# Solucionario Fundamentos Circuitos Eletr

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W
ith its objective to present circuit analysis in a manner that is clearer, more interesting, and easier to
understand than other texts, Fundamentals of Electric Circuits by Charles Alexander and Matthew Sadiku
has become the student choice for introductory electric circuits courses.
FiFth
Edition
FiFt h Edition
Building on the success of the previous editions, the ﬁfth edition features the latest updates and advances in the
ﬁeld, while continuing to present material with an unmatched pedagogy and communication style.
Fundamentals of
Pedagogical Features
Matched Example Problems and Extended Examples. Each illustrative example is immediately followed by a
practice problem and answer to test understanding of the preceding example. one extended example per
chapter shows an example problem worked using a detailed outline of the six-step method so students can
see how to practice this technique. Students follow the example step-by-step to solve the practice problem
without having to ﬂip pages or search the end of the book for answers.
■
Comprehensive Coverage of Material. not only is Fundamentals the most comprehensive text in terms of
material, but it is also self-contained in regards to mathematics and theory, which means that when students
have questions regarding the mathematics or theory they are using to solve problems, they can ﬁnd answers to
their questions in the text itself. they will not need to seek out other references.
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Computer tools. PSpice&reg; for Windows is used throughout the text with discussions and examples at the end of
each appropriate chapter. MAtLAB&reg; is also used in the book as a computational tool.
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new to the ﬁfth edition is the addition of 120 national instruments Multisim™ circuit ﬁles. Solutions for almost
all of the problems solved using PSpice are also available to the instructor in Multisim.
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We continue to make available KCidE for Circuits (a Knowledge Capturing integrated design Environment for
Circuits).
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An icon is used to identify homework problems that either should be solved or are more easily solved using
PSpice, Multisim, and/or KCidE. Likewise, we use another icon to identify problems that should be solved or are
more easily solved using MAtLAB.
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Electric Circuits
INSTRUCTOR
SOLUTIONS
MANUAL
MD DALIM 1167970 10/30/11 CYAN MAG YELO BLACK
■
Fundamentals of
Problem-Solving Methodology. A six-step method for solving circuits problems is introduced in Chapter 1 and
used consistently throughout the book to help students develop a systems approach to problem solving that
leads to better understanding and fewer mistakes in mathematics and theory.
Electric Circuits
■
the text also features a website of student and instructor resources. Check it out at www.mhhe.com/alexander.
Alexander
Charles K. Alexander | Matthew n. o. Sadiku
Chapter 1, Solution 1
(a) q = 6.482x1017 x [-1.602x10-19 C] = –103.84 mC
(b) q = 1. 24x1018 x [-1.602x10-19 C] = –198.65 mC
(c) q = 2.46x1019 x [-1.602x10-19 C] = –3.941 C
(d) q = 1.628x1020 x [-1.602x10-19 C] = –26.08 C
Chapter 1, Solution 2
(a)
(b)
(c)
(d)
(e)
i = dq/dt = 3 mA
i = dq/dt = (16t + 4) A
i = dq/dt = (-3e-t + 10e-2t) nA
i=dq/dt = 1200 cos 120 t pA
i =dq/dt =  e 4t (80 cos 50 t  1000 sin 50 t )  A
Chapter 1, Solution 3
(a) q(t)   i(t)dt  q(0)  (3t  1) C
(b) q(t)   (2t  s) dt  q(v)  (t 2  5t) mC
(c) q(t)   20 cos 10t   / 6   q(0)  (2sin(10t   / 6)  1)  C
10e -30t
q(t)   10e sin 40t  q(0) 
( 30 sin 40t - 40 cos t)
(d)
900  1600
  e - 30t (0.16cos40 t  0.12 sin 40t) C
-30t
Chapter 1, Solution 4
q = it = 7.4 x 20 = 148 C
Chapter 1, Solution 5
10
1
t 2 10
q   idt   tdt 
 25 C
2
4 0
0
Chapter 1, Solution 6
(a) At t = 1ms, i 
dq 30

 15 A
dt
2
(b) At t = 6ms, i 
dq
 0A
dt
(c) At t = 10ms, i 
dq  30

 –7.5 A
dt
4
Chapter 1, Solution 7
25A,
dq 
i
 - 25A,
dt 
 25A,
0t2
2t6
6t8
which is sketched below:
Chapter 1, Solution 8
q   idt 
10  1
 10  1  15 μC
2
Chapter 1, Solution 9
1
(a) q   idt   10 dt  10 C
0
3
5 1

q   idt  10  1  10 
  5 1
0
(b)
2 

 15  7.5  5  22.5C
5
(c) q   idt  10  10  10  30 C
0
Chapter 1, Solution 10
q = it = 10x103x15x10-6 = 150 mC
Chapter 1, Solution 11
q= it = 90 x10-3 x 12 x 60 x 60 = 3.888 kC
E = pt = ivt = qv = 3888 x1.5 = 5.832 kJ
Chapter 1, Solution 12
For 0 &lt; t &lt; 6s, assuming q(0) = 0,
t
t


0
0
q (t )  idt  q (0 )  3tdt  0  1.5t 2
At t=6, q(6) = 1.5(6)2 = 54
For 6 &lt; t &lt; 10s,
t
t


6
6
q (t )  idt  q (6 )  18 dt  54  18 t  54
At t=10, q(10) = 180 – 54 = 126
For 10&lt;t&lt;15s,
t
t


10
10
q (t )  idt  q (10 )  ( 12)dt  126  12t  246
At t=15, q(15) = -12x15 + 246 = 66
For 15&lt;t&lt;20s,
t

q (t )  0 dt  q (15) 66
15
Thus,

1.5t 2 C, 0 &lt; t &lt; 6s

 18 t  54 C, 6 &lt; t &lt; 10s
q (t )  
12t  246 C, 10 &lt; t &lt; 15s

66 C, 15 &lt; t &lt; 20s

The plot of the charge is shown below.
140
120
100
q(t)
80
60
40
20
0
0
5
10
t
15
20
Chapter 1, Solution 13
(a) i = [dq/dt] = 20πcos(4πt) mA
p = vi = 60πcos2(4πt) mW
At t=0.3s,
p = vi = 60πcos2(4π0.3) mW = 123.37 mW
(b) W =
W = 30π[0.6+(1/(8π))[sin(8π0.6)–sin(0)]] = 58.76 mJ
Chapter 1, Solution 14
1



(a) q   idt   0.02 1 - e -0.5t dt  0.02 t  2e -0.5t
0
(b)

1
0


 0.02 1  2e -0.5  2 = 4.261 mC
p(t) = v(t)i(t)
p(1) = 10cos(2)x0.02(1–e–0.5) = (–4.161)(0.007869)
= –32.74 mW
Chapter 1, Solution 15
 0.006 2t
q   idt   0.006e dt 
e
0
(a)
2
2

2
- 2t

0
 0.003 e  1 
2.945 mC
(b)
-4
10di
 0.012e - 2t (10)  0.12e - 2t V this leads to p(t) = v(t)i(t) =
dt
(-0.12e-2t)(0.006e-2t) = –720e–4t &micro;W
v
3
(c) w   pdt  -0.72 e
0
 720 -4t 6
dt 
e 10 = –180 &micro;J
4
0
3
- 4t
Chapter 1, Solution 16
(a)
 30t mA, 0 &lt; t &lt;2
i (t )  
120-30t mA, 2 &lt; t&lt;4
5 V, 0 &lt; t &lt;2
v(t )  
 -5 V, 2 &lt; t&lt;4
 150t mW, 0 &lt; t &lt;2
p(t )  
-600+150t mW, 2 &lt; t&lt;4
which is sketched below.
p(mW)
300
1
2
-300
(b) From the graph of p,
4
W   pdt  0 J
0
4
t (s)
Chapter 1, Solution 17
 p=0
 -205 + 60 + 45 + 30 + p 3 = 0
p 3 = 205 – 135 = 70 W
Thus element 3 receives 70 W.
Chapter 1, Solution 18
p 1 = 30(-10) = -300 W
p 2 = 10(10) = 100 W
p 3 = 20(14) = 280 W
p 4 = 8(-4) = -32 W
p 5 = 12(-4) = -48 W
Chapter 1, Solution 19
I = 8 –2 = 6 A
Calculating the power absorbed by each element means we need to find vi for
each element.
p 8 amp source = –8x9 = –72 W
p element with 9 volts across it = 2x9 = 18 W
p element with 3 bolts across it = 3x6 = 18 W
p 6 volt source = 6x6 = 36 W
One check we can use is that the sum of the power absorbed must equal zero
which is what it does.
Chapter 1, Solution 20
p 30 volt source = 30x(–6) = –180 W
p 12 volt element = 12x6 = 72 W
p 28 volt e.ement with 2 amps flowing through it = 28x2 = 56 W
p 28 volt element with 1 amp flowing through it = 28x1 = 28 W
p the 5Io dependent source = 5x2x(–3) = –30 W
Since the total power absorbed by all the elements in the circuit must equal zero,
or 0 = –180+72+56+28–30+p into the element with Vo or
p into the element with Vo = 180–72–56–28+30 = 54 W
Since p into the element with Vo = V o x3 = 54 W or V o = 18 V.
Chapter 1, Solution 21
p 60

 0.5 A
v 120
q = it = 0.5x24x60x60 = 43.2 kC
N e  qx6.24 x1018  2.696 x1023 electrons
p  vi


i
Chapter 1, Solution 22
q = it = 40x103x1.7x10–3 = 68 C
Chapter 1, Solution 23
W = pt = 1.8x(15/60) x30 kWh = 13.5kWh
C = 10cents x13.5 = \$1.35
Chapter 1, Solution 24
W = pt = 60 x24 Wh = 0.96 kWh = 1.44 kWh
C = 8.2 centsx0.96 = 11.808 cents
Chapter 1, Solution 25
Cost  1.5 kW 
3.5
hr  30  8.2 cents/kWh = 21.52 cents
60
Chapter 1, Solution 26
0.8A  h
 80 mA
10h
(b) p = vi = 6  0.08 = 0.48 W
(c) w = pt = 0.48  10 Wh = 0.0048 kWh
(a) i 
Chapter 1, Solution 27
(a) Let T  4h  4  3600
T
q   idt   3dt  3T  3  4  3600  43.2 kC
0
T
T
0. 5t 

( b) W   pdt   vidt   ( 3) 10 
dt
0
0
3600 

43600

0. 25t 2 

 310t 
3600  0

 475.2 kJ
( c)
 340  3600  0. 25  16  3600
W  475.2 kWs, (J  Ws)
475.2
kWh  9 cent  1.188 cents
Cost 
3600
Chapter 1, Solution 28
(a) i 
P 60

V 120
= 500 mA
(b) W  pt  60  365  24 Wh  525.6 kWh
Cost  \$0.095  525.6
= \$49.93
Chapter 1, Solution 29
(20  40  15  45)
 30 
hr  1.8 kW  hr
60
 60 
 2.4  0.9  3.3 kWh
Cost  12 cents  3.3  39.6 cents
w  pt  1. 2kW
Chapter 1, Solution 30
Monthly charge = \$6
First 250 kWh @ \$0.02/kWh = \$5
Remaining 2,436–250 kWh = 2,186 kWh @ \$0.07/kWh= \$153.02
Total = \$164.02
Chapter 1, Solution 31
Total energy consumed = 365(120x4 + 60x8) W
Cost = \$0.12x365x960/1000 = \$42.05
Chapter 1, Solution 32
i = 20 &micro;A
q = 15 C
t = q/i = 15/(20x10-6) = 750x103 hrs
Chapter 1, Solution 33
i
dq
 q   idt  2000  3  10  3  6 C
dt
Chapter 1, Solution 34
(a)
Energy =
 pt
= 200 x 6 + 800 x 2 + 200 x 10 + 1200 x 4 + 200 x 2
= 10 kWh
(b)
Average power = 10,000/24 = 416.7 W
Chapter 1, Solution 35
energy = (5x5 + 4x5 + 3x5 + 8x5 + 4x10)/60 = 2.333 MWhr
Chapter 1, Solution 36
160A  h
4A
40
160Ah 160, 000h
( b) t 

 6,667 days
0.001A 24h / day
(a)
i
Chapter 1, Solution 37
W = pt = vit = 12x 40x 60x60 = 1.728 MJ
Chapter 1, Solution 38
P = 10 hp = 7460 W
W = pt = 7460  30  60 J = 13.43  106 J
Chapter 1, Solution 39
W = pt = 600x4 = 2.4 kWh
C = 10cents x2.4 = 24 cents
Chapter 2, Solution 1.
Design a problem, complete with a solution, to help students to
better understand Ohm’s Law. Use at least two resistors and one voltage source. Hint,
you could use both resistors at once or one at a time, it is up to you. Be creative.
Although there is no correct way to work this problem, this is an example based on the
same kind of problem asked in the third edition.
Problem
The voltage across a 5-k resistor is 16 V. Find the current through the resistor.
Solution
v = iR
i = v/R = (16/5) mA = 3.2 mA
Chapter 2, Solution 2
p = v2/R 
R = v2/p = 14400/60 = 240 ohms
Chapter 2, Solution 3
For silicon,   6.4 x102 -m. A   r 2 . Hence,
R
L
A

L
 r2


r2 
 L 6.4 x102 x 4 x102

 0.033953
R
 x 240
r = 184.3 mm
Chapter 2, Solution 4
(a)
(b)
i = 40/100 = 400 mA
i = 40/250 = 160 mA
Chapter 2, Solution 5
n = 9;
l = 7; b = n + l – 1 = 15
Chapter 2, Solution 6
n = 12;
l = 8;
b = n + l –1 = 19
Chapter 2, Solution 7
6 branches and 4 nodes
Chapter 2, Solution 8.
Design a problem, complete with a solution, to help other students
to better understand Kirchhoff’s Current Law. Design the problem by specifying values
of i a , i b , and i c , shown in Fig. 2.72, and asking them to solve for values of i 1 , i 2 , and i 3 .
Be careful specify realistic currents.
Although there is no correct way to work this problem, this is an example based on the
same kind of problem asked in the third edition.
Problem
Use KCL to obtain currents i 1, i 2, and i 3 in the circuit shown in Fig. 2.72.
Solution
12 A
a
i1
b
8A
i2
i3
12 A
c
At node a,
At node c,
At node d,
9A d
8 = 12 + i 1
9 = 8 + i2
9 = 12 + i 3
i 1 = - 4A
i 2 = 1A
i 3 = -3A
Chapter 2, Solution 9
At A, 1+6–i 1 = 0 or i 1 = 1+6 = 7 A
At B, –6+i 2 +7 = 0 or i 2 = 6–7 = –1 A
At C, 2+i 3 –7 = 0 or i 3 = 7–2 = 5 A
Chapter 2, Solution 10
2
–8A
1
4A
i2
i1
3
–6A
At node 1,
–8–i 1 –6 = 0 or i 1 = –8–6 = –14 A
At node 2,
–(–8)+i 1 +i 2 –4 = 0 or i 2 = –8–i 1 +4 = –8+14+4 = 10 A
Chapter 2, Solution 11
V1  1  5  0
5  2  V2  0




V1  6 V
V2  3 V
Chapter 2, Solution 12
+ 30v –
loop 2
– 50v +
+
40v
-
+ 20v –
loop 1
+
v1
–
+ v2 –
loop 3
+
v3
–
For loop 1,
–40 –50 +20 + v 1 = 0 or v 1 = 40+50–20 = 70 V
For loop 2,
–20 +30 –v 2 = 0 or v 2 = 30–20 = 10 V
For loop 3,
–v 1 +v 2 +v 3 = 0 or v 3 = 70–10 = 60 V
Chapter 2, Solution 13
2A
I2
7A
1
I4
2
3
4
4A
I1
3A
At node 2,
3  7  I2  0


I 2  10 A
At node 1,
I1  I 2  2


I 1  2  I 2  12 A
At node 4,
2  I4  4


I 4  2  4  2 A
At node 3,
7  I4  I3


I3  7  2  5 A
Hence,
I 1  12 A,
I 2  10 A,
I 3  5 A,
I 4  2 A
I3
Chapter 2, Solution 14
+
3V
-
+
I3
4V
+
-
V3 -
+
I4
2V -
+
- V4
I2
+
V2
+
+
5V
I1
-
For mesh 1,
V4  2  5  0


V4  7V
For mesh 2,
4  V3  V4  0


V3  4  7  11V


V1  V3  3  8V


V2  V1  2  6V
For mesh 3,
3  V1  V3  0
For mesh 4,
V1  V2  2  0
Thus,
V1  8V ,
V1
V2  6V ,
V3  11V ,
V4  7V
Chapter 2, Solution 15
Calculate v and i x in the circuit of Fig. 2.79.
12 
+ v
10 V
+
_
+ 16 V –
–
ix
+
+
4V
_
_
Figure 2.79
For Prob. 2.15.
Solution
For loop 1, –10 + v +4 = 0, v = 6 V
For loop 2, –4 + 16 + 3i x =0, i x =
–4 A
3 ix
Chapter 2, Solution 16
Determine V o in the circuit in Fig. 2.80.
16 
14 

+
10 V
+
_
Vo
+
_
25 V
_

Figure 2.80
For Prob. 2.16.
Solution
Apply KVL,
–10 + (16+14)I + 25 = 0 or 30I = 10–25 = – or I = –15/30 = –500 mA
Also,
–10 + 16I + V o = 0 or V o = 10 – 16(–0.5) = 10+8 = 18 V
Chapter 2, Solution 17
Applying KVL around the entire outside loop we get,
–24 + v 1 + 10 + 12 = 0 or v 1 = 2V
Applying KVL around the loop containing v 2 , the 10-volt source, and the
12-volt source we get,
v 2 + 10 + 12 = 0 or v 2 = –22V
Applying KVL around the loop containing v 3 and the 10-volt source we
get,
–v 3 + 10 = 0 or v 3 = 10V
Chapter 2, Solution 18
Applying KVL,
-30 -10 +8 + I(3+5) = 0
8I = 32
I = 4A
-V ab + 5I + 8 = 0
V ab = 28V
Chapter 2, Solution 19
Applying KVL around the loop, we obtain
–(–8) – 12 + 10 + 3i = 0
Power dissipated by the resistor:
p 3 = i2R = 4(3) = 12W
Power supplied by the sources:
p 12V = 12 ((–2)) = –24W
p 10V = 10 (–(–2)) = 20W
p 8V = (–8)(–2) = 16W
i = –2A
Chapter 2, Solution 20
Determine i o in the circuit of Fig. 2.84.
io
54V
22 
+

+
–
Figure 2.84
For Prob. 2.20
Solution
Applying KVL around the loop,
–54 + 22i o + 5i o = 0
i o = 4A
5i o
Chapter 2, Solution 21
Applying KVL,
-15 + (1+5+2)I + 2 V x = 0
But V x = 5I,
-15 +8I + 10I =0,
I = 5/6
V x = 5I = 25/6 = 4.167 V
Chapter 2, Solution 22
Find V o in the circuit in Fig. 2.86 and the power absorbed by the dependent
source.
10 
+
10 
V1
Vo 
25A
2V o
Figure 2.86
For Prob. 2.22
Solution
At the node, KCL requires that [–V o /10]+[–25]+[–2V o ] = 0 or 2.1V o = –25
or V o = –11.905 V
The current through the controlled source is i = 2V 0 = –23.81 A
and the voltage across it is V 1 = (10+10) i 0 (where i 0 = –V 0 /10) = 20(11.905/10)
= 23.81 V.
Hence,
p dependent source = V 1 (–i) = 23.81x(–(–23.81)) = 566.9 W
Checking, (25–23.81)2(10+10) + (23.81)(–25) + 566.9 = 28.322 – 595.2 + 566.9
= 0.022 which is equal zero since we are using four places of accuracy!
Chapter 2, Solution 23
8//12 = 4.8, 3//6 = 2, (4 + 2)//(1.2 + 4.8) = 6//6 = 3
The circuit is reduced to that shown below.
1
ix
+
20A
vx
–
2
3
Applying current division,
i x = [2/(2+1+3)]20 = 6.667 and v x = 1x6.667 = 6.667 V
ix 
2
(6 A)  2 A,
2  1 3
v x  1i x  2V
The current through the 1.2-  resistor is 0.5i x = 3.333 A. The voltage across the
12-  resistor is 3.333 x 4.8 = 16V. Hence the power absorbed by the 12-ohm
resistor is equal to
(16)2/12 = 21.33 W
Chapter 2, Solution 24
(a)
I0 =
Vs
R1  R2
V0   I 0 R3 R4  = 
R 3R 4
Vs

R1  R 2 R 3  R 4
V0
 R3 R4

Vs  R1  R2  R3  R4 
(b)
If R 1 = R 2 = R 3 = R 4 = R,
V0
 R 

   10
VS
2R 2 4
 = 40
Chapter 2, Solution 25
V 0 = 5 x 10-3 x 10 x 103 = 50V
Using current division,
I 20 
5
(0.01x50)  0.1 A
5  20
V 20 = 20 x 0.1 kV = 2 kV
p 20 = I 20 V 20 = 0.2 kW
Chapter 2, Problem 26.
For the circuit in Fig. 2.90, i o = 3 A. Calculate i x and the total power absorbed by
the entire circuit.
ix
10 
io
8
4
2
16 
Figure 2.90
For Prob. 2.26.
Solution
If i 16 = i o = 3A, then v = 16x3 = 48 V and i 8 = 48/8 = 6A; i 4 = 48/4 = 12A; and
i 2 = 48/2 = 24A.
Thus,
i x = i 8 + i 4 + i 2 + i 16 = 6 + 12 + 24 + 3 = 45 A
p = (45)210 + (6)28 + (12)24 + (24)22 + (3)216 = 20,250 + 288 + 576 +1152 + 144
= 20250 + 2106 = 22.356 kW.
Chapter 2, Problem 27.
Calculate I o in the circuit of Fig. 2.91.
8
10V
+

Io
3
6
Figure 2.91
For Prob. 2.27.
Solution
The 3-ohm resistor is in parallel with the c-ohm resistor and can be replaced by a
[(3x6)/(3+6)] = 2-ohm resistor. Therefore,
I o = 10/(8+2) = 1 A.
Chapter 2, Solution 28
Design a problem, using Fig. 2.92, to help other students better
understand series and parallel circuits.
Although there is no correct way to work this problem, this is an example based on the
same kind of problem asked in the third edition.
Problem
Find v 1 , v 2 , and v 3 in the circuit in Fig. 2.92.
Solution
We first combine the two resistors in parallel
15 10  6 
We now apply voltage division,
v1 =
14
(40)  28 V
14  6
v2 = v3 =
Hence,
6
(40)  12 V
14  6
v 1 = 28 V, v 2 = 12 V, v s = 12 V
Chapter 2, Solution 29
All resistors in Fig. 2.93 are 5  each. Find R eq .
R eq
Figure 2.93
For Prob. 2.29.
Solution
R eq = 5 + 5||[5+5||(5+5)] = 5 + 5||[5+(5x10/(5+10))] = 5+5||(5+3.333) = 5 +
41.66/13.333
= 8.125 Ω
Chapter 2, Problem 30.
Find R eq for the circuit in Fig. 2.94.
25 
180 
60 
R eq
60 
Figure 2.94
For Prob. 2.30.
Solution
We start by combining the 180-ohm resistor with the 60-ohm resistor which in
turn is in parallel with the 60-ohm resistor or = [60(180+60)/(60+180+60)] = 48.
Thus,
R eq = 25+48 = 73 Ω.
Chapter 2, Solution 31
Req  3  2 // 4 //1  3 
1
 3.5714
1/ 2  1/ 4  1
i 1 = 200/3.5714 = 56 A
v 1 = 0.5714xi 1 = 32 V and i 2 = 32/4 = 8 A
i 4 = 32/1 = 32 A; i 5 = 32/2 = 16 A; and i 3 = 32+16 = 48 A
Chapter 2, Solution 32
Find i 1 through i 4 in the circuit in Fig. 2.96.
60 
i4
i2
200 
50 
40 
i3
i1
16 A
Figure 2.96
For Prob. 2.32.
Solution
We first combine resistors in parallel.
40 60 
40x 60
50x 200
 24  and 50 200 
 40 
100
250
Using current division principle,
24
40
i1  i 2 
(16)  6A, i 3  i 4  (16)  10A
24  40
64
i1 
200
50
(6)  –4.8 A and i 2 
(6)  –1.2 A
250
250
i3 
60
40
(10)  –6 A and i 4 
(10)  –4 A
100
100
Chapter 2, Solution 33
Combining the conductance leads to the equivalent circuit below
i
+
v
-
9A
6S 3S 
1S
4S
i
4S
6x3
 2S and 2S + 2S = 4S
9
Using current division,
i
1
1
1
2
(9)  6 A, v = 3(1) = 3 V
9A
+
v
-
1S
2S
Chapter 2, Solution 34
160//(60 + 80 + 20)= 80 ,
160//(28+80 + 52) = 80 
R eq = 20+80 = 100 Ω
I = 200/100 = 2 A or p = VI = 200x2 = 400 W.
Chapter 2, Solution 35
i
70 
200V
+
a
-
+
V1
i1 -
30 
Io
+
20 
i2
b
Vo 5 
-
Combining the resistors that are in parallel,
70 30 
i=
70x30
 21 ,
100
20 5 
20 x5
4 
25
200
8 A
21  4
v 1 = 21i = 168 V, v o = 4i = 32 V
v
v
i 1 = 1  2.4 A, i 2 = o  1.6 A
70
20
At node a, KCL must be satisfied
i1 = i2 + Io
2.4 = 1.6 + I o
I o = 0.8 A
Hence,
v o = 32 V and I o = 800 mA
Chapter 2, Solution 36
20//(30+50) = 16, 24 + 16 = 40, 60//20 = 15
R eq = 80+(15+25)40 = 80+20 = 100 Ω
i = 20/100 = 0.2 A
If i 1 is the current through the 24- resistor and i o is the current through the 50-
resistor, using current division gives
i 1 = [40/(40+40)]0.2 = 0.1 and i o = [20/(20+80)]0.1 = 0.02 A or
v o = 30i o = 30x0.02 = 600 mV.
Chapter 2, Solution 37
Applying KVL,
-20 + 10 + 10I – 30 = 0, I = 4
10  RI

 R
10
 2.5 
I
Chapter 2, Solution 38
20//80 = 80x20/100 = 16, 6//12 = 6x12/18 = 4
The circuit is reduced to that shown below.
2.5 
4
60 
15 
16 
R eq
(4 + 16)//60 = 20x60/80 = 15
R eq = 2.5+15||15 = 2.5+7.5 = 10 Ω and
i o = 35/10 = 3.5 A.
Chapter 2, Solution 39
(a) We note that the top 2k-ohm resistor is actually in parallel with the first 1k-ohm
resistor. This can be replaced (2/3)k-ohm resistor. This is now in series with the second
2k-ohm resistor which produces a 2.667k-ohm resistor which is now in parallel with the
second 1k-ohm resistor. This now leads to,
R eq = [(1x2.667)/3.667]k = 727.3 Ω.
(b) We note that the two 12k-ohm resistors are in parallel producing a 6k-ohm resistor.
This is in series with the 6k-ohm resistor which results in a 12k-ohm resistor which is in
parallel with the 4k-ohm resistor producing,
R eq = [(4x12)/16]k = 3 kΩ.
Chapter 2, Solution 40
Req = 8  4 (2  6 3)  8  2  10 
I=
15 15
  1.5 A
R eq 10
Chapter 2, Solution 41
Let R 0 = combination of three 12 resistors in parallel
1
1
1
1

 
R o 12 12 12
Ro = 4
R eq  30  60 (10  R 0  R )  30  60 (14  R )
50  30 
60(14  R )
74  R
or R = 16 
74 + R = 42 + 3R
Chapter 2, Solution 42
5x 20
 4
25
(a)
R ab = 5 (8  20 30)  5 (8  12) 
(b)
R ab = 2  4 (5  3) 8  5 10 4  2  4 4  5 2.857  2  2  1.8181  5.818 
Chapter 2, Solution 43
5x 20 400

 4  8  12 
25
50
(a)
R ab = 5 20  10 40 
(b)
1
1 
 1
60 20 30   
 
 60 20 30 
R ab = 80 (10  10) 
1

60
 10
6
80  20
 16 
100
Chapter 2, Solution 44
For the circuits in Fig. 2.108, obtain the equivalent resistance at terminals a-b.
5
20 
a
2
3
b
Figure 2.108
For Prob. 2.44
Solution
First we note that the 5 Ω and 20 Ω resistors are in parallel and can be replaced by
a 4 Ω [(5x20)/(5+20)] resistor which in now in series with the 2 Ω resistor and
can be replaced by a 6 Ω resistor in parallel with the 3 Ω resistor thus,
R ab = [(6x3)/(6+3)] = 2 Ω.
Chapter 2, Solution 45
(a) 10//40 = 8, 20//30 = 12, 8//12 = 4.8
Rab  5  50  4.8  59.8 
(b) 12 and 60 ohm resistors are in parallel. Hence, 12//60 = 10 ohm. This 10 ohm
and 20 ohm are in series to give 30 ohm. This is in parallel with 30 ohm to give
30//30 = 15 ohm. And 25//(15+10) = 12.5. Thus,
Rab  5  12.8  15  32.5
Chapter 2, Solution 46
R eq = 12 + 5||20 + [1/((1/15)+(1/15)+(1/15))] + 5 + 24||8
= 12 + 4 + 5 + 5 + 6 = 32 Ω
I = 80/32 = 2.5 A
Chapter 2, Solution 47
5 20 
6 3
5x 20
 4
25
6x3
 2
9
10 
8
a
b
4
R ab = 10 + 4 + 2 + 8 = 24 
2
Chapter 2, Solution 48
(a)
(b)
R 1 R 2  R 2 R 3  R 3 R 1 100  100  100

 30
R3
10
R a = R b = R c = 30 
Ra =
30x 20  30x50  20x 50 3100

 103.3
30
30
3100
3100
Rb 
 155, R c 
 62
20
50
R a = 103.3 , R b = 155 , R c = 62 
Ra 
Chapter 2, Solution 49
(a)
R1 =
RaRc
12 *12

 4
Ra  Rb  Rc
36
R1 = R2 = R3 = 4 
(b)
60x30
 18
60  30  10
60 x10
R2 
 6
100
30x10
R3 
 3
100
R1 
R 1 = 18, R 2 = 6, R 3 = 3
2.50 Design a problem to help other students better understand wye-delta transformations using
Fig. 2.114.
Although there is no correct way to work this problem, this is an example based on the same
kind of problem asked in the third edition.
Problem
What value of R in the circuit of Fig. 2.114 would cause the current source to deliver 800 mW to
the resistors.
Solution
Using R  = 3R Y = 3R, we obtain the equivalent circuit shown below:
R
30mA
3R
3R
3R
3R R 
30mA
R
3RxR 3
 R
4
4R
3
3Rx R
3 
3
3
2 =R
3R  R  R   3R R 
3
4 
2
4
3R  R
2
P = I2R
R = 889 
800 x 10-3 = (30 x 10-3)2 R
3R
3R/2
Chapter 2, Solution 51
(a)
30 30  15 and 30 20  30x 20 /(50)  12
R ab = 15 (12  12)  15x 24 /(39)  9.231 
a
a
30 
30 
30 
30 
b
12 
20 
15 
12 
20 
b
(b)
Converting the T-subnetwork into its equivalent  network gives
R a'b' = 10x20 + 20x5 + 5x10/(5) = 350/(5) = 70 
R b'c' = 350/(10) = 35, Ra'c' = 350/(20) = 17.5 
Also
30 70  30x 70 /(100)  21 and 35/(15) = 35x15/(50) = 10.5
R ab = 25 + 17.5 (21  10.5)  25  17.5 31.5
R ab = 36.25 
30 
30 
25 
10 
20 
a
a
5
b
15 
25 
a’
17.5 
b
70 
b’
35 
c’
15 
c’
Chapter 2, Solution 52
Converting the wye-subnetwork to delta-subnetwork, we obtain the circuit below.
9
3
9
3
9
3
3
3
3
6
3//1 = 3x1/4 = 0.75, 2//1 =2x1/3 = 0.6667. Combining these resistances leads to the
circuit below.
3
2.25 
3
2.25
9
3
2
We now convert the wye-subnetwork to the delta-subnetwork.
R a = [(2.25x3+2.25x3+2.25x2.25)/3] = 6.188 Ω
R b = R c = 18.562/2.25 = 8.25 Ω
This leads to the circuit below.
3
6.188
8.25
3
9
8.25
2
R = 9||6.188+8.25||2 = 3.667+1.6098 = 5.277
R eq = 3+3+8.25||5.277 = 9.218 Ω.
Chapter 2, Solution 53
(a)
Converting one  to T yields the equivalent circuit below:
30 
a’
20 
a
60 
b’
b
4
20 
c’
5
80 
40x10
10 x 50
40x 50
 4, R b 'n 
 5, R c 'n 
 20
40  10  50
100
100
R ab = 20 + 80 + 20 + (30  4) (60  5)  120  34 65
R a'n =
R ab = 142.32 
(b) We combine the resistor in series and in parallel.
30 (30  30) 
30 x 60
 20
90
We convert the balanced  s to Ts as shown below:
a
30 
30 
a
10 
30 
30 
20 
10 
30 
b
30 
10 
10 
b
R ab = 10 + (10  10) (10  20  10)  10  20  20 40
R ab = 33.33 
10 
10 
20 
Chapter 2, Solution 54
(a) Rab  50  100 / /(150  100  150 )  50  100 / /400  130 
(b) Rab  60  100 / /(150  100  150 )  60  100 / /400  140 
Chapter 2, Solution 55
We convert the T to  .
I0
a
24 V
+
-
I0
20 
40 
60 
10 
20 
50 
a
140 
24 V
+
35 
-
70 
b
R eq
b
R eq
R R  R 2 R 3  R 3 R 1 20 x 40  40 x10  10 x 20 1400


 35
R ab = 1 2
R3
40
40
R ac = 1400/(10) = 140, R bc = 1400/(20) = 70
70 70  35 and 140 160  140x60/(200) = 42
R eq = 35 (35  42)  24.0625
I 0 = 24/(R ab ) = 997.4mA
60 
70 
Chapter 2, Solution 56
We need to find R eq and apply voltage division. We first tranform the Y network to  .
30 
+
100 V
16 
15 
35 
12 
30 
16 
10 
20 
-
+
100 V
a
35 
R eq
R eq
15x10  10x12  12x15 450

 37.5
12
12
R ac = 450/(10) = 45, R bc = 450/(15) = 30
R ab =
Combining the resistors in parallel,
30||20 = (600/50) = 12 ,
37.5||30 = (37.5x30/67.5) = 16.667 
35||45 = (35x45/80) = 19.688 
R eq = 19.688||(12 + 16.667) = 11.672
By voltage division,
v =
11.672
100 = 42.18 V
11.672  16
37.5 
30 
45 
c
b
20 
Chapter 2, Solution 57
4 a
2
27 
1
18 
b
d
10 
36 
c
e
7
14 
28 
f
6 x12  12 x8  8x 6 216

 18 
12
12
R ac = 216/(8) = 27, R bc = 36 
4 x 2  2 x8  8x 4 56
R de =

7
8
8
R ef = 56/(4) = 14, R df = 56/(2) = 28 
R ab =
Combining resistors in parallel,
280
36 x 7
 7.368, 36 7 
 5.868
38
43
27 x 3
27 3 
 2 .7 
30
10 28 
4
4
18 
5.868 
7.568 
R cn
1.829 
3.977 
0.5964 
14 
7.568 
18x 2.7
18x 2.7

 1.829 
18  2.7  5.867 26.567
18x 5.868

 3.977 
26.567
5.868x 2.7

 0.5904 
26.567
R an 
R bn
2.7 
14 
R eq  4  1.829  (3.977  7.368) (0.5964  14)
 5.829  11.346 14.5964  12.21 
i = 20/(R eq ) = 1.64 A
Chapter 2, Solution 58
The resistance of the bulb is (120)2/60 = 240
40 
2A
+ 90 V - 0.5 A
VS
240 
+
-
1.5 A
+
120 V
80 
-
Once the 160 and 80 resistors are in parallel, they have the same voltage
120V. Hence the current through the 40 resistor is equal to 2 amps.
40(0.5 + 1.5) = 80 volts.
Thus
v s = 80 + 120 = 200 V.
Chapter 2, Solution 59
Three light bulbs are connected in series to a 120-V source as shown in Fig.
2.123. Find the current I through each of the bulbs. Each bulb is rated at 120
volts. How much power is each bulb absorbing? Do they generate much light?
Figure 2.123
For Prob. 2.59.
Solution
Using p = v2/R, we can calculate the resistance of each bulb.
R 30W = (120)2/30 = 14,400/30 = 480 Ω
R 40W = (120)2/40 = 14,400/40 = 360 Ω
R 50W = (120)2/50 = 14,400/50 = 288 Ω
The total resistance of the three bulbs in series is 480+360+288 = 1128 Ω.
The current flowing through each bulb is 120/1128 = 0.10638 A.
p 30 = (0.10638)2480 = 0.011317x480 = 5.432 W.
p 40 = (0.10638)2360 = 0.011317x360 = 4.074 W.
p 50 = (0.10638)2288 = 0.011317x288 = 3.259 W.
Clearly these values are well below the rated powers of each light bulb so we
would not expect very much light from any of them. To work properly, they need
to be connected in parallel.
Chapter 2, Solution 60
If the three bulbs of Prob. 2.59 are connected in parallel to the 120-V source,
calculate the current through each bulb.
Solution
Using p = v2/R, we can calculate the resistance of each bulb.
R 30W = (120)2/30 = 14,400/30 = 480 Ω
R 40W = (120)2/40 = 14,400/40 = 360 Ω
R 50W = (120)2/50 = 14,400/50 = 288 Ω
The current flowing through each bulb is 120/R.
i 30 = 120/480 = 250 mA.
i 40 = 120/360 = 333.3 mA.
i 30 = 120/288 = 416.7 mA.
Unlike the light bulbs in 2.59, the lights will glow brightly!
Chapter 2, Solution 61
There are three possibilities, but they must also satisfy the current range of 1.2 +
0.06 = 1.26 and 1.2 – 0.06 = 1.14.
(a)
Use R 1 and R 2 :
R = R 1 R 2  80 90  42.35
p = i2R = 70W
i2 = 70/42.35 = 1.6529 or i = 1.2857 (which is outside our range)
cost = \$0.60 + \$0.90 = \$1.50
(b)
Use R 1 and R 3 :
R = R 1 R 3  80 100  44.44 
i2 = 70/44.44 = 1.5752 or i = 1.2551 (which is within our range),
cost = \$1.35
(c)
Use R 2 and R 3 :
R = R 2 R 3  90 100  47.37
i2 = 70/47.37 = 1.4777 or i = 1.2156 (which is within our range),
cost = \$1.65
Note that cases (b) and (c) satisfy the current range criteria and (b) is the
cheaper of the two, hence the correct choice is:
R 1 and R 3
Chapter 2, Solution 62
p A = 110x8 = 880 W,
p B = 110x2 = 220 W
Energy cost = \$0.06 x 365 x10 x (880 + 220)/1000 = \$240.90
Chapter 2, Solution 63
Use eq. (2.61),
Im
2 x10 3 x100
Rn =
Rm 
 0.04
I  Im
5  2 x10 3
I n = I - I m = 4.998 A
p = I 2n R  (4.998) 2 (0.04)  0.9992  1 W
Chapter 2, Solution 64
When R x = 0, i x  10A
When R x is maximum, i x = 1A
i.e., R x = 110 - R = 99 
R x = 99 
Thus, R = 11 ,
R=
110
 11 
10
R  Rx 
110
 110 
1
Chapter 2, Solution 65
Rn 
Vfs
50
 Rm 
 1 k  4 k
I fs
10mA
Chapter 2, Solution 66
20 k/V = sensitivity =
1
I fs
1
k / V  50 A
20
V
The intended resistance R m = fs  10(20k / V)  200k
I fs
V
50 V
R n  fs  R m 
 200 k  800 k
(a)
i fs
50 A
i.e., I fs =
(b)
p = I fs2 R n  (50 A) 2 (800 k)  2 mW
Chapter 2, Solution 67
(a)
By current division,
i 0 = 5/(5 + 5) (2 mA) = 1 mA
V 0 = (4 k) i 0 = 4 x 103 x 10-3 = 4 V
(b)
4k 6k  2.4k. By current division,
5
(2mA)  1.19 mA
1  2.4  5
v '0  (2.4 k)(1.19 mA)  2.857 V
i '0 
v 0  v '0
1.143
x 100% =
x100  28.57%
v0
4
(c)
% error =
(d)
4k 36 k  3.6 k. By current division,
5
(2mA)  1.042mA
1  3.6  5
v '0 (3.6 k)(1.042 mA)  3.75V
i '0 
% error =
v  v '0
0.25x100
x100% 
 6.25%
v0
4
Chapter 2, Solution 68
(a)
40  24 60
4
 100 mA
16  24
4
i' 
 97.56 mA
16  1  24
0.1  0.09756
% error =
x100%  2.44%
0.1
i=
(b)
(c)
Chapter 2, Solution 69
With the voltmeter in place,
R2 Rm
V0 
VS
R1  R S  R 2 R m
where R m = 100 k without the voltmeter,
R2
V0 
VS
R1  R 2  R S
100
k
101
(a)
When R 2 = 1 k, R m R 2 
(b)
100
V 0 = 101 (40)  1.278 V (with)
100
 30
101
1
V0 =
(40)  1.29 V (without)
1  30
1000
When R 2 = 10 k, R 2 R m 
 9.091k
110
9.091
V0 =
(40)  9.30 V (with)
9.091  30
10
V0 =
(40)  10 V (without)
10  30
When R 2 = 100 k, R 2 R m  50k
(c)
50
(40)  25 V (with)
50  30
100
V0 =
(40)  30.77 V (without)
100  30
V0 
Chapter 2, Solution 70
(a) Using voltage division,
12
(25)  15V
12  8
10
vb 
(25)  10V
10  15
 va  vb  15  10  5V
va 
vab
(b)
c
8k 
+
25 V
–
15k 
a
12k 
b
10k 
v a = 0; v ac = –(8/(8+12))25 = –10V; v cb = (15/(15+10))25 = 15V.
v ab = v ac + v cb = –10 + 15 = 5V.
v b = –v ab = –5V.
Chapter 2, Solution 71
R1
iL
Vs +
−
RL
Given that v s = 30 V, R 1 = 20 , I L = 1 A, find R L .
v s  i L ( R1  R L )


RL 
vs
30
 R1 
 20  10
iL
1
Chapter 2, Solution 72
Converting the delta subnetwork into wye gives the circuit below.
1
⅓
1
⅓
Z in
10 V
+
_
1
1
1 1 1 4
Z in   (1  ) //(1  )   ( )  1 
3
3
3 3 2 3
Vo 
Z in
1
(10) 
(10)  5 V
1  Z in
11
⅓
1
Chapter 2, Solution 73
By the current division principle, the current through the ammeter will be
one-half its previous value when
R = 20 + R x
65 = 20 + R x
R x = 45 
Chapter 2, Solution 74
With the switch in high position,
6 = (0.01 + R 3 + 0.02) x 5
R 3 = 1.17 
At the medium position,
6 = (0.01 + R 2 + R 3 + 0.02) x 3
R 2 + R 3 = 1.97
or R 2 = 1.97 - 1.17 = 0.8 
At the low position,
6 = (0.01 + R 1 + R 2 + R 3 + 0.02) x 1
R 1 = 5.97 - 1.97 = 4 
R 1 + R 2 + R 3 = 5.97
Chapter 2, Solution 75
Converting delta-subnetworks to wye-subnetworks leads to the circuit below.
1
⅓
1
⅓
⅓
1
1
1
1
⅓
1
⅓
⅓
1
1
1 1 1 4
 (1  ) //(1  )   ( )  1
3
3
3 3 2 3
With this combination, the circuit is further reduced to that shown below.
1
1
1
1
1
1
1
1
1
Z ab  1   (1  ) //(1  )  1  1 = 2 
3
3
3
1
1
Chapter 2, Solution 76
Z ab = 1 + 1 = 2 
Chapter 2, Solution 77
(a)
5  = 10 10  20 20 20 20
i.e., four 20  resistors in parallel.
(b)
311.8 = 300 + 10 + 1.8 = 300 + 20 20  1.8
i.e., one 300 resistor in series with 1.8 resistor and
a parallel combination of two 20 resistors.
(c)
40k = 12k + 28k = (24 24k )  (56k 56k )
i.e., Two 24k resistors in parallel connected in series with two
56k resistors in parallel.
(a)
42.32k = 42l + 320
= 24k + 28k = 320
= 24k = 56k 56k  300  20
i.e., A series combination of a 20 resistor, 300 resistor,
24k resistor, and a parallel combination of two 56k
resistors.
Chapter 2, Solution 78
The equivalent circuit is shown below:
R
VS
V0 =
+
+
(1-)R
V0
-
-
(1  )R
1 
VS 
VS
R  (1  )R
2
V0 1  

VS 2  
Chapter 2, Solution 79
Since p = v2/R, the resistance of the sharpener is
R = v2/(p) = 62/(240 x 10-3) = 150
I = p/(v) = 240 mW/(6V) = 40 mA
Since R and R x are in series, I flows through both.
IR x = V x = 9 - 6 = 3 V
R x = 3/(I) = 3/(40 mA) = 3000/(40) = 75 
Chapter 2, Solution 80
The amplifier can be modeled as a voltage source and the loudspeaker as a resistor:
V
+
V
R1
-
R2
-
Case 1
V 2 p2 R1

,
Hence p 
R p1 R 2
+
Case 2
p2 
R1
10
p1  (12)  30 W
R2
4
Chapter 2, Solution 81
Let R 1 and R 2 be in k.
R eq  R 1  R 2 5
(1)
5 R2
V0

VS 5 R 2  R 1
(2)
From (1) and (2), 0.05 
From (1), 40 = R 1 + 2
5 R1
40
2 = 5 R2 
5R 2
or R 2 = 3.333 k
5 R2
R 1 = 38 k
Thus,
R 1 = 38 k, R 2 = 3.333 k
Chapter 2, Solution 82
(a)
10 
40 
10 
80 
1
2
R 12
R 12 = 80 + 10 (10  40)  80 
50
 88.33 
6
(b)
3
10 
10 
20 
40 
R 13
80 
1
R 13 = 80 + 10 (10  40)  20  100  10 50  108.33 
4
(c)
20 
10 
R 14
10 
80 
1
R 14 = 80  0 (10  40  10)  20  80  0  20  100 
40 
Chapter 2, Solution 83
The voltage across the fuse should be negligible when compared with 24
V (this can be checked later when we check to see if the fuse rating is
exceeded in the final circuit). We can calculate the current through the
devices.
p1 45mW

 5mA
V1
9V
p
480mW
I2 = 2 
 20mA
V2
24
I1 =
i 2 = 20 mA
I fuse
i R1
24 V
+
-
R1
i 1 = 5 mA
R2
i R2
Let R 3 represent the resistance of the first device, we can solve for its value from
knowing the voltage across it and the current through it.
R 3 = 9/0.005 = 1,800 Ω
This is an interesting problem in that it essentially has two unknowns, R 1 and R 2 but only
one condition that need to be met and that the voltage across R 3 must equal 9 volts.
Since the circuit is powered by a battery we could choose the value of R 2 which draws
the least current, R 2 = ∞. Thus we can calculate the value of R 1 that give 9 volts across
R3.
9 = (24/(R 1 + 1800))1800 or R 1 = (24/9)1800 – 1800 = 3 kΩ
This value of R 1 means that we only have a total of 25 mA flowing out of the battery
through the fuse which means it will not open and produces a voltage drop across it of
0.05V. This is indeed negligible when compared with the 24-volt source.
Chapter 3, Solution 1
Using Fig. 3.50, design a problem to help other students to better understand
nodal analysis.
R1
R2
Ix
12 V
+

R3
9V
+

Figure 3.50
For Prob. 3.1 and Prob. 3.39.
Solution
Given R 1 = 4 kΩ, R 2 = 2 kΩ, and R 3 = 2 kΩ, determine the value of I x using
nodal analysis.
Let the node voltage in the top middle of the circuit be designated as V x .
[(V x –12)/4k] + [(V x –0)/2k] + [(V x –9)/2k] = 0 or (multiply this by 4 k)
(1+2+2)V x = 12+18 = 30 or V x = 30/5 = 6 volts and
I x = 6/(2k) = 3 mA.
Chapter 3, Solution 2
At node 1,
v  v2
 v1 v1

 6 1
10
5
2
60 = - 8v 1 + 5v 2
(1)
At node 2,
v2
v  v2
 36 1
4
2
Solving (1) and (2),
v 1 = 0 V, v 2 = 12 V
36 = - 2v 1 + 3v 2
(2)
Chapter 3, Solution 3
Applying KCL to the upper node,
8
v0 vo vo
v


 20  0 = 0 or v 0 = –60 V
10 20 30
60
v0
v
 –6 A , i 2 = 0  –3 A,
10
20
v0
v
i3 =
 –2 A, i 4 = 0  1 A.
30
60
i1 =
Chapter 3, Solution 4
3A
v1
i1
20 
6A
i2
10 
v2
i3
40 
i4
40 
At node 1,
–6 – 3 + v 1 /(20) + v 1 /(10) = 0 or v 1 = 9(200/30) = 60 V
At node 2,
3 – 2 + v 2 /(10) + v 2 /(5) = 0 or v 2 = –1(1600/80) = –20 V
i 1 = v 1 /(20) = 3 A, i 2 = v 1 /(10) = 6 A,
i 3 = v 2 /(40) = –500 mA, i 4 = v 2 /(40) = –500 mA.
2A
Chapter 3, Solution 5
Apply KCL to the top node.
30  v 0 20  v 0 v 0


2k
5k
4k
v 0 = 20 V
Chapter 3, Solution 6.
Solve for V 1 using nodal analysis.
10 
10 V
–
+
5
10 
+
V1
10 
20 V
+

–
Figure 3.55
For Prob. 3.6.
Step 1.
The first thing to do is to select a reference node and to identify all the
unknown nodes. We select the bottom of the circuit as the reference node. The only
unknown node is the one connecting all the resistors together and we will call that node
V 1 . The other two nodes are at the top of each source. Relative to the reference, the one
at the top of the 10-volt source is –10 V. At the top of the 20-volt source is +20 V.
Step 2.
unknown).
Setup the nodal equation (there is only one since there is only one
Step 3.
Simplify and solve.
or
V 1 = –2 V.
The answer can be checked by calculating all the currents and see if they add up to zero.
The top two currents on the left flow right to left and are 0.8 A and 1.6 A respectively.
The current flowing up through the 10-ohm resistor is 0.2 A. The current flowing right to
left through the 10-ohm resistor is 2.2 A. Summing all the currents flowing out of the
node, V 1 , we get, +0.8+1.6 –0.2–2.2 = 0. The answer checks.
Chapter 3, Solution 7
2
Vx  0 Vx  0

 0.2Vx  0
10
20
0.35V x = 2 or V x = 5.714 V.
Substituting into the original equation for a check we get,
0.5714 + 0.2857 + 1.1428 = 1.9999 checks!
Chapter 3, Solution 8
6
i1
v1
i3
20
i2
+
V0
60V
4
+
–
+ 5V 0
–
–
20
i1 + i2 + i3 = 0
But
v1 ( v1  60)  0 v1  5v 0


0
10
20
20
2
v1 so that 2v 1 + v 1 – 60 + v 1 – 2v 1 = 0
5
or v 1 = 60/2 = 30 V, therefore v o = 2v 1 /5 = 12 V.
v0 
Chapter 3, Solution 9
Let V 1 be the unknown node voltage to the right of the 250-Ω resistor. Let the ground
reference be placed at the bottom of the 50-Ω resistor. This leads to the following nodal
equation:
V1  24 V1  0 V1  60I b  0


0
250
50
150
simplifying we get
3V1  72  15V1  5V1  300I b  0
But I b 
24  V1
. Substituting this into the nodal equation leads to
250
24.2V1  100.8  0 or V 1 = 4.165 V.
Thus,
I b = (24 – 4.165)/250 = 79.34 mA.
Chapter 3, Solution 10
v1
v2
v3
At node 1.
[(v 1 –0)/8] + [(v 1 –v 3 )/1] + 4 = 0
At node 2.
–4 + [(v 2 –0)/2] + 2i o = 0
At node 3.
–2i o + [(v 3 –0)/4] + [(v 3 –v 1 )/1] = 0
Finally, we need a constraint equation,
i o = v 1 /8
This produces,
1.125v 1 – v 3 = 4
(1)
0.25v 1 + 0.5v 2 = 4
(2)
–1.25v 1 + 1.25v 3 = 0 or v 1 = v 3
(3)
Substituting (3) into (1) we get (1.125–1)v 1 = 4 or v 1 = 4/0.125 = 32 volts. This leads to,
i o = 32/8 = 4 amps.
Chapter 3, Solution 11
Find V o and the power absorbed by all the resistors in the circuit of Fig. 3.60.
12 
60 V
Vo
+
_
12 
6
–
+
24 V
Figure 3.60
For Prob. 3.11.
Solution
At the top node, KCL produces
Vo  60 Vo  0 Vo  (24)


0
12
12
6
(1/3)V o = 1 or V o = 3 V.
P 12Ω = (3–60)2/1 = 293.9 W (this is for the 12 Ω resistor in series with the 60 V
source)
P 12Ω = (V o )2/12 = 9/12 = 750 mW (this is for the 12 Ω resistor connecting V o to
ground)
P 4Ω = (3–(–24))2/6 = 121.5 W.
Chapter 3, Solution 12
There are two unknown nodes, as shown in the circuit below.
20 
10 
V1
Vo
Ix
40 V
At node 1,
At node o,
+
_
20 
10 
4 Ix
V1  40 V1  0 V1  Vo


 0 or
20
20
10
(0.05+0.05+.1)V 1 – 0.1V o = 0.2V 1 – 0.1V o = 2
(1)
Vo  V1
V 0
 4I x  o
 0 and I x = V 1 /20
10
10
–0.1V 1 – 0.2V 1 + 0.2V o = –0.3V 1 + 0.2V o = 0 or
(2)
V 1 = (2/3)V o
(3)
Substituting (3) into (1),
0.2(2/3)V o – 0.1V o = 0.03333V o = 2 or
V o = 60 V.
Chapter 3, Solution 13
Calculate v 1 and v 2 in the circuit of Fig. 3.62 using nodal analysis.
10 V
15 A
Figure 3.62
For Prob. 3.13.
Solution
At node number 2, [((v 2 + 10) – 0)/10] + [(v 2 –0)/4] – 15 = 0 or
(0.1+0.25)v 2 = 0.35v 2 = –1+15 = 14 or
v 2 = 40 volts.
Next, I = [(v 2 + 10) – 0]/10 = (40 + 10)/10 = 5 amps and
v 1 = 8x5 = 40 volts.
Chapter 3, Solution 14
Using nodal analysis, find v o in the circuit of Fig. 3.63.
12.5 A
8
2
1
+
4
vo
100 V
+
–
50 V
–
+

Figure 3.63
For Prob. 3.14.
12.5 A
Solution
v0
v1
1
2
+
4
vo
100 V
+
–
8
50 V
–
+

At node 1,
[(v 1 –100)/1] + [(v 1 –v o )/2] + 12.5 = 0 or 3v 1 – v o = 200–25 = 175
At node o,
[(v o –v 1 )/2] – 12.5 + [(v o –0)/4] + [(v o +50)/8] = 0 or –4v 1 + 7v o = 50
(2)
(1)
4(1) + 3(2) = –4v o + 21v o = 700 + 150 or 17v o = 850 or
v o = 50 V.
Checking, we get v 1 = (175+v o )/3 = 75 V.
At node 1,
[(75–100)/1] + [(75–50)/2] + 12.5 = –25 + 12.5 + 12.5 = 0!
At node o,
[(50–75)/2] + [(50–0)/4] + [(50+50)/8] – 12.5 = –12.5 + 12.5 + 12.5 – 12.5 = 0!
Chapter 3, Solution 15
5A
v0
v1
1
2
4
40 V
8
20 V
–
+
+
–
Nodes 1 and 2 form a supernode so that v 1 = v 2 + 10
At the supernode, 2 + 6v 1 + 5v 2 = 3 (v 3 - v 2 )
At node 3, 2 + 4 = 3 (v 3 - v 2 )
2 + 6v 1 + 8v 2 = 3v 3
v3 = v2 + 2
Substituting (1) and (3) into (2),
2 + 6v 2 + 60 + 8v 2 = 3v 2 + 6
v 1 = v 2 + 10 =
v2 =
54
11
i 0 = 6v i = 29.45 A
2
P 65 =
v12
 54 
 v12 G    6  144.6 W
R
 11 
  56 
= v G
 5  129.6 W
 11 
2
P 55
2
2
P 35 = v L  v 3  G  (2) 2 3  12 W
2
(1)
 56
11
(2)
(3)
Chapter 3, Solution 16
2S
v2
v1
i0
2A
+
1S
v0
4S
8S
v3
13 V
–
+
–
At the supernode,
2 = v 1 + 2 (v 1 - v 3 ) + 8(v 2 – v 3 ) + 4v 2 , which leads to 2 = 3v 1 + 12v 2 - 10v 3 (1)
But
v 1 = v 2 + 2v 0 and v 0 = v 2 .
Hence
v 1 = 3v 2
v 3 = 13V
Substituting (2) and (3) with (1) gives,
v 1 = 18.858 V, v 2 = 6.286 V, v 3 = 13 V
(2)
(3)
Chapter 3, Solution 17
v1
i0
4
2
10 
v2
60 V
60 V
+
8
3i 0
–
60  v1 v1 v1  v 2


4
8
2
60  v 2 v1  v 2
At node 2, 3i 0 +

0
10
2
At node 1,
But i 0 =
120 = 7v 1 - 4v 2
(1)
60  v1
.
4
Hence
360  v1  60  v 2 v1  v 2


0
4
10
2
1020 = 5v 1 + 12v 2
Solving (1) and (2) gives v 1 = 53.08 V. Hence i 0 =
60  v1
 1.73 A
4
(2)
Chapter 3, Solution 18
–+
v2
v1
2
15A
v3
2
8
4
30 V
+
+
v1
v3
–
–
(a)
(b)
v 2  v1 v 2  v 3

–15 = 0 or –0.5v 1 + v 2 – 0.5v 3 = 15
(1)
2
2
v  v1 v 2  v 3 v1 v 3
At the supernode, 2

  = 0 and (v 1 /4) – 15 + (v 3 /8) = 0 or
2
2
4 8
2v 1 +v 3 = 120
(2)
At node 2, in Fig. (a),
From Fig. (b), – v 1 – 30 + v 3 = 0 or v 3 = v 1 + 30
Solving (1) to (3), we obtain,
v 1 = 30 V, v 2 = 60 V = v 3
(3)
Chapter 3, Solution 19
At node 1,
V1  V3 V1  V2 V1


2
8
4
At node 2,
5  3
V1  V2 V2 V2  V3


8
2
4
At node 3,
12  V3




0  V1  7V2  2V3
V1  V3 V2  V3

0
8
2
4
From (1) to (3),
3

 7  1  4  V1   16 

  

  1 7  2 V2    0 
4
2  7  V3    36 

Using MATLAB,
 10 
V  A 1 B   4.933 
12.267 
16  7V1  V2  4V3






(1)
(2)
 36  4V1  2V2  7V3 (3)
AV  B
V1  10 V, V2  4.933 V, V3  12.267 V
Chapter 3, Solution 20
Nodes 1 and 2 form a supernode; so do nodes 1 and 3. Hence
V1 V2 V3


0

 V1  4V2  V3  0
(1)
4
1
4
.
V1
V2
.
4
2
1
Between nodes 1 and 3,
 V1  12  V3  0

 V3  V1  12
Similarly, between nodes 1 and 2,
V1  V2  2i
But i  V3 / 4 . Combining this with (2) and (3) gives
. V2
V3
 6  V1 / 2
4
(2)
(3)
(4)
Solving (1), (2), and (4) leads to
V1  3V, V2  4.5V, V3  15V
Chapter 3, Solution 21
4 k
v1
2 k
v3
3v 0
+
3v 0
v2
+
v0
3 mA
–
1 k
+
+
+
v3
v2
–
–
(b)
(a)
Let v 3 be the voltage between the 2k resistor and the voltage-controlled voltage source.
At node 1,
v  v 2 v1  v 3
3x10 3  1
12 = 3v 1 - v 2 - 2v 3
(1)

4000
2000
At node 2,
v1  v 2 v1  v 3 v 2


4
2
1
3v 1 - 5v 2 - 2v 3 = 0
(2)
Note that v 0 = v 2 . We now apply KVL in Fig. (b)
- v 3 - 3v 2 + v 2 = 0
From (1) to (3),
v 1 = 1 V, v 2 = 3 V
v 3 = - 2v 2
(3)
Chapter 3, Solution 22
At node 1,
12  v 0 v1
v  v0

3 1
2
4
8
At node 2, 3 +
24 = 7v 1 - v 2
(1)
v 1  v 2 v 2  5v 2

8
1
But, v 1 = 12 - v 1
Hence, 24 + v 1 - v 2 = 8 (v 2 + 60 + 5v 1 ) = 4 V
456 = 41v 1 - 9v 2
Solving (1) and (2),
v 1 = - 10.91 V, v 2 = - 100.36 V
(2)
Chapter 3, Solution 23
We apply nodal analysis to the circuit shown below.
1
30 V
+
_
4
Vo
2
2 Vo
+
+
Vo
V1
–
16 
_
3A
At node o,
Vo  30 Vo  0 Vo  (2Vo  V1 )
 0  1.25Vo  0.25V1  30


1
2
4
(1)
At node 1,
(2Vo  V1 )  Vo V1  0

 3  0  5V1  4Vo  48
4
16
From (1), V 1 = 5V o – 120. Substituting this into (2) yields
29V o = 648 or V o = 22.34 V.
(2)
Chapter 3, Solution 24
Consider the circuit below.
8
+ Vo _
4A
V1
1
4
V2
2A
V3
2
V4
2
V1  0
V  V4
4 1
 0  1.125V1  0.125V4  4
8
1
V  0 V2  V3
4 2

 0  0.75V2  0.25V3  4
2
4
V3  V2 V3  0

 2  0  0.25V2  0.75V3  2
4
2
V  V1 V4  0
2 4

 0  0.125V1  1.125V4  2
8
1
1
(1)
(2)
(3)
(4)
0
0
 0.125
 1.125
4
 0

  4
0.75  0.25
0 

V 
 0
  2
 0.25 0.75
0 


 
0
0
1.125 
 0.125
2
Now we can use MATLAB to solve for the unknown node voltages.
&gt;&gt; Y=[1.125,0,0,-0.125;0,0.75,-0.25,0;0,-0.25,0.75,0;-0.125,0,0,1.125]
Y=
1.1250
0
0 -0.1250
0 0.7500 -0.2500
0
0 -0.2500 0.7500
0
-0.1250
0
0 1.1250
&gt;&gt; I=[4,-4,-2,2]'
I=
4
-4
-2
2
&gt;&gt; V=inv(Y)*I
V=
3.8000
-7.0000
-5.0000
2.2000
V o = V 1 – V 4 = 3.8 – 2.2 = 1.6 V.
Chapter 3, Solution 25
Consider the circuit shown below.
20

4


10
1  
1
2

10



3 

30
8
4
20
At node 1.
V  V2 V1  V4
4 1


 80  21V1  20V2  V4
1
20
At node 2,
V1  V2 V2 V2  V3




0  80V1  98V2  8V3
1
8
10
At node 3,
V2  V3 V3 V3  V4



 0  2V2  5V3  2V4
10
20
10
At node 4,
V1  V4 V3  V4 V4



 0  3V1  6V3  11V4
20
10
30
Putting (1) to (4) in matrix form gives:
(1)
(2)
(3)
(4)
1   V1 
 80   21 20 0
  
 
 0    80 98 8 0  V2 
0  0
2 5 2  V3 
  
 
0
6 11 V4 
0  3
B =A V
V = A-1 B
V 1 = 25.52 V,
V 2 = 22.05 V, V 3 = 14.842 V, V 4 = 15.055 V
Chapter 3, Solution 26
At node 1,
V  V3 V1  V2
15  V1
 3 1



 45  7V1  4V2  2V3
20
10
5
At node 2,
V1  V2 4 I o  V2 V2  V3


5
5
5
V1  V3
But I o 
. Hence, (2) becomes
10
0  7V1  15V2  3V3
At node 3,
V  V3  10  V3 V2  V3
3 1


0

 70  3V1  6V2  11V3
10
15
5
Putting (1), (3), and (4) in matrix form produces
 7  4  2  V1    45 

  

 7  15 3  V2    0 
  3  6 11  V   70 

 3  

  7.19 


V  A 1B    2.78 
 2.89 




AV  B
Thus,
V 1 = –7.19V; V 2 = –2.78V; V 3 = 2.89V.
(1)
(2)
(3)
(4)
Chapter 3, Solution 27
At node 1,
2 = 2v 1 + v 1 – v 2 + (v 1 – v 3 )4 + 3i 0 , i 0 = 4v 2 . Hence,
2 = 7v 1 + 11v 2 – 4v 3
(1)
At node 2,
v 1 – v 2 = 4v 2 + v 2 – v 3
0 = – v 1 + 6v 2 – v 3
(2)
At node 3,
2v 3 = 4 + v 2 – v 3 + 12v 2 + 4(v 1 – v 3 )
– 4 = 4v 1 + 13v 2 – 7v 3
or
(3)
In matrix form,
7 11  4  v 1   2 
1  6 1   v    0 
 2   

4 13  7  v 3    4
7
11
  1 6
4
13
7
2
4
2
11
1  176,  1  0
6
7
4
4
7
2  1 0
1  66,
4 4 7
v1 =
13
11
4
1  110
7
2
 3  1  6 0  286
4 13  4
 1 110

66

 0.625V, v 2 = 2 
 0.375V


176
176
v3 =
3
286

 1.625V.
176

v 1 = 625 mV, v 2 = 375 mV, v 3 = 1.625 V.
Chapter 3, Solution 28
At node c,
V d  V c V c  Vb V c



 0  5Vb  11Vc  2Vd
(1)
10
4
5
At node b,
Va  90  Vb Vc  Vb Vb




 90  Va  4Vb  2Vc (2)
8
4
8
At node a,
Va  60  Vd Va Va  90  Vb


0


60  7Va  2Vb  4Vd (3)
4
16
8
At node d,
Va  60  Vd Vd Vd  Vc



 300  5Va  2Vc  8Vd (4)
4
20
10
In matrix form, (1) to (4) become
 0  5 11  2  Va   0 

  

 1  4 2 0  Vb    90 


AV  B
 7  2 0  4  V    60 
c

  

 5 0 2  8  V   300 

 d  

We use MATLAB to invert A and obtain
  10.56 


 20.56 
1
VA B
1.389 


  43.75 


Thus,
V a = –10.56 V; V b = 20.56 V; V c = 1.389 V; VC d = –43.75 V.
Chapter 3, Solution 29
At node 1,
5  V1  V4  2V1  V1  V2  0


 5  4V1  V2  V4
At node 2,
V1  V2  2V2  4(V2  V3 )  0

 0  V1  7V2  4V3
At node 3,
6  4(V2  V3 )  V3  V4

 6  4V2  5V3  V4
At node 4,
2  V3  V4  V1  V4  3V4

 2  V1  V3  5V4
In matrix form, (1) to (4) become
 4  1 0  1 V1    5 
   

  1 7  4 0 V2   0 
AV  B


 0  4 5  1 V    6 
3
   

  1 0  1 5 V   2 
 4   

Using MATLAB,
(1)
(2)
(3)
(4)
  0.7708 


 1.209 
1
V  A B
2.309 


 0.7076 


i.e.
V1  0.7708 V, V 2  1.209 V, V 3  2.309 V, V 4  0.7076 V
Chapter 3, Solution 30
i0
v1
10 
+
80 V
–+
96 V
20 
v0
1
2
4v 0
–
40 
+
–
2i 0
80 
At node 1,
[(v 1 –80)/10]+[(v 1 –4v o )/20]+[(v 1 –(v o –96))/40] = 0 or
(0.1+0.05+0.025)v 1 – (0.2+0.025)v o =
0.175v 1 – 0.225v o = 8–2.4 = 5.6
(1)
At node 2,
–2i o + [((v o –96)–v 1 )/40] + [(v o –0)/80] = 0 and i o = [(v 1 –(v o –96))/40]
–2[(v 1 –(v o –96))/40] + [((v o –96)–v 1 )/40] + [(v o –0)/80] = 0
–3[(v 1 –(v o –96))/40] + [(v o –0)/80] = 0 or
–0.0.075v 1 + (0.075+0.0125)v o = 7.2 =
–0.075v 1 + 0.0875v o = 7.2
(2)
Using (1) and (2) we get,
 0.175  0.225  v1  5.6
 0.075 0.0875  v   7.2 or

 o   
0.0875 0.225
0.0875 0.225


 v1 
0.075 0.175 5.6  0.075 0.175 5.6

v   0.0153125  0.016875 7.2   0.0015625 7.2
 
 
 o
v 1 = –313.6–1036.8 = –1350.4
v o = –268.8–806.4 = –1.0752 kV
and i o = [(v 1 –(v o –96))/40] = [(–1350.4–(–1075.2–96))/40] = –4.48 amps.
Chapter 3, Solution 31
1
+ v0 –
v2
v1
2v 0
v3
2
i0
1A
4
1
10 V
4
–
At the supernode,
1 + 2v 0 =
v1 v 2 v1  v 3


4
1
1
(1)
But v o = v 1 – v 3 . Hence (1) becomes,
4 = -3v 1 + 4v 2 +4v 3
(2)
At node 3,
2v o +
or
v3
10  v 3
 v1  v 3 
4
2
20 = 4v 1 + 0v 2 – v 3
At the supernode, v 2 = v 1 + 4i o . But i o =
(3)
v3
. Hence,
4
v2 = v1 + v3
Solving (2) to (4) leads to,
v 1 = 4.97V, v 2 = 4.85V, v 3 = –0.12V.
+
(4)
Chapter 3, Solution 32
5 k
v1
v3
v2
+
10 k
v1
10 V
20 V
–+
+–
12 V
–
4 mA
+
loop 1
+
loop 2
–
v3
–
(b)
(a)
We have a supernode as shown in figure (a). It is evident that v 2 = 12 V, Applying KVL
to loops 1and 2 in figure (b), we obtain,
-v 1 – 10 + 12 = 0 or v 1 = 2 and -12 + 20 + v 3 = 0 or v 3 = -8 V
Thus,
v 1 = 2 V, v 2 = 12 V, v 3 = -8V.
Chapter 3, Solution 33
(a) This is a planar circuit. It can be redrawn as shown below.
5
3
1
2
4
6
2A
(b) This is a planar circuit. It can be redrawn as shown below.
4
3
5
12 V
+
–
1
2
Chapter 3, Solution 34
(a)
This is a planar circuit because it can be redrawn as shown below,
7
2
1
3
6
10 V
5
+
–
4
(b)
This is a non-planar circuit.
Chapter 3, Solution 35
30 V
20 V
+
–
+
–
i1
2 k
+
i2
v0
4 k
–
5 k
Assume that i 1 and i 2 are in mA. We apply mesh analysis. For mesh 1,
-30 + 20 + 7i 1 – 5i 2 = 0 or 7i 1 – 5i 2 = 10
(1)
For mesh 2,
-20 + 9i 2 – 5i 1 = 0 or -5i 1 + 9i 2 = 20
Solving (1) and (2), we obtain, i 2 = 5.
v 0 = 4i 2 = 20 volts.
(2)
Chapter 3, Solution 36
10 V
4
+–
i1
12 V
i2
+
I1
–
6
i3
I2
2
Applying mesh analysis gives,
10I 1 – 6I 2 = 12 and –6I 1 + 8I 2 = –10
or
4 3


 6   5  3  I1   6   I1  3 5  6 
 5   3 4  I    5 or I   11  5
 
 2     2 
  
I 1 = (24–15)/11 = 0.8182 and I 2 = (18–25)/11 = –0.6364
i 1 = –I 1 = –818.2 mA; i 2 = I 1 – I 2 = 0.8182+0.6364 = 1.4546 A; and
i 3 = I 2 = –636.4 mA.
Chapter 3, Solution 37
6
60 V +
+
v0
20
4
–
i1
i2
–
5v 0
+
–
20
Applying mesh analysis to loops 1 and 2, we get,
30i 1 – 20i 2 + 60 = 0 which leads to i 2 = 1.5i 1 + 3
(1)
–20i 1 + 40i 2 – 60 + 5v 0 = 0
(2)
But, v 0 = –4i 1
(3)
Using (1), (2), and (3) we get –20i 1 + 60i 1 + 120 – 60 – 20i 1 = 0 or
20i 1 = –60 or i 1 = –3 amps and i 2 = 7.5 amps.
Therefore, we get,
v 0 = –4i 1 = 12 volts.
Chapter 3, Solution 38
Consider the circuit below with the mesh currents.
4
+
_
60 V
3
I3
I4
1
10 A
2
2
Io
1
1
I1
I2
+
_
22.5V
4
5A
I 1 = –5 A
(1)
1(I 2 –I 1 ) + 2(I 2 –I 4 ) + 22.5 + 4I 2 = 0
7I 2 – I 4 = –27.5
(2)
–60 + 4I 3 + 3I 4 + 1I 4 + 2(I 4 –I 2 ) + 2(I 3 – I 1 ) = 0 (super mesh)
–2I 2 + 6 I 3 + 6I 4 = +60 – 10 = 50
(3)
But, we need one more equation, so we use the constraint equation –I 3 + I 4 = 10. This
now gives us three equations with three unknowns.
0  1 I 2   27.5
7
 2 6 6   I    50 

 3  

 0  1 1  I 4   10 
We can now use MATLAB to solve the problem.
&gt;&gt; Z=[7,0,-1;-2,6,6;0,-1,0]
Z=
7 0 -1
-2 6 6
0 -1 0
&gt;&gt; V=[–27.5,50,10]'
V=
–27.5
50
10
&gt;&gt; I=inv(Z)*V
I=
–1.3750
–10.0000
17.8750
I o = I 1 – I 2 = –5 – 1.375 = –6.375 A.
Check using the super mesh (equation (3)):
–2I 2 + 6 I 3 + 6I 4 = 2.75 – 60 + 107.25 = 50!
Chapter 3, Solution 39
Using Fig. 3.50 from Prob. 3.1, design a problem to help other students to better
understand mesh analysis.
Solution
Given R 1 = 4 kΩ, R 2 = 2 kΩ, and R 3 = 2 kΩ, determine the value of I x using
mesh analysis.
R1
R2
Ix
12 V
+

I2
I1
R3
9V
+

Figure 3.50
For Prob. 3.1 and 3.39.
For loop 1 we get –12 +4kI 1 + 2k(I 1 –I 2 ) = 0 or 6I 1 – 2I 2 = 0.012 and at
loop 2 we get 2k(I 2 –I 1 ) + 2kI 2 + 9 = 0 or –2I 1 + 4I 2 = –0.009.
Now 6I 1 – 2I 2 = 0.012 + 3[–2I 1 + 4I 2 = –0.009] leads to,
10I 2 = 0.012 – 0.027 = –0.015 or I 2 = –1.5 mA and I 1 = (–0.003+0.012)/6 = 1.5
mA.
Thus,
I x = I 1 –I 2 = (1.5+1.5) mA = 3 mA.
Chapter 3, Solution 40
2 k
6 k
6 k
56V
+
–
i2
2 k
i1
i3
4 k
4 k
Assume all currents are in mA and apply mesh analysis for mesh 1.
–56 + 12i 1 – 6i 2 – 4i 3 = 0 or 6i 1 – 3i 2 – 2i 3 = 28
(1)
for mesh 2,
–6i 1 + 14i 2 – 2i 3 = 0 or –3i 1 + 7i 2 – i 3
=0
(2)
=0
(3)
for mesh 3,
–4i 1 – 2i 2 + 10i 3 = 0 or –2i 1 – i 2 + 5i 3
Solving (1), (2), and (3) using MATLAB, we obtain,
i o = i 1 = 8 mA.
Chapter 3, Solution 41
10 
i1
6V
2
+–
1
i2
4
5
i3
8V
+
–
i
i2
i3
0
For loop 1,
6 = 12i 1 – 2i 2
3 = 6i 1 – i 2
(1)
For loop 2,
-8 = – 2i 1 +7i 2 – i 3
(2)
For loop 3,
-8 + 6 + 6i 3 – i 2 = 0
2 = – i 2 + 6i 3
We put (1), (2), and (3) in matrix form,
6  1 0  i1   3
 2  7 1  i    8 

 2   
0  1 6 i 3  2
6
1 0
6 3 0
  2  7 1  234,  2  2 8 1  240
0
1 6
0 2 6
(3)
6
1 3
 3  2  7 8  38
0 1 2
At node 0, i + i 2 = i 3 or i = i 3 – i 2 =
3  2
 38  240

= 1.188 A

 234
Chapter 3, Solution 42
Although there are many ways to work this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
Determine the mesh currents in the circuit of Fig. 3.88.
Figure 3.88
Solution
For mesh 1,
 12  50I 1  30I 2  0

 12  50I1  30I 2
For mesh 2,
 8  100 I 2  30 I 1  40 I 3  0


8  30 I 1  100 I 2  40 I 3
For mesh 3,
 6  50 I 3  40 I 2  0


6  40 I 2  50 I 3
Putting eqs. (1) to (3) in matrix form, we get
0  I 1  12 
 50  30
   

  30 100  40  I 2    8 
 0
 40 50  I 3   6 



AI  B
Using Matlab,
 0.48 


I  A B   0.40 
 0.44 


1
i.e. I 1 = 480 mA, I 2 = 400 mA, I 3 = 440 mA
(1)
(2)
(3)
Chapter 3, Solution 43
20 
a
80 V
+
i1
–
30 
+
30 
i3
20 
80 V
+
i2
–
30 
20 
V ab
–
b
For loop 1,
80 = 70i 1 – 20i 2 – 30i 3
8 = 7i 1 – 2i 2 – 3i 3
(1)
80 = 70i 2 – 20i 1 – 30i 3
8 = -2i 1 + 7i 2 – 3i 3
(2)
0 = -30i 1 – 30i 2 + 90i 3
0 = i 1 + i 2 – 3i 3
For loop 2,
For loop 3,
(3)
Solving (1) to (3), we obtain i 3 = 16/9
I o = i 3 = 16/9 = 1.7778 A
V ab = 30i 3 = 53.33 V.
Chapter 3, Solution 44
90 V
+
2
4
i3
i2
1
180V +
–
5
i1
45 A
i1
i2
Loop 1 and 2 form a supermesh. For the supermesh,
6i 1 + 4i 2 – 5i 3 + 180 = 0
(1)
For loop 3,
–i 1 – 4i 2 + 7i 3 + 90 = 0
(2)
Also,
i 2 = 45 + i 1
(3)
Solving (1) to (3), i 1 = –46, i 3 = –20;
i o = i 1 – i 3 = –26 A
Chapter 3, Solution 45
4
30V
+
–
8
i3
i4
2
6
i1
3
i2
1
For loop 1,
30 = 5i 1 – 3i 2 – 2i 3
(1)
For loop 2,
10i 2 - 3i 1 – 6i 4 = 0
(2)
For the supermesh,
6i 3 + 14i 4 – 2i 1 – 6i 2 = 0
(3)
But
i 4 – i 3 = 4 which leads to i 4 = i 3 + 4
Solving (1) to (4) by elimination gives i = i 1 = 8.561 A.
(4)
Chapter 3, Solution 46
For loop 1,
 12  3i1  8(i1  i 2 )  12  11i1  8i 2  0
For loop 2,
8(i 2  i1 )  6i 2  2v o  8i1  14i 2  2v o  0
But vo  3i1 ,


11i1  8i 2  12
 8i1  14i2  6i1  0

 i1  7i2
Substituting (2) into (1),
77i2  8i2  12

 i 2  0.1739 A and i1  7i2  1.217 A
(1)
(2)
Chapter 3, Solution 47
First, transform the current sources as shown below.
- 6V +
2
8
V1
4
V2
I3
V3
4
2
I1
8
I2
+
20V
-
+
12V
-
For mesh 1,
 20  14 I 1  2 I 2  8I 3  0
For mesh 2,
10  7 I 1  I 2  4 I 3
(1)
 6   I1  7 I 2  2I 3
(2)
 6  14 I 3  4 I 2  8I 1  0

 3  4 I 1  2 I 2  7 I 3
Putting (1) to (3) in matrix form, we obtain
(3)
12  14 I 2  2 I 1  4 I 3  0
For mesh 3,




 7  1  4  I 1   10 
   

  1 7  2  I 2     6 
  4  2 7  I   3 
 3   

Using MATLAB,
 2 
I  A 1 B  0.0333
1.8667 
But


AI  B

 I 1  2.5, I 2  0.0333, I 3  1.8667
I1 
20  V
4


V1  20  4 I 1  10 V
V 2  2( I 1  I 2 )  4.933 V
Also,
I2 
V3  12
8


V3  12  8I 2  12.267 V.
Chapter 3, Solution 48
We apply mesh analysis and let the mesh currents be in mA.
3k 
I4
4k 
2k 
Io
1k 
I3
I2
I1
+
6V
-
5k 
+
4V
-
10k 
For mesh 1,
 6  8  5I1  I 2  4I 4  0

 2  5I1  I 2  4I 4
For mesh 2,
 4  13I 2  I1  10I 3  2I 4  0

 4  I1  13I 2  10I 3  2I 4
For mesh 3,
 3  15I 3  10I 2  5I 4  0

 3  10I 2  15I 3  5I 4
For mesh 4,
 4 I 1  2 I 2  5I 3  14 I 4  0
Putting (1) to (4) in matrix form gives
1
 4  I1   2 
0
 5

   
  1 13  10  2  I 2   4 

 AI B
 0  10 15  5  I    3 
3

   
  4  2  5 14  I   0 

 4   
Using MATLAB,
 3.608 


 4.044 
1
IA B
0.148
3.896 


 3 


The current through the 10k  resistor is I o = I 2 – I 3 = 148 mA.
3V
+
(1)
(2)
(3)
(4)
Chapter 3, Solution 49
3
i3
2
1
2
i1
27 V
i2
+
–
2i 0
i1
i2
0
(a)
2
1
2
i1
+
+
v0
or
v0
–
i2
27V
+
–
(b)
For the supermesh in figure (a),
3i 1 + 2i 2 – 3i 3 + 27 = 0
(1)
At node 0,
i 2 – i 1 = 2i 0 and i 0 = –i 1 which leads to i 2 = –i 1
(2)
For loop 3,
–i 1 –2i 2 + 6i 3 = 0 which leads to 6i 3 = –i 1
(3)
Solving (1) to (3), i 1 = (–54/3)A, i 2 = (54/3)A, i 3 = (27/9)A
i 0 = –i 1 = 18 A, from fig. (b), v 0 = i 3 –3i 1 = (27/9) + 54 = 57 V.
Chapter 3, Solution 50
i1
4
2
i3
10 
8
35 V
+
–
i2
3i 0
i2
For loop 1,
i3
16i 1 – 10i 2 – 2i 3 = 0 which leads to 8i 1 – 5i 2 – i 3 = 0
(1)
For the supermesh, –35 + 10i 2 – 10i 1 + 10i 3 – 2i 1 = 0
or
–6i 1 + 5i 2 + 5i 3 = 17.5
Also, 3i 0 = i 3 – i 2 and i 0 = i 1 which leads to 3i 1 = i 3 – i 2
Solving (1), (2), and (3), we obtain i 1 = 1.0098 and
i 0 = i 1 = 1.0098 A
(2)
(3)
Chapter 3, Solution 51
5A
i1
8
2
i3
1
+
i2
40 V
4
+
v0
20V
–
+

–
For loop 1,
i 1 = 5A
(1)
For loop 2,
-40 + 7i 2 – 2i 1 – 4i 3 = 0 which leads to 50 = 7i 2 – 4i 3
(2)
For loop 3,
-20 + 12i 3 – 4i 2 = 0 which leads to 5 = - i 2 + 3 i 3
(3)
Solving with (2) and (3),
And,
i 2 = 10 A, i 3 = 5 A
v 0 = 4(i 2 – i 3 ) = 4(10 – 5) = 20 V.
Chapter 3, Solution 52
+
v0 2 
i2
–
VS
+
–
8
3A
i2
i1
i3
4
i3
+
–
2V 0
For mesh 1,
2(i 1 – i 2 ) + 4(i 1 – i 3 ) – 12 = 0 which leads to 3i 1 – i 2 – 2i 3 = 6
(1)
For the supermesh, 2(i 2 – i 1 ) + 8i 2 + 2v 0 + 4(i 3 – i 1 ) = 0
But v 0 = 2(i 1 – i 2 ) which leads to -i 1 + 3i 2 + 2i 3 = 0
(2)
For the independent current source, i 3 = 3 + i 2
Solving (1), (2), and (3), we obtain,
i 1 = 3.5 A, i 2 = -0.5 A, i 3 = 2.5 A.
(3)
Chapter 3, Solution 53
–12 + 4kI 1 – 3kI 2 – 1kI 3 = 0
–3kI 1 + 7kI 2 – 4kI 4 = 0
–3kI 1 + 7kI 2 = –12
–1kI 1 + 15kI 3 – 8kI 4 – 6kI 5 = 0
–1kI 1 + 15kI 3 – 6k = –24
I 4 = –3mA
–6kI 3 – 8kI 4 + 16kI 5 = 0
–6kI 3 + 16kI 5 = –24
Putting these in matrix form (having substituted I 4
(1)
(2)
(3)
(4)
(5)
= 3mA in the above),
 4  3  1 0   I1   12 
 3 7
0
0  I 2    12 


k
  1 0 15  6  I 3   24

   

0  6 16   I 5   24
0
ZI = V
Using MATLAB,
&gt;&gt; Z = [4,-3,-1,0;-3,7,0,0;-1,0,15,-6;0,0,-6,16]
Z=
4 -3 -1 0
-3 7 0 0
-1 0 15 -6
0 0 -6 16
&gt;&gt; V = [12,-12,-24,-24]'
V=
12
-12
-24
-24
We obtain,
&gt;&gt; I = inv(Z)*V
I=
1.6196 mA
–1.0202 mA
–2.461 mA
3 mA
–2.423 mA
Chapter 3, Solution 54
Let the mesh currents be in mA. For mesh 1,
 12  10  2 I 1  I 2  0


2  2I1  I 2
For mesh 2,
 10  3I 2  I 1  I 3  0

 10   I 1  3I 2  I 3
For mesh 3,
 12  2 I 3  I 2  0

 12   I 2  2 I 3
Putting (1) to (3) in matrix form leads to
 2  1 0  I 1   2 
   

  1 3  1 I 2   10 
 0  1 2  I  12 
 3   

Using MATLAB,
 5.25 
I  A B   8.5 
10.25
1


AI  B

 I 1  5.25 mA, I 2  8.5 mA, I 3  10.25 mA
I 1 = 5.25 mA, I 2 = 8.5 mA, and I 3 = 10.25 mA.
(1)
(2)
(3)
Chapter 3, Solution 55
10 V
I2
b
i1
4A
c
+
1A
I2
6
1A
2
i2
i3
I4
4A
I3
d
I1
12 
4
+–
a
I3
0
I4
8V
It is evident that I 1 = 4
For mesh 4,
(2)
12(I 4 – I 1 ) + 4(I 4 – I 3 ) – 8 = 0
For the supermesh
At node c,
(1)
6(I 2 – I 1 ) + 10 + 2I 3 + 4(I 3 – I 4 ) = 0
or -3I 1 + 3I 2 + 3I 3 – 2I 4 = -5
I2 = I3 + 1
(3)
(4)
Solving (1), (2), (3), and (4) yields, I 1 = 4A, I 2 = 3A, I 3 = 2A, and I 4 = 4A
At node b,
i 1 = I 2 – I 1 = -1A
At node a,
i 2 = 4 – I 4 = 0A
At node 0,
i 3 = I 4 – I 3 = 2A
Chapter 3, Solution 56
+ v1 –
2
2
i2
2
2
12 V
+
i1
–
2
i3
+
v2
–
For loop 1, 12 = 4i 1 – 2i 2 – 2i 3 which leads to 6 = 2i 1 – i 2 – i 3
(1)
For loop 2, 0 = 6i 2 –2i 1 – 2 i 3 which leads to 0 = -i 1 + 3i 2 – i 3
(2)
For loop 3, 0 = 6i 3 – 2i 1 – 2i 2 which leads to 0 = -i 1 – i 2 + 3i 3
(3)
In matrix form (1), (2), and (3) become,
 2  1  1  i1  6
  1 3  1 i   0
 2   

  1  1 3  i 3  0
2
1 1
2
6 1
 =  1 3  1  8,  2 =  1 3  1  24
1 1 3
1 0 3
2
1 6
 3 =  1 3 0  24 , therefore i 2 = i 3 = 24/8 = 3A,
1 1 0
v 1 = 2i 2 = 6 volts, v = 2i 3 = 6 volts
Chapter 3, Solution 57
Assume R is in kilo-ohms.
V2  4kx15mA  60V,
V1  90  V2  90  60  30V
Current through R is
3
3

 30 
(15)R
iR 
i o,
V1  i R R
3 R
3 R
This leads to R = 90/15 = 6 kΩ.
Chapter 3, Solution 58
30 
i2
30 
10 
i1
10 
30 
i3
+
–
120 V
For loop 1, 120 + 40i 1 – 10i 2 = 0, which leads to -12 = 4i 1 – i 2
(1)
For loop 2, 50i 2 – 10i 1 – 10i 3 = 0, which leads to -i 1 + 5i 2 – i 3 = 0
For loop 3, -120 – 10i 2 + 40i 3 = 0, which leads to 12 = -i 2 + 4i 3
Solving (1), (2), and (3), we get, i 1 = -3A, i 2 = 0, and i 3 = 3A
(2)
(3)
Chapter 3, Solution 59
40 
–+
I0
96 V
i2
10 
20 
+
80V
i1
+
4v 0
–
i3
+
–
80 
v0
–
2I 0
i2
i3
For loop 1, -80 + 30i 1 – 20i 2 + 4v 0 = 0, where v 0 = 80i 3
or 4 = 1.5i 1 – i 2 + 16i 3
(1)
For the supermesh, 60i 2 – 20i 1 – 96 + 80i 3 – 4 v 0 = 0, where v 0 = 80i 3
or 4.8 = -i 1 + 3i 2 – 12i 3
(2)
Also, 2I 0 = i 3 – i 2 and I 0 = i 2 , hence, 3i 2 = i 3
(3)
From (1), (2), and (3),
3
 = 1
0
2
32
 3  2 32 
  1 3  12


3
 1 
 0
3
8
 i1   8 
i   4.8
 2  
i3   0 
32
3
2
3
 12  5,  2 =  1 4.8  12  22.4,  3 =  1
3
3
1
3
0
0
1
I 0 = i 2 =  2 / = -28/5 = –4.48 A
v 0 = 8i 3 = (-84/5)80 = –1.0752 kvolts
0
8
4.8  67.2
0
Chapter 3, Solution 60
0.5i 0
4
56 V
8
v1
v2
1
56 V
+
2
–
i0
At node 1, [(v 1 –0)/1] + [(v 1 –56)/4] + 0.5[(v 1 –0)/1] = 0 or 1.75v 1 = 14 or v 1 = 8 V
At node 2, [(v 2 –56)/8] – 0.5[8/1] + [(v 2 –0)/2] = 0 or 0.625v 2 = 11 or v 2 = 17.6 V
P 1 = (v 1 )2/1 = 64 watts, P 2 = (v 2 )2/2 = 154.88 watts,
P 4 = (56 – v 1 )2/4 = 576 watts, P 8 = (56 – v 2 )2/8 = 1.84.32 watts.
Chapter 3, Solution 61
v1
20 
v2
10 
i0
is
+
v0
–
30 
–
+ 5v 0
At node 1, i s = (v 1 /30) + ((v 1 – v 2 )/20) which leads to 60i s = 5v 1 – 3v 2
But v 2 = -5v 0 and v 0 = v 1 which leads to v 2 = -5v 1
Hence, 60i s = 5v 1 + 15v 1 = 20v 1 which leads to v 1 = 3i s , v 2 = -15i s
i 0 = v 2 /50 = -15i s /50 which leads to i 0 /i s = -15/50 = –0.3
40 
(1)
Chapter 3, Solution 62
4 k
100V +
–
A
i1
8 k
i2
B
2 k
i3
+
–
40 V
We have a supermesh. Let all R be in k, i in mA, and v in volts.
For the supermesh, -100 +4i 1 + 8i 2 + 2i 3 + 40 = 0 or 30 = 2i 1 + 4i 2 + i 3 (1)
At node A,
i1 + 4 = i2
(2)
At node B,
i 2 = 2i 1 + i 3
(3)
Solving (1), (2), and (3), we get i 1 = 2 mA, i 2 = 6 mA, and i 3 = 2 mA.
Chapter 3, Solution 63
10 
A
5
50 V
+
–
i1
i2
+
–
4i x
For the supermesh, -50 + 10i 1 + 5i 2 + 4i x = 0, but i x = i 1 . Hence,
50 = 14i 1 + 5i 2
At node A, i 1 + 3 + (v x /4) = i 2 , but v x = 2(i 1 – i 2 ), hence, i 1 + 2 = i 2
Solving (1) and (2) gives i 1 = 2.105 A and i 2 = 4.105 A
v x = 2(i 1 – i 2 ) = –4 volts and i x = i 2 – 2 = 2.105 amp
(1)
(2)
Chapter 3, Solution 64
i1
50 
A
i0
i1
i 2 10 
+

v0
10 
i2
+
–
4i 0
i3
40 
250V +
–
5A
0.2V 0
i1
For mesh 2,
B
i3
20i 2 – 10i 1 + 4i 0 = 0
(1)
But at node A, i o = i 1 – i 2 so that (1) becomes i 1 = (16/6)i 2
(2)
For the supermesh, –250 + 50i 1 + 10(i 1 – i 2 ) – 4i 0 + 40i 3 = 0
or
28i 1 – 3i 2 + 20i 3 = 125
(3)
At node B,
But,
i 3 + 0.2v 0 = 2 + i 1
v 0 = 10i 2 so that (4) becomes i 3 = 5 + (2/3)i 2
Solving (1) to (5), i 2 = 0.2941 A,
v 0 = 10i 2 = 2.941 volts, i 0 = i 1 – i 2 = (5/3)i 2 = 490.2mA.
(4)
(5)
Chapter 3, Solution 65
For mesh 1,
–12 + 12I 1 – 6I 2 – I 4 = 0 or
12  12 I 1  6 I 2  I 4
(1)
–6I 1 + 16I 2 – 8I 3 – I 4 – I 5 = 0
(2)
–8I 2 + 15I 3 – I 5 – 9 = 0 or
9 = –8I 2 + 15I 3 – I 5
(3)
For mesh 2,
For mesh 3,
For mesh 4,
–I 1 – I 2 + 7I 4 – 2I 5 – 6 = 0 or
6 = –I 1 – I 2 + 7I 4 – 2I 5
(4)
For mesh 5,
–I 2 – I 3 – 2I 4 + 8I 5 – 10 = 0 or
10   I 2  I 3  2 I 4  8 I 5
Casting (1) to (5) in matrix form gives
1
0  I1  12 
 12  6 0

   
  6 16  8  1  1  I 2   0 
 0  8 15 0  1  I    9 



 3   
7  2  I 4   6 
 1 1 0
 0  1  1  2 8  I  10 

 5   
(5)
AI  B
Using MATLAB we input:
Z=[12,-6,0,-1,0;-6,16,-8,-1,-1;0,-8,15,0,-1;-1,-1,0,7,-2;0,-1,-1,-2,8]
and V=[12;0;9;6;10]
&gt;&gt; Z=[12,-6,0,-1,0;-6,16,-8,-1,-1;0,-8,15,0,-1;-1,-1,0,7,-2;0,-1,-1,-2,8]
Z=
12
-6
0
-1
0
-6 0 -1 0
16 -8 -1 -1
-8 15 0 -1
-1 0 7 -2
-1 -1 -2 8
&gt;&gt; V=[12;0;9;6;10]
V=
12
0
9
6
10
&gt;&gt; I=inv(Z)*V
I=
2.1701
1.9912
1.8119
2.0942
2.2489
Thus,
I = [2.17, 1.9912, 1.8119, 2.094, 2.249] A.
Chapter 3, Solution 66
The mesh equations are obtained as follows.
12  24  30I1  4I2  6I3  2I4  0
or
30I 1 – 4I 2 – 6I 3 – 2I 4 = –12
24  40  4I1  30I2  2I4  6I5  0
or
–4I 1 + 30I 2 – 2I 4 – 6I 5 = –16
(1)
(2)
–6I 1 + 18I 3 – 4I 4 = 30
(3)
–2I 1 – 2I 2 – 4I 3 + 12I 4 –4I 5 = 0
(4)
–6I 2 – 4I 4 + 18I 5 = –32
(5)
Putting (1) to (5) in matrix form
 30  4  6  2 0    12 
 4 30 0  2  6   16 

 

  6 0 18  4 0  I   30 

 

 2  2  4 12  4  0 
 0  6 0  4 18   32
ZI = V
Using MATLAB,
&gt;&gt; Z = [30,-4,-6,-2,0;
-4,30,0,-2,-6;
-6,0,18,-4,0;
-2,-2,-4,12,-4;
0,-6,0,-4,18]
Z=
30 -4 -6 -2 0
-4 30 0 -2 -6
-6 0 18 -4 0
-2 -2 -4 12 -4
0
-6
0
-4
18
&gt;&gt; V = [-12,-16,30,0,-32]'
V=
-12
-16
30
0
-32
&gt;&gt; I = inv(Z)*V
I=
-0.2779 A
-1.0488 A
1.4682 A
-0.4761 A
-2.2332 A
Chapter 3, Solution 67
Consider the circuit below.
5A
V1
4
V2
2
V3
+ Vo 3 Vo
10 
5
10 A
0 
 0.35  0.25
 5  3Vo 
 0.25 0.95  0.5 V  

0




 0
 15 
 0.5
0.5 
Since we actually have four unknowns and only three equations, we need a constraint
equation.
Vo = V2 – V3
Substituting this back into the matrix equation, the first equation becomes,
0.35V 1 – 3.25V 2 + 3V 3 = –5
This now results in the following matrix equation,
3 
 5
 0.35  3.25
 0.25 0.95  0.5 V   0 
 


 15 
 0
0.5 
 0.5
Now we can use MATLAB to solve for V.
&gt;&gt; Y=[0.35,-3.25,3;-0.25,0.95,-0.5;0,-0.5,0.5]
Y=
0.3500 -3.2500 3.0000
-0.2500 0.9500 -0.5000
0 -0.5000 0.5000
&gt;&gt; I=[–5,0,15]'
I=
–5
0
15
&gt;&gt; V=inv(Y)*I
V=
–410.5262
–194.7368
–164.7368
V o = V 2 – V 3 = –77.89 + 65.89 = –30 V.
Let us now do a quick check at node 1.
–3(–30) + 0.1(–410.5) + 0.25(–410.5+194.74) + 5 =
90–41.05–102.62+48.68+5 = 0.01; essentially zero considering the
accuracy we are using. The answer checks.
Chapter 3, Solution 68
Although there are many ways to work this problem, this is an example based on the
same kind of problem asked in the third edition.
Problem
Find the voltage V o in the circuit of Fig. 3.112.
3A
10 
25 
+
4A
40 
20 
Vo
+
_
24 V
+
_
24 V
_
Figure 3.112
For Prob. 3.68.
Solution
Consider the circuit below. There are two non-reference nodes.
3A
V1
10 
Vo
25 
+
4A
40 
Vo
_
20 
0.125  0.1
 43   7 
V

  0.1 0.19 
 3  24 / 25   2.04



 

Using MATLAB, we get,
&gt;&gt; Y=[0.125,-0.1;-0.1,0.19]
Y=
0.1250 -0.1000
-0.1000 0.1900
&gt;&gt; I=[7,-2.04]'
I=
7.0000
-2.0400
&gt;&gt; V=inv(Y)*I
V=
81.8909
32.3636
Thus, V o = 32.36 V.
We can perform a simple check at node V o ,
3 + 0.1(32.36–81.89) + 0.05(32.36) + 0.04(32.36–24) =
3 – 4.953 + 1.618 + 0.3344 = – 0.0004; answer checks!
Chapter 3, Solution 69
Assume that all conductances are in mS, all currents are in mA, and all voltages are in
volts.
G 11 = (1/2) + (1/4) + (1/1) = 1.75, G 22 = (1/4) + (1/4) + (1/2) = 1,
G 33 = (1/1) + (1/4) = 1.25, G 12 = -1/4 = -0.25, G 13 = -1/1 = -1,
G 21 = -0.25, G 23 = -1/4 = -0.25, G 31 = -1, G 32 = -0.25
i 1 = 20, i 2 = 5, and i 3 = 10 – 5 = 5
The node-voltage equations are:
 1   v 1   20
 1.75  0.25
  0.25
1
 0.25 v 2    5 

  1
 0.25 1.25  v 3   5 
Chapter 3, Solution 70
 4I x  20 
3 0
0 5  V    4 I  7 


x


With two equations and three unknowns, we need a constraint equation,
I x = 2V 1 , thus the matrix equation becomes,
  5 0
 20 
 8 5 V    7 


 
This results in V 1 = 20/(–5) = –4 V and
V 2 = [–8(–4) – 7]/5 = [32 – 7]/5 = 5 V.
Chapter 3, Solution 71
 9  4  5  30 
 4 7  1 I   15

 

  5  1 9   0 
We can now use MATLAB solve for our currents.
&gt;&gt; R=[9,–4,–5;–4,7,–1;–5,–1,9]
R=
9 –4 –5
–4 7 –1
–5 –1 9
&gt;&gt; V=[30,–15,0]'
V=
30
–15
0
&gt;&gt; I=inv(R)*V
I=
6.255 A
1.9599 A
3.694 A
Chapter 3, Solution 72
R 11 = 5 + 2 = 7, R 22 = 2 + 4 = 6, R 33 = 1 + 4 = 5, R 44 = 1 + 4 = 5,
R 12 = -2, R 13 = 0 = R 14 , R 21 = -2, R 23 = -4, R 24 = 0, R 31 = 0,
R 32 = -4, R 34 = -1, R 41 = 0 = R 42 , R 43 = -1, we note that R ij = R ji for
all i not equal to j.
v 1 = 8, v 2 = 4, v 3 = -10, and v 4 = -4
Hence the mesh-current equations are:
0   i1   8 
 7 2 0
 2 6  4 0  i   4 

 2   

 0  4 5  1  i 3    10

  

0  1 5  i4    4 
 0
Chapter 3, Solution 73
R 11 = 2 + 3 +4 = 9, R 22 = 3 + 5 = 8, R 33 = 1+1 + 4 = 6, R 44 = 1 + 1 = 2,
R 12 = -3, R 13 = -4, R 14 = 0, R 23 = 0, R 24 = 0, R 34 = -1
v 1 = 6, v 2 = 4, v 3 = 2, and v 4 = -3
Hence,
 9  3  4 0   i1   6 
 3 8
0
0   i 2   4 


 4 0
6  1  i 3   2 

   
0  1 2   i 4    3
 0
Chapter 3, Solution 74
R 11 = R 1 + R 4 + R 6 , R 22 = R 2 + R 4 + R 5 , R 33 = R 6 + R 7 + R 8 ,
R 44 = R 3 + R 5 + R 8 , R 12 = -R 4 , R 13 = -R 6 , R 14 = 0, R 23 = 0,
R 24 = -R 5 , R 34 = -R 8 , again, we note that R ij = R ji for all i not equal to j.
 V1 
 V 
2
The input voltage vector is = 
 V3 


  V4 
 R 1  R4  R6

 R4


 R6

0

 R4
R2  R4  R5
0
 R5
 R6
0
R6  R7  R8
 R8
  i1   V1 
 i   V 
 2    2 
  i 3   V3 
  

R3  R5  R8   i 4    V4 
0
 R5
 R8
Chapter 3, Solution 75
* Schematics Netlist *
R_R4
R_R2
R_R1
R_R3
R_R5
V_V4
v_V3
v_V2
v_V1
\$N_0002 \$N_0001 30
\$N_0001 \$N_0003 10
\$N_0005 \$N_0004 30
\$N_0003 \$N_0004 10
\$N_0006 \$N_0004 30
\$N_0003 0 120V
\$N_0005 \$N_0001 0
0 \$N_0006 0
0 \$N_0002 0
i3
i1
i2
Clearly, i 1 = –3 amps, i 2 = 0 amps, and i 3 = 3 amps, which agrees with the answers in
Problem 3.44.
Chapter 3, Solution 76
* Schematics Netlist *
I_I2
R_R1
R_R3
R_R2
F_F1
VF_F1
R_R4
R_R6
I_I1
R_R5
0 \$N_0001 DC 4A
\$N_0002 \$N_0001 0.25
\$N_0003 \$N_0001 1
\$N_0002 \$N_0003 1
\$N_0002 \$N_0001 VF_F1 3
\$N_0003 \$N_0004 0V
0 \$N_0002 0.5
0 \$N_0001 0.5
0 \$N_0002 DC 2A
0 \$N_0004 0.25
Clearly, v 1 = 625 mVolts, v 2 = 375 mVolts, and v 3 = 1.625 volts, which agrees with
the solution obtained in Problem 3.27.
1
Chapter 3, Solution 77
As a check we can write the nodal equations,
 1.7  0.2
5
 1.2 1.2  V   2


 
Solving this leads to V 1 = 3.111 V and V 2 = 1.4444 V. The answer checks!
Chapter 3, Solution 78
The schematic is shown below. When the circuit is saved and simulated the node
voltages are displayed on the pseudocomponents as shown. Thus,
V1  3V, V2  4.5V, V3  15V,
.
Chapter 3, Solution 79
The schematic is shown below. When the circuit is saved and simulated, we obtain the
node voltages as displayed. Thus,
V a = –10.556 volts; V b = 20.56 volts; V c = 1.3889 volts; and V d = –43.75 volts.
1.3889 V
R3
10
–43.75 V
R6
4
R4
5
R5
4
R1
R2
20
8
20.56 V
R7
16
R8
8
V1
V2
60Vdc
90Vdc
–10.556 V
Chapter 3, Solution 80
* Schematics Netlist *
H_H1
VH_H1
I_I1
V_V1
R_R4
R_R1
R_R2
R_R5
R_R3
\$N_0002 \$N_0003 VH_H1 6
0 \$N_0001 0V
\$N_0004 \$N_0005 DC 8A
\$N_0002 0 20V
0 \$N_0003 4
\$N_0005 \$N_0003 10
\$N_0003 \$N_0002 12
0 \$N_0004 1
\$N_0004 \$N_0001 2
Clearly, v 1 = 84 volts, v 2 = 4 volts, v 3 = 20 volts, and v 4 = -5.333 volts
Chapter 3, Solution 81
Clearly, v 1 = 26.67 volts, v 2 = 6.667 volts, v 3 = 173.33 volts, and v 4 = –46.67 volts
which agrees with the results of Example 3.4.
This is the netlist for this circuit.
* Schematics Netlist *
R_R1
R_R2
R_R3
R_R4
R_R5
I_I1
V_V1
E_E1
0 \$N_0001 2
\$N_0003 \$N_0002 6
0 \$N_0002 4
0 \$N_0004 1
\$N_0001 \$N_0004 3
0 \$N_0003 DC 10A
\$N_0001 \$N_0003 20V
\$N_0002 \$N_0004 \$N_0001 \$N_0004 3
Chapter 3, Solution 82
2i 0
+ v0 –
3 k
1
2 k
3v 0
2
3
6 k
4
+
4A
4 k
8 k
100V +
–
0
This network corresponds to the Netlist.
Chapter 3, Solution 83
The circuit is shown below.
1
20 V
+
20 
70 
2
50 
2A
3
30 
–
0
When the circuit is saved and simulated, we obtain v 2 = –12.5 volts
Chapter 3, Solution 84
From the output loop, v 0 = 50i 0 x20x103 = 106i 0
(1)
From the input loop, 15x10-3 + 4000i 0 – v 0 /100 = 0
(2)
From (1) and (2) we get, i 0 = 2.5 A and v 0 = 2.5 volt.
Chapter 3, Solution 85
The amplifier acts as a source.
Rs
+
Vs
-
RL
For maximum power transfer,
R L  Rs  9
Chapter 3, Solution 86
Let v 1 be the potential across the 2 k-ohm resistor with plus being on top. Then,
Since i = [(0.047–v 1 )/1k]
[(v 1 –0.047)/1k] – 400[(0.047–v 1 )/1k] + [(v 1 –0)/2k] = 0 or
401[(v 1 –0.047)] + 0.5v 1 = 0 or 401.5v 1 = 401x0.047 or
v 1 = 0.04694 volts and i = (0.047–0.04694)/1k = 60 nA
Thus,
v 0 = –5000x400x60x10-9 = –120 mV.
Chapter 3, Solution 87
v 1 = 500(v s )/(500 + 2000) = v s /5
v 0 = -400(60v 1 )/(400 + 2000) = -40v 1 = -40(v s /5) = -8v s ,
Therefore, v 0 /v s = –8
Chapter 3, Solution 88
Let v 1 be the potential at the top end of the 100-ohm resistor.
(v s – v 1 )/200 = v 1 /100 + (v 1 – 10-3v 0 )/2000
(1)
For the right loop, v 0 = -40i 0 (10,000) = -40(v 1 – 10-3)10,000/2000,
or, v 0 = -200v 1 + 0.2v 0 = -4x10-3v 0
(2)
Substituting (2) into (1) gives, (v s + 0.004v 1 )/2 = -0.004v 0 + (-0.004v 1 – 0.001v 0 )/20
This leads to 0.125v 0 = 10v s or (v 0 /v s ) = 10/0.125 = –80
Chapter 3, Solution 89
Consider the circuit below.
_
0.7 V
C
+
100 k
|
|
+ IC
V CE
_
2.25 V
15 V
+
_
1 k
E
For the left loop, applying KVL gives
–2.25 – 0.7 + 105I B + V BE = 0 but V BE = 0.7 V means 105I B = 2.25 or
I B = 22.5 &micro;A.
For the right loop, –V CE + 15 – I C x103 = 0. Addition ally, I C = βI B = 100x22.5x10–6 =
2.25 mA. Therefore,
V CE = 15–2.25x10–3x103 = 12.75 V.
Chapter 3, Solution 90
1 k
10 k
vs
+
-
i1
IB
i2
+
V CE
+
V BE –
–
18V
500 
+
IE
–
+
-
V0
For loop 1, -v s + 10k(I B ) + V BE + I E (500) = 0 = -v s + 0.7 + 10,000I B + 500(1 + )I B
which leads to v s + 0.7 = 10,000I B + 500(151)I B = 85,500I B
But, v 0 = 500I E = 500x151I B = 4 which leads to I B = 5.298x10-5
Therefore, v s = 0.7 + 85,500I B = 5.23 volts
Chapter 3, Solution 91
We first determine the Thevenin equivalent for the input circuit.
R Th = 6||2 = 6x2/8 = 1.5 k and V Th = 2(3)/(2+6) = 0.75 volts
5 k
IC
1.5 k
+
0.75 V
-
i1
IB
i2
+
V CE
+
V BE –
–
9V
400 
+
IE
–
+
-
V0
For loop 1, -0.75 + 1.5kI B + V BE + 400I E = 0 = -0.75 + 0.7 + 1500I B + 400(1 + )I B
I B = 0.05/81,900 = 0.61 A
v 0 = 400I E = 400(1 + )I B = 49 mV
For loop 2, -400I E – V CE – 5kI C + 9 = 0, but, I C = I B and I E = (1 + )I B
V CE = 9 – 5kI B – 400(1 + )I B = 9 – 0.659 = 8.641 volts
Chapter 3, Solution 92
Although there are many ways to work this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
Find I B and V C for the circuit in Fig. 3.128. Let  = 100, V BE = 0.7V.
Figure 3.128
Solution
I1
5 k
10 k
VC
IB
IC
+
V CE
+
V BE –
–
4 k
IE
12V
+
V0
–
I 1 = I B + I C = (1 + )I B and I E = I B + I C = I 1
+
-
Applying KVL around the outer loop,
4kI E + V BE + 10kI B + 5kI 1 = 12
12 – 0.7 = 5k(1 + )I B + 10kI B + 4k(1 + )I B = 919kI B
I B = 11.3/919k = 12.296 A
Also, 12 = 5kI 1 + V C which leads to V C = 12 – 5k(101)I B = 5.791 volts
Chapter 3, Solution 93
1
4
v1
i1
24V
+
3v 0
i
2
2
2
v2 i3
+
8
i
i2
–
3v 0
4
+
+
+
v0
v1
v2
–
–
–
(a)
+
(b)
From (b), -v 1 + 2i – 3v 0 + v 2 = 0 which leads to i = (v 1 + 3v 0 – v 2 )/2
At node 1 in (a), ((24 – v 1 )/4) = (v 1 /2) + ((v 1 +3v 0 – v 2 )/2) + ((v 1 – v 2 )/1), where v 0 = v 2
or 24 = 9v 1 which leads to v 1 = 2.667 volts
At node 2, ((v 1 – v 2 )/1) + ((v 1 + 3v 0 – v 2 )/2) = (v 2 /8) + v 2 /4, v 0 = v 2
v 2 = 4v 1 = 10.66 volts
Now we can solve for the currents, i 1 = v 1 /2 = 1.333 A, i 2 = 1.333 A, and
i 3 = 2.6667 A.
Chapter 4, Solution 1.
5
30V
+

25  i
o
i
40 
15 
40 (25  15)  20 , i = [30/(5+20)] = 1.2 and i o = i20/40 = 600 mA.
Since the resistance remains the same we get can use linearity to find the new
value of the voltage source = (30/0.6)5 = 250 V.
4.2 Using Fig. 4.70, design a problem to help other students better understand linearity.
Although there are many ways to work this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
Find v o in the circuit of Fig. 4.70. If the source current is reduced to 1 A, what is v o ?
Figure 4.70
Solution
6 (4  2)  3, i1  i 2 
io 
1
A
2
1
1
i1  , v o  2i o  0.5V
2
4
5
4
i1
io
i2
1A
If i s = 1A, then v o = 0.5V
8
6
2
Chapter 4, Solution 3.
R
3R
io
3R
Vs
3R
+

+
R
vo
1V
(a) We transform the Y sub-circuit to the equivalent  .
3
3
3
3R 2 3
 R, R  R  R
4
4
2
4R
4
vs
independent of R
2
i o = v o /(R)
vo 
When v s = 1V, v o = 0.5V, i o = 0.5A
(b)
(c)
3R

(a)
R 3R 
+

When v s = 10V, v o = 5V, i o = 5A
When v s = 10V and R = 10,
v o = 5V, i o = 10/(10) = 500mA
(b)
1.5R
Chapter 4, Solution 4.
If I o = 1, the voltage across the 6 resistor is 6V so that the current through the 3
resistor is 2A.
2
2
2A
1A
3A
3A
i1
+
3
6
4
Is
2
4
v1

(a)
3 6  2 , v o = 3(4) = 12V, i1 
(b)
vo
 3A.
4
Hence I s = 3 + 3 = 6A
If
I s = 6A
I s = 9A
Io = 1
I o = 9/6 = 1.5A
Is
Chapter 4, Solution 5.
2
Vs
If v o = 1V,
If v s =
10
3
Then v s = 15
3
v1
+

6
1
V1     1  2V
3
10
2
Vs  2   v1 
3
3
vo = 1
vo =
3
x15  4.5V
10
vo
6
6
Chapter 4, Solution 6.
Due to linearity, from the first experiment,
1
Vo  Vs
3
Applying this to other experiments, we obtain:
Experiment
2
3
4
Vs
Vo
48
1V
-6 V
16 V
0.333 V
-2V
Chapter 4, Solution 7.
If V o = 1V, then the current through the 2- and 4- resistors is &frac12; = 0.5. The voltage
across the 3- resistor is &frac12; (4 + 2) = 3 V. The total current through the 1- resistor is
0.5 +3/3 = 1.5 A. Hence the source voltage
vs  1x1.5  3  4.5 V
If vs  4.5

 1V
Then vs  4


1
x4  0.8889 V = 888.9 mV.
4.5
1
Chapter 4, Solution 8.
Let V o = V 1 + V 2 , where V 1 and V 2 are due to 9-V and 3-V sources respectively. To
find V 1 , consider the circuit below.
V1
3
9
1
+
_
9  V1 V1 V1
 
3
9 1
9V

 V1  27 /13  2.0769
To find V 2 , consider the circuit below.
V1
9
V2 V2 3  V2


9
3
1
3

 V2  27 /13  2.0769
V o = V 1 + V 2 = 4.1538 V
+
_
3V
Chapter 4. Solution 9.
Given that I = 4 amps when V s = 40 volts and I s = 4 amps and I = 1 amp when V s = 20 volts and
I s = 0, determine the value of I when V s = 60 volts and I s = –2 amps.
VS
+

I
IS
At first this appears to be a difficult problem. However, if you take it one step at a time then it is
not as hard as it seems. The important thing to keep in mind is that it is linear!
If I = 1 amp when V s = 20 and I s = 0 then I = 2 amps when V s = 40 volts and I s = 0 (linearity).
This means that I is equal to 2 amps (4–2) when I s = 4 amps and V s = 0 (superposition). Thus,
I = (60/20)1 + (–2/4)2 = 3–1 = 2 amps.
Chapter 4, Solution 10.
Using Fig. 4.78, design a problem to help other students better understand superposition. Note,
the letter k is a gain you can specify to make the problem easier to solve but must not be zero.
Although there are many ways to work this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
For the circuit in Fig. 4.78, find the terminal voltage V ab using superposition.
Figure 4.78
For Prob. 4.10.
Solution
Let v ab = v ab1 + v ab2 where v ab1 and v ab2 are due to the 4-V and the 2-A sources respectively.
3v ab1
10 
10 
+
3v ab2
+
+
4V
+

v ab1
+
2A

(a)

(b)
For v ab1 , consider Fig. (a). Applying KVL gives,
- v ab1 – 3 v ab1 + 10x0 + 4 = 0, which leads to v ab1 = 1 V
For v ab2 , consider Fig. (b). Applying KVL gives,
v ab2
–v ab2 – 3v ab2 + 10x2 = 0, which leads to v ab2 = 5
v ab = 1 + 5 = 6 V
Chapter 4, Solution 11.
Let v o = v 1 + v 2 , where v 1 and v 2 are due to the 6-A and 80-V sources respectively. To
find v 1 , consider the circuit below.
I1
va
+ V1 _
40 
6A
20 
10 
vb
4 i1
At node a,
6
va va  vb

40
10

 240  5va  4vb
(1)
At node b,
–I 1 – 4I 1 + (v b – 0)/20 = 0 or v b = 100I 1
But
i1 
va  vb
10
which leads to 100(v a –v b )10 = v b or v b = 0.9091v a
Substituting (2) into (1),
5v a – 3.636v a = 240 or v a = 175.95 and v b = 159.96
However,
v 1 = v a – v b = 15.99 V.
To find v 2 , consider the circuit below.
(2)
io
10 
+ v2 _
40 
20 
vc
 
4 io
–
+
30 V
0  vc
(30  vc )
 4io 
0
50
20
(0  vc )
But io 
50
5vc (30  vc )

0


50
20
0  vc 0  10 1
i2 


50
50
5
v2  10i2  2 V

vc  10 V
v o = v 1 + v 2 =15.99 + 2 = 17.99 V and i o = v o /10= 1.799 A.
Chapter 4, Solution 12.
Let v o = v o1 + v o2 + v o3 , where v o1 , v o2 , and v o3 are due to the 2-A, 12-V, and 19-V sources
respectively. For v o1 , consider the circuit below.
2A
2A
5
4
+ v o1 
6
io 5 
+ v o1 
12 
3
5
6||3 = 2 ohms, 4||12 = 3 ohms. Hence,
i o = 2/2 = 1, v o1 = 5io = 5 V
For v o2 , consider the circuit below.
6
5
4
6
+ v o2 
12V
+

3
12 
12V
+

5
+
+ v o2 
v1
3
3

3||8 = 24/11, v 1 = [(24/11)/(6 + 24/11)]12 = 16/5
v o2 = (5/8)v 1 = (5/8)(16/5) = 2 V
For v o3 , consider the circuit shown below.
5
4
+ v o3 
6
3
12 
5
+

+ v o3 
19V
2
12 
4
+
v2
+
 19V

7||12 = (84/19) ohms, v 2 = [(84/19)/(4 + 84/19)]19 = 9.975
v = (-5/7)v2 = -7.125
v o = 5 + 2 – 7.125 = -125 mV
Chapter 4, Solution 13.
Let vo  v1  v2  v 3 , where v 1 , v 2 , and v 3 are due to the independent sources. To
find v 1 , consider the circuit below.
8
+
5
10 
2A
v1
_
10
x2  4.3478
10  8  5
To find v 2 , consider the circuit below.
v1  5 x
4A
8
+
10 
5
v2
_
8
x4  6.9565
8  10  5
To find v 3 , consider the circuit below.
v2  5 x
8
12 V
+ –
10 
5
+
v3
_
5


v3  12 
  2.6087
 5  10  8 
vo  v1  v2  v 3  8.6956 V =8.696V.
Chapter 4, Solution 14.
Let v o = v o1 + v o2 + v o3 , where v o1 , v o2 , and v o3 , are due to the 20-V, 1-A, and 2-A sources
respectively. For v o1 , consider the circuit below.
6
4
2
+
+
 20V
3
v o1

6||(4 + 2) = 3 ohms, v o1 = (&frac12;)20 = 10 V
For v o2 , consider the circuit below.
6
4
6
4V
2
2
+
+
1A
4
+
3
v o2
v o2


3||6 = 2 ohms, v o2 = [2/(4 + 2 + 2)]4 = 1 V
For v o3 , consider the circuit below.
6
2A
4
2A
2
3
+
v o3
3

3
 v o3 +
6||(4 + 2) = 3, v o3 = (-1)3 = –3
v o = 10 + 1 – 3 = 8 V
3
Chapter 4, Solution 15.
Let i = i 1 + i 2 + i 3 , where i 1 , i 2 , and i 3 are due to the 20-V, 2-A, and 16-V sources.
For i 1 , consider the circuit below.
io
20V
+

1
i1
2
4
3
4||(3 + 1) = 2 ohms, Then i o = [20/(2 + 2)] = 5 A, i 1 = i o /2 = 2.5 A
For i 3 , consider the circuit below.
+
2
vo’
1
4
i3

+
3
16V

2||(1 + 3) = 4/3, v o ’ = [(4/3)/((4/3) + 4)](-16) = -4
i 3 = v o ’/4 = -1
For i 2 , consider the circuit below.
2
1
1
2A
2A
(4/3)
i2
4
3
2||4 = 4/3, 3 + 4/3 = 13/3
i2
3
Using the current division principle.
i 2 = [1/(1 + 13/2)]2 = 3/8 = 0.375
i = 2.5 + 0.375 - 1 = 1.875 A
p = i2R = (1.875)23 = 10.55 watts
Chapter 4, Solution 16.
Let i o = i o1 + i o2 + i o3 , where i o1 , i o2 , and i o3 are due to
the 12-V, 4-A, and 2-A sources. For i o1 , consider the circuit below.
12V
3
4
i o1
+

10 
2
5
10||(3 + 2 + 5) = 5 ohms, i o1 = 12/(5 + 4) = (12/9) A
4A
For i o2 , consider the circuit below.
3
i o2
4
2
5
10
i1
2 + 5 + 4||10 = 7 + 40/14 = 69/7
i 1 = [3/(3 + 69/7)]4 = 84/90, i o2 =[-10/(4 + 10)]i 1 = -6/9
For i o3 , consider the circuit below.
3
i o3
2
i2
4
10 
5 2A
3 + 2 + 4||10 = 5 + 20/7 = 55/7
i 2 = [5/(5 + 55/7)]2 = 7/9, i o3 = [-10/(10 + 4)]i 2 = -5/9
i o = (12/9) – (6/9) – (5/9) = 1/9 = 111.11 mA
Chapter 4, Solution 17.
Let v x = v x1 + v x2 + v x3 , where v x1 ,v x2 , and v x3 are due to the 90-V, 6-A, and 40-V
sources. For v x1 , consider the circuit below.
30 
+
90V
+

20 
10 
60 
v x1

i o 10 
+

v x1
30 
20 
3A
12 
20||30 = 12 ohms, 60||30 = 20 ohms
By using current division,
i o = [20/(22 + 20)]3 = 60/42, v x1 = 10i o = 600/42 = 14.286 V
For v x2 , consider the circuit below.
10  i ’
o
+
30 
10  i ’
o
v x2 
+ v x2 
60  6A
30 
20 
6A
20 
12 
i o ’ = [12/(12 + 30)]6 = 72/42, v x2 = –10i o ’ = –17.143 V
For v x3 , consider the circuit below.
10 
+
30 
60 
v x3
20 
10 

30 
+
40V
+

20 
v x3
io

12
4A
i o ” = [12/(12 + 30)]2 = 24/42, v x3 = -10i o ” = -5.714= [12/(12 + 30)]2 = 24/42,
v x3 = -10i o ” = -5.714
= [12/(12 + 30)]2 = 24/42, v x3 = -10i o ” = -5.714
v x = 14.286 – 17.143 – 5.714 = -8.571 V
Chapter 4, Solution 18.
Let V o = V 1 + V 2, where V 1 and V 2 are due to 10-V and 2-A sources respectively. To
find V 1 , we use the circuit below.
1
0.5 V 1
2
+
10 V
+
_
V1
_
2
1
0.5 V 1
- +
+
10 V
i
+
_
4
-10 + 7i – 0.5V 1 = 0
But V 1 = 4i
`10  7i  2i  5i

 i  2,
V1  8 V
V1
_
To find V 2 , we use the circuit below.
1
0.5 V 2
2
+
4
2A
2
V2
_
1
0.5 V 2
- +
+
4V
+
_
- 4 + 7i – 0.5V 2 =0
But V 2 = 4i
4  7i  2 i  5 i

 i  0.8,
i
4
V2  4i  3.2
V o = V 1 + V 2 = 8 +3.2 =11.2 V
V2
_
Chapter 4, Solution 19.
Let v x = v 1 + v 2 , where v 1 and v 2 are due to the 4-A and 6-A sources respectively.
v1
ix
ix
v2
+
2
4A
8
v1
+
2
6A 8

+
+
4i x
v2

4i x
(a)
(b)
To find v 1 , consider the circuit in Fig. (a).
v 1 /8 – 4 + (v 1 – (–4i x ))/2 = 0 or (0.125+0.5)v 1 = 4 – 2i x or v 1 = 6.4 – 3.2i x
But,
i x = (v 1 – (–4i x ))/2 or i x = –0.5v 1 . Thus,
v 1 = 6.4 + 3.2(0.5v 1 ), which leads to v 1 = –6.4/0.6 = –10.667
To find v 2 , consider the circuit shown in Fig. (b).
v 2 /8 – 6 + (v 2 – (–4i x ))/2 = 0 or v 2 + 3.2i x = 9.6
But i x = –0.5v 2 . Therefore,
v 2 + 3.2(–0.5v 2 ) = 9.6 which leads to v 2 = –16
Hence,
v x = –10.667 – 16 = –26.67V.
Checking,
i x = –0.5v x = 13.333A
Now all we need to do now is sum the currents flowing out of the top node.
13.333 – 6 – 4 + (–26.67)/8 = 3.333 – 3.333 = 0
Chapter 4, Solution 20.
Convert the voltage sources to current sources and obtain the circuit shown below.
3A
10 
0.6
1
1
1
1



 0.1 0.05  0.025  0.175
Req 10 20 40
20 
0.4
40 
= 5.7143
5.714 Ω
eq 

 RReq
I eq = 3 + 0.6 + 0.4 = 4
Thus, the circuit is reduced as shown below. Please note, we that this is merely an
exercise in combining sources and resistors. The circuit we have is an equivalent circuit
which has no real purpose other than to demonstrate source transformation. In a practical
situation, this would need some kind of reference and a use to an external circuit to be of
real value.
5.714 
18.285 V
4A
5.714 
+
_
4.21 Using Fig. 4.89, design a problem to help other students to better understand source
transformation.
Although there are many ways to work this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
Apply source transformation to determine v o and i o in the circuit in Fig. 4.89.
Figure 4.89
Solution
To get i o , transform the current sources as shown in Fig. (a).
io
6
3
i
+
 12V
+

6V 2 A
6
3
+
vo 2 A

(a)
From Fig. (a),
(b)
-12 + 9i o + 6 = 0, therefore i o = 666.7 mA
To get v o , transform the voltage sources as shown in Fig. (b).
i = [6/(3 + 6)](2 + 2) = 8/3
v o = 3i = 8 V
Chapter 4, Solution 22.
We transform the two sources to get the circuit shown in Fig. (a).
5

+ 10V
5
4
10
2A
(a)
i
1A
10
4
10
2A
(b)
We now transform only the voltage source to obtain the circuit in Fig. (b).
10||10 = 5 ohms, i = [5/(5 + 4)](2 – 1) = 5/9 = 555.5 mA
Chapter 4, Solution 23
If we transform the voltage source, we obtain the circuit below.
8
10 
6
3
5A
3A
3//6 = 2-ohm. Convert the current sources to voltages sources as shown below.
10 
8
2
+
10V
-
+
30V
-
Applying KVL to the loop gives
 30  10  I (10  8  2)  0

 I = 1 A
p  VI  I 2 R  8 W
Chapter 4, Solution 24.
Transform the two current sources in parallel with the resistors into their voltage source
equivalents yield,
a 30-V source in series with a 10-Ω resistor and a 20V x -V sources in
series with a 10-Ω resistor.
We now have the following circuit,
8
+
40 V
10 
Vx –
– +
30 V
+
_
I
10 
20V x
+
–
We now write the following mesh equation and constraint equation which will lead to a
solution for V x ,
28I – 70 + 20V x = 0 or 28I + 20V x = 70, but V x = 8I which leads to
28I + 160I = 70 or I = 0.3723 A or V x = 2.978 V.
Chapter 4, Solution 25.
Transforming only the current source gives the circuit below.
18 V
9
+
–
+
12V
5
i
4
+
vo
2


+
30 V
+
30 V
Applying KVL to the loop gives,
–(4 + 9 + 5 + 2)i + 12 – 18 – 30 – 30 = 0
20i = –66 which leads to i = –3.3
v o = 2i = –6.6 V
Chapter 4, Solution 26.
Transforming the current sources gives the circuit below.
2
15 V
5
io
4
– +
12 V
+
_
–12 + 11i o –15 +20 = 0 or 11i o = 7 or i o = 636.4 mA.
+
_
20 V
Chapter 4, Solution 27.
Transforming the voltage sources to current sources gives the circuit in Fig. (a).
10||40 = 8 ohms
Transforming the current sources to voltage sources yields the circuit in Fig. (b).
Applying KVL to the loop,
-40 + (8 + 12 + 20)i + 200 = 0 leads to i = -4
v x 12i = -48 V
12 
+ vx 
5A
10
40
8A
20
2A
(a)
8
+

12 
+ vx 
40V
i
(b)
20 
+

200V
Chapter 4, Solution 28.
Convert the dependent current source to a dependent voltage source as shown below.
1
4
io
3
+ Vo _
8V
+
_
–
+
Applying KVL,
8  io(1 4  3)  Vo  0
But Vo  4io
8  8io  4io  0

 io  2 A
Vo
Chapter 4, Solution 29.
Transform the dependent voltage source to a current source as shown in Fig. (a). 2||4 =
(4/3) k ohms
4 k
2 k
2v o
(4/3) k
+
1.5v o
+
3 mA
1 k
3 mA
i
1 k
+
vo

vo

(a)
(b)
It is clear that i = 3 mA which leads to v o = 1000i = 3 V
If the use of source transformations was not required for this problem, the actual answer
could have been determined by inspection right away since the only current that could
have flowed through the 1 k ohm resistor is 3 mA.
Chapter 4, Solution 30
Transform the dependent current source as shown below.
ix
24 
60 
+
12V
-
10 
+
30 
7i x
-
Combine the 60-ohm with the 10-ohm and transform the dependent source as shown
below.
ix
24 
+
12V
-
30 
70 
0.1i x
Combining 30-ohm and 70-ohm gives 30//70 = 70x30/100 = 21-ohm. Transform the
dependent current source as shown below.
24 
21 
ix
+
12V
-
+
2.1i x
-
Applying KVL to the loop gives
45i x  12  2.1i x  0


ix 
12
= 254.8 mA.
47.1
Chapter 4, Solution 31.
Transform the dependent source so that we have the circuit in
Fig. (a). 6||8 = (24/7) ohms. Transform the dependent source again to get the circuit in
Fig. (b).
3
+
12V
+

vx

8
6
v x /3
(a)
3
+
12V
+

vx

i
+

(8/7)v x
(b)
From Fig. (b),
v x = 3i, or i = v x /3.
Applying KVL,
-12 + (3 + 24/7)i + (24/21)v x = 0
12 = [(21 + 24)/7]v x /3 + (8/7)v x , leads to v x = 84/23 = 3.652 V
Chapter 4, Solution 32.
As shown in Fig. (a), we transform the dependent current source to a voltage source,
15 
10 
5i x
+
60V
+

50 
40 
(a)
15 
60V
+

50 
50 
0.1i x
(b)
ix
60V
15 
+

25 
ix

+
(c)
In Fig. (b), 50||50 = 25 ohms. Applying KVL in Fig. (c),
-60 + 40i x – 2.5i x = 0, or i x = 1.6 A
2.5i x
Chapter 4, Solution 33.
Determine the Thevenin equivalent circuit as seen by the 5-ohm resistor. Then calculate
the current flowing through the 5-ohm resistor.
10 
10 
4A
5
Solution
Step 1.
We need to find V oc and I sc . To do this, we will need two circuits, label
the appropriate unknowns and solve for V oc , I sc , and then R eq which is equal to V oc /I sc .
V1
10 
V2
10 
+
4A
10 
V oc
4A
10 
I sc
–
Note, in the first case V 1 = V oc and the nodal equation at 1 produces –4+(V 1 –0)/10 = 0.
In the second case, I sc = (V 2 –0)/10 where the nodal equation at 2 produces,
–4+[(V 2 –0)/10]+[(V 2 –0)/10] = 0.
Step 2.
0.1V 1 = 4 or V 1 = 40 V = V oc = V Thev . Next, (0.1+0.1)V 2 = 4 or 0.2V 2 =
4 or V 2 = 20 V. Thus, I sc = 20/10 = 2 A. This leads to R eq = 40/2 = 20 Ω. We can check
our results by using source transformation. The 4-amp current source in parallel with the
10-ohm resistor can be replaced by a 40-volt voltage source in series with a 10-ohm
resistor which in turn is in series with the other 10-ohm resistor yielding the same
Thevenin equivalent circuit. Once the 5-ohm resistor is connected to the Thevenin
equivalent circuit, we now have 40 V across 25 Ω producing a current of 1.6 A.
4.34 Using Fig. 4.102, design a problem that will help other students better understand Thevenin
equivalent circuits.
Although there are many ways to work this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
Find the Thevenin equivalent at terminals a-b of the circuit in Fig. 4.102.
Figure 4.102
Solution
To find R Th , consider the circuit in Fig. (a).
3A
10 
10 
20 
20 
v1
v2
+
R Th
40 
+

(a)
40V
V Th
40 
(b)
R Th = 20 + 10||40 = 20 + 400/50 = 28 ohms
To find V Th , consider the circuit in Fig. (b).
At node 1,
(40 – v 1 )/10 = 3 + [(v 1 – v 2 )/20] + v 1 /40, 40 = 7v 1 – 2v 2
(1)
At node 2,
3 + (v1- v2)/20 = 0, or v1 = v2 – 60
(2)
Solving (1) and (2),
v 1 = 32 V, v 2 = 92 V, and V Th = v 2 = 92 V
Chapter 4, Solution 35.
To find R Th , consider the circuit in Fig. (a).
R Th = R ab = 6||3 + 12||4 = 2 + 3 =5 ohms
To find V Th , consider the circuit shown in Fig. (b).
R Th
a
6
b
12 
3
4
(a)
2A
6
v1
v2 4 
+ V Th
+
+

12V v 1
+
3
12

v2
+

19V

(b)
At node 1,
2 + (12 – v 1 )/6 = v 1 /3, or v 1 = 8
At node 2,
(19 – v 2 )/4 = 2 + v 2 /12, or v 2 = 33/4
But,
-v 1 + V Th + v 2 = 0, or V Th = v 1 – v 2 = 8 – 33/4 = -0.25
a +
vo

b
10 
R Th = 5 
+
V Th = (-
v o = V Th /2 = -0.25/2 = –125 mV
Chapter 4, Solution 36.
Remove the 30-V voltage source and the 20-ohm resistor.
a
R Th
10
a
10
+
+

40
V Th
40
50V
b
b
(a)
(b)
From Fig. (a),
R Th = 10||40 = 8 ohms
From Fig. (b),
V Th = (40/(10 + 40))50 = 40V
8
+

i
a
12 
40V
+

30V
b
(c)
The equivalent circuit of the original circuit is shown in Fig. (c). Applying KVL,
30 – 40 + (8 + 12)i = 0, which leads to i = 500mA
Chapter 4, Solution 37
R N is found from the circuit below.
20 
a
40 
12 
b
R N  12 //( 20  40)  10 Ω.
I N is found from the circuit below.
2A
20 
a
40 
+
120V
-
12 
IN
b
Applying source transformation to the current source yields the circuit below.
20 
40 
+ 80 V -
+
120V
-
Applying KVL to the loop yields
 120  80  60 I N  0


IN
I N  40 / 60  666.7 mA.
Chapter 4, Solution 38
We find Thevenin equivalent at the terminals of the 10-ohm resistor. For R Th , consider
the circuit below.
1
4
5
R Th
16 
RTh  1  5 //( 4  16)  1  4  5
For V Th , consider the circuit below.
V1
4
1
V2
5
3A
+
16 
V Th
+
12 V
-
At node 1,
V V  V2
3 1  1


48  5V1  4V2
16
4
At node 2,
V1  V2 12  V2

0


48  5V1  9V2
4
5
Solving (1) and (2) leads to
VTh  V2  19.2
-
(1)
(2)
Thus, the given circuit can be replaced as shown below.
5
+
19.2V
-
+
Vo
-
10 
Using voltage division,
Vo 
10
(19.2) = 12.8 V.
10  5
Chapter 4, Solution 39.
We obtain R Th using the circuit below.
10 
16
5
10 
R Th
R Thev = 16 + (20||5) = 16 + (20x5)/(20+5) = 20 Ω
To find V Th , we use the circuit below.
3A
10
16
V1
+
10 
24
V2
+
+
_
V2
5
V Th
_
At node 1,
24  V1
V  V2
3 1

 54  2V1  V2
10
10
At node 2,
V1  V2
V
3 2

 60  2V1  6V2
10
5
_
(1)
(2)
Substracting (1) from (2) gives
6 = -5V 2 or V 2 = -1.2 V
But
-V 2 + 16x3 + V Thev = 0 or V Thev = -(48 + 1.2) = –49.2 V
Chapter 4, Solution 40.
To obtain V Th , we apply KVL to the loop.
70  (10  20)kI  4Vo  0
But Vo  10kI
70  70kI

 I  1mA
70  10kI  VTh  0

 VTh  60 V
To find R Th , we remove the 70-V source and apply a 1-V source at terminals a-b, as
shown in the circuit below.
a

I2 
–
Vo
I1
10 k
1V
+
+
_
b
We notice that V o = -1 V.
1 20kI1  4Vo  0

 I1  0.25 mA
I2  I1 
1V
 0.35 mA
10k
RTh 
1V
1

k  2.857 k
I2
0.35
20 
+
–
4 Vo
Chapter 4, Solution 41
To find R Th , consider the circuit below
14 
a
6
5
b
RTh  5 //(14  6)  4  R N
Applying source transformation to the 1-A current source, we obtain the circuit below.
6
- 14V +
14 
V Th
a
+
6V
3A
5
b
At node a,
14  6  VTh
V
 3  Th
6  14
5
IN 


VTh  8 V
VTh
 (8) / 4  2 A
RTh
Thus,
RTh  RN  4 ,
VTh  8V,
I N  2 A
Chapter 4, Solution 42.
To find R Th , consider the circuit in Fig. (a).
20 
10 
30 
20 
a
a
b
30  30 
b
10
10 
10 
10 
(a)
10 
(b)
20||20 = 10 ohms. Transform the wye sub-network to a delta as shown in Fig. (b).
10||30 = 7.5 ohms. R Th = R ab = 30||(7.5 + 7.5) = 10 ohms.
To find V Th , we transform the 20-V (to a current source in parallel with the 20 Ω resistor and
then back into a voltage source in series with the parallel combination of the two 20 Ω resistors)
and the 5-A sources. We obtain the circuit shown in Fig. (c).
a
10 
+
+
10 
b
10 
i1
30V
10 V
10 
+

i2
50V
10 
+

(c)
For loop 1,
-30 + 50 + 30i 1 – 10i 2 = 0, or -2 = 3i 1 – i 2
(1)
For loop 2,
-50 – 10 + 30i 2 – 10i 1 = 0, or 6 = -i 1 + 3i 2
(2)
Solving (1) and (2),
i 1 = 0, i 2 = 2 A
Applying KVL to the output loop, -v ab – 10i 1 + 30 – 10i 2 = 0, v ab = 10 V
V Th = v ab = 10 volts
Chapter 4, Solution 43.
To find R Th , consider the circuit in Fig. (a).
R Th
a
10
b
5
10
(a)
10 
a
+
+

20V v a
b
+ V Th
10 
+
vb

5

(b)
R Th = 10||10 + 5 = 10 ohms
To find V Th , consider the circuit in Fig. (b).
v b = 2x5 = 10 V, v a = 20/2 = 10 V
But,
-v a + V Th + v b = 0, or V Th = v a – v b = 0 volts
2A
Chapter 4, Solution 44.
(a)
For R Th , consider the circuit in Fig. (a).
R Th = 1 + 4||(3 + 2 + 5) = 3.857 ohms
For V Th , consider the circuit in Fig. (b). Applying KVL gives,
10 – 24 + i(3 + 4 + 5 + 2), or i = 1
V Th = 4i = 4 V
1
3
a
+
3
1
a
+

24V
+

R Th
4
2
2
i
b
5
b
10V
5
(b)
(a)
(b)
V Th
4
For R Th , consider the circuit in Fig. (c).
3
1
4
3
b
24V
2
R Th
5
1
4
vo
+

b
+
2
5
2A
c
(c)
c
(d)
R Th = 5||(2 + 3 + 4) = 3.214 ohms
V Th
To get V Th , consider the circuit in Fig. (d). At the node, KCL gives,
[(24 – vo)/9] + 2 = vo/5, or vo = 15
V Th = vo = 15 V
Chapter 4, Solution 45.
For R N , consider the circuit in Fig. (a).
6
6
6
4
RN
4A
(a)
6
4
IN
(b)
R N = (6 + 6)||4 = 3 Ω
For I N , consider the circuit in Fig. (b). The 4-ohm resistor is shorted so that 4-A current is
equally divided between the two 6-ohm resistors. Hence,
I N = 4/2 = 2 A
Chapter 4, Solution 46.
Using Fig. 4.113, design a problem to help other students better understand Norton
equivalent circuits.
Although there are many ways to work this problem, this is an example based on the
same kind of problem asked in the third edition.
Problem
Find the Norton equivalent at terminals a-b of the circuit in Fig. 4.113.
10 
 a
20 
10 
4A
b
Figure 4.113 For Prob. 4.46.
Solution
R N is found using the circuit below.
10 
 a
10 
20 
RN
b
R N = 20//(10+10) = 10 
To find I N , consider the circuit below.
10 
4A
10 
The 20- resistor is short-circuited and can be ignored.
IN = &frac12; x 4 = 2 A
20 

IN
Chapter 4, Solution 47
Since V Th = V ab = V x , we apply KCL at the node a and obtain
30  VTh VTh

 2VTh

 VTh  150 / 126  1.1905 V
12
60
To find R Th , consider the circuit below.
12 
Vx
a
60 
2V x
1A
At node a, KCL gives
V V
1  2V x  x  x

 V x  60 / 126  0.4762
60 12
V
V
RTh  x  0.4762,
I N  Th  1.19 / 0.4762  2.5
1
RTh
Thus,
V Thev = 1.1905 V, R eq = 476.2 mΩ, and I N = 2.5 A.
Chapter 4, Solution 48.
To get R Th , consider the circuit in Fig. (a).
10I o
10I o
+
2
+
+
Io
+
Io
4
V Th
4
V

1A

2A
(a)
From Fig. (a),
2
(b)
I o = 1,
6 – 10 – V = 0, or V = -4
R eq = V/1 = -4 ohms
Note that the negative value of R eq indicates that we have an active device in the circuit since we
cannot have a negative resistance in a purely passive circuit.
To solve for I N we first solve for V Th , consider the circuit in Fig. (b),
I o = 2, V Th = -10I o + 4I o = -12 V
I N = V Th /R Th = 3A
Chapter 4, Solution 49.
R N = R Th = 28 ohms
To find I N , consider the circuit below,
3A
10 
vo
20 
io
40V
At the node,
+

40 
I sc = I N
(40 – v o )/10 = 3 + (v o /40) + (v o /20), or v o = 40/7
i o = v o /20 = 2/7, but I N = I sc = i o + 3 = 3.286 A
Chapter 4, Solution 50.
From Fig. (a), R N = 6 + 4 = 10 ohms
6
6
I sc = I N
4
4
2A
(a)
From Fig. (b),
+
12V 
(b)
2 + (12 – v)/6 = v/4, or v = 9.6 V
-I N = (12 – v)/6 = 0.4, which leads to I N = -0.4 A
Combining the Norton equivalent with the right-hand side of the original circuit produces
the circuit in Fig. (c).
i
0.4A
10 
5
4A
(c)
i = [10/(10 + 5)] (4 – 0.4) = 2.4 A
Chapter 4, Solution 51.
(a)
From the circuit in Fig. (a),
R N = 4||(2 + 6||3) = 4||4 = 2 ohms
R Th
6
V Th
+
4
6
3
2
120V
4
+

3
(a)
6A
2
(b)
For I N or V Th , consider the circuit in Fig. (b). After some source transformations, the
circuit becomes that shown in Fig. (c).
+ V Th
2
40V
+

4
i
2
12V
+

(c)
Applying KVL to the circuit in Fig. (c),
-40 + 8i + 12 = 0 which gives i = 7/2
V Th = 4i = 14 therefore I N = V Th /R N = 14/2 = 7 A
(b)
To get R N , consider the circuit in Fig. (d).
R N = 2||(4 + 6||3) = 2||6 = 1.5 ohms
6
2
4
i
+
3
2
(d)
RN
V Th
12V
+

(e)
To get I N , the circuit in Fig. (c) applies except that it needs slight modification as in
Fig. (e).
i = 7/2, V Th = 12 + 2i = 19, I N = V Th /R N = 19/1.5 = 12.667 A
Chapter 4, Solution 52.
For the transistor model in Fig. 4.118, obtain the Thevenin equivalent at terminals a-b.
12 V
Figure 4.118
For Prob. 4.52.
Solution
Step 1.
To find the Thevenin equivalent for this circuit we need to find v oc and i sc .
Then V Thev = v oc and R eq = v oc /i sc .
3 k
12
+

+
v oc
2 k
I
20I o
–
b
For v oc , I o = (12–0)/3k = 4 mA and 20I o + (v oc –0)/2k = 0.
For i sc , i sc = –20I o .
Step 2.
v oc = –2k(20I o ) = –40x4 = –160 volts = V Thev
i sc = –20x4x10–3 = –80 mA or
R eq = –160/(80x10–3) = 2 kΩ.
i sc
Chapter 4, Solution 53.
To get R Th , consider the circuit in Fig. (a).
0.25v o
0.25v o
2
2
a
+
6
3
+
vo
2
1A

1/2
a
1/2
vo
v ab


b
b
(a)
(b)
From Fig. (b),
v o = 2x1 = 2V, -v ab + 2x(1/2) +v o = 0
v ab = 3V
R N = v ab /1 = 3 ohms
To get I N , consider the circuit in Fig. (c).
0.25v o
6
2
a
+
18V
+

3
vo
I sc =

b
(c)
[(18 – v o )/6] + 0.25v o = (v o /2) + (v o /3) or v o = 4V
But,
(v o /2) = 0.25v o + I N , which leads to I N = 1 A
+
1A
Chapter 4, Solution 54
To find V Th =V x , consider the left loop.
 3  1000io  2V x  0


For the right loop,
V x  50 x 40i o  2000io
Combining (1) and (2),
3  1000io  4000io  3000io
V x  2000io  2


3  1000io  2V x
(1)
(2)


io  1mA
VTh  2
To find R Th , insert a 1-V source at terminals a-b and remove the 3-V independent
source, as shown below.
1 k
ix
.
io
+
2V x
-
2V x
 2mA
1000
V
1
i x  40io  x  80mA  A  -60mA
50
50
V x  1,
RTh 
io  
1
 1 / 0.060   16.67
ix
40i o
+
Vx
-
50 
+
1V
-
Chapter 4, Solution 55.
To get R N , apply a 1 mA source at the terminals a and b as shown in Fig. (a).
8 k
a
I
+
2V
+

v ab /100
80I
+

50 k
IN
v ab

b
(b)
We assume all resistances are in k ohms, all currents in mA, and all voltages in volts. At
node a,
(1)
(v ab /50) + 80I = 1
Also,
(2)
-8I = (v ab /1000), or I = -v ab /8000
From (1) and (2),
(v ab /50) – (80v ab /8000) = 1, or v ab = 100
R N = v ab /1 = 100 k ohms
To get I N , consider the circuit in Fig. (b).
a
I
+
v ab /100
8 k
80I
+

50 k
v ab

b
(a)
Since the 50-k ohm resistor is shorted,
I N = -80I, v ab = 0
Hence,
8i = 2 which leads to I = (1/4) mA
I N = -20 mA
1mA
Chapter 4, Solution 56.
We remove the 1-k resistor temporarily and find Norton equivalent across its terminals.
R eq is obtained from the circuit below.
12 k
2 k
10 k

RN
24 k

R eq = 10 + 2 + (12//24) = 12+8 = 20 k
I N is obtained from the circuit below.
12 k
36 V
+
_
2k
10 k
3 mA
24 k

IN
We can use superposition theorem to find I N . Let I N = I 1 + I 2 , where I 1 and I 2 are due
to 16-V and 3-mA sources respectively. We find I 1 using the circuit below.
12 k
36 V
+
_
24 k
2k
10 k

I1
Using source transformation, we obtain the circuit below.
12 k
3 mA
24 k
12 k

I1

I2
12//24 = 8 k
8
(3mA)  1.2 mA
8  12
To find I 2 , consider the circuit below.
I1 
2k
24 k
12 k
3 mA
2k + 12k//24 k = 10 k
I 2 =0.5(-3mA) = -1.5 mA
I N = 1.2 –1.5 = -0.3 mA
The Norton equivalent with the 1-k resistor is shown below
a
+
In
20 k
10 k
Vo
1 k
–
b
V o = 1k(20/(20+1))(-0.3 mA) = -285.7 mV.
Chapter 4, Solution 57.
To find R Th , remove the 50V source and insert a 1-V source at a – b, as shown in Fig. (a).
2
B
a
A
i
+
3
6
vx
10 
0.5v x

+

1V
b
(a)
We apply nodal analysis. At node A,
i + 0.5v x = (1/10) + (1 – v x )/2, or i + v x = 0.6
(1)
At node B,
(1 – v o )/2 = (v x /3) + (v x /6), and v x = 0.5
From (1) and (2),
(2)
i = 0.1 and
R Th = 1/i = 10 ohms
To get V Th , consider the circuit in Fig. (b).
3
2
v1
v2
a
+
50V
+

vx
+
6
0.5v x

(b)
10  V Th

b
At node 1,
(50 – v 1 )/3 = (v 1 /6) + (v 1 – v 2 )/2, or 100 = 6v 1 – 3v 2
(3)
At node 2,
0.5v x + (v 1 – v 2 )/2 = v 2 /10, v x = v 1 , and v 1 = 0.6v 2
(4)
From (3) and (4),
v 2 = V Th = 166.67 V
I N = V Th /R Th = 16.667 A
R N = R Th = 10 ohms
Chapter 4, Solution 58.
This problem does not have a solution as it was originally stated. The reason for this is
that the load resistor is in series with a current source which means that the only
equivalent circuit that will work will be a Norton circuit where the value of R N =
infinity. I N can be found by solving for I sc .
ib
VS
R1
 ib
vo
+

R2
I sc
Writing the node equation at node vo,
i b + i b = v o /R 2 = (1 + )i b
But
i b = (V s – v o )/R 1
vo = Vs – ibR1
V s – i b R 1 = (1 + )R 2 i b , or i b = V s /(R 1 + (1 + )R 2 )
I sc = I N = -i b = -V s /(R 1 + (1 + )R 2 )
Chapter 4, Solution 59.
R Th = (10 + 20)||(50 + 40) 30||90 = 22.5 ohms
To find V Th , consider the circuit below.
i1
i2
10 
20 
+ V Th
8A
50 
40 
i 1 = i 2 = 8/2 = 4, 10i 1 + V Th – 20i 2 = 0, or V Th = 20i 2 –10i 1 = 10i 1 = 10x4
V Th = 40V, and I N = V Th /R Th = 40/22.5 = 1.7778 A
Chapter 4, Solution 60.
The circuit can be reduced by source transformations.
2A
18 V
12 V
+ 
+ 
10 V
10 
5
+ 
2A
10 
a
b
3A
5
2A
3A
a
3.333
Norton Equivalent Circuit
b
a
3.333
10 V
+ 
Thevenin Equivalent Circuit
b
Chapter 4, Solution 61.
To find R Th , consider the circuit in Fig. (a).
Let
R Th = 2R||R = (2/3)R = 1.2 ohms.
R = 2||18 = 1.8 ohms,
To get V Th , we apply mesh analysis to the circuit in Fig. (d).
2
a
6
6
2
2
6
b
(a)
2
a
18 
1.8 
2
18 
a
2
18 
1.8 
1.8 
R Th
b
b
(b)
(c)
2
a
6
12V
6
i3
+

12V
+
+

V Th
6
2
i1
i2

+
2
12V
b
(d)
-12 – 12 + 14i 1 – 6i 2 – 6i 3 = 0, and 7 i 1 – 3 i 2 – 3i 3 = 12
(1)
12 + 12 + 14 i 2 – 6 i 1 – 6 i 3 = 0, and -3 i 1 + 7 i 2 – 3 i 3 = -12
(2)
14 i 3 – 6 i 1 – 6 i 2 = 0, and
(3)
-3 i 1 – 3 i 2 + 7 i 3 = 0
This leads to the following matrix form for (1), (2) and (3),
 7  3  3  i1   12 
 3 7  3 i    12

 2  

 3  3 7  i 3   0 
7
  3
3 3
 3  100 ,
7
3 3
7
7
12
3
 2   3  12  3  120
3
0
7
i 2 = / 2 = -120/100 = -1.2 A
V Th = 12 + 2i 2 = 9.6 V, and I N = V Th /R Th = 8 A
Chapter 4, Solution 62.
Since there are no independent sources, V Th = 0 V
To obtain R Th , consider the circuit below.
0.1i o
ix
2
+
vo
10 

v1
1
io
40 
VS
+

20 
2v o
+ 
At node 2,
i x + 0.1i o = (1 – v 1 )/10, or 10i x + i o = 1 – v 1
(1)
(v 1 /20) + 0.1i o = [(2v o – v 1 )/40] + [(1 – v 1 )/10]
(2)
At node 1,
But i o = (v 1 /20) and v o = 1 – v 1 , then (2) becomes,
1.1v 1 /20 = [(2 – 3v 1 )/40] + [(1 – v 1 )/10]
2.2v 1 = 2 – 3v 1 + 4 – 4v 1 = 6 – 7v 1
v 1 = 6/9.2
or
(3)
From (1) and (3),
10i x + v 1 /20 = 1 – v 1
10i x = 1 – v 1 – v 1 /20 = 1 – (21/20)v 1 = 1 – (21/20)(6/9.2)
i x = 31.52 mA, R Th = 1/i x = 31.73 ohms.
Chapter 4, Solution 63.
Because there are no independent sources, I N = I sc = 0 A
R N can be found using the circuit below.
10 
+
vo

Applying KCL at node 1,
v1
io
20 
0.5v o
+

1V
v 1 = 1, and v o = (20/30)v 1 = 2/3
i o = (v 1 /30) – 0.5v o = (1/30) – 0.5x2/3 = 0.03333 –
0.33333 = – 0.3 A.
Hence,
R N = 1/(–0.3) = –3.333 ohms
Chapter 4, Solution 64.
With no independent sources, V Th = 0 V. To obtain R Th , consider the circuit shown
below.
4
1
vo
io
ix
+
–
2
+

1V
10i x
i x = [(1 – v o )/1] + [(10i x – v o )/4], or 5v o = 4 + 6i x
But i x = v o /2. Hence,
5v o = 4 + 3v o , or v o = 2, i o = (1 – v o )/1 = -1
Thus, R Th = 1/i o = –1 ohm
(1)
Chapter 4, Solution 65
At the terminals of the unknown resistance, we replace the circuit by its Thevenin equivalent.
12
Req  2  (4 || 12)  2  3  5,
VTh 
(32)  24 V
12  4
Thus, the circuit can be replaced by that shown below.
5
Io
+
24 V
-
+
Vo
-
Applying KVL to the loop,
 24  5I o  Vo  0


V o = 24 – 5I o .
Chapter 4, Solution 66.
We first find the Thevenin equivalent at terminals a and b. We find R Th using the circuit
in Fig. (a).
2
10V
 +
3
2
a
b
+
3
a
V Th
5
b
R Th
+

5
20V

+
i
30V
(a)
(b)
R Th = 2||(3 + 5) = 2||8 = 1.6 ohms
By performing source transformation on the given circuit, we obatin the circuit in (b).
We now use this to find V Th .
10i + 30 + 20 + 10 = 0, or i = –6
V Th + 10 + 2i = 0, or V Th = 2 V
p = V Th 2/(4R Th ) = (2)2/[4(1.6)] = 625 m watts
Chapter 4, Solution 67.
We first find the Thevenin equivalent. We find R Th using the circuit below.
80 
20 
R Th
10 
90 
RTh  20 // 80  90 //10  16  9  25 
We find V Th using the circuit below. We apply mesh analysis.
80 
I1
20 
40 V
+–
10 
I2
(80  20)i1  40  0
(10  90)i2  40  0
90i2  20i1  VTh  0
+
V TH
90 

 i1  0.4

 i2  0.4

 VTh  28 V
(a) R = R Th = 25 
V2
(28)2
(b) Pmax  Th 
 7.84 W
4RTh 100
_
Chapter 4, Solution 68.
This is a challenging problem in that the load is already specified. This now becomes a
&quot;minimize losses&quot; style problem. When a load is specified and internal losses can be
adjusted, then the objective becomes, reduce R Thev as much as possible, which will result
in maximum power transfer to the load.
R
12 V
10 
+
20 
+
-
8V
Removing the 10 ohm resistor and solving for the Thevenin Circuit results in:
R Th = (Rx20/(R+20)) and a V oc = V Th = 12x(20/(R +20)) + (-8)
As R goes to zero, R Th goes to zero and V Th goes to 4 volts, which produces the
maximum power delivered to the 10-ohm resistor.
P = vi = v2/R = 4x4/10 = 1.6 watts
Notice that if R = 20 ohms which gives an R Th = 10 ohms, then V Th becomes -2 volts
and the power delivered to the load becomes 0.1 watts, much less that the 1.6 watts.
It is also interesting to note that the internal losses for the first case are 122/20 = 7.2 watts
and for the second case are = to 12 watts. This is a significant difference.
Chapter 4, Solution 69.
We need the Thevenin equivalent across the resistor R. To find R Th , consider the circuit
below.
22 k v 1
+
10 k
40 k
vo
30 k
0.003v o

1mA
Assume that all resistances are in k ohms and all currents are in mA.
10||40 = 8, and 8 + 22 = 30
1 + 3v o = (v 1 /30) + (v 1 /30) = (v 1 /15)
15 + 45v o = v 1
But v o = (8/30)v 1 , hence,
15 + 45x(8v 1 /30) v 1 , which leads to v 1 = 1.3636
R Th = v 1 /1 = –1.3636 k ohms
R Th being negative indicates an active circuit and if you now make R equal to 1.3636 k
ohms, then the active circuit will actually try to supply infinite power to the resistor. The
2
2
VTh
V 


pR = 
 1363.6   Th  1363.6 =
 0 
  1363.6  1363.6 
∞
It may still be instructive to find V Th . Consider the circuit below.
10 k v o 22 k
v1
+
100V
+

vo

+
40 k
0.003v o
(100 – v o )/10 = (v o /40) + (v o – v 1 )/22
30 k
V Th

(1)
[(v o – v 1 )/22] + 3v o = (v 1 /30)
Solving (1) and (2),
v 1 = V Th = -243.6 volts
(2)
Chapter 4, Solution 70
We find the Thevenin equivalent across the 10-ohm resistor. To find V Th , consider the
circuit below.
3V x
5
5
+
+
15 
4V
-
V Th
6
–
+
Vx
-
From the figure,
15
(4)  3V
15  5
consider the circuit below:
V x  0,
To find R eq,
VTh 
3V x
5
5
V1
+
4V
–
15 
+
I sc
6
Vx
-
At node 1,
[(V 1 –V x )/15] + [(V 1 –(4+V x ))/5] + [(V 1 –0)/5] + 3V x = 0 or
0.4667V 1 + 2.733V x = 0.8
(1)
At node x,
[(V x –0)/6] + [((V x +4)–V 1 )/5] + [(V x –V 1 )/15] = 0 or
–(0.2667)V 1 + 0.4333V x = –0.8
(2)
(0.4667–0.2667)V 1 + (2.733+0.4333)V x = 0 or V 1 = –(3.166/0.2)V x = –15.83V x
Now we can put this into (1) and we get,
0.4667(–15.83V x ) + 2.733V x = 0.8 = (–7.388+2.733)V x = –4.655V x or V x = –0.17186
V.
I sc = –V x /6 = 0.02864 and R eq = 3/(0.02864) = 104.75 Ω
An alternate way to find R eq is replace I sc with a 1 amp current source flowing up and
setting the 4 volts source to zero. We then find the voltage across the 1 amp current
source which is equal to R eq . First we note that V x = 6 volts ;
V 1 = 6+3.75 = –9.75; V 2 = 19x5 + V 1 = 95+9.75 = 104.75 or R eq = 104.75 Ω.
Clearly setting the load resistance to 104.75 Ω means that the circuit will deliver
maximum power to it. Therefore,
p max = [3/(2x104.75)]2x104.75 = 21.48 mW
Chapter 4, Solution 71.
We need R Th and V Th at terminals a and b. To find R Th , we insert a 1-mA source at the
terminals a and b as shown below.
10 k
+
3 k
vo

+
1 k

120v o
a
40 k
1mA
b
Assume that all resistances are in k ohms, all currents are in mA, and all voltages are in
volts. At node a,
1 = (v a /40) + [(v a + 120v o )/10], or 40 = 5v a + 480v o
(1)
The loop on the left side has no voltage source. Hence, v o = 0. From (1), v a = 8 V.
R Th = v a /1 mA = 8 kohms
To get V Th , consider the original circuit. For the left loop,
v o = (1/4)8 = 2 V
For the right loop,
v R = V Th = (40/50)(-120v o ) = -192
The resistance at the required resistor is
R = R Th = 8 kΩ
p = V Th 2/(4R Th ) = (-192)2/(4x8x103) = 1.152 watts
Chapter 4, Solution 72.
(a)
R Th and V Th are calculated using the circuits shown in Fig. (a) and (b)
respectively.
From Fig. (a),
R Th = 2 + 4 + 6 = 12 ohms
From Fig. (b),
-V Th + 12 + 8 + 20 = 0, or V Th = 40 V
4
2
6
4
12V
6
 +
+
2
R Th
+

V Th
8V
20V
(a)
(b)
+ 
(b)
i = V Th /(R Th + R) = 40/(12 + 8) = 2A
(c)
For maximum power transfer,
R L = R Th = 12 ohms
(d)
p = V Th 2/(4R Th ) = (40)2/(4x12) = 33.33 watts.

Chapter 4, Solution 73
Find the Thevenin’s equivalent circuit across the terminals of R.
10 
25 
R Th
20 
5
RTh  10 // 20  25 // 5  325 / 30  10.833
10 
+
60 V
-
25 
+
V Th -
+
+
Va
Vb
20 
5
-
20
(60)  40,
30
 Va  VTh  Vb  0
Va 
-
5
(60)  10
30

 VTh  Va  Vb  40  10  30 V
Vb 
2
p max
V
30 2
= 20.77 W.
 Th 
4 RTh 4 x10.833
Chapter 4, Solution 74.
When R L is removed and V s is short-circuited,
R Th = R 1 ||R 2 + R 3 ||R 4 = [R 1 R 2 /( R 1 + R 2 )] + [R 3 R 4 /( R 3 + R 4 )]
R L = R Th = (R 1 R 2 R 3 + R 1 R 2 R 4 + R 1 R 3 R 4 + R 2 R 3 R 4 )/[( R 1 + R 2 )( R 3 + R 4 )]
When R L is removed and we apply the voltage division principle,
V oc = V Th = v R2 – v R4
= ([R 2 /(R 1 + R 2 )] – [R 4 /(R 3 + R 4 )])V s = {[(R 2 R 3 ) – (R 1 R 4 )]/[(R 1 + R 2 )(R 3 + R 4 )]}V s
p max = V Th 2/(4R Th )
= {[(R 2 R 3 ) – (R 1 R 4 )]2/[(R 1 + R 2 )(R 3 + R 4 )]2}V s 2[( R 1 + R 2 )( R 3 + R 4 )]/[4(a)]
where a = (R 1 R 2 R 3 + R 1 R 2 R 4 + R 1 R 3 R 4 + R 2 R 3 R 4 )
p max =
[(R 2 R 3 ) – (R 1 R 4 )]2V s 2/[4(R 1 + R 2 )(R 3 + R 4 ) (R 1 R 2 R 3 + R 1 R 2 R 4 + R 1 R 3 R 4 + R 2 R 3 R 4 )]
Chapter 4, Solution 75.
We need to first find R Th and V Th .
R
R
R
R
R
R
vo
+
R Th
1V
+

2V
+

+

3V
V Th

(a)
(b)
Consider the circuit in Fig. (a).
(1/R eq ) = (1/R) + (1/R) + (1/R) = 3/R
R eq = R/3
From the circuit in Fig. (b),
((1 – v o )/R) + ((2 – v o )/R) + ((3 – v o )/R) = 0
v o = 2 = V Th
For maximum power transfer,
R L = R Th = R/3
P max = [(V Th )2/(4R Th )] = 3 mW
R Th = [(V Th )2/(4P max )] = 4/(4xP max ) = 1/P max = R/3
R = 3/(3x10-3) = 1 kΩ
1 kΩ, 3 mW
Chapter 4, Solution 76.
Follow the steps in Example 4.14. The schematic and the output plots are shown below.
From the plot, we obtain,
V = 92 V [i = 0, voltage axis intercept]
R = Slope = (120 – 92)/1 = 28 ohms
Chapter 4, Solution 77.
(a)
The schematic is shown below. We perform a dc sweep on a current source, I1,
connected between terminals a and b. We label the top and bottom of source I1 as 2 and 1
respectively. We plot V(2) – V(1) as shown.
V Th = 4 V [zero intercept]
R Th = (7.8 – 4)/1 = 3.8 ohms
(b)
Everything remains the same as in part (a) except that the current source, I1, is connected
between terminals b and c as shown below. We perform a dc sweep on I1 and obtain the
plot shown below. From the plot, we obtain,
V = 15 V [zero intercept]
R = (18.2 – 15)/1 = 3.2 ohms
Chapter 4, Solution 78.
The schematic is shown below. We perform a dc sweep on the current source, I1,
connected between terminals a and b. The plot is shown. From the plot we obtain,
V Th = -80 V [zero intercept]
R Th = (1920 – (-80))/1 = 2 k ohms
Chapter 4, Solution 79.
After drawing and saving the schematic as shown below, we perform a dc sweep on I1
connected across a and b. The plot is shown. From the plot, we get,
V = 167 V [zero intercept]
R = (177 – 167)/1 = 10 ohms
Chapter 4, Solution 80.
The schematic in shown below. We label nodes a and b as 1 and 2 respectively. We
perform dc sweep on I1. In the Trace/Add menu, type v(1) – v(2) which will result in the
plot below. From the plot,
V Th = 40 V [zero intercept]
R Th = (40 – 17.5)/1 = 22.5 ohms [slope]
Chapter 4, Solution 81.
The schematic is shown below. We perform a dc sweep on the current source, I2,
connected between terminals a and b. The plot of the voltage across I2 is shown below.
From the plot,
V Th = 10 V [zero intercept]
R Th = (10 – 6.7)/1 = 3.3 ohms. Note that this is in good agreement with the exact
value of 3.333 ohms.
Chapter 4, Solution 82.
V Th = V oc = 12 V, I sc = 20 A
R Th = V oc /I sc = 12/20 = 0.6 ohm.
0.6 
i
12V
i = 12/2.6 ,
+

2
p = i2R = (12/2.6)2(2) = 42.6 watts
Chapter 4, Solution 83.
V Th = V oc = 12 V, I sc = I N = 1.5 A
R Th = V Th /I N = 8 ohms, V Th = 12 V, R Th = 8 ohms
Chapter 4, Solution 84
Let the equivalent circuit of the battery terminated by a load be as shown below.
R Th
IL
+
+
V Th
VL
-
RL
-
For open circuit,
R L  ,

 VTh  Voc  VL  10.8 V
When R L = 4 ohm, V L =10.5,
IL 
VL
 10.8 / 4  2.7
RL
But
VTh  VL  I L RTh


RTh 
VTh  V L 12  10.8

 0.4444
2.7
IL
= 444.4 mΩ.
Chapter 4, Solution 85
(a) Consider the equivalent circuit terminated with R as shown below.
R Th
a
+
V Th
-
+
V ab
-
R
b
Vab
R

VTh
R  RTh


10
6
VTh
10  RTh
or
60  6 RTh  10VTh
where R Th is in k-ohm.
(1)
Similarly,
30


VTh
30  RTh
Solving (1) and (2) leads to
12 
360  12 RTh  30VTh
VTh  24 V, RTh  30k
(b)
V ab 
20
( 24)  9.6 V
20  30
(2)
Chapter 4, Solution 86.
We replace the box with the Thevenin equivalent.
R Th
+
V Th
+

i
R
v

V Th = v + iR Th
When i = 1.5, v = 3, which implies that V Th = 3 + 1.5R Th
(1)
When i = 1, v = 8, which implies that V Th = 8 + 1xR Th
(2)
From (1) and (2), R Th = 10 ohms and V Th = 18 V.
(a)
When R = 4, i = V Th /(R + R Th ) = 18/(4 + 10) = 1.2857 A
(b)
For maximum power, R = R TH
Pmax = (V Th )2/4R Th = 182/(4x10) = 8.1 watts
Chapter 4, Solution 87.
(a)
i m = 9.876 mA
i m = 9.975 mA
+
Is
vm
Rs Rm
Is
Rs
Rs Rm

(a)
(b)
From Fig. (a),
v m = R m i m = 9.975 mA x 20 = 0.1995 V
I s = 9.975 mA + (0.1995/R s )
(1)
From Fig. (b),
v m = R m i m = 20x9.876 = 0.19752 V
I s = 9.876 mA + (0.19752/2k) + (0.19752/R s )
= 9.975 mA + (0.19752/R s )
Solving (1) and (2) gives,
R s = 8 k ohms,
I s = 10 mA
(b)
i m ’ = 9.876
Is
Rs
Rs Rm
(b)
8k||4k = 2.667 k ohms
i m ’ = [2667/(2667 + 20)](10 mA) = 9.926 mA
(2)
Chapter 4, Solution 88
To find R Th, consider the circuit below.
5k 
R Th
A
B
30k 
20k 
10k 
RTh  30  10  20 // 5  44k
To find V Th , consider the circuit below.
5k 
A
B
io
30k 
20k 
+
4mA
60 V
-
10k 
V A  30 x 4  120,
VB 
20
(60)  48,
25
VTh  V A  VB  72 V
The Thevenin equivalent circuit is shown below.
44k 
I
Ri
+
72 V
-
2k 
72
mA
44  2  Ri
assuming R i is in k-ohm.
I
(a) When R i =500  ,
I
72
 1.548 mA
44  2  0.5
(b) When R i = 0  ,
I
72
 1.565 mA
44  2  0
Chapter 4, Solution 89
It is easy to solve this problem using Pspice.
(a) The schematic is shown below. We insert IPROBE to measure the desired ammeter
reading. We insert a very small resistance in series IPROBE to avoid problem. After the
circuit is saved and simulated, the current is displaced on IPROBE as 99.99A .
(b) By interchanging the ammeter and the 12-V voltage source, the schematic is shown
below. We obtain exactly the same result as in part (a).
Chapter 4, Solution 90.
R x = (R 3 /R 1 )R 2 = (4/2)R 2 = 42.6, R 2 = 21.3
which is (21.3ohms/100ohms)% = 21.3%
Chapter 4, Solution 91.
R x = (R 3 /R 1 )R 2
(a)
Since 0 &lt; R 2 &lt; 50 ohms, to make 0 &lt; R x &lt; 10 ohms requires that when R 2 = 50
ohms, R x = 10 ohms.
10 = (R 3 /R 1 )50 or R 3 = R 1 /5
so we select R 1 = 100 ohms and R 3 = 20 ohms
(b)
For 0 &lt; R x &lt; 100 ohms
100 = (R 3 /R 1 )50, or R 3 = 2R 1
So we can select R 1 = 100 ohms and R 3 = 200 ohms
Chapter 4, Solution 92.
For a balanced bridge, v ab = 0. We can use mesh analysis to find v ab . Consider the
circuit in Fig. (a), where i 1 and i 2 are assumed to be in mA.
2 k
3 k
220V
+

a
i1
+
6 k
i2
b
v ab
5 k
10 k
0
(a)
220 = 2i 1 + 8(i 1 – i 2 ) or 220 = 10i 1 – 8i 2 (1)
0 = 24i 2 – 8i 1 or i 2 = (1/3)i 1
(2)
From (1) and (2),
i 1 = 30 mA and i 2 = 10 mA
Applying KVL to loop 0ab0 gives
5(i 2 – i 1 ) + v ab + 10i 2 = 0 V
Since v ab = 0, the bridge is balanced.
When the 10 k ohm resistor is replaced by the 18 k ohm resistor, the gridge becomes
unbalanced. (1) remains the same but (2) becomes
0 = 32i 2 – 8i 1 , or i 2 = (1/4)i 1
Solving (1) and (3),
i 1 = 27.5 mA, i 2 = 6.875 mA
v ab = 5(i 1 – i 2 ) – 18i 2 = -20.625 V
V Th = v ab = -20.625 V
(3)
To obtain R Th , we convert the delta connection in Fig. (b) to a wye connection shown in
Fig. (c).
2 k
3 k
6 k
a
5 k
R Th
R2
6 k
R1
b
18 k
a R Th
R3
(b)
b
18 k
(c)
R 1 = 3x5/(2 + 3 + 5) = 1.5 k ohms, R 2 = 2x3/10 = 600 ohms,
R 3 = 2x5/10 = 1 k ohm.
R Th = R 1 + (R 2 + 6)||(R 3 + 18) = 1.5 + 6.6||9 = 6.398 k ohms
R L = R Th = 6.398 k ohms
P max = (V Th )2/(4R Th ) = (20.625)2/(4x6.398) = 16.622 mWatts
Chapter 4, Solution 93.
Rs
ix
VS
+

Ro
R o i x
ix
+

-V s + (R s + R o )i x + R o i x = 0
i x = V s /(R s + (1 + )R o )
Chapter 4, Solution 94.
(a)
V o /V g = R p /(R g + R s + R p )
(1)
R eq = R p ||(R g + R s ) = R g
R g = R p (R g + R s )/(R p + R g + R s )
RgRp + Rg2 + RgRs = RpRg + RpRs
R p R s = R g (R g + R s )
From (1),
(2)
R p / = R g + R s + R p
R g + R s = R p ((1/) – 1) = R p (1 - )/
(1a)
Combining (2) and (1a) gives,
R s = [(1 - )/]R eq
= (1 – 0.125)(100)/0.125 = 700 ohms
From (3) and (1a),
R p (1 - )/ = R g + [(1 - )/]R g = R g /
R p = R g /(1 - ) = 100/(1 – 0.125) = 114.29 ohms
(b)
R Th
I
V Th
+

RL
V Th = V s = 0.125V g = 1.5 V
R Th = R g = 100 ohms
I = V Th /(R Th + R L ) = 1.5/150 = 10 mA
(3)
Chapter 4, Solution 95.
Let 1/sensitivity = 1/(20 k ohms/volt) = 50 A
For the 0 – 10 V scale,
R m = V fs /I fs = 10/50 A = 200 k ohms
For the 0 – 50 V scale,
R m = 50(20 k ohms/V) = 1 M ohm
R Th
V Th
+

Rm
V Th = I(R Th + R m )
(a)
I = (4/10)I fs = 0.4x50 A = 20 A
V Th = 20 A R Th + 20 A 250 k ohms
= 4 + 20 A R Th
(b)
(1)
I = (5/50)I fs = 0.1 x 50 A = 5 A
V Th = 5 A x R Th + 5 A x 1 M ohm
V Th = 5 + 5 A R Th
(2)
From (1) and (2)
0 = -1 + 15 A R Th which leads to R Th = 66.67 k ohms
From (1),
V Th = 4 + 20x10-6x(1/(15x10-6)) = 5.333 V
Chapter 4, Solution 96.
(a)
The resistance network can be redrawn as shown in Fig. (a),
10 
10 
8
R Th
9V
+

i1
40
+
i2
60 
8
10 
V Th

R
+
V Th
+

(a)
Vo
R

(b)
R Th = 10 + 10 + [60||(8 + 8 + [10||40])] = 20 + (60||24) = 37.14 ohms
Using mesh analysis,
-9 + 50i 1 - 40i 2 = 0
116i 2 – 40i 1 = 0 or i 1 = 2.9i 2
From (1) and (2),
(1)
(2)
i 2 = 9/105 = 0.08571
V Th = 60i 2 = 5.143 V
From Fig. (b),
V o = [R/(R + R Th )]V Th = 1.8 V
R/(R + 37.14) = 1.8/5.143 = 0.35 or R = 0.35R + 13 or R = (13)/(1–0.35)
which leads to R = 20 Ω (note, this is just for the V o = 1.8 V)
(b)
Asking for the value of R for maximum power would lead to R = R Th = 37.14 Ω.
However, the problem asks for the value of R for maximum current. This happens when
the value of resistance seen by the source is a minimum thus R = 0 is the correct value.
I max = V Th /(R Th ) = 5.143/(37.14) = 138.48 mA.
Chapter 4, Solution 97.
6 k
12V
+

B
+
4 k
V Th

E
R Th = R 1 ||R 2 = 6||4 = 2.4 k ohms
V Th = [R 2 /(R 1 + R 2 )]v s = [4/(6 + 4)](12) = 4.8 V
Chapter 4, Solution 98.
The 20-ohm, 60-ohm, and 14-ohm resistors form a delta connection which needs to be
connected to the wye connection as shown in Fig. (b),
20 
30 
30 
R2
R1
14 
60 
a
b
R Th
a
b
R3
(a)
R Th
(b)
R 1 = 20x60/(20 + 60 + 14) = 1200/94 = 12.766 ohms
R 2 = 20x14/94 = 2.979 ohms
R 3 = 60x14/94 = 8.936 ohms
R Th = R 3 + R 1 ||(R 2 + 30) = 8.936 + 12.766||32.98 = 18.139 ohms
To find V Th , consider the circuit in Fig. (c).
IT
30 
20 
I1
14 
b
60 
a
+
IT
16 V
+ 
(c)
V Th
I T = 16/(30 + 15.745) = 349.8 mA
I 1 = [20/(20 + 60 + 14)]I T = 74.43 mA
V Th = 14I 1 + 30I T = 11.536 V
I 40 = V Th /(R Th + 40) = 11.536/(18.139 + 40) = 198.42 mA
P 40 = I 40 2R = 1.5748 watts
Chapter 5, Solution 1.
(a)
(b)
(c)
R in = 1.5 M
R out = 60 
A = 8x104
Therefore A dB = 20 log 8x104 = 98.06 dB
Chapter 5, Solution 2.
v 0 = Av d = A(v 2 - v 1 )
= 105 (20-10) x 10-6 = 1V
Chapter 5, Solution 3.
v 0 = Av d = A(v 2 - v 1 )
= 2 x 105 (30 + 20) x 10-6 = 10V
Chapter 5, Solution 4.
v 0 = Av d = A(v 2 - v 1 )
v
4
 2V
v2 - v1 = 0 
A 2 x10 6
v 2 - v 1 = -2 &micro;V = –0.002 mV
1 mV - v 1 = -0.002 mV
v 1 = 1.002 mV
Chapter 5, Solution 5.
I
R0
R in
vd
+
vi
Av d
+
v0
-
+
-
-v i + Av d + (R i + R 0 ) I = 0
But
+
-
(1)
v d = R i I,
-v i + (R i + R 0 + R i A) I = 0
I=
vi
R 0  (1  A)R i
(2)
-Av d - R 0 I + v 0 = 0
v 0 = Av d + R 0 I = (R 0 + R i A)I =
(R 0  R i A) v i
R 0  (1  A)R i
v0
R 0  RiA
100  10 4 x10 5


 10 4
v i R 0  (1  A)R i 100  (1  10 5 )

10 9
100,000
 10 4 
 0.9999990
5
100,001
1  10


Chapter 5, Solution 6.
vi
+ -
R0
I
R in
vd
+
-
+
Av d
+
vo
-
(R 0 + R i )R + v i + Av d = 0
But
v d = R i I,
v i + (R 0 + R i + R i A)I = 0
I=
 vi
R 0  (1  A)R i
(1)
-Av d - R 0 I + v o = 0
v o = Av d + R 0 I = (R 0 + R i A)I
Substituting for I in (1),
 R 0  R iA 
 v i
v 0 =  
 R 0  (1  A)R i 
50  2x10 6 x 2 x10 5  10 3
= 
50  1  2x10 5 x 2 x10 6





 200,000 x 2 x10
mV
200,001x 2 x10 6
6
v 0 = -0.999995 mV
Chapter 5, Solution 7.
100 k
10 k
VS
+
–
R out = 100 
1
2
+
Vd
R in
–
+
AV d
–
At node 1,
+
V out
–
(V S – V 1 )/10 k = [V 1 /100 k] + [(V 1 – V 0 )/100 k]
10 V S – 10 V 1 = V 1 + V 1 – V 0
which leads to V 1 = (10V S + V 0 )/12
At node 2,
(V 1 – V 0 )/100 k = (V 0 – (–AV d ))/100
But V d = V 1 and A = 100,000,
V 1 – V 0 = 1000 (V 0 + 100,000V 1 )
0= 1001V 0 + 99,999,999[(10V S + V 0 )/12]
0 = 83,333,332.5 V S + 8,334,334.25 V 0
which gives us (V 0 / V S ) = –10 (for all practical purposes)
If V S = 1 mV, then V 0 = –10 mV
Since V 0 = A V d = 100,000 V d , then V d = (V 0 /105) V = –100 nV
Chapter 5, Solution 8.
(a)
If v a and v b are the voltages at the inverting and noninverting terminals of the op amp.
va = vb = 0
1mA =
0  v0
2k
v 0 = –2 V
(b)
10 k
2V
+
ia
va
2V
+
vb
1V
+
vo
+
+2 k
-
-
10 k
+
va
-
+
ia
vo
-
(b)
(a)
Since v a = v b = 1V and i a = 0, no current flows through the 10 k resistor. From Fig. (b),
-v a + 2 + v 0 = 0
v 0 = v a – 2 = 1 – 2 = –1V
Chapter 5, Solution 9.
(a)
Let v a and v b be respectively the voltages at the inverting and noninverting terminals of
the op amp
v a = v b = 4V
At the inverting terminal,
1mA =
4  v0
2k
v 0 = 2V
1V
(b)
+-
+
+
vb
vo
-
-
Since v a = v b = 3V,
-v b + 1 + v o = 0
v o = v b – 1 = 2V
Chapter 5, Solution 10.
Since no current enters the op amp, the voltage at the input of the op amp is v s . Hence
 10  v o
vs = vo 

 10  10  2
vo
=2
vs
5.11 Using Fig. 5.50, design a problem to help other students to better understand how ideal op
amps work.
Although there are many ways to work this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
Find v o and i o in the circuit in Fig. 5.50.
Figure 5.50 for Prob. 5.11
Solution
8 k
2 k
3V
vb =
+

5 k
a
b

+
10 k
+
4 k
vo

10
(3)  2V
10  5
At node a,
3  va va  vo

2
8
io
12 = 5v a – v o
But v a = v b = 2V,
12 = 10 – v o
–i o =
v o = –2V
va  vo 0  vo 2  2 2


  1mA
8
4
8
4
i o = –1mA
Chapter 5, Solution 12.
Step 1.
Label the unknown nodes in the op amp circuit. Next we write the node
equations and then apply the constraint, V a = V b . Finally, solve for V o in terms of V s .
25 k
5 k
VS
+

a
b

+
10 k
Step 2.
+
Vo

[(V a -V s )/5k] + [(V a -V o )/25k] + 0 = 0 and
[(V b -0)/10k] + 0 = 0 or V b = 0 = V a ! Thus,
[(-V s )/5k] + [(-V o )/25k] = 0 or,
V o = ( –25/5)V s or V o /V s = –5.
Chapter 5, Solution 13.
10 k
a
b
1V
+

+

io
100 k i 2
10 k
90 k
50 k
By voltage division,
va =
90
(1)  0.9V
100
vb =
v
50
vo  o
150
3
But v a = v b
io = i1 + i2 =
v0
 0.9
3
i1
v o = 2.7V
vo
v
 o  0.27mA + 0.018mA = 288 A
10k 150k
+
vo

Chapter 5, Solution 14.
Transform the current source as shown below. At node 1,
10  v1 v1  v 2 v1  v o


5
20
10
10 k
vo
10 k
5 k
20 k
v1
10V

+
v2
+

+
vo

But v 2 = 0. Hence 40 – 4v 1 = v 1 + 2v 1 – 2v o
At node 2,
v1  v 2 v 2  v o

,
20
10
v 2  0 or v 1 = –2v o
From (1) and (2), 40 = –14v o - 2v o
v o = –2.5V
40 = 7v 1 – 2v o
(1)
(2)
Chapter 5, Solution 15
(a) Let v 1 be the voltage at the node where the three resistors meet. Applying
KCL at this node gives
 1
v1 v1  vo
1  vo
 

 v1 

R2
R3
 R2 R3  R3
At the inverting terminal,
is 
0  v1

 v1  i s R1
R1
Combining (1) and (2) leads to

v
R
R 
i s  1  1  1    o


R2 R3 
R3

(1)
is 
(2)

vo
RR
  R1  R3  1 3
is
R2

(b) For this case,
vo
20 x 40 

  20  40 
 k  - 92 k
is
25 

= –92 kΩ



Chapter 5, Solution 16
Using Fig. 5.55, design a problem to help students better understand inverting op amps.
Although there are many ways to work this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
Obtain i x and i y in the op amp circuit in Fig. 5.55.
Figure 5.55
Solution
10k 
5k 
ix
va
vb
+
0.5V
-
iy
+
2k 
8k 
vo
Let currents be in mA and resistances be in k  . At node a,
0.5  v a v a  v o


 1  3v a  vo
5
10
(1)
But
8
10
vo

 vo  v a
(2)
82
8
Substituting (2) into (1) gives
10
8
1  3v a  v a

 v a 
8
14
Thus,
0.5  v a
ix 
 1 / 70 mA   14.28 A
5
v  vb vo  va
10
0.6 8
iy  o

 0.6(v o  v a )  0.6( v a  v a ) 
x mA
2
10
8
4 14
v a  vb 
= 85.71 &micro;A
Chapter 5, Solution 17.
(a)
(b)
(c)
G=
R
vo
12
  f    –2.4
vi
Ri
5
vo
80

= –16
vi
5
vo
2000

 –400
vi
5
(a) –2.4, (b) –16, (c) –400
Chapter 5, Solution 18.
For the circuit, shown in Fig. 5.57, solve for the Thevenin equivalent circuit
looking into terminals A and B.
10 k
10 k 
a
b
7.5 V
+


+
c
A
2.5 
B
Figure 5.57
For Prob. 5.18.
Write a node equation at a. Since node b is tied to ground, v b = 0. We cannot write a
node equation at c, we need to use the constraint equation, v a = v b . Once, we know v c ,
we then proceed to solve for V open circuit and I short circuit . This will lead to V Thev (t) = V open
circuit and R equivalent = V open circuit /I short circuit .
[(v a – 7.5)/10k] + [(v a – v c )/10k] + 0 = 0
v a = v b = 0 or v c = –7.5 volts
This is also the open circuit voltage (note, the op-amp keeps the output voltage at –5 volts
in spite of any connection between A and B. Since this means that even a short from A to
B would theoretically then produce an infinite current, R equivalent = 0. In real life, the
short circuit current will be limited to whatever the op-amp can put out into a short
circuited output.
V Thev = –7.5 volts; R equivalent = 0-ohms.
Chapter 5, Solution 19.
We convert the current source and back to a voltage source.
24
(4/3) k
4 k
4
3
10 k
0V
(1.5/3)V
+


+
vo
2 k
10k  1.5 
   –937.5 mV.
4  3 

 4  k
3

v
v 0
 –562.5 &micro;A.
io  o  o
2k 10k
vo  
Chapter 5, Solution 20.
8 k
2 k
4 k
9V
a
+

4 k
vs
b

+
+

+
vo

At node a,
9  va va  vo va  vb


4
8
4
18 = 5v a – v o – 2v b
(1)
At node b,
va  vb vb  vo

4
2
v a = 3v b – 2v o
But v b = v s = 2 V; (2) becomes v a = 6 –2v o and (1) becomes
–18 = 30–10v o – v o – 4
v o = –44/(–11) = 4 V.
(2)
Chapter 5, Solution 21.
Let the voltage at the input of the op amp be v a .
va  1 V,
3-v a va  vo

4k
10k


3-1 1 vo

4
10
v o = –4 V.
Chapter 5, Solution 22.
A v = -R f /R i = -15.
If R i = 10k, then R f = 150 k.
Chapter 5, Solution 23
At the inverting terminal, v=0 so that KCL gives
vs  0
0 0  vo


R1
R2
Rf

vo
vs

Rf
R1
Chapter 5, Solution 24
v1
Rf
R1
R2
- vs +
+
+
R4
R3
vo
-
v2
We notice that v 1 = v 2 . Applying KCL at node 1 gives
v1 (v1  v s ) v1  vo


0
R1
R2
Rf


 1

  1  1 v1  v s  vo
R R
R2 R f
R f 
2
 1
Applying KCL at node 2 gives
R3
v1 v1  v s

0

 v1 
vs
R3
R4
R3  R4
Substituting (2) into (1) yields
 R
R
R  R3  1 
   v s
vo  R f  3  3  4 
 R1 R f R2  R3  R4  R2 
i.e.
 R
R
R  R3  1 
  
k  R f  3  3  4 
 R1 R f R2  R3  R4  R2 
(2)
(1)
Chapter 5, Solution 25.
This is a voltage follower. If v 1 is the output of the op amp,
v 1 = 3.7 V
v o = [20k/(20k+12k)]v 1 = [20/32]3.7 = 2.312 V.
Chapter 5, Solution 26
Using Fig. 5.64, design a problem to help other students better understand noninverting op amps.
Although there are many ways to work this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
Determine i o in the circuit of Fig. 5.64.
Figure 5.64
Solution
+
vb
+
0.4V
-
-
io
+
2k 
8k 
vo
-
vb  0.4 
8
vo  0.8vo
82


vo  0.4 / 0.8  0.5 V
Hence,
io 
vo
0.5

 0.1 mA
5k 5k
5k 
Chapter 5, Solution 27.
This is a voltage follower.
v 1 = [24/(24+16)]7.5 = 4.5 V; v 2 = v 1 = 4.5 V; and
v o = [12/(12+8)]4.5 = 2.7 V.
Chapter 5, Solution 28.
50 k
v1

+
va
10 k
At node 1,
+

10 V
0  v1 v1  v o

10k
50k
But v 1 = 10V,
–5v 1 = v 1 – v o , leads to
v o = 6v 1 = 60V
Alternatively, viewed as a noninverting amplifier,
v o = (1 + (50/10)) (10V) = 60V
i o = v o /(20k) = 60/(20k) = 3 mA.
vo
20 k
Chapter 5, Solution 29
R1
va
vb
+
vi
-
+
-
R2
+
R2
vo
R1
-
va 
R2
vi ,
R1  R2
But v a  vb
vb 


R1
vo
R1  R2
R2
R1
vi 
vo
R1  R2
R1  R2
Or
v o R2

vi
R1
Chapter 5, Solution 30.
The output of the voltage becomes
v o = v i = 1.2 V
(30k 20k )  12k
By voltage division,
vx 
12
(1.2)  0.2V
12  60
ix 
vx
20
0.2


 10A
20k 20k 2 x10 6
v 2x 0.04

 2W.
p
R
20k
Chapter 5, Solution 31.
After converting the current source to a voltage source, the circuit is as shown below:
12 k
3 k
1
6 k v
o
v1
12 V
+

+

2
vo
6 k
At node 1,
12  v1 v1  v o v1  v o


3
6
12
48 = 7v 1 - 3v o
(1)
At node 2,
v1  v o v o  0

 ix
6
6
From (1) and (2),
48
11
v
i x  o  727.2μA
6k
vo 
v 1 = 2v o
(2)
Chapter 5, Solution 32.
Let v x = the voltage at the output of the op amp. The given circuit is a non-inverting
amplifier.
 50 
v x  1   (4 mV) = 24 mV
 10 
60 30  20k
By voltage division,
v
20
v x  x  12mV
20  20
2
vx
24mV

 600 A
ix =
20  20k 40k
vo =
p=
v o2 144x10 6

 204 W.
R
60x10 3
Chapter 5, Solution 33.
After transforming the current source, the current is as shown below:
1 k
4 k
vi
+

va
4V
+

2 k
vo
3 k
This is a noninverting amplifier.
3
 1
v o  1   v i  v i
2
 2
Since the current entering the op amp is 0, the source resistor has a 0 V potential drop.
Hence v i = 4V.
vo 
3
(4)  6V
2
Power dissipated by the 3k resistor is
v o2 36

 12mW
R 3k
ix 
12mW, –2mA
va  vo 4  6

 –2mA.
R
1k
Chapter 5, Solution 34
v1  vin v1  vin

0
R1
R2
(1)
R3
vo
R3  R 4
(2)
but
va 
Combining (1) and (2),
v1  va 
R1
R
v 2  1 va  0
R2
R2

R 
R
v a 1  1   v1  1 v 2
R2
 R2 
R 3v o 
R 
R
1  1   v1  1 v 2
R3  R 4  R 2 
R2
vo 

R3  R 4 
R
 v1  1 v 2 
R2 

R 
R 3 1  1  
 R2 
vO =
R3  R4
(v1 R2  v2 )
R3 ( R1  R2 )
Chapter 5, Solution 35.
Av 
vo
R
 1  f  7.5
vi
Ri
R f = 6.5R i
If R i = 60 k, R f = 390 k.
Chapter 5, Solution 36
VTh  Vab
But
R1
Vab . Thus,
R1  R2
R
R  R2
 1
v s  (1  2 )v s
R1
R1
vs 
VTh  Vab
To get R Th , apply a current source I o at terminals a-b as shown below.
v1
v2
+
-
a
+
R2
vo
io
R1
b
Since the noninverting terminal is connected to ground, v 1 = v 2 =0, i.e. no current passes
through R 1 and consequently R 2 . Thus, v o =0 and
RTh 
vo
0
io
Chapter 5, Solution 37.
R

R
R
v o    f v1  f v 2  f v 3 
R2
R3 
 R1
30
30
 30

   (2)  (2)  (4.5)
20
30
 10

v o = 1.5 V.
Chapter 5, Solution 38.
Using Fig. 5.75, design a problem to help other students better understand summing amplifiers.
Although there are many ways to work this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
Calculate the output voltage due to the summing amplifier shown in Fig. 5.75.
Figure 5.75
Solution
R

R
R
R
v o    f v1  f v 2  f v 3  f v 4 
R2
R3
R4 
 R1
50
50
50
 50

   (10)  (20)  (50)  (100)
20
10
50
 25

= -120mV
Chapter 5, Solution 39
This is a summing amplifier.
Rf
Rf 
 Rf
50
50
 50

vo  
v1 
v2 
v3    (2)  v 2  (1)   9  2.5v 2
R2
R3 
20
50
 10

 R1
Thus,
v o  16.5  9  2.5v 2


v2  3 V
Chapter 5, Solution 40
Determine V o in terms of V 1 and V 2 .
200 k
100 k
100 k
V1
Va
+

V2
+


+
Vc
10 
+
40 
Vb
Vo

Step 1.
Label the reference and node voltages in the circuit, see above.
Note we now can consider nodes a and b, we cannot write a node equation at c
without introducing another unknown. The node equation at a is [(V a –V 1 )/105] +
[(V a –V 2 )/105] + 0 + [(V a –V c )/2x105] = 0. At b it is clear that V b = 0. Since we
have two equations and three unknowns, we need another equation. We do get
that from the constraint equation, V a = V b . After we find V c in terms of V 1 and
V 2 , we then can determine V o which is equal to [(V c –0)/50] times 40.
Step 2.
Letting V a = V b = 0, the first equation can be simplified to,
[–V 1 /105] + [–V 2 /105] + [–V c /2x105] = 0
Taking V c to the other side of the equation and multiplying everything by 2x105,
we get,
V c = –2V 1 – 2V 2
Now we can find V o which is equal to (40/50)V c = 0.8[–2V 1 –2V 2 ]
V o = –1.6V 1 –1.6V 2 .
Chapter 5, Solution 41.
R f /R i = 1/(4)
R i = 4R f = 40k
The averaging amplifier is as shown below:
R 1 = 40 k
10 k
v1
R 2 = 40 k
v2
R 3 = 40 k
v3
R 4 = 40 k
v4

+
vo
Chapter 5, Solution 42
Since the average of three numbers is the sum of those numbers divided by three, the
value of the feedback resistor needs to be equal to one-third of the input resistors or,
1
R f  R 1  25 kΩ.
3
Chapter 5, Solution 43.
In order for

R
R
R
R
v o   f v1  f v 2  f v 3  f v 4 
R2
R3
R4 
 R1
to become
1
v 1  v 2  v 3  v 4 
4
Rf 1
R 80k

Rf  i 
 20 k.
Ri 4
4
4
vo  
Chapter 5, Solution 44.
R4
R3
a
R1
v1

+
b
R2
vo
v2
At node b,
v b  v1 v b  v 2

0
R1
R2
At node a,
0  va va  vo

R3
R4
v1 v 2

R1 R 2
vb 
1
1

R1 R 2
(1)
vo
1 R4 / R3
(2)
va 
But v a = v b . We set (1) and (2) equal.
vo
R v  R 1v 2
 2 1
1 R4 / R3
R1  R 2
or
vo =
R3  R4  
R2 v 1  R1 v 2 
R3  R1  R2 
Chapter 5, Solution 45.
This can be achieved as follows:
 R
 v1   R v 2 
v o  
R/2 
R / 3
R

R
   f  v1   f v 2 
R2 
 R1
i.e. R f = R, R 1 = R/3, and R 2 = R/2
Thus we need an inverter to invert v 1 , and a summer, as shown below (R&lt;100k).
R
R
R
v1

+
R/3
-v 1
R/2
v2

+
vo
Chapter 5, Solution 46.
v1 1
R
R
R
1
 (  v 2 )  v 3  f v1  x (  v 2 )  f v 3
3 3
2
R1
R2
R3
i.e. R 3 = 2R f , R 1 = R 2 = 3R f . To get -v 2 , we need an inverter with R f = R i . If R f =
10k, a solution is given below.
 vo 
30 k
10 k
v1
10 k
v2

+
10 k
30 k
-v 2
20 k
v3

+
vo
Chapter 5, Solution 47.
Using eq. (5.18), R1  2k, R2  30k, R3  2k, R4  20k
vo 
30(1 2 / 30)
30
32
v2 
V1 
(2)  15(1)  14.09 V
2(1 2 / 20)
2
2.2
= 14.09 V.
Chapter 5, Solution 48.
We can break this problem up into parts. The 5 mV source separates the lower
circuit from the upper. In addition, there is no current flowing into the input of
the op amp which means we now have the 40-kohm resistor in series with a
parallel combination of the 60-kohm resistor and the equivalent 100-kohm
resistor.
+10 mV
Thus,
40k + (60x100k)/(160) = 77.5k
which leads to the current flowing through this part of the circuit,
i = 10 m/77.5k = 129.03x10–9 A
The voltage across the 60k and equivalent 100k is equal to,
v = ix37.5k = 4.839 mV
We can now calculate the voltage across the 80-kohm resistor.
v 80 = 0.8x4.839 m = 3.87 mV
which is also the voltage at both inputs of the op amp and the voltage between the
20-kohm and 80-kohm resistors in the upper circuit. Let v 1 be the voltage to the
left of the 20-kohm resistor of the upper circuit and we can write a node equation
at that node.
(v 1 –10m)/(10k) + v 1 /30k + (v 1 –3.87m)/20k = 0
or
6v 1 – 60m + 2v 1 + 3v 1 – 11.61m = 0
or
v 1 = 71.61/11 = 6.51 mV.
The current through the 20k-ohm resistor, left to right, is,
i 20 = (6.51m–3.87m)/20k = 132 x10–9 A
thus,
v o = 3.87m – 132 x10–9x80k = –6.69 mV.
Chapter 5, Solution 49.
R 1 = R 3 = 20k, R 2 /(R 1 ) = 4
i.e.
R 2 = 4R 1 = 80k = R 4
vo 
Verify:
4
R 2 1  R1 / R 2
R
v 2  2 v1
R1 1  R 3 / R 4
R1
(1  0.25)
v 2  4v1  4v 2  v1 
1  0.25
Thus, R 1 = R 3 = 20 k, R 2 = R 4 = 80 k.
Chapter 5, Solution 50.
(a) We use a difference amplifier, as shown below:
R1
R2
v1

+
R1
vo
R2
v2
vo 
R2
v 2  v1   2.5v 2  v1 , i.e. R 2 /R 1 = 2.5
R1
If R 1 = 100 k then R 2 = 250k
(b) We may apply the idea in Prob. 5.35.
v 0  2.5v1  2.5v 2
 R
 v1   R v 2 
 
R/2 
R / 2
R

R
   f  v1   f v 2 
R2 
 R1
i.e. R f = R, R 1 = R/2.5 = R 2
We need an inverter to invert v 1 and a summer, as shown below. We may let R =
100 k.
R
R
R
v1

+
R/2.5
-v 1
R/2.5
v2

+
vo
Chapter 5, Solution 51.
We achieve this by cascading an inverting amplifier and two-input inverting summer as
shown below:
R
R
R
v1

+
R
va
R
v2
Verify:
But
v o = -v a - v 2
v a = -v 1 . Hence
vo = v1 - v2.

+
vo
Chapter 5, Solution 52
Design an op amp circuit such that
v o = 4v 1 + 6v 2 – 3v 3 – 5v 4
Let all the resistors be in the range of 20 to 200 k.
Solution
A summing amplifier shown below will achieve the objective. An inverter is
inserted to invert v 2 . Since the smallest resistance must be at least 20 kΩ, then let
R/6 = 20kΩ therefore let R = 120 kΩ.
R/4
R
v1
R/6
v2
R

+
R
R/3
v3
R/5
v4

+
Chapter 5, Solution 53.
(a)
R1
R2
v1
va
vb

+
R1
vo
R2
v2
At node a,
At node b,
v1  v a v a  v o

R1
R2
R2
vb 
v2
R1  R 2
va 
R 2 v1  R 1 v o
R1  R 2
(1)
(2)
But v a = v b . Setting (1) and (2) equal gives
R v  R 1v o
R2
v2  2 1
R1  R 2
R1  R 2
R
v 2  v1  1 v o  v i
R2
vo R 2

vi
R1
(b)

v1
vi
+ v2
R 1 /2 v
A
R 1 /2
R2
va
Rg
R 1 /2
R 1 /2
vB
vb

+
R2
+
vo

At node A,
v1  v A v B  v A v A  v a


R1 / 2
Rg
R1 / 2
or
v1  v A 
At node B,
v2  vB vB  vA vB  vb


R1 / 2
R1 / 2
Rg
or
v2  vB 
R1
v B  v A   v A  v a
2R g
R1
(v B  v A )  v B  v b
2R g
(1)
(2)
Subtracting (1) from (2),
v 2  v1  v B  v A 
2R 1
v B  v A   v B  v A  v b  v a
2R g
R
v 2  v1 
 1 1
 2R
2
g


v B  v A   v i

2

Since, v a = v b ,
or
vB  vA 
vi

2
1
R
1 1
2R g
(3)
But for the difference amplifier,
R2
v B  v A 
R1 / 2
R
vB  vA  1 vo
2R 2
vo 
or
Equating (3) and (4),
R1
v
vo  i 
2R 2
2
vo R 2


vi
R1
(4)
1
R
1 1
2R g
1
R
1 1
2R g
(c)
At node a,
At node b,
v1  v a v a  v A

R1
R2 /2
2R 1
2R 1
v1  v a 
va 
vA
R2
R2
2R 1
2R 1
v2  vb 
vb 
vB
R2
R2
(1)
(2)
Since v a = v b , we subtract (1) from (2),
v
 2R 1
(v B  v A )  i
R2
2
 R2
vB  vA 
vi
2R 1
v 2  v1 
or
(3)
At node A,
va  vA vB  vA vA  vo


R2 /2
Rg
R/2
va  vA 
At node B,
R2
v B  v A   v A  v o
2R g
(4)
vb  vB vB  vA vB  0


R/2
Rg
R/2
vb  vB 
R2
v B  v A   v B
2R g
(5)
Subtracting (5) from (4),
v B v A 
R2
v B  v A   v A  v B  v o
Rg

R 
2v B  v A 1  2    v o
 2R 
g 

Combining (3) and (6),
 R 2 
R 
v i 1  2   v o
 2R 
R1
g 

v o R2 
R 
1 2 

vi
R1 
2 R g 
(6)
Chapter 5, Solution 54.
The first stage is a summer (please note that we let the output of the first stage be v 1 ).
R 
R
v1   v s  v o  = –v s – v o
R 
R
The second stage is a noninverting amplifier
v o = (1 + R/R)v 1 = 2v 1 = 2(–v s – v o ) or 3v o = –2v s
v o /v s = –0.6667.
Chapter 5, Solution 55.
Let A 1 = k, A 2 = k, and A 3 = k/(4)
A = A 1 A 2 A 3 = k3/(4)
20Log10 A  42
Log10 A  2.1
A = 102 1 = 125.89
k3 = 4A = 503.57
k = 3 503.57  7.956
Thus
A 1 = A 2 = 7.956, A 3 = 1.989
Chapter 5, Solution 56.
Using Fig. 5.83, design a problem to help other students better understand cascaded op
amps.
Although there are many ways to work this problem, this is an example based on the
same kind of problem asked in the third edition.
Problem
Calculate the gain of the op amp circuit shown in Fig. 5.83.
10 k
40 k
1 k
20 k
–
+
+
vi
–
Figure 5.83
–
+
For Prob. 5.56.
Solution
Each stage is an inverting amplifier. Hence,
vo
10
40
 ( )( )  20
vs
1
20
Chapter 5, Solution 57.
Let v 1 be the output of the first op amp and v 2 be the output of the second op amp.
The first stage is an inverting amplifier.
50
v1  
vs1  2vs1
25
The second state is a summer.
v 2 = –(100/50)v s2 – (100/100)v 1 = –2v s2 + 2v s1
The third state is a noninverting amplifier
vo  (1
100
)v2  3v2  6vs1  6vs2
50
Chapter 5, Solution 58.
Looking at the circuit, the voltage at the right side of the 5-kΩ resistor must be at 0V if
the op amps are working correctly. Thus the 1-kΩ is in series with the parallel
combination of the 3-kΩ and the 5-kΩ. By voltage division, the input to the voltage
follower is:
v1 
35
1 3 5
(0.6)  0.3913V = to the output of the first op amp.
Thus,
v o = –10((0.3913/5)+(0.3913/2)) = –2.739 V.
io 
0  vo
 684.8 &micro;A.
4k
Chapter 5, Solution 59.
The first stage is a noninverting amplifier. If v 1 is the output of the first op amp,
v 1 = (1 + 2R/R)v s = 3v s
The second stage is an inverting amplifier
v o = –(4R/R)v 1 = –4v 1 = –4(3v s ) = –12v s
v o /v s = –12.
Chapter 5, Solution 60.
The first stage is a summer. Let V 1 be the output of the first stage.
10
10
vi  vo

 v1  2vi  2.5vo
5
4
By voltage division,
10
5
v1 
vo  vo
10  2
6
Combining (1) and (2),
5
10
vo  2v1  2.5v0


v0  2vi
6
3
v1  
vo
 6 /10  0.6
vi
(1)
(2)
Chapter 5, Solution 61.
The first op amp is an inverter. If v 1 is the output of the first op amp,
V 1 = –(200/100)(0.4) = –0.8 V
The second op amp is a summer
V o = –(40/10)(–0.2) – (40/20)(–0.8) = 0.8 + 1.6
= 2.4 V.
Chapter 5, Solution 62.
Let v 1 = output of the first op amp
v 2 = output of the second op amp
The first stage is a summer
v1  
R2
R
vi – 2 vo
R1
Rf
(1)
The second stage is a follower. By voltage division
vo  v2 
R4
v1
R3  R4
v1 
R3  R4
vo
R4
(2)
From (1) and (2),
 R3 
R
R
 v o   2 v i  2 v o
1 
Rf
R1
 R4 
 R3 R2 
R
 v o   2 v i
1 

R1
 R4 Rf 
 R2 R4 R f
vo
R
1
 2 

R
R
R1 R2 R4  R3 R f  R4 R f
vi
R1
1 3  2
R4 Rf


Chapter 5, Solution 63.
The two op amps are summers. Let v 1 be the output of the first op amp. For the first
stage,
v1  
R2
R
vi  2 vo
R1
R3
(1)
For the second stage,
vo  
R4
R
v1  4 v i
R5
R6
(2)
Combining (1) and (2),
 R2 
R R 
R
 v i  4  2  v o  4 v i

R5  R3 
R6
 R1 
 R R  R R
R 
v o 1  2 4    2 4  4  v i
 R 3 R 5   R 1R 5 R 6 
vo 
R4
R5
R2 R4 R4

v o R1 R5 R6

R R
vi
1 2 4
R3 R5
Chapter 5, Solution 64
G4
G
G3
G1
1
+
0V
G
+
v
2
0V +
G2
vs
-
-
At node 1, v 1 =0 so that KCL gives
G1v s  G4 vo  Gv
(1)
At node 2,
G2 v s  G3 v o  Gv
From (1) and (2),
G1v s  G4 v o  G2 v s  G3 vo
or
+
vo
(2)


(G1  G2 )v s  (G3  G4 )vo
vo G1  G2

v s G3  G4
Chapter 5, Solution 65
The output of the first op amp (to the left) is 6 mV. The second op amp is an inverter so
that its output is
30
(6mV)  -18 mV
10
The third op amp is a noninverter so that
vo '  
vo ' 
40
vo
40  8


vo 
48
v o '   21.6 mV
40
Chapter 5, Solution 66.
We can start by looking at the contributions to v o from each of the sources and the fact
that each of them go through inverting amplifiers.
The 6 V source contributes –[100k/25k]6; the 4 V source contributes
–[40k/20k][–(100k/20k)]4; and the 2 V source contributes –[100k/10k]2 or
vo 
40  100 
100
 100
(6)   
(2)
(4) 
25
20  20 
10
 24  40  20  –4V
Chapter 5, Solution 67.
80  80 
80
  (0.3)  (0.7)
20
40  20 
 4.8  2.8  2 V.
vo = 
Chapter 5, Solution 68.
If R q = , the first stage is an inverter.
Va  
15
(15)  45 mV
5
when V a is the output of the first op amp.
The second stage is a noninverting amplifier.
 6
v o  1   v a  (1  3)(45)  –180mV.
 2
Chapter 5, Solution 69.
In this case, the first stage is a summer
va  
15
15
(15)  v o  45  1.5v o
5
10
For the second stage,
 6
v o  1   v a  4v a  4 45  1.5v o 
 2
180
7 v o  180 v o  
 –25.71 mV.
7
Chapter 5, Solution 70.
The output of amplifier A is
vA  
30
30
(1)  (2)  9
10
10
The output of amplifier B is
vB  
20
20
(3)  (4)  14
10
10
40 k
20 k
vA
a
60 k

+
vB
b
vo
10 k
vb 
10
(14)  2V
60  10
At node a,
vA  va va  vo

20
40
But v a = v b = -2V, 2(-9+2) = -2-v o
Therefore,
v o = 12V
Chapter 5, Solution 71
20k 
5k 
100k 
40k 
+
+
1.5 V
–
v2
10k 
80k 
+
20k 
+
vo
-
+
10k 
v1
+
2.25V
–
v1  2.25,
+
-
30k 
v2  
v3
50k 
20
50
(1.5)  6,
v 3  (1  ) v1  6
5
30
100 
 100
v o  
v2 
v 3   (15  7.5)  7.5 V.
80 
 40
Chapter 5, Solution 72.
Since no current flows into the input terminals of ideal op amp, there is no voltage
drop across the 20 k resistor. As a voltage summer, the output of the first op
amp is
v 01 = 1.8 V
The second stage is an inverter
v2  
250
v 01
100
 2.5(1.8)  –4.5 V.
Chapter 5, Solution 73.
The first stage is a noninverting amplifier. The output is
50
vo1  (1.8)  1.8  10.8V
10
The second stage is another noninverting amplifier whose output is
vL  v01  10.8V
Chapter 5, Solution 74.
Let v 1 = output of the first op amp
v 2 = input of the second op amp.
The two sub-circuits are inverting amplifiers
100
(0.9)  9V
10
32
v2  
(0.6)  12V
1.6
 9  12
v  v2

io  1
 150 A.
20k
20k
v1  
Chapter 5, Solution 75.
The schematic is shown below. Pseudo-components VIEWPOINT and IPROBE are involved as
shown to measure v o and i respectively. Once the circuit is saved, we click Analysis | Simulate.
The values of v and i are displayed on the pseudo-components as:
i = 200 A
(v o /v s ) = -4/2 = –2
The results are slightly different than those obtained in Example 5.11.
Chapter 5, Solution 76.
The schematic is shown below. IPROBE is inserted to measure i o . Upon simulation, the value
of i o is displayed on IPROBE as
i o = –562.5 A
11.25V
–19.358uV
0.750V
375mV
750 mV
–936.8mV
2 kΩ
–11.25V
Chapter 5, Solution 77.
The schematic for the PSpice solution is shown below.
Note that the output voltage, –6.686 mV, agrees with the answer to problem, 5.48.
6.510mV
3.872mV
–6.686mV
3.872mV
0.0100V
4.838mV
Chapter 5, Solution 78.
The circuit is constructed as shown below. We insert a VIEWPOINT to display v o . Upon
simulating the circuit, we obtain,
v o = 667.75 mV
Chapter 5, Solution 79.
The schematic is shown below.
v o = –4.992 V
R3
R4
R5
20k
10k
40k
V5
V3
20Vdc
1Vdc
0
1.000V
1
OS1
6
2.000V OUT
5
OS2
7
V-
4
1.000V
+
3
7
U1
+
OS2
OUT
2
uA741
R1
R2
20k
10k
-
4
0
0
20.00V
0V
V+
0V
OS1
V-
-20.00V
V4
20Vdc
1.666V
V6
5Vdc
3
V+
0
V2
20Vdc
-
U2
uA741
0V
2
R6
100k
5
6
1
20.00V
V1
20Vdc
-4.992V
1.666V
5.000V
0
Checking using nodal analysis we get,
For the first op-amp we get v a1 = [5/(20+10)]10 = 1.6667 V = v b1 .
For the second op-amp, [(v b1 – 1)/20] + [(v b1 – v c2 )/10] = 0 or v c2 = 10[1.6667–1)/20] +
1.6667 = 2 V;
[(v a2 – v c2 )/40] + [(v a2 – v c1 )/100] = 0; and v b2 = 0 = v a2 . This leads to v c1 = –2.5v c2 .
Thus,
= –5 V.
Chapter 5, Solution 80.
The schematic is as shown below. After it is saved and simulated, we obtain
v o = 2.4 V.
Chapter 5, Solution 81.
The schematic is shown below. We insert one VIEWPOINT and one IPROBE to measure v o
and i o respectively. Upon saving and simulating the circuit, we obtain,
v o = 343.4 mV
i o = 24.51 A
Chapter 5, Solution 82.
The maximum voltage level corresponds to
11111 = 25 – 1 = 31
Hence, each bit is worth
(7.75/31) = 250 mV
Chapter 5, Solution 83.
The result depends on your design. Hence, let R G = 10 k ohms, R 1 = 10 k ohms, R 2 = 20 k
ohms, R 3 = 40 k ohms, R 4 = 80 k ohms, R 5 = 160 k ohms, R 6 = 320 k ohms, then,
-v o = (R f /R 1 )v 1 + --------- + (R f /R 6 )v 6
= v 1 + 0.5v 2 + 0.25v 3 + 0.125v 4 + 0.0625v 5 + 0.03125v 6
(a)
|v o | = 1.1875 = 1 + 0.125 + 0.0625 = 1 + (1/8) + (1/16) which implies,
[v 1 v 2 v 3 v 4 v 5 v 6 ] = [100110]
(b)
|v o | = 0 + (1/2) + (1/4) + 0 + (1/16) + (1/32) = (27/32) = 843.75 mV
(c)
This corresponds to [1 1 1 1 1 1].
|v o | = 1 + (1/2) + (1/4) + (1/8) + (1/16) + (1/32) = 63/32 = 1.96875 V
Chapter 5, Solution 84.
(a)
The easiest way to solve this problem is to use superposition and to solve for each
term letting all of the corresponding voltages be equal to zero. Also, starting with
each current contribution (i k ) equal to one amp and working backwards is easiest.
2R
v1
+

R
R
R
2R
2R
2R
ik
v2
+

v3
+

v4
2R
+

For the first case, let v 2 = v 3 = v 4 = 0, and i 1 = 1A.
Therefore,
v 1 = 2R volts or i 1 = v 1 /(2R).
Second case, let v 1 = v 3 = v 4 = 0, and i 2 = 1A.
2R
R
R
2R
R
2R
2R
i2
v2
Simplifying, we get,
+

2R
2R
R
1A
2R
v2
+

Therefore,
v 2 = 1xR + (3/2)(2R) = 4R volts or i 2 = v 2 /(4R) or i 2 =
0.25v 2 /R. Clearly this is equal to the desired 1/4th.
Now for the third case, let v 1 = v 2 = v 4 = 0, and i 3 = 1A.
2R
5R/3
1.5
2R
v3
+

The voltage across the 5R/3-ohm resistor is 5R/2 volts. The current through the
2R resistor at the top is equal to (5/4) A and the current through the 2R-ohm
resistor in series with the source is (3/2) + (5/4) = (11/4) A. Thus,
v 3 = (11/2)R + (5/2)R = (16/2)R = 8R volts or i 3 = v 3 /(8R) or 0.125v 3 /R. Again,
we have the desired result.
For the last case, v 1 = v 2 = v 3 and i 4 = 1A. Simplifying the circuit we get,
R
1A
R
2R
R
2R
v4
5R/3
1.5A
2R
2R
+

R
2R
2R
v4
+

2R
21R/11
11/4A
2R
v4
+

2R
Since the current through the equivalent 21R/11-ohm resistor is (11/4) amps, the
voltage across the 2R-ohm resistor on the right is (21/4)R volts. This means the
current going through the 2R-ohm resistor is (21/8) A. Finally, the current going
through the 2R resistor in series with the source is ((11/4)+(21/8)) = (43/8) A.
Now, v 4 = (21/4)R + (86/8)R = (128/8)R = 16R volts or i 4 = v 4 /(16R) or
0.0625v 4 /R. This is just what we wanted.
(b)
If R f = 12 k ohms and R = 10 k ohms,
-v o = (12/20)[v 1 + (v 2 /2) + (v 3 /4) + (v 4 /8)]
= 0.6[v 1 + 0.5v 2 + 0.25v 3 + 0.125v 4 ]
For
[v 1 v 2 v 3 v 4 ] = [1 0 11],
|v o | = 0.6[1 + 0.25 + 0.125] = 825 mV
For
[v 1 v 2 v 3 v 4 ] = [0 1 0 1],
|v o | = 0.6[0.5 + 0.125] = 375 mV
Chapter 5, Solution 85.
This is a noninverting amplifier.
v o = (1 + R/40k)v s = (1 + R/40k)2
The power being delivered to the 10-kΩ give us
P = 10 mW = (v o )2/10k or v o = 10  2 x10 4 = 10V
Returning to our first equation we get
10 = (1 + R/40k)2 or R/40k = 5 – 1 = 4
Thus,
R = 160 kΩ.
Chapter 5, Solution 86.
Design a voltage controlled ideal current source (within the operating limits of the op
amp) where the output current is equal to 200v s (t) &micro;A.
The easiest way to solve this problem is to understand that the op amp creates an output
voltage so that the current through the feedback resistor remains equal to the input
current.
In the following circuit, the op amp wants to keep the voltage at a equal to zero. So, the
input current is v s /R = 200v s (t) &micro;A = v s (t)/5k.
Thus, this circuit acts like an ideal voltage controlled current source no matter what
(within the operational parameters of the op amp) is connected between a and b. Note,
you can change the direction of the current between a and b by sending v s (t) through an
inverting op amp circuit.
b
a
5kΩ
v s (t)
+
–

+
Chapter 5, Solution 87.
The output, v a , of the first op amp is,
Also,
v a = (1 + (R 2 /R 1 ))v 1
(1)
v o = (-R 4 /R 3 )v a + (1 + (R 4 /R 3 ))v 2
(2)
Substituting (1) into (2),
v o = (-R 4 /R 3 ) (1 + (R 2 /R 1 ))v 1 + (1 + (R 4 /R 3 ))v 2
Or,
If
v o = (1 + (R 4 /R 3 ))v 2 – (R 4 /R 3 + (R 2 R 4 /R 1 R 3 ))v 1
R 4 = R 1 and R 3 = R 2 , then,
v o = (1 + (R 4 /R 3 ))(v 2 – v 1 )
which is a subtractor with a gain of (1 + (R 4 /R 3 )).
Chapter 5, Solution 88.
We need to find V Th at terminals a – b, from this,
v o = (R 2 /R 1 )(1 + 2(R 3 /R 4 ))V Th = (500/25)(1 + 2(10/2))V Th
= 220V Th
Now we use Fig. (b) to find V Th in terms of v i .
a
a
30 k
20 k
20 k
vi
vi
30 k
+
40 k
80 k
40 k
b
80 k
b
(a)
(b)
v a = (3/5)v i , v b = (2/3)v i
V Th = v b – v a (1/15)v i
(v o /v i ) = A v = -220/15 = -14.667
Chapter 5, Solution 89.
A summer with v o = –v 1 – (5/3)v 2 where v 2 = 6-V battery and an inverting amplifier
with v 1 = –12v s .
Chapter 5, Solution 90.
The op amp circuit in Fig. 5.107 is a current amplifier. Find the current gain i o /i s of the
amplifier.
Figure 5.107
For Prob. 5.90.
Solution
Transforming the current source to a voltage source produces the circuit below,
v b = (2/(2 + 4))v o = v o /3
At node b,
20 k
5 k a
5i s
+

b

+
4 k
io
2 k
+
vo

At node a,
(5i s – v a )/5 = (v a – v o )/20
But v a = v b = v o /3.
20i s – (4/3)v o = (1/3)v o – v o , or i s = v o /30
i o = [(2/(2 + 4))/2]v o = v o /6
i o /i s = (v o /6)/(v o /30) = 5
Chapter 5, Solution 91.

+
vo
R2
R1
is
i2
i1
But
io
io = i1 + i2
(1)
i1 = is
(2)
R 1 and R 2 have the same voltage, v o , across them.
R 1 i 1 = R 2 i 2 , which leads to i 2 = (R 1 /R 2 )i 1
Substituting (2) and (3) into (1) gives,
i o = i s (1 + R 1 /R 2 )
i o /i s = 1 + (R 1 /R 2 ) = 1 + 8/1 = 9
(3)
Chapter 5, Solution 92
The top op amp circuit is a non-inverter, while the lower one is an inverter. The output
at the top op amp is
v 1 = (1 + 60/30)v i = 3v i
while the output of the lower op amp is
v 2 = -(50/20)v i = -2.5v i
Hence,
v o = v 1 – v 2 = 3v i + 2.5v i = 5.5v i
v o /v i = 5.5
Chapter 5, Solution 93.
R3
R1 v
a
vb

+
io
+
R4
vi
+

R2
iL
vL
RL
+
vo


At node a,
(v i – v a )/R 1 = (v a – v o )/R 3
v i – v a = (R 1 /R 2 )(v a – v o )
v i + (R 1 /R 3 )v o = (1 + R 1 /R 3 )v a
(1)
But v a = v b = v L . Hence, (1) becomes
v i = (1 + R 1 /R 3 )v L – (R 1 /R 3 )v o
(2)
i o = v o /(R 4 + R 2 ||R L ), i L = (R 2 /(R 2 + R L ))i o = (R 2 /(R 2 + R L ))(v o /( R 4 + R 2 ||R L ))
Or,
v o = i L [(R 2 + R L )( R 4 + R 2 ||R L )/R 2
(3)
But,
vL = iLRL
(4)
Substituting (3) and (4) into (2),
v i = (1 + R 1 /R 3 ) i L R L – R 1 [(R 2 + R L )/(R 2 R 3 )]( R 4 + R 2 ||R L )i L
= [((R 3 + R 1 )/R 3 )R L – R 1 ((R 2 + R L )/(R 2 R 3 )(R 4 + (R 2 R L /(R 2 + R L ))]i L
= (1/A)i L
Thus,
A =
1

 R  RL
R 
 1  1  R L  R 1  2
R3 

 R 2R 3

R 2RL
 R 4 
R2  RL




Please note that A has the units of mhos. An easy check is to let every resistor equal 1ohm and v i equal to one amp. Going through the circuit produces i L = 1A. Plugging into
the above equation produces the same answer so the answer does check.
Chapter 6, Solution 1.
iC


dv
 7.5 2e 3t  6te 3t  15(1 – 3t)e-3t A
dt
p = vi = 15(1–3t)e–3t  2t e–3t = 30t(1 – 3t)e–6t W.
15(1 – 3t)e-3t A, 30t(1 – 3t)e–6t W
Chapter 6, Solution 2.
w(t) = (1/2)C(v(t))2 or (v(t))2 = 2w(t)/C = (20cos2(377t))/(50x10–6) =
0.4x106cos2(377t) or v(t) = &plusmn;632.5cos(377t) V. Let us assume that v(t) =
632.5cos(377t) V, which leads to i(t) = C(dv/dt) = 50x10–6(632.5)(–377sin(377t))
= –11.923sin(377t) A.
Please note that if we had chosen the negative value for v,
then i(t) would have been positive.
Chapter 6, Solution 3.
Design a problem to help other students to better understand how capacitors work.
Although there are many ways to work this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
In 5 s, the voltage across a 40-mF capacitor changes from 160 V to
220 V. Calculate the average current through the capacitor.
Solution
i=C
dv
220  160
 40x10 3
 480 mA
dt
5
Chapter 6, Solution 4.
v
1 t
idt  v(0)
C o
t
1 t
 0.8

  4 sin( 4t )dt  1   
cos(4t )   1  0.2 cos(4t )  0.2  1
0
5
 4
0
= [1.2 – 0.2 cos(4t)] V.
Chapter 6, Solution 5.
 5000 t , 0  t  2ms

v =  20  5000 t , 2  t  6ms
 40  5000 t , 6  t  8ms

dv 4 x10 6

iC
dt
10 3
 5,

5,

 5,
0  t  2ms  20 mA,

2  t  6ms  20 mA,
6  t  8ms  20 mA,
0  t  2ms
2  t  6ms
6  t  8ms
Chapter 6, Solution 6.
dv
 55 x10 6 times the slope of the waveform.
dt
For example, for 0 &lt; t &lt; 2,
iC
dv
10

dt 2x10 3
dv
10
 (55 x106 )
 275mA
dt
2 x103
Thus the current i(t) is sketched below.
i= C
i(t) (mA)
275
4
2
–275
8
6
t (msec)
10
12
Chapter 6, Solution 7.
v
=
1
1
idt  v(t o ) 

25 x10 3
C
t
 5tx10
o
2.5t 2
10  [0.1t2 + 10] V.
25
3
dt  10
Chapter 6, Solution 8.
(a) i  C
dv
 100 ACe 100t  600 BCe 600t
dt
i (0)  2  100 AC  600 BC


(1)
5   A  6B
v (0  )  v (0  )

 50  A  B
Solving (2) and (3) leads to
A=61, B=-11
(b) Energy 
(2)
(3)
1 2
1
Cv (0)  x 4 x10 3 x 2500  5 J
2
2
(c ) From (1),
i  100 x 61 x 4 x10 3 e 100 t  600 x11 x 4 x10 3 e 600 t   24.4e 100 t  26.4e 600 t A
Chapter 6, Solution 9.
v(t) =




t
1 t
6 1  e  t dt  0  12 t  e  t V = 12(t + e-t) – 12

12 o
0
v(2) = 12(2 + e-2) – 12 = 13.624 V
p = iv = [12 (t + e-t) – 12]6(1-e-t)
p(2) = [12 (2 + e-2) – 12]6(1-e-2) = 70.66 W
Chapter 6, Solution 10
iC
dv
dv
 5 x10 3
dt
dt
 16t , 0  t  1s

v   16, 1  t  3 s
64 - 16t, 3  t  4s

 16 x10 6 , 0  t  1s
dv 
  0, 1  t  3 s
dt 
6
- 16x10 , 3  t  4 s
0  t  1s
 80 kA,

i ( t )   0, 1  t  3 s
- 80 kA,
3  t  4s

Chapter 6, Solution 11.
t
t
1
1
v   idt  v(0)  10 
i(t)dt
C0
4 x10 3 0
t
For 0&lt;t &lt;2, i(t)=15mA, V(t)= 10+
v  10 
103
15dt  10  3.76t
4 x10 3 0
v(2) = 10+7.5 =17.5
For 2 &lt; t &lt;4, i(t) = –10 mA
t
t
1
10 x10 3
v(t) 
i(t)dt  v(2)  
dt  17.5  22.5  2.5t
4 x10 3 2
4 x10 3 2
v(4)=22.5-2.5x4 =12.5
t
v(t) 
For 4&lt;t&lt;6, i(t) = 0,
1
0dt  v(4) 12.5
4 x103 2
For 6&lt;t&lt;8, i(t) = 10 mA
t
v(t) 
10 x103
dt  v(6) 2.5(t  6)  12.5  2.5t  2.5
4 x103 4
Hence,
 10  3.75t V,
22.5  2.5t V,

v(t) = 
 12.5 V,
 2.5t  2.5 V,
which is sketched below.
v(t)
0  t  2s
2  t  4s
4  t  6s
6  t  8s
20
15
10
5
t (s)
0
2
4
6
8
Chapter 6, Solution 12.
i R = V/R = (30/12)e–2000t = 2.5 e–2000t and i C = C(dv/dt) = 0.1x30(–2000) e–2000t
= –6000 e–2000t A. Thus, i = i R + i C = –5,997.5 e–2000t. The power is equal to:
vi = –179.925 e–4000t W.
Chapter 6, Solution 13.
Under dc conditions, the circuit becomes that shown below:
i1
10 
i2
20 
+
70 
50 
+
v2
v1

60V
+

i 2 = 0, i 1 = 60/(70+10+20) = 0.6 A
v 1 = 70i 1 = 42 V, v 2 = 60–20i 1 = 48 V
Thus, v 1 = 42 V, v 2 = 48 V.

Chapter 6, Solution 14.
20 pF is in series with 60pF = 20*60/80=15 pF
30-pF is in series with 70pF = 30x70/100=21pF
15pF is in parallel with 21pF = 15+21 = 36 pF
Chapter 6, Solution 15.
Arranging the capacitors in parallel results in circuit shown in Fig. (1) (It should
be noted that the resistors are in the circuits only to limit the current surge as the
capacitors charge. Once the capacitors are charged the current through the
resistors are obviously equal to zero.):
v 1 = v 2 = 100
R
R
+
100V
+

C1
v1
+
v2

C2
100V
+

+
C1
+
v2

(1)
v1 

(2)
1 2 1
Cv  x 25 x10 6 x100 2  125 mJ
2
2
1
w 30 = x75 x10 6 x100 2  375 mJ
2
w 20 =
(b)
Arranging the capacitors in series results in the circuit shown in Fig. (2):
v1 
C2
75
V
x100  75 V, v 2 = 25 V
C1  C2
100
w 25 =
1
x 25 x10 6 x75 2  70.31 mJ
2
w 75 =
1
x75 x10 6 x 25 2  23.44 mJ.
2
(a) 125 mJ, 375 mJ (b) 70.31 mJ, 23.44 mJ
C2
Chapter 6, Solution 16
C eq  14 
Cx 80
 30
C  80


C  20 F
Chapter 6, Solution 17.
(a)
(b)
(c)
4F in series with 12F = 4 x 12/(16) = 3F
3F in parallel with 6F and 3F = 3+6+3 = 12F
4F in series with 12F = 3F
i.e. C eq = 3F
C eq = 5 + [6x(4 + 2)/(6+4+2)] = 5 + (36/12) = 5 + 3 = 8F
3F in series with 6F = (3 x 6)/9 = 2F
1
1 1 1
   1
C eq 2 6 3
C eq = 1F
Chapter 6, Solution 18.
4 F in parallel with 4 F = 8F
4 F in series with 4 F = 2 F
2 F in parallel with 4 F = 6 F
Hence, the circuit is reduced to that shown below.
8F
6 F
6 F
C eq
1
1 1 1
    0.4583
Ceq 6 6 8

 Ceq  2.1818 F
Chapter 6, Solution 19.
We combine 10-, 20-, and 30-  F capacitors in parallel to get 60  F. The 60 -  F
capacitor in series with another 60-  F capacitor gives 30  F.
30 + 50 = 80  F, 80 + 40 = 120  F
The circuit is reduced to that shown below.
12
120
12
80
120-  F capacitor in series with 80  F gives (80x120)/200 = 48
48 + 12 = 60
60-  F capacitor in series with 12  F gives (60x12)/72 = 10 &micro;F
Chapter 6, Solution 20.
Consider the circuit shown below.
C1
C2
C3
C1  1 1  2  F
C2  2  2  2  6 F
C3  4 x3  12  F
1/C eq = (1/C 1 ) + (1/C 2 ) + (1/C 3 ) = 0.5 + 0.16667 + 0.08333 = 0.75x106
C eq = 1.3333 &micro;F.
Chapter 6, Solution 21.
4F in series with 12F = (4x12)/16 = 3F
3F in parallel with 3F = 6F
6F in series with 6F = 3F
3F in parallel with 2F = 5F
5F in series with 5F = 2.5F
Hence C eq = 2.5F
Chapter 6, Solution 22.
Combining the capacitors in parallel, we obtain the equivalent circuit shown below:
a
b
40 F
60 F
30 F
20 F
Combining the capacitors in series gives C1eq , where
1
1
1
1
1




1
C eq 60 20 30 10
Thus
C eq = 10 + 40 = 50 F
C1eq = 10F
Chapter 6, Solution 23.
Using Fig. 6.57, design a problem to help other students better understand how capacitors work
together when connected in series and parallel.
Although there are many ways to work this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
For the circuit in Fig. 6.57, determine:
(a) the voltage across each capacitor,
(b) the energy stored in each capacitor.
Figure 6.57
Solution
(a)
(b)
3F is in series with 6F
v 4F = 1/2 x 120 = 60V
v 2F = 60V
3
v 6F =
(60)  20V
63
v 3F = 60 - 20 = 40V
3x6/(9) = 2F
Hence w = 1/2 Cv2
w 4F = 1/2 x 4 x 10-6 x 3600 = 7.2mJ
w 2F = 1/2 x 2 x 10-6 x 3600 = 3.6mJ
w 6F = 1/2 x 6 x 10-6 x 400 = 1.2mJ
w 3F = 1/2 x 3 x 10-6 x 1600 = 2.4mJ
Chapter 6, Solution 24.
20F is series with 80F = 20x80/(100) = 16F
14F is parallel with 16F = 30F
(a) v 30F = 90V
v 60F = 30V
v 14F = 60V
80
v 20F =
x 60  48V
20  80
v 80F = 60 - 48 = 12V
1 2
Cv
2
w 30F = 1/2 x 30 x 10-6 x 8100 = 121.5mJ
w 60F = 1/2 x 60 x 10-6 x 900 = 27mJ
w 14F = 1/2 x 14 x 10-6 x 3600 = 25.2mJ
w 20F = 1/2 x 20 x 10-6 x (48)2 = 23.04mJ
w 80F = 1/2 x 80 x 10-6 x 144 = 5.76mJ
(b) Since w =
Chapter 6, Solution 25.
(a) For the capacitors in series,
Q1 = Q2
vs = v1 + v2 =
Similarly, v 1 
v1 C 2

v 2 C1
C1v1 = C2v2
C2
C  C2
v2  v2  1
v2
C1
C1
C2
vs
C1  C 2
(b) For capacitors in parallel
Q1 Q 2

C1 C 2
C  C2
C
Qs = Q1 + Q2 = 1 Q 2  Q 2  1
Q2
C2
C2
v1 = v2 =
or
C2
C1  C 2
C1
Q1 
Qs
C1  C 2
Q2 =
i=
dQ
dt
i1 
C1
is ,
C1  C 2
i2 
C2
is
C1  C 2
v2 
C1
vs
C1  C 2
Chapter 6, Solution 26.
(a)
C eq = C 1 + C 2 + C 3 = 35F
(b)
Q 1 = C 1 v = 5 x 150C = 0.75mC
Q 2 = C 2 v = 10 x 150C = 1.5mC
Q 3 = C 3 v = 20 x 150 = 3mC
(c)
w=
1
1
C eq v 2  x35x150 2 J = 393.8mJ
2
2
Chapter 6, Solution 27.
If they are all connected in parallel, we get CT  4 x4  F  16 F
If they are all connected in series, we get
1
4


 CT  1 F
CT 4  F
All other combinations fall within these two extreme cases. Hence,
C min = 1 &micro;F, C max = 16 &micro;F
Chapter 6, Solution 28.
We may treat this like a resistive circuit and apply delta-wye transformation, except that
R is replaced by 1/C.
Cb
50 F
Cc
20 F
Ca
 1  1   1  1   1  1 
          
1
10 40
10 30
30 40
         
1
Ca
30
3
1
1
2
=
 

40 10 40 10
C a = 5F
1
1
1


1
2
 400 300 1200 
1
Cb
30
10
C b = 15F
1
1
1


1
4
 400 300 1200 
1
Cc
15
40
C c = 3.75F
C b in parallel with 50F = 50 + 15 = 65F
C c in series with 20F = 23.75F
65x 23.75
65F in series with 23.75F =
 17.39F
88.75
17.39F in parallel with C a = 17.39 + 5 = 22.39F
Hence C eq = 22.39F
Chapter 6, Solution 29.
(a)
C in series with C = C/(2)
C/2 in parallel with C = 3C/2
3C
in series with C =
2
3
3C
2  3C
C
5
5
2
Cx
C
C
in parallel with C = C + 3  1.6 C
5
5
(b)
2C
C eq
2C
1
1
1
1



C eq 2C 2C C
C eq = 1 C
Chapter 6, Solution 30.
1 t
idt  i(0)
C o
For 0 &lt; t &lt; 1, i = 90t mA,
10 3 t
vo 
90tdt  0  15t 2 kV
3x10  6 o
v o (1) = 15 kV
vo =
For 1&lt; t &lt; 2, i = (180 – 90t) mA,
10 3 t
vo =
(180  90t )dt  vo (1)
3x10t 6 1
= [60t – 15t2 ] 1  15kV
= [60t – 15t2 – (60–15) + 15] kV = [60t – 15t2 – 30] kV
15t 2 kV ,
0t 1
vo ( t )  
2
[60t  15t  30]kV , 1  t  2
Chapter 6, Solution 31.
0  t 1
30tmA,

is (t )  30mA,
1 t  3
 75  15t , 3  t  5
C eq = 4 + 6 = 10F
1 t
v
idt  v(0)
C eq o
For 0 &lt; t &lt; 1,
v
10 3
10 x10 6
t
 30t dt + 0 = 1.5t
o
2
kV
For 1 &lt; t &lt; 3,
103 t
v
20dt  v(1)  [3(t  1)  1.5]kV
10 1
 [3t  1.5]kV
For 3 &lt; t &lt; 5,
103 t
v
15(t  5)dt  v(3)
10 3
 t2

 1.5  7.5t  t3 7.5kV  [0.75t 2  7.5t  23.25]kV
 2

1.5t 2 kV ,
0  t  1s

v ( t )  [3t  1.5]kV ,
1  t  3s
[0.75t 2  7.5t  23.25]kV , 3  t  5 s

dv
dv
 6x10 6
dt
dt
0  t  1s
18tmA,

i1  18mA,
1  t  3s
[9t  45]mA, 3  t  5 s
i 1  C1
i2  C2
dv
dv
 4x10 6
dt
dt
0  t  1s
12tmA,

i2  12mA,
1  t  3s
[6t  30]mA, 3  t  5 s
Chapter 6, Solution 32.
(a) C eq = (12x60)/72 = 10  F
103
v1 
12 x10 6
103
v2 
60 x10 6
t
 50e
 2t
dt  v1 (0)   2083e  2 t
t
0
 50   2083e  2 t  2133V
0
t
 50e
 2t
dt  v2 (0)   416.7e  2 t
t
0
 20   416.7e  2 t  436.7V
0
(b) At t=0.5s,
v1  2083e 1  2133  1366.7,
w12 F 
1
x12 x10  6 x(1366.7 )2  11.207 J
2
1
x 20 x10 6 x ( 283.4)2  803.2 mJ
2
1
 x 40 x10 6 x ( 283.4)2  1.6063 J
2
w20 F 
w40 F
v2  416.7e 1  436.7  283.4
Chapter 6, Solution 33
Because this is a totally capacitive circuit, we can combine all the capacitors using
the property that capacitors in parallel can be combined by just adding their
values and we combine capacitors in series by adding their reciprocals. However,
for this circuit we only have the three capacitors in parallel.
3 F + 2 F = 5 F (we need this to be able to calculate the voltage)
C Th = C eq = 5+3+2 = 10 F
The voltage will divide equally across the two 5 F capacitors. Therefore, we get:
V Th = 15 V, C Th = 10 F.
15 V, 10 F
Chapter 6, Solution 34.
i = 10e–t/2
di
1
v  L  10 x10 3 (10) e t / 2
dt
2
-t/2
= –50e mV
v(3) = –50e–3/2 mV = –11.157 mV
p = vi = –500e–t mW
p(3) = –500e–3 mW = –24.89 mW.
Chapter 6, Solution 35.
di
vL
dt
v
160 x10 3

 L

 6.4 mH
di / dt (100  50)x10 3
2 x10 3
Chapter 6, Solution 36.
Design a problem to help other students to better understand how inductors work.
Although there are many ways to work this problem, this is an example based on the
same kind of problem asked in the third edition.
Problem
The current through a 12-mH inductor is i(t)  30te2t A, t  0. Determine: (a) the
voltage across the inductor, (b) the power being delivered to the inductor at t = 1 s, (c)
the energy stored in the inductor at t = 1 s.
Solution
di
 12 x10 3(30e2t  60te2 t )  (0.36  0.72t)e2t V
dt
(b) p  vi  (0.36  0.72 x1)e2 x30 x1e2  0.36 x30e4  0.1978 W
1
(c) w  Li 2 = 0.5x12x10–3(30x1xe–2)2 = 98.9 mJ.
2
(a) v  L
Chapter 6, Solution 37.
di
 12 x10 3 x 4(100) cos100t
dt
= 4.8 cos (100t) V
vL
p = vi = 4.8 x 4 sin 100t cos 100t = 9.6 sin 200t
w=
t
11 / 200
o
o
 pdt  
9.6 sin 200 t
9.6
/ 200
cos 200t 11
J
o
200
 48(cos   1)mJ  96 mJ

Please note that this problem could have also been done by using (&frac12;)Li2.
Chapter 6, Solution 38.
vL


di
 40x10 3 e  2 t  2te  2 t dt
dt
= 40(1  2t )e 2 t mV , t  0
Chapter 6, Solution 39
vL
i
di
1

 i   0t idt  i(0)
dt
L
1
200x10
t
(3t 2
3  0
 5( t 3  t 2  4t )
t
0
 2 t  4)dt  1
1
i(t) = [5t3 + 5t2 + 20t + 1] A
Chapter 6, Solution 40.
 5t, 0  t  2ms

i   10, 2  t  4ms
30  5t, 4  t  6ms

 5,
di 5 x10 3 
vL

 0,
dt
10 3 
5,
0  t  2ms
 25, 0  t  2ms

2  t  4ms   0, 2  t  4ms
4  t  6ms  25, 4  t  6ms
At t = 1ms, v = 25 V
At t = 3ms, v = 0 V
At t = 5ms, v= –25 V
Chapter 6, Solution 41.
i


1 t
1 t
vdt  C     20 1  e  2 t dt  C

L 0
 2 o
 1

= 10 t  e 2 t  ot  C  10t  5e 2 t  4.7A
2


Note, we get C = –4.7 from the initial condition for i needing to be 0.3 A.
We can check our results be solving for v = Ldi/dt.
v = 2(10 – 10e–2t)V which is what we started with.
At t = l s, i = 10 + 5e-2 – 4.7 = 10 + 0.6767 – 4.7 = 5.977 A
w
1 2
L i = 35.72J
2
Chapter 6, Solution 42.
1 t
1 t
vdt  i(0)   v( t )dt  1

5 o
L o
10 t
For 0 &lt; t &lt; 1, i   dt  1  2t  1 A
5 0
i
For 1 &lt; t &lt; 2, i = 0 + i(1) = 1A
1
10dt  i(2)  2t 2t 1
5
= 2t - 3 A
For 2 &lt; t &lt; 3, i =
For 3 &lt; t &lt; 4, i = 0 + i(3) = 3 A
1 t
10dt  i(4)  2 t 4t 3

4
5
= 2t - 5 A
For 4 &lt; t &lt; 5, i =
Thus,
 2t  1 A,
1 A,

i ( t )   2t  3 A,

 3 A,
 2t  5,
0t 1
1 t  2
2t 3
3t 4
4t 5
Chapter 6, Solution 43.
w = L
t

idt 
1 2
1 2
Li ( t )  Li ()
2
2


2
1
x80x10 3 x 60x10 3  0
2
= 144 J.

Chapter 6, Solution 44.
di
 100 x10 3(400)x50 x10 3 e400t  2e400t V
dt
(b) Since R and L are in parallel, vR  vL  2e400t V
(c) No
1
(d) w  Li 2 = 0.5x100x10–3(0.05)2 = 125 &micro;J.
2
(a) vL  L
Chapter 6, Solution 45.
i(t) =
1 t
v( t )  i(0)
L o
For 0 &lt; t &lt; 1, v = 5t
i
1
10x10 3
t
 5t dt + 0
o
= 250t2 A
For 1 &lt; t &lt; 2, v = -10 + 5t
i
1
10x10 3
t
 (10  5t )dt  i(1)
1
t
  (0.5t  1)dt  0.25kA
1
= [1 – t + 0.25t2 ] kA
 250t 2 A,
0  t  1s
i(t )  
2
1  t  2s
[1  t  0.25t ] kA,
Chapter 6, Solution 46.
Under dc conditions, the circuit is as shown below:
2
iL
+
vC
4
3A

By current division,
iL 
4
(3)  2A, v c = 0V
42
wL 
1 2 11 2
L i L   (2)  1J
2
22
wc 
1
1
C v c2  (2)( v)  0J
2
2
Chapter 6, Solution 47.
Under dc conditions, the circuit is equivalent to that shown below:
R
+
vC

2
5A
iL 
2
10
10R
, v c  Ri L 
(5) 
R2
R2
R2
1 2
100R 2
6
w c  Cv c  80x10 x
2
(R  2) 2
1
100
w L  Li12  2x10 3 x
2
(R  2) 2
If w c = w L ,
80x10
6
x
100R 2
(R  2)
2

2x10 3 x100
(R  2)
2
R = 5
80 x 10-3R2 = 2
iL
Chapter 6, Solution 48.
Under steady-state, the inductor acts like a short-circuit, while the capacitor acts like
an open circuit as shown below.
i
+
5 mA
30k
v
20 k
–
Using current division,
i = (30k/(30k+20k))(5mA) = 3 mA
v = 20ki = 60 V
Chapter 6, Solution 49.
Converting the wye-subnetwork to its equivalent delta gives the circuit below.
30 mH
30mH
5mH
30 mH
30//0 = 0, 30//5 = 30x5/35=4.286
Leq  30 // 4.286 
30 x4.286
 3.75 mH
34.286
Chapter 6, Solution 50.
16mH in series with 14 mH = 16+14=30 mH
24 mH in series with 36 mH = 24+36=60 mH
30mH in parallel with 60 mH = 30x60/90 = 20 mH
Chapter 6, Solution 51.
1
1
1
1
1




L 60 20 30 10
L eq  10 25  10 
L = 10 mH
10x35
45
= 7.778 mH
Chapter 6, Solution 52.
Using Fig. 6.74, design a problem to help other students better understand how inductors
behave when connected in series and when connected in parallel.
Although there are many ways to work this problem, this is an example based on the
same kind of problem asked in the third edition.
Problem
Find L eq in the circuit of Fig. 6.74.
10 H
4H
6H
5H
3H
L eq
7H
Figure 6.74
For Prob. 6.52.
Solution
Leq  5 //(7  3  10 //(4  6))  5 //(7  3  5)) 
5 x15
 3.75 H
20
Chapter 6, Solution 53.
L eq  6  10  8 5 (8  12)  6 (8  4)
 16  8 (4  4)  16  4
L eq = 20 mH
Chapter 6, Solution 54.
L eq  4  (9  3) 10 0  6 12
 4  12 (0  4)  4  3
L eq = 7H
Chapter 6, Solution 55.
(a) L//L = 0.5L, L + L = 2L
Leq  L  2 L // 0.5L  L 
2 Lx0.5 L
 1.4 L = 1.4 L.
2 L  0.5L
(b) L//L = 0.5L, L//L + L//L = L
L eq = L//L = 500 mL
Chapter 6, Solution 56.
1 L

3 3
L
Hence the given circuit is equivalent to that shown below:
LLL
L
L/3
L/3
L
L eq
5
Lx L
2 

3  5L
 L L  L 
5
8
3 

L L
3
Chapter 6, Solution 57.
Let v  L eq
di
dt
v  v1  v 2  4
i = i1 + i2
v2  3
(1)
di
 v2
dt
i2 = i – i1
di1 di1 v 2
or

dt
dt
3
(2)
(3)
(4)
and
di
di
5 2  0
dt
dt
di
di
v2  2  5 2
dt
dt
 v2  2
Incorporating (3) and (4) into (5),
v2  2
di
v
di
di
di
5 5 1  7 5 2
dt
dt
dt
dt
3
di
 5
v 2 1    7
dt
 3
21 di
v2 
8 dt
Substituting this into (2) gives
v4

di 21 di

dt 8 dt
53 di
8 dt
Comparing this with (1),
L eq 
53
 6.625 H
8
(5)
Chapter 6, Solution 58.
vL
di
di
 3  3 x slope of i(t).
dt
dt
Thus v is sketched below:
v(t) (V)
6
t (s)
1
-6
2
3
4
5
6
7
Chapter 6, Solution 59.
(a) v s  L1  L 2 
di
dt
vs
di

dt L1  L 2
di
di
v 1  L1 , v 2  L 2
dt
dt
v1 
(b)
v i  v 2  L1
L1
L2
vs , vL 
vs
L1  L2
L1  L2
di1
di
 L2 2
dt
dt
i s  i1  i 2
di s di1 di 2
L  L 2 
v
v




v 1
dt
dt
dt
L1 L 2
L1 L 2
L1 L 2
1
1
vdt 


L1 L1  L 2
L1
L1 L 2
1
1
i2 
vdt 


L2
L 2 L1  L 2
i1 
L2
di s
dt 
is
L1  L2
dt
L1
di s
dt 
is
L1  L2
dt
Chapter 6, Solution 60
15
8
di 15 d
vo  Leq

4e  2t   15e 2t
dt 8 dt
Leq  3 // 5 

t

t
I
1
io   vo (t )dt  io (0)  2   (15)e  2t dt  2  1.5e  2t
L0
50
i o = (0.5 + 1.5e–2t) A
t
0
Chapter 6, Solution 61.
(a) Leq  20 //(4  6)  20 x10 / 30  6.667 mH
Using current division,
i1(t) 
10
i s  et mA
10  20
i2(t)  2e t mA
(b) vo  Leq
(c ) w 
dis 20

x10 3(3e t x10 3 )  20e t  V
dt
3
1 2 1
Li1  x20 x10 3 xe2 x10 6  1.3534 nJ
2
2
Chapter 6, Solution 62.
(a)
Leq  25  20 // 60  25 
v  Leq
di
dt
20 x60
 40 mH
80
1
10 3
(
)
(
0
)
v
t
dt

i

12e 3t dt  i (0)  0.1(e 3t  1)  i (0)
Leq 
40 x10 3 0
t


i
Using current division and the fact that all the currents were zero when the circuit was put
together, we get,
60
3
1
i1 
i  i, i 2  i
80
4
4
3
i1 (0)  i (0)

 0.75i (0)  0.01 
 i (0)  0.01333
4
i2 
1
(0.1e 3t  0.08667) A  - 25e -3t  21.67 mA
4
i 2 (0)  25  21.67   3.33 mA
(b) i1 
3
( 0.1e  3 t  0.08667 ) A  - 75e - 3t  65 mA
4
i 2  - 25e -3t  21.67 mA
Chapter 6, Solution 63.
We apply superposition principle and let
vo  v1  v 2
where v 1 and v 2 are due to i 1 and i 2 respectively.
di1
di  2, 0  t  3
2 1 
dt
dt  2, 3  t  6
0t 2
 4,
di2
di2 
2
  0,
2t 4
v2  L
dt
dt 
4t6
 4,
v1
v1  L
v2
2
4
0
3
6
t
0
-2
2
4
-4
Adding v 1 and v 2 gives v o , which is shown below.
v o (t) V
6
2
0
2
-2
-6
3
4
6
t (s)
6
t
Chapter 6, Solution 64.
(a) When the switch is in position A,
i= –6 = i(0)
When the switch is in position B,
i ()  12 / 4  3,
  L / R  1/ 8
i (t )  i ()  [i (0)  i ()]e  t / 
i(t) = (3 – 9e–8t) A
(b) -12 + 4i(0) + v=0, i.e. v = 12 – 4i(0) = 36 V
(c) At steady state, the inductor becomes a short circuit so that
v=0V
Chapter 6, Solution 65.
1
1
L1i12  x5x (4) 2  40 J
2
2
1
w 20  (20)(2) 2  40 J
2
(b) w = w 5 + w 20 = 80 J
1 t
(c) i1 
 50e  200 t dt  i1 (0) 

0
L1
= [5x10-5(e-200t – 1) + 4] A
(a)
w5 


t
1 1 
 200 t
x10  3  4

 50e
0
5  200 


1 t
1  1 
 200 t
 200 t
3 t
50
e
dt
i
(
0
)



50
e
x
10
2


2
L2  0
20  200 
0
= [1.25x10-5 (e-200t – 1) – 2] A
i2 
(d)
i = i 1 + i 2 = [6.25x10-5 (e-200t– 1) + 2] A
Chapter 6, Solution 66.
If v=i, then
di
dt di
iL



dt
L
i
Integrating this gives
 i 
t
  i = C o et/L
 ln(i)  ln(C o )  ln
L
 Co 
i(0) = 2 = C o
i(t) = 2et/0.02 = 2e50t A.
Chapter 6, Solution 67.
1
vi dt, RC = 50 x 103 x 0.04 x 10-6 = 2 x 10-3
RC 
 10 3
10 sin 50t dt
vo 
2 
v o = 100cos(50t) mV
vo  
Chapter 6, Solution 68.
1
vi dt + v(0), RC = 50 x 103 x 100 x 10-6 = 5
RC 
1 t
v o =   10dt  0  2t
5 o
The op amp will saturate at v o =  12
vo  
-12 = -2t
t = 6s
Chapter 6, Solution 69.
RC = 4 x 106 x 1 x 10-6 = 4
vo  
1
1
v i dt    v i dt

4
RC
For 0 &lt; t &lt; 1, v i = 20, v o  
1 t
20dt  -5t mV
4 o
1 t
10dt  v(1)  2.5( t  1)  5
4 1
= -2.5t - 2.5mV
For 1 &lt; t &lt; 2, v i = 10, v o  
1 t
20dt  v(2)  5( t  2)  7.5
4 2
= 5t - 17.5 mV
For 2 &lt; t &lt; 4, v i = - 20, v o  
1 t
10dt  v(4)  2.5( t  4)  2.5
4 4
= 2.5t - 7.5 mV
For 4 &lt; t &lt; 5m, v i = -10, v o 
1 t
20dt  v(5)  5( t  5)  5
4 5
= - 5t + 30 mV
For 5 &lt; t &lt; 6, v i = 20, v o  
Thus v o (t) is as shown below:
v(t) (V)
5
2.5
t (s)
0
1
-2.5
-5
-7.5
2
3
4
5
6
7
Chapter 6, Solution 70.
One possibility is as follows:
1
 50
RC
Let R = 100 k, C 
1
 0.2 F
50 x100 x10 3
Chapter 6, Solution 71.
By combining a summer with an integrator, we have the circuit below:
R1
v1
C
R2
v2
R3

+
vo
v3
vo  
1
1
1
v1dt 
v 2 dt 
v 2 dt


R 1C
R 2C
R 2C 
For the given problem, C = 2F,
R1C = 1
R 2 C = 1/(4)
R 3 C = 1/(10)
R 1 = 1/(C) = 106/(2) = 500 k
R 2 = 1/(4C) = 500k/(4) = 125 k
R 3 = 1/(10C) = 50 k
Chapter 6, Solution 72.
The output of the first op amp is
v1  
t
1
1
100t
v i dt = 
v i dt  


RC
2
10x10 3 x 2 x10 6 o
= - 50t
vo  
1
1
v i dt = 
3

RC
20x10 x 0.5x10 6
= 2500t2
At t = 1.5ms,
v o  2500(1.5) 2 x10 6  5.625 mV
t
 (50t )dt
o
Chapter 6, Solution 73.
Consider the op amp as shown below:
Let v a = v b = v
At node a,
0  v v  vo

R
R
2v - v o = 0
(1)
R
R
v
a
R

+
R
v
vo
b
vi
At node b,
+

C
vi  v v  vo
dv

C
R
R
dt
dv
v i  2v  v o  RC
dt
Combining (1) and (2),
v i v o v o 
RC dv o
2 dt
or
vo 
+
2
v i dt
RC 
showing that the circuit is a noninverting integrator.

(2)
Chapter 6, Solution 74.
RC = 0.01 x 20 x 10-3 sec
v o   RC
dv i
dv
 0.2 m sec
dt
dt
  2V,
v o  2V,
 2V,
0  t 1
1 t  3
3 t 4
Thus v o (t) is as sketched below:
v o (t) (V)
2
t (ms)
1
-2
2
3
Chapter 6, Solution 75.
v 0   RC
dv i
, RC  250 x10 3 x10x10 6  2.5
dt
v o  2.5
d
(12t )  –30 mV
dt
Chapter 6, Solution 76.
dv i
, RC = 50 x 103 x 10 x 10-6 = 0.5
dt
 10, 0  t  5
dv
v o  0.5 i  
5  t  15
dt
5,
v o   RC
The input is sketched in Fig. (a), while the output is sketched in Fig. (b).
v o (t) (V)
v i (t) (mV)
100
5
t (ms)
0
5
10
t (ms)
15
0
5
10
(a)
-10
(b)
15
Chapter 6, Solution 77.
i = iR + iC
vi  0 0  v0
d

 C 0  v o 
R
RF
dt
R F C  10 6 x10 6  1
dv 

Hence v i   v o  o 
dt 

Thus v i is obtained from v o as shown below:
–dv o (t)/dt
– v o (t) (V)
4
4
t (s)
t (s)
0
1
2
3
-4
-4
v i (t) (V)
8
t (s)
-4
-8
0
4
1
2
3
4
1
2
3
4
Chapter 6, Solution 78.
d 2 vo
2dv o
 10 sin 2 t 
 vo
dt
dt
Thus, by combining integrators with a summer, we obtain the appropriate analog
computer as shown below:
2v o
 +
C
C
R
2
d v o /dt
2
t=0
R
R

+
R

+
-dv o /dt

+
vo
R
2
R

+
R/2
dv o /dt
R
R
sin2t
+


+
d2v o /dt
R/10
-sin2t
Chapter 6, Solution 79.
We can write the equation as
dy
 f (t )  4 y (t )
dt
which is implemented by the circuit below.
1V
t=0
C
R
R
R
R/4
dy/dt
+
+
-y
R
f(t)
R
+
dy/dt
Chapter 6, Solution 80.
From the given circuit,
d 2vo
1000k
1000k dv o
 f (t) 
vo 
2
5000k
200k dt
dt
or
d 2vo
dv
 5 o  2v o  f ( t )
2
dt
dt
Chapter 6, Solution 81
We can write the equation as
d 2v
 5v  2 f (t )
dt 2
which is implemented by the circuit below.
C
C
R
R
2
2
d v/dt
+
R
R/5
-
-dv/dt
+
v
+
R/2
f(t)
d2v/dt2
Chapter 6, Solution 82
The circuit consists of a summer, an inverter, and an integrator. Such circuit is shown
below.
10R
R
R
R
+
+
R
C=1/(2R)
R
+
vs
-
+
vo
Chapter 6, Solution 83.
Since two 10F capacitors in series gives 5F, rated at 600V, it requires 8 groups in
parallel with each group consisting of two capacitors in series, as shown below:
+
600

Answer: 8 groups in parallel with each group made up of 2 capacitors in series.
Chapter 6, Solution 84.
v = L(di/dt) = 8x10–3x5x2πsin(πt)cos(πt)10–3 = 40πsin(2πt) &micro;V
p = vi = 40πsin(2πt)5sin2(πt)10–9 W, at t=0 p = 0W
w
1 2 1
Li  x8 x10 3 x[5 sin2( / 2)x10 3 ]2  4 x25 x10 9  100 nJ
2
2
= 100 ηJ
Chapter 6, Solution 85.
It is evident that differentiating i will give a waveform similar to v. Hence,
di
vL
dt
4t ,0  t  1ms
i
8  4 t ,1  t  2ms
 di 4000L,0  t  1ms 
v  L  

 dt  4000L,1  t  2ms
But,
5V,0  t  1ms
v
 5V,1  t  2ms
Thus, 4000L = 5
L = 1.25 mH in a 1.25 mH inductor
Chapter 6, Solution 86.
v  vR  vL  Ri  L
di
 12 x2te10t  200 x10 3 x(20te10t  2e10t )  (0.4  20t)e10t V
dt
Chapter 7, Solution 1.
(a)
=RC = 1/200
For the resistor, V=iR= 56e200 t  8Re 200t x10 3
C


R
56
 7 k
8
1
1

 0.7143  F
200R 200 X7 X103
(b)
 =1/200= 5 ms
(c) If value of the voltage at = 0 is 56 .
1
x56  56e200t
2


200to  ln2
e200 t  2


to 
1
ln2  3.466 ms
200
Chapter 7, Solution 2.
  R th C
where R th is the Thevenin equivalent at the capacitor terminals.
R th  120 || 80  12  60 
  60  200  10 -3  12 s.
Chapter 7, Solution 3.
R = 10 +20//(20+30) =10 + 40x50/(40 + 50)=32.22 k
  RC  32.22 X103 X100 X10 12  3.222  S
Chapter 7, Solution 4.
For t&lt;0, v(0-)=40 V.
For t &gt;0. we have a source-free RC circuit.
  RC  2 x103 x10 x10 6  0.02
v(t)  v(0)et /   40e50t V
Chapter 7, Solution 5.
Using Fig. 7.85, design a problem to help other students to better understand source-free
RC circuits.
Although there are many ways to work this problem, this is an example based on the
same kind of problem asked in the third edition.
Problem
For the circuit shown in Fig. 7.85, find i(t), t&gt;0.
2
t=0
i
+
5
4Ω
24 V 4 
_
1/3 F
Figure 7.85
For Prob. 7.5.
Solution
Let v be the voltage across the capacitor.
For t &lt;0,
4
v(0  ) 
(24)  16 V
24
For t &gt;0, we have a source-free RC circuit as shown below.
i
5
+
v
–
1
3
  RC  (4  5)  3 s
1/3 F
4
v(t)  v(0)et /   16e t / 3 V
dv
1 1
i(t)  C
  ( )16e t / 3  1.778et / 3 A
dt
3 3
Chapter 7, Solution 6.
v o  v ( 0) 
2
(40)  6.667 V
10  2
v( t )  v o e  t /  ,   RC  40 x10 6 x 2 x10 3 
v( t )  6.667e 12.5 t V
2
25
Chapter 7, Solution 7.
Assuming that the switch in Fig. 7.87 has been in position A for a long time and is
moved to position B at t=0. Then at t = 1second, the switch moves from B to C.
Find v C (t) for t  0.
10 k
12V
+

A
B
500 
C
1 k
2 mF
Figure 7.87
For Prob. 7.7
Solution
Step 1.
Determine the initial voltage on the capacitor. Clearly it charges to
12 volts when the switch is at position A because the circuit has reached steady state.
This then leaves us with two simple circuits, the first a 500 Ω resistor in series with a
2 mF capacitor and an initial charge on the capacitor of 12 volts. The second circuit
which exists from t = 1 sec to infinity. The initial condition for the second circuit
will be v C (1) from the first circuit. The time constant for the first circuit is
(500)(0.002) = 1 sec and the time constant for the second circuit is (1,000)(0.002) = 2
sec. v C (∞) = 0 for both circuits.
Step 1.
v C (t) = 12e-t volts for 0 &lt; t &lt; 1 sec and = 12e-1e-2(t-1) at t = 1 sec, and
= 4.415e-2(t-1) volts for 1 sec &lt; t &lt; ∞.
12e-t volts for 0 &lt; t &lt; 1 sec, 4.415e-2(t-1) volts for 1 sec &lt; t &lt; ∞.
Chapter 7, Solution 8.
(a)
  RC 
dv
dt
-4t
- 0.2 e  C (10)(-4) e-4t 
 C  5 mF
1
R
 50 
4C
1
  RC   0.25 s
4
1
1
w C (0)  CV02  (5  10 -3 )(100)  250 mJ
2
2
1 1
1
w R   CV02  CV02 1  e -2t 0  
2 2
2
1
 e -8t 0 
0.5  1  e -8t 0 
2
8t 0
e 2
or
1
t 0  ln (2)  86.6 ms
8
-i  C
(b)
(c)
(d)
1
4
Chapter 7, Solution 9.
For t &lt; 0, the switch is closed so that
4
vo(0) 
(6)  4 V
24
For t &gt;0, we have a source-free RC circuit.
  RC  3 x103 x4 x103  12 s
v o (t) = v o (0)e–t/τ = 4e–t/12 V.
Chapter 7, Solution 10.
3
(36V )  9 V
39
For t&gt;0, we have a source-free RC circuit
  RC  3 x103 x20 x106  0.06 s
For t&lt;0,
v(0  ) 
v o (t) = 9e–16.667t V
Let the time be t o .
3 = 9e–16.667to or e16.667to = 9/3 = 3
t o = ln(3)/16.667 = 65.92 ms.
Chapter 7, Solution 11.
For t&lt;0, we have the circuit shown below.
4
4H
4
24 V
8
+
_
4H
io
4
6A
4
4||4= 4x4/8=2
i o (0–) = [2/(2+8)]6 = 1.2 A
For t &gt;0, we have a source-free RL circuit.
L
4
 
 1/ 3 thus,
R 48
i o (t) = 1.2e–3t A.
8
Chapter 7, Solution 12.
Using Fig. 7.92, design a problem to help other students better understand source-free RL
circuits.
Although there are many ways to work this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
The switch in the circuit in Fig. 7.90 has been closed for a long time. At t = 0, the switch is
opened. Calculate i(t) for t &gt; 0.
Figure 7.90
Solution
When t &lt; 0, the switch is closed and the inductor acts like a short circuit to dc. The 4  resistor
is short-circuited so that the resulting circuit is as shown in Fig. (a).
3
12 V
i(0-)
+

4
(a)
12
4A
3
Since the current through an inductor cannot change abruptly,
i(0)  i(0  )  i(0  )  4 A
i (0  ) 
2H
(b)
When t &gt; 0, the voltage source is cut off and we have the RL circuit in Fig. (b).
L 2
    0.5
R 4
Hence,
i( t )  i(0) e - t   4 e -2t A
Chapter 7, Solution 13.
(a)  
1
 1ms
103
= 1 ms.
v(t) = i(t)R = 80e–1000t V = R5e–1000tx10–3 or R = 80,000/5 = 16 kΩ.
But τ = L/R = 1/103 or L = 16x103/103 = 16 H.
(b) The energy dissipated in the resistor is
= 200(1–e–1)x10–6 = 126.42 &micro;J.
(a) 16 kΩ, 16 H, 1 ms
(b) 126.42 &micro;J
Chapter 7, Solution 14.
60 x40
 24 k
100
5 x10 3
  L/R 
 0.2083  s
24 x103
RTh  (40  20)//(10  30) 
Chapter 7, Solution 15
(a) R Th  2  10 // 40  10,
(b) RTh  40 // 160  48  40,
L
 5 / 10  0.5s
R Th
L

 (20 x10 3 ) / 80  0.25 ms
RTh

(a) 10 Ω, 500 ms
(b) 40 Ω, 250 &micro;s
Chapter 7, Solution 16.

(a)
L eq
R eq
L eq  L and R eq  R 2 

(b)
R 1R 3
R 2 (R 1  R 3 )  R 1 R 3

R1  R 3
R1  R 3
L( R 1  R 3 )
R 2 (R 1  R 3 )  R 1 R 3
R 3 (R 1  R 2 )  R 1 R 2
L1 L 2
R 1R 2
and R eq  R 3 

L1  L 2
R1  R 2
R1  R 2
L1L 2 (R 1  R 2 )

(L 1  L 2 ) ( R 3 ( R 1  R 2 )  R 1 R 2 )
where L eq 
Chapter 7, Solution 17.
i( t )  i(0) e - t  ,

14 1
L


R eq
4 16
i (t )  6 e -16t
vo (t )  3i  L
di
 18 e -16t  (1 4)(-16) 6 e -16t
dt
v o ( t )  –6e–16tu(t) V
Chapter 7, Solution 18.
If v( t )  0 , the circuit can be redrawn as shown below.
+
0.4 H
R eq
i(t)
6
L 2 5 1
,
   
5
R 5 6 3
-t 
-3t
i (t )  i (0) e  5e
di - 2
vo (t )  -L  (-3)5 e -3t  6 e -3t V
dt 5
R eq  2 || 3 
v o (t)

Chapter 7, Solution 19.
i
1V
 +
10 
i1
i1
i2
i/2
i2
40 
To find R th we replace the inductor by a 1-V voltage source as shown above.
10 i1  1  40 i 2  0
But
i  i2  i 2
and
i  i1
i.e.
i1  2 i 2  i
1
10 i  1  20 i  0 
 i 
30
1
R th   30 
i
L
6


 0.2 s
R th 30
i( t )  6 e -5t u (t ) A
Chapter 7, Solution 20.
(a)
(b)
(c)
L
1


 R  50L
R 50
di
v  L
dt
-50t
90 e   L(30)(-50) e -50t 
 L  60 mH
R  50L  3 Ω
L
1
 
 20 ms
R 50
1
1
w  L i 2 (0)  (0.06)(30) 2  27 J
2
2
The value of the energy remaining at 10 ms is given by:

w 10 = 0.03(30e–0.5)2 = 0.03(18.196)2 = 9.933 J.
So, the fraction of the energy dissipated in the first 10 ms is given by:
(27–9.933)/27 = 0.6321 or 63.21%.
Chapter 7, Solution 21.
The circuit can be replaced by its Thevenin equivalent shown below.
R th
V th
+

2H
80
(60)  40 V
80  40
80
R
R th  40 || 80  R 
3
Vth
40
I  i(0)  i() 

R th 80 3  R
Vth 

 1
3
40
40
1 
 R 
3
R  80 3
1  40
1
w  L I 2  (2)
2  R  80
2
2
R  13.333 Ω
Chapter 7, Solution 22.
i( t )  i(0) e - t  ,

L
R eq
R eq  5 || 20  1  5  ,

2
5
i( t )  10e–2.5t A
Using current division, the current through the 20 ohm resistor is
5
-i
io 
(-i)   -2 e -2.5t
5  20
5
v( t )  20 i o  –40e–2.5t V
Chapter 7, Solution 23.
Since the 2  resistor, 1/3 H inductor, and the (3+1)  resistor are in parallel,
they always have the same voltage.
10 10

 7.5 
 i (0)  -7.5
2 3 1
The Thevenin resistance R th at the inductor’s terminals is
13 1
L
4



R th  2 || (3  1)  ,
3
R th 4 3 4
-i 
i (t )  i (0) e - t   -7.5 e -4t , t  0
di
v L  vo  L  -7.5(-4)(1/3) e - 4t
dt
-4t
v o  10e V , t  0
1
vx 
v L  2.5 e-4t V , t  0
3 1
Chapter 7, Solution 24.
(a) v( t )  - 5 u(t)
(b) i( t )  -10  u ( t )  u ( t  3)  10 u ( t  3)  u ( t  5)
= - 10 u(t )  20 u(t  3)  10 u(t  5)
(c) x ( t )  ( t  1)  u ( t  1)  u ( t  2)   u ( t  2)  u ( t  3)
 (4  t )  u ( t  3)  u ( t  4)
= ( t  1) u ( t  1)  ( t  2) u ( t  2)  ( t  3) u ( t  3)  ( t  4) u ( t  4)
= r(t  1)  r(t  2)  r(t  3)  r(t  4)
(d) y( t )  2 u (-t )  5  u ( t )  u ( t  1)
= 2 u(-t )  5 u(t )  5 u(t  1)
Chapter 7, Solution 25.
Design a problem to help other students to better understand singularity functions.
Although there are many ways to work this problem, this is an example based on the
same kind of problem asked in the third edition.
Problem
Sketch each of the following waveforms.
(a) i(t) = [u(t–2)+u(t+2)] A
(b) v(t) = [r(t) – r(t–3) + 4u(t–5) – 8u(t–8)] V
Solution
The waveforms are sketched below.
(a)
i(t) (A)
2
1
-2 -1 0
1
2
3
4
t
(b)
v(t) (V)
7
3
0
–1
1
2
3
4
5
6
7
8
t
Chapter 7, Solution 26.
(a)
(b)
(c)
(d)
v1 ( t )  u ( t  1)  u ( t )   u ( t  1)  u ( t )
v1 ( t )  u(t  1)  2 u(t )  u(t  1)
v 2 ( t )  ( 4  t )  u ( t  2)  u ( t  4) 
v 2 ( t )  -( t  4) u ( t  2)  ( t  4) u ( t  4)
v 2 ( t )  2 u(t  2)  r(t  2)  r(t  4)
v 3 ( t )  2  u(t  2)  u(t  4)  4  u(t  4)  u(t  6)
v 3 ( t )  2 u(t  2)  2 u(t  4)  4 u(t  6)
v 4 ( t )  -t  u ( t  1)  u ( t  2)  -t u(t  1)  t u ( t  2)
v 4 ( t )  (-t  1  1) u ( t  1)  ( t  2  2) u ( t  2)
v 4 ( t )  - r(t  1)  u(t  1)  r(t  2)  2 u(t  2)
Chapter 7, Solution 27.
v(t)= [5u(t+1)+10u(t)–25u(t–1)+15u(t-2)] V
Chapter 7, Solution 28.
i(t) is sketched below.
i(t)
1
0
-1
1
2
3
4
t
Chapter 7, Solution 29
x(t)
(a)
3.679
0
(b)
1
t
y(t)
27.18
0
(c)
t
z (t )  cos 4t (t  1)  cos 4 (t  1)  0.6536 (t  1) , which is sketched below.
z(t)
1
0
t
–0.653δ(t–1)
Chapter 7, Solution 30.

(a)
  4t
(b)
- 4t

2
2
( t  1) dt  4t 2 t 1  4
cos(2t ) ( t  0.5) dt  4t 2 cos(2t ) t 0.5  cos   - 1
Chapter 7, Solution 31.
(a)
(b)
 e  112  10
  e (t  2) dt  e
  5 (t )  e (t )  cos 2t (t ) dt   5  e  cos(2t )

-

-
- 4t 2
- 4t 2
-t
t2
-16
-9
-t
t 0
 5 11  7
Chapter 7, Solution 32.
t
(a)
t
t
1
1
 u ( )d   1d  
1
4
1
 t 1
4
t2
(b)  r (t  1)dt   0dt   (t  1)dt   t 14  4.5
2
0
0
1
5
(c )
 (t  6)
1
2
 (t  2)dt  (t  6) 2
t 2
 16
Chapter 7, Solution 33.
i( t ) 
i (t ) 
1 t
 v(t ) dt  i(0)
L 0
t
10 -3
15 (t  2) dt  0
-3 0
10  10
i( t )  1.5 u( t  2) A
Chapter 7, Solution 34.
(a)
d
u(t  1) u(t  1)   (t  1)u(t  1) 
dt
u( t  1) ( t  1)   ( t  1)1  0 ( t  1)   ( t  1)
(b)
d
r (t  6) u(t  2)  u(t  6)u(t  2) 
dt
r ( t  6) ( t  2)  u( t  6)1  0 ( t  2)  u( t  6)
(c)
d
sin 4t u(t  3)  4 cos 4t u(t  3)  sin 4t (t  3)
dt
 4 cos 4t u( t  3)  sin 4 x 3 ( t  3)
 4 cos 4t u( t  3)  0.5366 ( t  3)
Chapter 7, Solution 35.
(a)
v  Ae2t ,
v(0)  A  1
v(t) = –e–2tu(t) V
i  Ae3 t / 2 ,
i(0)  A  2
(b)
i(t) = 2e–1.5tu(t) A
Chapter 7, Solution 36.
(a)
(b)
v( t )  A  B e-t , t  0
A  1,
v(0)  0  1  B
v( t ) 
1  e -t V , t  0
v( t )  A  B e t 2 , t  0
A  -3 ,
v(0)  -6  -3  B
v( t )  - 3  1  e
t 2
 V,
t0
or
B  -1
or
B  -3
Chapter 7, Solution 37.
Let v = v h + v p , v p =10.

1
vh  4 v
h
0


v h  Ae t / 4
v  10  Ae 0.25t
v(0)  2  10  A
v  10  8e 0.25t
(a)   4 s
(b) v ()  10 V


(c ) v  10  8e 0.25 t u( t ) V


A  8
Chapter 7, Solution 38.
Let i = i p +i h

i h  3ih  0
Let i p  ku (t ),

ip  0,


3ku (t )  2u (t )
ip 
ih  Ae 3t u (t )


2
u (t )
3
2
i  ( Ae 3t  )u (t )
3
If i(0) =0, then A + 2/3 = 0, i.e. A=-2/3. Thus,
i
2
(1  e  3 t )u( t )
3
k
2
3
Chapter 7, Solution 39.
(a)
Before t = 0,
v( t ) 
1
(20)  4 V
4 1
After t = 0,
v( t )  v()   v(0)  v() e - t 
  RC  (4)(2)  8 , v(0)  4 ,
v()  20
v( t )  20  (4  20) e -t 8
v( t )  20  16 e - t 8 V
Before t = 0, v  v1  v 2 , where v1 is due to the 12-V source and v 2 is
due to the 2-A source.
v1  12 V
To get v 2 , transform the current source as shown in Fig. (a).
v 2  -8 V
Thus,
v  12  8  4 V
After t = 0, the circuit becomes that shown in Fig. (b).
(b)
4
2F
+
v2
2F

+

8V
12 V
+

3
3
(a)
(b)
v( t )  v()   v(0)  v() e - t 
v()  12 ,
v(0)  4 ,
  RC  (2)(3)  6
-t 6
v( t )  12  (4  12) e
v( t )  12  8 e -t 6 V
Chapter 7, Solution 40.
(a)
Before t = 0, v  12 V .
After t = 0, v( t )  v()   v(0)  v() e - t 
v()  4 ,
v(0)  12 ,
  RC  (2)(3)  6
v( t )  4  (12  4) e - t 6
v( t )  4  8 e - t 6 V
(b)
Before t = 0, v  12 V .
After t = 0, v( t )  v()   v(0)  v() e - t 
After transforming the current source, the circuit is shown below.
t=0
2
12 V
v(0)  12 ,
v  12 V
+

v()  12 ,
4
5F
  RC  (2)(5)  10
Chapter 7, Solution 41.
Using Fig. 7.108, design a problem to help other students to better understand the step response
of an RC circuit.
Although there are many ways to work this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
For the circuit in Fig. 7.108, find v(t) for t &gt; 0.
Figure 7.108
Solution
v(0)  0 ,
v ( ) 
R eq C  (6 || 30)(1) 
30
(12)  10
36
(6)(30)
5
36
v( t )  v()   v(0)  v() e - t 
v( t )  10  (0  10) e - t 5
v( t )  10 (1  e -0.2t ) u ( t )V
Chapter 7, Solution 42.
(a)
v o ( t )  v o ()   v o (0)  v o () e - t 
4
v o () 
(12)  8
v o (0)  0 ,
42
4
  R eq C eq , R eq  2 || 4 
3
4
  (3)  4
3
v o ( t )  8  8 e -t 4
v o ( t )  8 ( 1  e -0.25t ) V
(b)
For this case, v o ()  0 so that
v o ( t )  v o (0) e - t 
4
(12)  8 ,
v o (0) 
42
  RC  (4)(3)  12
v o ( t )  8 e -t 12 V
Chapter 7, Solution 43.
Before t = 0, the circuit has reached steady state so that the capacitor acts like an open
circuit. The circuit is equivalent to that shown in Fig. (a) after transforming the voltage
source.
0.5i
vo
i
40 
2A
80 
0.5i
(a)
vo
vo
i
,
40
80
vo
1 vo
320
 2

 v o 
 64
Hence,
2 80
40
5
vo
i
 0.8 A
80
0.5i  2 
After t = 0, the circuit is as shown in Fig. (b).
0.5i
vC
i
3 mF
80 
0.5i
(b)
v C ( t )  v C (0) e - t  ,
  R th C
To find R th , we replace the capacitor with a 1-V voltage source as shown in Fig. (c).
0.5i
vC
i
1V
+

0.5i
(c)
80 
vC
1
0.5

,
i o  0.5 i 
80 80
80
1 80
R th  
 160  ,
  R th C  480
i o 0.5
v C (0)  64 V
i
v C ( t )  64 e - t 480
dv C
 1 
 64 e - t 480
0.5 i  -i C  -C
 -3 
 480 
dt
i( t )  800 e - t 480 u( t ) mA
Chapter 7, Solution 44.
R eq  6 || 3  2  ,
  RC  4
v( t )  v()   v(0)  v() e - t 
Using voltage division,
3
v(0) 
(60)  20 V ,
36
v ( ) 
3
(24)  8 V
3 6
Thus,
v(t )  8  (20  8) e -t 4  8  12 e -t 4
dv
 -1
i (t )  C
 (2)(12)   e -t 4  - 6 e -0.25t A
dt
4
Chapter 7, Solution 45.
To find R Th , consider the circuit shown below.
20 k
10 k
40 k
RTh  10  20 // 40  10 
  RThC 
R Th
20 x40 70

k
60
3
70
x103 x3 x10 6  0.07
3
To find vo(), consider the circuit below.
20 k
10 k
+
30V
+
_
40 k
vo
–
v o (∞) = [40/(40+20)]30 = 20 V
v o (t) = v o (∞) + [v o (0)– v o (∞)]e–t/0.07 = [20 –15e–14.286t]u(t) V.
Chapter 7, Solution 46.
  RTh C  (2  6) x0.25  2s,
v(0)  0,
v()  6i s  6 x5  30
v ( t )  v ( )  [v (0)  v ( )]e  t /   30(1  e  t / 2 ) u( t ) V
Chapter 7, Solution 47.
For t &lt; 0, u ( t )  0 ,
u ( t  1)  0 ,
v(0)  0
For 0 &lt; t &lt; 1,   RC  (2  8)(0.1)  1
v(0)  0 ,
v()  (8)(3)  24
v( t )  v()   v(0)  v() e - t 
v( t )  24 1  e - t 
For t &gt; 1,
v(1)  24 1  e -1   15.17
- 6  v() - 24  0 
 v()  30
v( t )  30  (15.17  30) e -(t-1)
v( t )  30  14.83 e -(t-1)
Thus,


 24 1  e - t V ,
0t1
v( t )  
-(t -1)
V,
t 1
 30  14.83 e
Chapter 7, Solution 48.
For t &lt; 0,
u (-t)  1 ,
For t &gt; 0,
u (-t)  0 ,
R th  20  10  30 ,
v()  0
  R th C  (30)(0.1)  3
v( t )  v()   v(0)  v() e - t 
v( t )  10 e -t 3 V
i( t )  C
 - 1
dv
 (0.1)  10 e - t 3
3
dt
i( t ) 
- 1 -t 3
e A
3
Chapter 7, Solution 49.
For 0 &lt; t &lt; 1, v(0)  0 ,
R eq  4  6  10 ,
v()  (2)(4)  8
  R eq C  (10)(0.5)  5
v( t )  v()   v(0)  v() e - t 
v( t )  8  1  e - t 5  V
v(1)  8  1  e -0.2   1.45 ,
For t &gt; 1,
v( t )  v()   v(1)  v() e -( t 1) 
v( t )  1.45 e -( t 1) 5 V
Thus,


 8 1  e -t 5 V , 0  t  1
v( t )  
- ( t 1 ) 5
V,
t 1
 1.45 e
v()  0
Chapter 7, Solution 50.
For the capacitor voltage,
v( t )  v()   v(0)  v() e- t 
v(0)  0
For t &gt; 0, we transform the current source to a voltage source as shown in Fig. (a).
1 k
1 k
+
30 V
+

2 k
v

(a)
2
(30)  15 V
2 11
R th  (1  1) || 2  1 k
1
1
  R th C  10 3   10 -3 
4
4
-4t
v( t )  15  1  e  , t  0
v() 
We now obtain i x from v(t). Consider Fig. (b).
i T 1 k
v
ix
30 mA
1 k
1/4 mF
(b)
But
i x  30 mA  i T
dv
v
C
iT 
dt
R3
i T ( t )  7.5  1  e -4t  mA 
i T ( t )  7.5  1  e -4t  mA
1
 10 -3 (-15)(-4) e -4t A
4
Thus,
i x ( t )  30  7.5  7.5 e -4t mA
i x ( t )  7.5  3  e -4t  mA , t  0
2 k
Chapter 7, Solution 51.
Consider the circuit below.
t=0
R
+
VS
+

i
L
v

After the switch is closed, applying KVL gives
di
VS  Ri  L
dt

VS 
di

or
L  -R  i 

dt
R
di
-R

dt
i  VS R
L
Integrating both sides,

V  i( t ) - R
ln  i  S  I 0 
t

R
L
 i  VS R  - t

ln 
 I0  VS R  
or
i  VS R
 e- t 
I0  VS R
i( t ) 
VS 
VS  -t 
e
  I0 
R 
R
which is the same as Eq. (7.60).
Chapter 7, Solution 52.
Using Fig. 7.118, design a problem to help other students to better understand the step response
of an RL circuit.
Although there are many ways to work this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
For the circuit in Fig. 7.118, find i(t) for t &gt; 0.
Figure 7.118
Solution
20
i()  2 A
 2 A,
10
i( t )  i()   i(0)  i() e- t 
i(0) 
i( t )  2 A
Chapter 7, Solution 53.
25
5A
3 2
i( t )  i(0) e - t 
After t = 0,
L 4
    2,
i(0)  5
R 2
i( t )  5 e - t 2 u( t ) A
i
(a)
Before t = 0,
(b)
Before t = 0, the inductor acts as a short circuit so that the 2  and 4 
resistors are short-circuited.
i( t )  6 A
After t = 0, we have an RL circuit.
L 3
i( t )  i(0) e - t  ,
 
R 2
-2t 3
u( t ) A
i( t )  6 e
Chapter 7, Solution 54.
(a)
Before t = 0, i is obtained by current division or
4
(2)  1 A
i( t ) 
44
After t = 0,
i( t )  i()   i(0)  i() e- t 
L

,
Req  4  (4 || 12)  7 
R eq

3.5 1

7
2
i(0)  1 ,
i ( ) 
(4 || 12)
3
6
(2) 
(2) 
4  (4 || 12)
43
7
6 
6
  1   e -2 t
7 
7
1
i( t )   6  e - 2t  A
7
10
2A
Before t = 0, i( t ) 
23
Req  3  (6 || 2)  4.5
After t = 0,
i( t ) 
(b)
L
2
4


R eq 4.5 9
i(0)  2
To find i() , consider the circuit below, at t = when the inductor becomes a
short circuit,

v
i
10 V
2
+

24 V
6
+

2H
3
10  v 24  v v
v



 v  9 i()   3 A and
2
6
3
3
i( t )  3  (2  3) e -9 t 4
i( t )  3  e - 9 t 4 A
Chapter 7, Solution 55.
For t &lt; 0, consider the circuit shown in Fig. (a).
0.5 H
io
3
i
io
+

24 V
0.5 H
+
4i o
+

v
2

(a)
3i o  24  4i o  0 
 i o  24
v
v( t )  4i o  96 V
i   48 A
2
For t &gt; 0, consider the circuit in Fig. (b).
i( t )  i()   i(0)  i() e- t 
i(0)  48 ,
i ( )  0
L
0.5 1
R th  2  ,  


R th
2
4
i( t )  (48) e -4t
v( t )  2 i( t )  96 e -4t u( t )V
8
20 V
+
v
+


(b)
2
Chapter 7, Solution 56.
R eq  6  20 || 5  10  ,

L
 0.05
R
i( t )  i()   i(0)  i() e- t 
i(0) is found by applying nodal analysis to the following circuit.
5
vx
12 
2A
20 
i
6
+
0.5 H
v

20  v x v x v x v x



5
12 20 6
v
i(0)  x  2 A
6
2

 v x  12
Since 20 || 5  4 ,
4
(4)  1.6
i() 
46
i( t )  1.6  (2  1.6) e- t 0.05  1.6  0.4 e-20t
di 1
v( t )  L  (0.4) (-20) e -20t
dt 2
v( t )  - 4 e -20t V
+

20 V
Chapter 7, Solution 57.
At t  0  , the circuit has reached steady state so that the inductors act like short circuits.
6
30 V
i
+

i1
i2
5
20 
20
30
30

 3,
i1 
(3)  2.4 ,
25
6  (5 || 20) 10
i 1 ( 0 )  2 .4 A ,
i 2 ( 0 )  0 .6 A
i
i 2  0 .6
For t &gt; 0, the switch is closed so that the energies in L1 and L 2 flow through the closed
switch and become dissipated in the 5  and 20  resistors.
L
2.5 1
i1 ( t )  i1 (0) e - t 1 ,
1  1 

R1
5
2
i1 ( t )  2.4e–2tu(t) A
i 2 ( t )  i 2 (0) e - t  2 ,
2 
L2
4 1


R 2 20 5
i 2 ( t )  600e–5tu(t) mA
Chapter 7, Solution 58.
For t &lt; 0,
v o (t)  0
For t &gt; 0,
i(0)  10 ,
R th  1  3  4  ,
20
5
1 3
L 14 1



R th
4 16
i() 
i( t )  i()   i(0)  i() e- t 
i( t )  5  1  e-16t  A
di
1
 15  1  e-16t   (-16)(5) e-16t
dt
4
-16t
v o ( t )  15  5 e V
vo (t )  3i  L
Chapter 7, Solution 59.
Let i(t) be the current through the inductor.
i(0)  0
For t &lt; 0,
vs  0 ,
For t &gt; 0,
R eq  4  (6 || 3)  6 Ω and  
L 1.5

 0.25 sec.
R eq
6
At t = ∞, the inductor becomes a short and the current delivered by the 18 volts
source is I s = 18/[6+(3||4)] = 18/7.714 = 2.333 amps. The voltage across the 4ohm resistor is equal to 18–6(2.333) = 18–14 = 4 volts. Therefore the current
through the inductor is equal to i(∞) = 4/4 = 1 amp.
i( t )  i()   i(0)  i() e- t 
i( t )  1(1  e -4t ) amps.
v o (t)  L
di
 (1.5)(1)(-4)(-e- 4t )
dt
v o ( t )  [6e–4t]u(t) volts.
Chapter 7, Solution 60.
Let I be the inductor current.
u(t)  0 
 i(0)  0
For t &lt; 0,
For t &gt; 0,
R eq  5 || 20  4  ,
i()  4
i( t )  i()   i(0)  i() e- t 
i( t )  4  1  e - t 2 
 - 1
di
 (8)(-4)  e - t 2
2
dt
v( t )  16 e -0.5t V
v( t )  L

L
8
 2
R eq 4
Chapter 7, Solution 61.
The current source is transformed as shown below.
4
20u(-t) + 40u(t)
+

0.5 H
L 12 1
i(0)  5 ,

 ,
R
4
8
i( t )  i()   i(0)  i() e - t 

i()  10
i( t )  (10 – 5e–8t)u(t) A
v( t )  L
di  1 
  (-5)(-8) e -8t
dt  2 
v( t )  20e–8tu(t) V
Chapter 7, Solution 62.

L
2

1
R eq 3 || 6
For 0 &lt; t &lt; 1, u ( t  1)  0 so that
1
i(0)  0 ,
i() 
6
1
i( t )   1  e - t 
6
1
 1  e -1   0.1054
6
1 1 1
i()   
3 6 2
i( t )  0.5  (0.1054  0.5) e-(t -1)
i( t )  0.5  0.3946 e-(t -1)
For t &gt; 1,
i(1) 
Thus,

1
 1  e -t  A
0t1
i( t )  
6
 0.5  0.3946 e -(t -1) A
t1
Chapter 7, Solution 63.
10
2
5
For t &lt; 0,
u (- t )  1 ,
i(0) 
For t &gt; 0,
u (-t)  0 ,
i()  0
L
0.5 1



R th
4 8
R th  5 || 20  4  ,
i( t )  i()   i(0)  i() e - t 
i( t )  2e–8tu(t) A
v( t )  L
di  1 
  (-8)(2) e -8t
dt  2 
v( t )  –8e–8tu(t) V
2e–8tu(t) A, –8e–8tu(t) V
Chapter 7, Solution 64
Determine the value of i L (t) and the total energy dissipated by the circuit from t =
0 sec to t = ∞ sec. The value of v in (t) is equal to [40–40u(t)] volts.
40 
v in (t)
+

40 
R eq
v 1 i L (t)
10 H
v Thev (t
+

i L (t)
10 H
Solution
Step 1.
Determine the Thevenin equivalent circuit to the left of the inductor. This
means we need to find v oc (t) and i sc (t) which gives us v Thev (t) = v oc (t) and R eq =
v oc (t)/i sc (t) (note, this only works for resistor networks in the time domain). This leads to
the second circuit shown above.
Now, with this circuit, we can use the generalized solution to a first order differential
equation, i L (t) = Ae–(t–0)/τ+B where, t 0 = 0, τ = L/R, A+B = i L (0) and 0+B = i L (∞).
Finally, we can use w = (1/2)Li L (t)2 to calculate the energy dissipated by the circuit (w =
[(1/2)Li L (∞)2–(1/2)Li L (0)2].
Step 2.
We now determine the Thevenin equivalent circuit. First we need to pick
a reference node and mark the unknown voltages, as seen above. With the inductor out
of the circuit, the node equation is simply [(v 1 –v in (t))/40] + [(v 1 –0)/40] + 0 (since the
inductor is out of the circuit, there is an open circuit where it was) = 0. This leads to
[(1/40)+(1/40)]v 1 = (1/40)v in (t) or 2v 1 = v in (t) or v 1 = 0.5v in (t) = [20–20u(t)] v oc (t) =
v Thev (t). Now to short the open circuit which produces v 1 = 0 and i sc = –[(0–v in (t))/40] =
v in (t)/40 = 0.025v in (t) A.
Step 3.
Now, everything comes together, R eq = v oc (t)/i sc (t) = 0.5v in (t)/[0.025v in (t)]
= 0.5/0.025 = 20 Ω. Next we find τ = L/R eq = 10/20 = (1/2) sec. At t = 0–, v in (0–) = [10–
0] V (note u(t) = 0 until t = 0). Since it has been at this value for a very long time, the
inductor can be considered a short and the value of the current is equal to 20/20 or i L (0–)
= 1 amp. Since you cannot change the current instantaneously, i L (0) = 1 amp = A+B.
Since v Thev (t) = 20–20 = 0 for all t &gt; 0, all the energy in the inductor will be dissipated by
the circuit and i L (∞) = 0 = B which means that A = 1 and i L (t) = [e–2t] u(t) amps. The
total energy dissipated from t = 0 to ∞ sec is equal to [(1/2)Li L (0)2–(1/2)Li L (∞)2] =
(0.5)10(1)2–0 = 5 J.
Chapter 7, Solution 65.
Since v s  10  u ( t )  u ( t  1) , this is the same as saying that a 10 V source is turned on at
t = 0 and a -10 V source is turned on later at t = 1. This is shown in the figure below.
vs
10
1
t
-10
For 0 &lt; t &lt; 1, i(0)  0 ,
R th  5 || 20  4 ,
10
2
5
L
2 1

 
R th 4 2
i() 
i( t )  i()   i(0)  i() e- t 
i( t )  2  1  e -2t  A
i(1)  2  1  e-2   1.729
For t &gt; 1,
i()  0
since vs  0
i( t )  i(1) e- ( t 1) 
i( t )  1.729 e-2( t 1) A
Thus,
 2  1  e - 2t  A 0  t  1
i( t )  
t1
 1.729 e - 2( t 1) A
Chapter 7, Solution 66.
Using Fig. 7.131, design a problem to help other students to better understand first-order
op amp circuits.
Although there are many ways to work this problem, this is an example based on the
same kind of problem asked in the third edition.
Problem
For the op-amp circuit of Fig. 7.131, find v o . Assume that v s changes abruptly from
0 to 1 V at t=0. Find v o .
50 k
0.5 F
20 k
–
+
vs
+
vo
+
_
–
Figure 7.131 For Prob. 7.66.
Solution
For t&lt;0-, v s =0 so that v o (0)=0
:Let v be the capacitor voltage
For t&gt;0, v s =1. At steady state, the capacitor acts like an open circuit so that we
have an inverting amplifier
v o (∞) = –(50k/20k)(1V) = –2.5 V
τ = RC = 50x103x0.5x10–6 = 25 ms
v o (t) = v o (∞) + (v o (0) – v o (∞))e–t/0.025 = 2.5(e–40t – 1) V.
Chapter 7, Solution 67.
The op amp is a voltage follower so that v o  v as shown below.
R
R

+
vo
v1
vo
+
R
vo
C

At node 1,
v o  v1 v1  0 v1  v o


R
R
R

 v1 
2
v
3 o
At the noninverting terminal,
dv
v  v1
C o  o
0
dt
R
dv
2
1
 RC o  v o  v1  v o  v o  v o
dt
3
3
dv o
v
 o
dt
3RC
v o ( t )  VT e - t 3RC
VT  v o (0)  5 V ,
  3RC  (3)(10  103 )(1  10- 6 ) 
v o ( t )  5e–100t/3u(t) V
3
100
Chapter 7, Solution 68.
This is a very interesting problem which has both an ideal solution as well as a realistic solution.
Let us look at the ideal solution first. Just before the switch closes, the value of the voltage
across the capacitor is zero which means that the voltage at both terminals input of the op amp
are each zero. As soon as the switch closes, the output tries to go to a voltage such that both
inputs to the op amp go to 4 volts. The ideal op amp puts out whatever current is necessary to
reach this condition. An infinite (impulse) current is necessary if the voltage across the capacitor
is to go to 8 volts in zero time (8 volts across the capacitor will result in 4 volts appearing at the
negative terminal of the op amp). So v o will be equal to 8 volts for all t &gt; 0.
What happens in a real circuit? Essentially, the output of the amplifier portion of the op amp
goes to whatever its maximum value can be. Then this maximum voltage appears across the
output resistance of the op amp and the capacitor that is in series with it. This results in an
exponential rise in the capacitor voltage to the steady-state value of 8 volts.
vC(t) = V op amp max (1 – e-t/(RoutC)) volts, for all values of vC less than 8 V,
= 8 V when t is large enough so that the 8 V is reached.
Chapter 7, Solution 69.
Let v x be the capacitor voltage.
For t &lt; 0,
v x ( 0)  0
For t &gt; 0, the 20 k and 100 k resistors are in series and together, they are in
parallel with the capacitor since no current enters the op amp terminals.
As t   , the capacitor acts like an open circuit so that
4
v o ( ) 
(20  100)  48
10
R th  20  100  120 k ,
  R th C  (120  103 )(25  10-3 )  3000
v o ( t )  v o ()   v o (0)  v o () e - t 


v o ( t )  48 1  e - t 3000 V = 48(e–t/3000–1)u(t)V
Chapter 7, Solution 70.
Let v = capacitor voltage.
For t &lt; 0, the switch is open and v(0)  0 .
For t &gt; 0, the switch is closed and the circuit becomes as shown below.
1
+

2
vS
+
+

vo
v

C
R
v1  v 2  v s
0  vs
dv
C
R
dt
 v o  v s  v
where v  v s  v o 
From (1),
dv v s

0
dt RC
- t vs
-1
v
v s dt  v(0) 

RC
RC
Since v is constant,
RC  (20  10 3 )(5  10 -6 )  0.1
- 20 t
v
mV  -200 t mV
0.1
From (3),
v o  v s  v  20  200 t
v o  20 ( 1  10t ) mV
(1)
(2)
(3)
Chapter 7, Solution 71.
We temporarily remove the capacitor and find the Thevenin equivalent at its
terminals. To find R Th , we consider the circuit below.
Ro
20 k
R Th
Since we are assuming an ideal op amp, R o = 0 and R Th =20k . The op amp circuit
is a noninverting amplifier. Hence,
10
)vs  2vs  6V
10
The Thevenin equivalent is shown below.
20 k
VTh  (1
+
6V
+
_
v
10 F
–
Thus,
v(t)  6(1 e t /  ) , t  0
where   RTH C  20 x10 3 x10 x10 6  0.2
v(t)  6(1 e5t ), t  0 V
Chapter 7, Solution 72.
The op amp acts as an emitter follower so that the Thevenin equivalent circuit is
shown below.
C
+
3u(t)
Hence,
+

v

io
R
v( t )  v()   v(0)  v() e - t 
v(0)  -2 V , v()  3 V ,   RC  (10  10 3 )(10  10 -6 )  0.1
v( t )  3  (-2 - 3) e -10t  3  5 e -10t
dv
 (10  10 -6 )(-5)(-10) e -10t
dt
i o  0.5 e -10t mA , t  0
io  C
Chapter 7, Solution 73.
Consider the circuit below.
Rf
v1
R1
v2
+
v1
+

C
v
v3


+
+
vo

At node 2,
v1  v 2
dv
C
R1
dt
At node 3,
dv v 3  v o
C

dt
Rf
But v 3  0 and v  v 2  v 3  v 2 . Hence, (1) becomes
v1  v
dv
C
dt
R1
dv
v 1  v  R 1C
dt
v1
dv
v
or


dt R 1C R 1C
which is similar to Eq. (7.42). Hence,

vT
t0
v( t )  
-t 
t0
 v1   v T  v1  e
where v T  v(0)  1 and v1  4
  R 1C  (10  10 3 )(20  10 -6 )  0.2
 1
t0
v( t )  
-5t
t0
4  3 e
From (2),
dv
 (20  10 3 )(20  10 -6 )(15 e -5t )
dt
v o  -6 e -5t , t  0
v o  -R f C
(1)
(2)
v o  - 6 e -5t u(t ) V
Chapter 7, Solution 74.
Let v = capacitor voltage. For t &lt; 0, v(0)  0
10 kΩ
2 &micro;F
is

+
50 kΩ
is
+
vo

For t &gt; 0,
i s  10 A .
Since the current through the feedback resistor is i s , then
v o = –i s x104 volts = –10–5x104 = –100 mV.
It is interesting to look at the capacitor voltage.
dv v

dt R
v( t )  v()   v(0)  v() e - t 
is  C
It is evident that
  RC  (2  10 6 )(50  10 3 )  0.1
At steady state, the capacitor acts like an open circuit so that i s passes through R.
Hence,
v()  i s R  (10  10 6 )(50  10 3 )  0.5 V
Then the voltage across the capacitor is,
v(t) = 500(1–e–10t) mV.
Chapter 7, Solution 75.
Let v1 = voltage at the noninverting terminal.
Let v 2 = voltage at the inverting terminal.
For t &gt; 0,
v1  v 2  v s  4
0  vs
 i o , R 1  20 k
R1
vo  -ioR
Also, i o 
i.e.
dv
v
C ,
dt
R2
(1)
R 2  10 k , C  2 F
- vs
v
dv

C
R1 R 2
dt
(2)
This is a step response.
v( t )  v()   v(0)  v() e - t  ,
where   R 2 C  (10  10 3 )(2  10 -6 ) 
v(0)  1
1
50
At steady state, the capacitor acts like an open circuit so that i o passes through
R 2 . Hence, as t  
- vs
v()
 io 
R1
R2
-R2
- 10
(4)  -2
i.e.
vs 
v() 
20
R1
v( t )  -2  (1  2) e -50t
v( t )  -2  3 e -50t
But
v  vs  vo
or
v o  v s  v  4  2  3 e -50 t
v o  6  3 e -50 t u( t )V
- vs
-4

 -0.2 mA
R 1 20k
v
dv
io 
C
 - 0.2 mA
R2
dt
io 
or
Chapter 7, Solution 76.
The schematic is shown below. For the pulse, we use IPWL and enter the corresponding
values as attributes as shown. By selecting Analysis/Setup/Transient, we let Print Step =
25 ms and Final Step = 2 s since the width of the input pulse is 1 s. After saving and
simulating the circuit, we select Trace/Add and display –V(C1:2). The plot of V(t) is
shown below.
Chapter 7, Solution 77.
The schematic is shown below. We click Marker and insert Mark Voltage Differential at
the terminals of the capacitor to display V after simulation. The plot of V is shown
below. Note from the plot that V(0) = 12 V and V() = -24 V which are correct.
Chapter 7, Solution 78.
(a)
When the switch is in position (a), the schematic is shown below. We insert IPROBE to
display i. After simulation, we obtain,
i(0) = 7.714 A
from the display of IPROBE.
(b)
When the switch is in position (b), the schematic is as shown below. For inductor I1, we
let IC = 7.714. By clicking Analysis/Setup/Transient, we let Print Step = 25 ms and Final Step =
2 s. After Simulation, we click Trace/Add in the probe menu and display I(L1) as shown below.
Note that i() = 12A, which is correct.
Chapter 7, Solution 79.
When the switch is in position 1, i o (0) = 12/3 = 4A. When the switch is in position 2,
4
L
i o ( )  
 0.5 A,
R Th  (3  5) // 4  8 / 3,

 3 / 80
53
R Th
i o ( t )  i o ()  [i o (0)  i o ()]e  t /    0.5  4.5e 80 t / 3 u(t)A
Chapter 7, Solution 80.
(a) When the switch is in position A, the 5-ohm and 6-ohm resistors are short-circuited so
that
i 1 ( 0 )  i 2 ( 0)  v o ( 0)  0
but the current through the 4-H inductor is i L (0) =30/10 = 3A.
(b) When the switch is in position B,
R Th  3 // 6  2,

L
 4 / 2  2 sec
R Th
i L ( t )  i L ( )  [i L (0)  i L ( )]e  t /   0  3e  t / 2  3e  t / 2 A
(c) i1 () 
30
 2 A,
10  5
v o (t )  L
3
i 2 ( )   i L ( )  0 A
9
di L
dt


v o ( )  0 V
Chapter 7, Solution 81.
The schematic is shown below. We use VPWL for the pulse and specify the attributes as shown.
In the Analysis/Setup/Transient menu, we select Print Step = 25 ms and final Step = 3 S. By
inserting a current marker at one terminal of LI, we automatically obtain the plot of i after
simulation as shown below.
2.0A
1.5A
1.0A
0.5A
0A
0s
0.5s
1.0s
1.5s
-I(L1)
Time
2.0s
2.5s
3.0s
Chapter 7, Solution 82.
3  10 -3

  RC 
 R  
 30 
C 100  10 -6
Chapter 7, Solution 83.
v()  120,
v(0)  0,
  RC  34 x10 6 x15 x10 6  510s
v(t )  v()  [v(0)  v()]e t / 
Solving for t gives


85.6  120(1  e t / 510 )
t  510 ln 3.488  637.16 s
speed = 4000m/637.16s = 6.278m/s
Chapter 7, Solution 84.
Let I o be the final value of the current. Then
i(t )  I o (1  e  t /  ),
0.6 I o  I o (1  e  50 t )
  R / L  0.16 / 8  1 / 50


t
1
1
ln
 18.33 ms.
50 0.4
Chapter 7, Solution 85.
(a)
The light is on from 75 volts until 30 volts. During that time we essentially have
a 120-ohm resistor in parallel with a 6-&micro;F capacitor.
v(0) = 75, v(∞) = 0, τ = 120x6x10-6 = 0.72 ms
v(t 1 ) = 75 e  t1 /   30 which leads to t 1 = –0.72ln(0.4) ms = 659.7 &micro;s of lamp on
time.
(b)
  RC  (4  106 )(6  10-6 )  24 s
Since v( t )  v()   v(0)  v() e - t 
v( t 1 )  v()   v(0)  v() e - t1 
v( t 2 )  v()   v(0)  v() e- t 2 
Dividing (1) by (2),
v( t1 )  v()
 e( t 2  t1 ) 
v( t 2 )  v()
 v( t )  v() 

t 0  t 2  t1   ln  1
 v( t 2 )  v() 
 75  120 
  24 ln (2)  16.636 s
t 0  24 ln 
 30  120 
(1)
(2)
Chapter 7, Solution 86.
v( t )  v()   v(0)  v() e- t 
v()  12 ,
v(0)  0
-t 
v( t )  12  1  e 
v( t 0 )  8  12  1  e- t 0  
8
1
 1  e- t 0  
 e- t 0  
12
3
t 0   ln (3)
For R  100 k ,
  RC  (100  103 )(2  10-6 )  0.2 s
t 0  0.2 ln (3)  0.2197 s
For R  1 M ,
  RC  (1  106 )(2  10-6 )  2 s
t 0  2 ln (3)  2.197 s
Thus,
0.2197 s  t 0  2.197 s
Chapter 7, Solution 87.
Let i be the inductor current.
For t &lt; 0,
i (0  ) 
120
 1.2 A
100
For t &gt; 0, we have an RL circuit
L
50
 
 0.1 ,
i()  0
R 100  400
i( t )  i()   i(0)  i() e - t 
i( t )  1.2 e -10t
At t = 100 ms = 0.1 s,
i(0.1)  1.2 e -1  441mA
which is the same as the current through the resistor.
Chapter 7, Solution 88.
(a)
(b)
  RC  (300  10 3 )(200  10 -12 )  60 s
As a differentiator,
T  10   600 s  0.6 ms
i.e.
Tmin  0.6 ms
  RC  60 s
As an integrator,
T  0.1  6 s
i.e.
Tmax  6 s
Chapter 7, Solution 89.
Since   0.1 T  1 s
L
 1 s
R
L  R  10 -6  (200  10 3 )(1  10 -6 )
L  200 mH
Chapter 7, Solution 90.
We determine the Thevenin equivalent circuit for the capacitor C s .
Rs
v th 
v,
R th  R s || R p
Rs  Rp i
R th
V th
+

Cs
The Thevenin equivalent is an RC circuit. Since
Rs
1
1
v th  v i 


10
10 R s  R p
Rs 
6 2
1
R p   M
9 3
9
Also,
  R th C s  15 s
6 (2 3)
 0.6 M
where R th  R p || R s 
62 3

15  10 -6

 25 pF
Cs 
R th 0.6  10 6
Chapter 7, Solution 91.
12
 240 mA ,
i()  0
50
i( t )  i()   i(0)  i() e - t 
i( t )  240 e - t 
i o (0) 
L 2

R R
i( t 0 )  10  240 e - t 0


e t 0   24 
 t 0   ln (24)
t0
2
5


 1.573 
R
ln (24) ln (24)
2
R
 1.271 
1.573
Chapter 7, Solution 92.
 10

-3
dv
 4  10 -9   2  10
iC
- 10
dt

 5  10 -6
0  t  tR
tR  t  tD
 20 A
0  t  2 ms
i( t )  
- 8 mA 2 ms  t  2 ms  5 s
which is sketched below.
i(t)
5 s
20 A
t
2 ms
-8 mA
(not to scale)
Chapter 8, Solution 1.
(a)
At t = 0-, the circuit has reached steady state so that the equivalent circuit is
shown in Figure (a).
6
VS
+

6
6
+
+
vL
10 H

(a)
10 F
v

(b)
i(0-) = 12/6 = 2A, v(0-) = 12V
At t = 0+, i(0+) = i(0-) = 2A, v(0+) = v(0-) = 12V
(b)
For t &gt; 0, we have the equivalent circuit shown in Figure (b).
v L = Ldi/dt or di/dt = v L /L
Applying KVL at t = 0+, we obtain,
v L (0+) – v(0+) + 10i(0+) = 0
v L (0+) – 12 + 20 = 0, or v L (0+) = -8
Hence,
di(0+)/dt = -8/2 = -4 A/s
Similarly,
i C = Cdv/dt, or dv/dt = i C /C
i C (0+) = -i(0+) = -2
dv(0+)/dt = -2/0.4 = -5 V/s
(c)
As t approaches infinity, the circuit reaches steady state.
i() = 0 A, v() = 0 V
Chapter 8, Solution 2.
Using Fig. 8.63, design a problem to help other students better understand finding initial and
final values.
Although there are many ways to work this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
In the circuit of Fig. 8.63, determine:
(a) i R (0+), i L (0+), and i C (0+),
(b) di R (0+)/dt, di L (0+)/dt, and di C (0+)/dt,
(c) i R (), i L (), and i C ().
Figure 8.63
Solution
(a)
At t = 0-, the equivalent circuit is shown in Figure (a).
25 k
20 k
iR
80V
+

+
60 k v

(a)
iL
25 k
20 k
iL
iR
+

80V
(b)
60||20 = 15 kohms, i R (0-) = 80/(25 + 15) = 2mA.
By the current division principle,
i L (0-) = 60(2mA)/(60 + 20) = 1.5 mA
v C (0-) = 0
At t = 0+,
v C (0+) = v C (0-) = 0
i L (0+) = i L (0-) = 1.5 mA
80 = i R (0+)(25 + 20) + v C (0-)
i R (0+) = 80/45k = 1.778 mA
iR = iC + iL
But,
1.778 = i C (0+) + 1.5 or i C (0+) = 0.278 mA
(b)
v L (0+) = v C (0+) = 0
But,
v L = Ldi L /dt and di L (0+)/dt = v L (0+)/L = 0
di L (0+)/dt = 0
Again, 80 = 45i R + v C
0
But,
= 45di R /dt + dv C /dt
dv C (0+)/dt = i C (0+)/C = 0.278 mamps/1 F = 278 V/s
Hence,
di R (0+)/dt = (-1/45)dv C (0+)/dt = -278/45
di R (0+)/dt = -6.1778 A/s
Also, i R = i C + i L
di R (0+)/dt = di C (0+)/dt + di L (0+)/dt
-6.1788 = di C (0+)/dt + 0, or di C (0+)/dt = -6.1788 A/s
(c)
As t approaches infinity, we have the equivalent circuit in Figure (b).
i R () = i L () = 80/45k = 1.778 mA
i C () = Cdv()/dt = 0.
Chapter 8, Solution 3.
At t = 0-, u(t) = 0. Consider the circuit shown in Figure (a). i L (0-) = 0, and v R (0-) =
0. But, -v R (0-) + v C (0-) + 10 = 0, or v C (0-) = -10V.
(a)
At t = 0+, since the inductor current and capacitor voltage cannot change abruptly,
the inductor current must still be equal to 0A, the capacitor has a voltage equal to
–10V. Since it is in series with the +10V source, together they represent a direct
short at t = 0+. This means that the entire 2A from the current source flows
through the capacitor and not the resistor. Therefore, v R (0+) = 0 V.
(b)
At t = 0+, v L (0+) = 0, therefore Ldi L (0+)/dt = v L (0+) = 0, thus, di L /dt = 0A/s,
i C (0+) = 2 A, this means that dv C (0+)/dt = 2/C = 8 V/s. Now for the value of
dv R (0+)/dt. Since v R = v C + 10, then dv R (0+)/dt = dv C (0+)/dt + 0 = 8 V/s.
40 
40 
+
+
vC
+
vR
10 
+


2A
vR
+


10V
iL
vC

10 
+

10V
(b)
(a)
As t approaches infinity, we end up with the equivalent circuit shown in
(c)
Figure (b).
i L () = 10(2)/(40 + 10) = 400 mA
v C () = 2[10||40] –10 = 16 – 10 = 6V
v R () = 2[10||40] = 16 V
Chapter 8, Solution 4.
(a)
At t = 0-, u(-t) = 1 and u(t) = 0 so that the equivalent circuit is shown in Figure (a).
i(0-) = 40/(3 + 5) = 5A, and v(0-) = 5i(0-) = 25V.
i(0+) = i(0-) = 5A
Hence,
v(0+) = v(0-) = 25V
3
i
40V
+
+

v
5

(a)
3
i
40V
0.25 H
+ vL  iC
+

0.1F
iR
4A
5
(b)
(b)
i C = Cdv/dt or dv(0+)/dt = i C (0+)/C
For t = 0+, 4u(t) = 4 and 4u(-t) = 0. The equivalent circuit is shown in Figure (b). Since i
and v cannot change abruptly,
i R = v/5 = 25/5 = 5A, i(0+) + 4 =i C (0+) + i R (0+)
5 + 4 = i C (0+) + 5 which leads to i C (0+) = 4
dv(0+)/dt = 4/0.1 = 40 V/s
Similarly,
v L = Ldi/dt which leads to di(0+)/dt = v L (0+)/L
3i(0+) + v L (0+) + v(0+) = 0
15 + v L (0+) + 25 = 0 or v L (0+) = -40
di(0+)/dt = -40/0.25 = –160 A/s
(c)
As t approaches infinity, we have the equivalent circuit in Figure (c).
3
i
+
4A
v
5

(c)
i() = -5(4)/(3 + 5) = –2.5 A
v() = 5(4 – 2.5) = 7.5 V
Chapter 8, Solution 5.
(a)
For t &lt; 0, 4u(t) = 0 so that the circuit is not active (all initial conditions = 0).
i L (0-) = 0 and v C (0-) = 0.
For t = 0+, 4u(t) = 4. Consider the circuit below.
A
iL
i
4A
+
4  vC
1H
iC +
vL 
0.25F
+
6

v

Since the 4-ohm resistor is in parallel with the capacitor,
i(0+) = v C (0+)/4 = 0/4 = 0 A
Also, since the 6-ohm resistor is in series with the inductor,
v(0+) = 6i L (0+) = 0V.
(b)
di(0+)/dt = d(v R (0+)/R)/dt = (1/R)dv R (0+)/dt = (1/R)dv C (0+)/dt
= (1/4)4/0.25 A/s = 4 A/s
v = 6i L or dv/dt = 6di L /dt and dv(0+)/dt = 6di L (0+)/dt = 6v L (0+)/L = 0
Therefore dv(0+)/dt = 0 V/s
(c)
As t approaches infinity, the circuit is in steady-state.
i() = 6(4)/10 = 2.4 A
v() = 6(4 – 2.4) = 9.6 V
Chapter 8, Solution 6.
(a)
Let i = the inductor current. For t &lt; 0, u(t) = 0 so that
i(0) = 0 and v(0) = 0.
For t &gt; 0, u(t) = 1. Since, v(0+) = v(0-) = 0, and i(0+) = i(0-) = 0.
v R (0+) = Ri(0+) = 0 V
Also, since v(0+) = v R (0+) + v L (0+) = 0 = 0 + v L (0+) or v L (0+) = 0 V.
(1)
(b)
Since i(0+) = 0,
But,
i C (0+) = V S /R S
i C = Cdv/dt which leads to dv(0+)/dt = V S /(CR S )
(2)
From (1),
(3)
dv(0+)/dt = dv R (0+)/dt + dv L (0+)/dt
v R = iR or dv R /dt = Rdi/dt
(4)
But,
v L = Ldi/dt, v L (0+) = 0 = Ldi(0+)/dt and di(0+)/dt = 0
(5)
From (4) and (5),
dv R (0+)/dt = 0 V/s
From (2) and (3),
dv L (0+)/dt = dv(0+)/dt = V s /(CR s )
(c)
As t approaches infinity, the capacitor acts like an open circuit, while the inductor
acts like a short circuit.
v R () = [R/(R + R s )]V s
v L () = 0 V
Chapter 8, Solution 7.
α = [R/(2L)] = 20x103/(2x0.2x10–3) = 50x106
ω o = [1/(LC)0.5] = 1/(0.2x10–3x5x10–6)0.5 = 3.162 x104
  o
overdamped


overdamped
Chapter 8, Solution 8.
Design a problem to help other students better understand source-free RLC circuits.
Although there are many ways to work this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
The branch current in an RLC circuit is described by the differential equation
d 2i
di
 6  9i  0
2
dt
dt
and the initial conditions are i(0) = 0, di(0)/dt = 4. Obtain the characteristic equation and
determine i(t) for t &gt; 0.
Solution
s2 + 6s + 9 = 0, thus s 1,2 =
 6  6 2  36
= -3, repeated roots.
2
i(t) = [(A + Bt)e-3t], i(0) = 0 = A
di/dt = [Be-3t] + [-3(Bt)e-3t]
di(0)/dt = 4 = B.
Therefore, i(t) = [4te-3t] A
Chapter 8, Solution 9.
s2 + 10s + 25 = 0, thus s 1,2 =
 10  10  10
= –5, repeated roots.
2
i(t) = [(A + Bt)e-5t], i(0) = 10 = A
di/dt = [Be-5t] + [-5(A + Bt)e-5t]
di(0)/dt = 0 = B – 5A = B – 50 or B = 50.
Therefore, i(t) = [(10 + 50t)e-5t] A
Chapter 8, Solution 10.
s2 + 5s + 4 = 0, thus s 1,2 =
 5  25  16
= –4, –1.
2
v(t) = (Ae-4t + Be-t), v(0) = 0 = A + B, or B = -A
dv/dt = (-4Ae-4t - Be-t)
dv(0)/dt = 10 = – 4A – B = –3A or A = –10/3 and B = 10/3.
Therefore, v(t) = (–(10/3)e-4t + (10/3)e-t) V
Chapter 8, Solution 11.
s2 + 2s + 1 = 0, thus s 1,2 =
2 44
= –1, repeated roots.
2
v(t) = [(A + Bt)e-t], v(0) = 10 = A
dv/dt = [Be-t] + [-(A + Bt)e-t]
dv(0)/dt = 0 = B – A = B – 10 or B = 10.
Therefore, v(t) = [(10 + 10t)e-t] V
Chapter 8, Solution 12.
(a)
Overdamped when C &gt; 4L/(R2) = 4x1.5/2500 = 2.4x10-3, or
C &gt; 2.4 mF
(b)
Critically damped when C = 2.4 mF
(c)
Underdamped when C &lt; 2.4 mF
Chapter 8, Solution 13.
Let R||60 = R o . For a series RLC circuit,
o =
1
LC
=
1
0.01x 4
= 5
For critical damping,  o =  = R o /(2L) = 5
or R o = 10L = 40 = 60R/(60 + R)
which leads to R = 120 ohms
Chapter 8, Solution 14.
When the switch is in position A, v(0-)= 0 and i L (0) = 80/40 = 2 A. When the
switch is in position B, we have a source-free series RCL circuit.
R
10


 1.25
2L 2 x4
1
1
o 

1
LC
1
x4
4
When the switch is in position A, v(0–)= 0. When the switch is in position B, we
have a source-free series RCL circuit.
R
10


 1.25
2L 2 x4
1
1
o 

1
LC
1
x4
4
Since   o , we have overdamped case.
–2 64, 0.9336
s1,2     2  o2  1.25  1.5625–0.5
 1 and
1.56
v(t) = Ae–2t + Be–0.5t
(1)
v(0) = 0 = A + B
(2)
i C (0) = C(dv(0)/dt) = –2 or dv(0)/dt = –2/C = –8.
But
dv( t )
 2Ae  2 t  0.5Be  0.5t
dt
dv(0)
 2 A  0.5B  8
dt
Solving (2) and (3) gives A= 1.3333 and B = –1.3333
v(t) = 5.333e–2t–5.333e–0.5t V.
(3)
Chapter 8, Solution 15.
Given that s 1 = -10 and s 2 = -20, we recall that
s 1,2 =     2  o2 = -10, -20
Clearly,
s 1 + s 2 = -2 = -30 or  = 15 = R/(2L) or R = 60L
(1)
s 1 =  15  15 2   o2 = -10 which leads to 152 –  o 2 = 25
or  o =
225  25 =
200  1
LC , thus LC = 1/200
Since we have a series RLC circuit, i L = i C = Cdv C /dt which gives,
i L /C = dv C /dt = [200e-20t – 300e-30t] or i L = 100C[2e-20t – 3e-30t]
But,
i is also = 20{[2e-20t – 3e-30t]x10-3} = 100C[2e-20t – 3e-30t]
Therefore,
C = (0.02/102) = 200 F
L = 1/(200C) = 25 H
R = 30L = 750 ohms
(2)
Chapter 8, Solution 16.
At t = 0, i(0) = 0, v C (0) = 40x30/50 = 24V
For t &gt; 0, we have a source-free RLC circuit.
 = R/(2L) = (40 + 60)/5 = 20 and  o =
1
LC
=
1
10 3 x 2.5
 o =  leads to critical damping
i(t) = [(A + Bt)e-20t], i(0) = 0 = A
di/dt = {[Be-20t] + [-20(Bt)e-20t]},
but di(0)/dt = -(1/L)[Ri(0) + v C (0)] = -(1/2.5)[0 + 24]
Hence,
B = -9.6 or i(t) = [–9.6te–20t] A
= 20
Chapter 8, Solution 17.
i (0)  I 0  0, v(0)  V0  4 x5  20
di (0)
1
  ( RI 0  V0 )  4(0  20)  80
dt
L
1
1
o 

 10
LC
1 1
4 25
R 10


 20, which is  o .
2L 2 1
4
s     2  o2  20  300  20  10 3  2.679,  37.32
i (t )  A1e  2.679t  A2 e 37.32t
di (0)
 2.679 A1  37.32 A2  80
dt
This leads to A1  2.309   A2
i (0)  0  A1  A2 ,

i (t )  2.309 e 37.32t  e  2.679t
Since, v(t ) 

1 t
i (t )dt  20, we get
C 0
v(t) = [21.55e-2.679t – 1.55e-37.32t] V
Chapter 8, Solution 18.
When the switch is off, we have a source-free parallel RLC circuit.
o 
  o
1
LC

1
0.25 x1


 2,

1
 0.5
2 RC
underdamped case  d   o   2  4  0.25  1.936
2
I o (0) = i(0) = initial inductor current = 100/5 = 20 A
V o (0) = v(0) = initial capacitor voltage = 0 V
v(t )  e t ( A1 cos(d t )  A2 sin(d t ))  e 0.5t ( A1 cos(1.936t )  A2 sin(1.936t ))
v(0) =0 = A 1
dv
 e 0.5t (0.5)( A1 cos(1.936t )  A2 sin(1.936t ))  e 0.5t (1.936 A1 sin(1.936t )  1.936 A2 cos(1.936t ))
dt
(V  RI o )
(0  20)
dv(0)
 o

 20  0.5 A1  1.936 A2
1
dt
RC
Thus,
v ( t )  [10.333e 0.5 t sin(1.936t )]volts


A2  10.333
Chapter 8, Solution 19.
For t &lt; 0, the equivalent circuit is shown in Figure (a).
10 
i
+
120V
+

i
+
v
L
v
C


(b)
(a)
i(0) = 120/10 = 12, v(0) = 0
For t &gt; 0, we have a series RLC circuit as shown in Figure (b) with R = 0 = .
o =
1
LC
=
1
4
= 0.5 =  d
i(t) = [Acos0.5t + Bsin0.5t], i(0) = 12 = A
v = -Ldi/dt, and -v/L = di/dt = 0.5[-12sin0.5t + Bcos0.5t],
which leads to -v(0)/L = 0 = B
Hence,
i(t) = 12cos0.5t A and v = 0.5
However, v = -Ldi/dt = -4(0.5)[-12sin0.5t] = 24sin(0.5t) V
Chapter 8, Solution 20.
For t &lt; 0, the equivalent circuit is as shown below.
2
i
30
+ 

vC
+
v(0) = –30 V and i(0) = 30/2 = 15 A
For t &gt; 0, we have a series RLC circuit.
 = R/(2L) = 2/(2x0.5) = 2
 o = 1/ LC  1 / 0.5x 1 4  2 2
Since  is less than  o , we have an under-damped response.
d  o2   2  8  4  2
i(t) = (Acos(2t) + Bsin(2t))e-2t
i(0) = 15 = A
di/dt = –2(15cos(2t) + Bsin(2t))e-2t + (–2x15sin(2t) + 2Bcos(2t))e-t
di(0)/dt = –30 + 2B = –(1/L)[Ri(0) + v C (0)] = –2[30 – 30] = 0
Thus, B = 15 and i(t) = (15cos(2t) + 15sin(2t))e-2t A
Chapter 8, Solution 21.
By combining some resistors, the circuit is equivalent to that shown below.
60||(15 + 25) = 24 ohms.
12 
6
t=0
i
3H
24V
+

24 
+
(1/27)F
v

At t = 0-,
i(0) = 0, v(0) = 24x24/36 = 16V
For t &gt; 0, we have a series RLC circuit.
R = 30 ohms, L = 3 H, C = (1/27) F
 = R/(2L) = 30/6 = 5
 o  1 / LC  1 / 3x1 / 27 = 3, clearly  &gt;  o (overdamped response)
s 1,2 =     2  o2  5  5 2  3 2 = -9, -1
v(t) = [Ae-t + Be-9t], v(0) = 16 = A + B
(1)
i = Cdv/dt = C[-Ae-t - 9Be-9t]
i(0) = 0 = C[-A – 9B] or A = -9B
From (1) and (2),
B = -2 and A = 18.
Hence,
v(t) = (18e-t – 2e-9t) V
(2)
Chapter 8, Solution 22.
Compare the characteristic equation with eq. (8.8), i.e.
R
1
s2  s 
0
L
LC
we obtain
R
R
2000
 100

 L

 20H
L
100 100
1
 106
LC
 C
1
10 6

 50 nF
20
106 L
Chapter 8, Solution 23.
Let C o = C + 0.01. For a parallel RLC circuit,
 = 1/(2RC o ),  o = 1/ LC o
 = 1 = 1/(2RC o ), we then have C o = 1/(2R) = 1/20 = 50 mF
 o = 1/ 0.02 x0.05 = 141.42 &gt;  (underdamped)
C o = C + 10 mF = 50 mF or C = 40 mF
Chapter 8, Solution 24.
When the switch is in position A, the inductor acts like a short circuit so
i(0  )  4
When the switch is in position B, we have a source-free parallel RCL circuit
1
1


5
2RC 2 x10 x10 x10 3
1
1
o 

 20
LC
1
x10 x10 3
4
Since   o , we have an underdamped case.
s1,2  5  25  400  5  j19.365
i(t)  e5t  A1 cos19.365t  A2 sin19.365t 
i(0)  4  A1
di
vL


dt
di(0) v(0)

0
dt
L
di
 e5t  5 A1 cos19.365t  5 A2 sin19.365t  19.365 A1 sin19.365t  19.365 A2 cos19.365t 
dt
0 = [di(0)/dt] = –5A 1 + 19.365A 2 or A 2 = 20/19.365 = 1.0328
i(t) = e–5t[4cos(19.365t) + 1.0328sin(19.365t)] A
Chapter 8, Solution 25.
Using Fig. 8.78, design a problem to help other students to better understand source-free RLC
circuits.
Although there are many ways to work this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
In the circuit in Fig. 8.78, calculate i o (t) and v o (t) for t &gt; 0.
Figure 8.78
Solution
In the circuit in Fig. 8.76, calculate i o (t) and v o (t) for t&gt;0.
2
30V
+

1H
i o (t
+
t=0, note this is a
make before break
switch so the
inductor current is
not interrupted.
Figure 8.78
8
For Problem 8.25.
At t = 0-, v o (0) = (8/(2 + 8)(30) = 24
For t &gt; 0, we have a source-free parallel RLC circuit.
 = 1/(2RC) = &frac14;
 o = 1/ LC  1 / 1x 1 4  2
(1/4)F
v o (t)

Since  is less than  o , we have an under-damped response.
d  o2   2  4  (1 / 16)  1.9843
v o (t) = (A 1 cos d t + A 2 sin d t)e-t
v o (0) = 30(8/(2+8)) = 24 = A 1 and i o (t) = C(dv o /dt) = 0 when t = 0.
dv o /dt = -(A 1 cos d t + A 2 sin d t)e-t + (- d A 1 sin d t +  d A 2 cos d t)e-t
at t = 0, we get dv o (0)/dt = 0 = -A 1 +  d A 2
Thus, A 2 = (/ d )A 1 = (1/4)(24)/1.9843 = 3.024
v o (t) = (24cos1.9843t + 3.024sin1.9843t)e-t/4 volts.
i 0 (t) = Cdv/dt = 0.25[–24(1.9843)sin1.9843t + 3.024(1.9843)cos1.9843t –
0.25(24cos1.9843t) – 0.25(3.024sin1.9843t)]e–t/4
= [– 12.095sin1.9843t]e–t/4 A.
Chapter 8, Solution 26.
s2 + 2s + 5 = 0, which leads to s 1,2 =
 2  4  20
= –1j4
2
These roots indicate an underdamped circuit which has the generalized solution
given as:
i(t) = I s + [(A 1 cos(4t) + A 2 sin(4t))e-t],
At t = ∞, (di(t)/dt) = 0 and (d2i(t)/dt2) = 0 so that
I s = 10/5 = 2 (from (d2i(t)/dt2)+2(di(t)/dt)+5 = 10)
i(0) = 2 = 2 + A 1 , or A 1 = 0
di/dt = [(4A 2 cos(4t))e-t] + [(-A 2 sin(4t))e-t] = 4 = 4A 2 , or A 2 = 1
i(t) = [2 + sin(4te-t)] amps
Chapter 8, Solution 27.
s2 + 4s + 8 = 0 leads to s =
 4  16  32
 2  j2
2
v(t) = V s + (A 1 cos2t + A 2 sin2t)e-2t
8V s = 24 means that V s = 3
v(0) = 0 = 3 + A 1 leads to A 1 = -3
dv/dt = -2(A 1 cos2t + A 2 sin2t)e-2t + (-2A 1 sin2t + 2A 2 cos2t)e-2t
0 = dv(0)/dt = -2A 1 +2A 2 or A 2 = A 1 = -3
v(t) = [3 – 3(cos(2t) + sin(2t))e–2t] volts.
Chapter 8, Solution 28.
The characteristic equation is
1
Ls2  Rs   0


C
s1,2 
1 2
1
s  4s 
0
2
0.2

 s2  8 s  10  0
8  64  40
–6.45 and
–1.5505
 0.838,
7.162
2
i( t )  i s  Ae 6.45t  Be 1.5505t
But
[i s /C] = 10 or i s = 0.2x10 = 2
i (t )  2  Ae 6.45t  Be 1.5505t
i(0) = 1 = 2 + A + B or A + B = –1 or A = –1–B
(1)
di (t )
 6.45 Ae 6.45t  1.5505Be 1.5505t
dt
di (0)
but
 0  6.45 A  1.5505B
dt
(2)
Solving (1) and (2) gives –6.45(–1–B) – 1.5505B = 0 or (6.45–1.5505)B = –6.45
B = –6.45/(4.9) = –1.3163 and A = –1–1.3163 = –2.3163
A= –2.3163, B= –1.3163
Hence,
i(t) = [2–2.3163e–6.45t –1.3163e–1.5505t] A.
Chapter 8, Solution 29.
(a)
s2 + 4 = 0 which leads to s 1,2 = j2 (an undamped circuit)
v(t) = V s + Acos2t + Bsin2t
4V s = 12 or V s = 3
v(0) = 0 = 3 + A or A = -3
dv/dt = -2Asin2t + 2Bcos2t
dv(0)/dt = 2 = 2B or B = 1, therefore v(t) = (3 – 3cos2t + sin2t) V
(b)
s2 + 5s + 4 = 0 which leads to s 1,2 = -1, -4
i(t) = (I s + Ae-t + Be-4t)
4I s = 8 or I s = 2
i(0) = -1 = 2 + A + B, or A + B = -3
(1)
di/dt = -Ae-t - 4Be-4t
di(0)/dt = 0 = -A – 4B, or B = -A/4
From (1) and (2) we get A = -4 and B = 1
i(t) = (2 – 4e-t + e-4t) A
(c)
s2 + 2s + 1 = 0, s 1,2 = -1, -1
v(t) = [V s + (A + Bt)e-t], V s = 3.
v(0) = 5 = 3 + A or A = 2
dv/dt = [-(A + Bt)e-t] + [Be-t]
dv(0)/dt = -A + B = 1 or B = 2 + 1 = 3
v(t) = [3 + (2 + 3t)e-t] V
(2)
(d)
s2 + 2s +5 = 0, s 1,2 = -1 + j2, -1 – j2
i(t) = [I s + (Acos2t + Bsin2t)e-t], where 5I s = 10 or I s = 2
i(0) = 4 = 2 + A or A = 2
di/dt = [-(Acos2t + Bsin2t)e-t] + [(-2Asin2t + 2Bcos2t)e-t]
di(0)/dt = -2 = -A + 2B or B = 0
i(t) = [2 + (2cos2t)e-t] A
Chapter 8, Solution 30.
The step responses of a series RLC circuit are
v C (t) = [40–10e–2000t–10e–4000t] volts, t &gt; 0 and
i L (t) = [3e–2000t+6e–4000t ] m A, t &gt; 0.
(a) Find C. (b) Determine what type of damping exhibited by the circuit.
Solution
Step 1.
For a series RLC circuit, i R (t) = i L (t) = i C (t).
We can determine C from i C (t) = i L (t) = C(dv C /dt) and we can determine that the
circuit is overdamped since the exponent value are real and negative.
Step 2.
C(dv C /dt) = C[20,000e–2000t+40,000e–4000t] = 0.003e–2000t+0.006e–4000t or
C = 0.003/20,000 = 150 F.
Chapter 8, Solution 31.
For t = 0-, we have the equivalent circuit in Figure (a). For t = 0+, the equivalent
circuit is shown in Figure (b). By KVL,
v(0+) = v(0-) = 40, i(0+) = i(0-) = 1
By KCL, 2 = i(0+) + i 1 = 1 + i 1 which leads to i 1 = 1. By KVL, -v L + 40i 1 +
v(0+) = 0 which leads to v L (0+) = 40x1 + 40 = 80
v L (0+) = 80 V,
40 
i 1 40 
10 
+
i
v C (0+) = 40 V
+
+
v
50V

+

(a)
v
vL

10 
50V

0.5H
(b)
+

Chapter 8, Solution 32.
For t = 0-, the equivalent circuit is shown below.
2A
i
+
v

6
i(0-) = 0, v(0-) = -2x6 = -12V
For t &gt; 0, we have a series RLC circuit with a step input.
 = R/(2L) = 6/2 = 3,  o = 1/ LC  1 / 0.04
s =  3  9  25  3  j4
Thus, v(t) = V f + [(Acos4t + Bsin4t)e-3t]
where V f = final capacitor voltage = 50 V
v(t) = 50 + [(Acos4t + Bsin4t)e-3t]
v(0) = -12 = 50 + A which gives A = -62
i(0) = 0 = Cdv(0)/dt
dv/dt = [-3(Acos4t + Bsin4t)e-3t] + [4(–Asin4t + Bcos4t)e–3t]
0 = dv(0)/dt = -3A + 4B or B = (3/4)A = –46.5
v(t) = {50 + [(–62cos4t – 46.5sin4t)e–3t]} V
Chapter 8, Solution 33.
We may transform the current sources to voltage sources. For t = 0-, the equivalent
circuit is shown in Figure (a).
10 
i
i
1H
+
30V
+

v
+
5

5
v
20V
4F
+


(a)
(b)
i(0) = 30/15 = 2 A, v(0) = 5x30/15 = 10 V
For t &gt; 0, we have a series RLC circuit, shown in (b).
 = R/(2L) = 5/2 = 2.5
 o  1 / LC  1 / 4 = 0.5, clearly  &gt;  o (overdamped response)
s 1,2 =     2   2o  2.5  6.25  0.25 = –4.949, –0.0505
v(t) = V s + [A 1 e–4.949t + A 2 e–0.0505t], V s = 20.
v(0) = 10 = 20 + A 1 + A 2 or
A 2 = –10 – A 1
(1)
i(0) = Cdv(0)/dt or dv(0)/dt = –2/4 = –1/2
Hence,
–0.5 = – 4.949A 1 – 0.0505A 2
From (1) and (2),
–0.5 = –4.949A 1 + 0.0505(10 + A 1 ) or
–4.898A 1 = –0.5–0.505 = –1.005
A 1 = 0.2052, A 2 = –10.205
v(t) = [20 + 0.2052e–4.949t – 10.205e-0.05t] V.
(2)
Chapter 8, Solution 34.
Before t = 0, the capacitor acts like an open circuit while the inductor behaves like
a short circuit.
i(0) = 0, v(0) = 50 V
For t &gt; 0, the LC circuit is disconnected from the voltage source as shown below.
+
Vx

i
(1/16)F
(&frac14;) H
This is a lossless, source-free, series RLC circuit.
 = R/(2L) = 0,  o = 1/ LC = 1/
1
1

= 8, s = j8
16 4
Since  is equal to zero, we have an undamped response. Therefore,
i(t) = A 1 cos(8t) + A 2 sin(8t) where i(0) = 0 = A 1
di(0)/dt = (1/L)v L (0) = –(1/L)v(0) = –4x50 = –200
However, di/dt = 8A 2 cos(8t), thus, di(0)/dt = –200 = 8A 2 which leads to A 2 = –25
Now we have
i(t) = –25sin(8t) A
Chapter 8, Solution 35.
Using Fig. 8.83, design a problem to help other students to better understand the step response of
series RLC circuits.
Although there are many ways to work this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
Determine v(t) for t &gt; 0 in the circuit in Fig. 8.83.
Figure 8.83
Solution
At t = 0-, i L (0) = 0, v(0) = v C (0) = 8 V
For t &gt; 0, we have a series RLC circuit with a step input.
 = R/(2L) = 2/2 = 1,  o = 1/ LC = 1/ 1 / 5 =
5
s 1,2 =     2   2o  1  j2
v(t) = V s + [(Acos2t + Bsin2t)e-t], V s = 12.
v(0) = 8 = 12 + A or A = -4, i(0) = Cdv(0)/dt = 0.
But dv/dt = [-(Acos2t + Bsin2t)e-t] + [2(-Asin2t + Bcos2t)e-t]
0
= dv(0)/dt = -A + 2B or 2B = A = -4 and B = -2
v(t) = {12 – (4cos2t + 2sin2t)e-t V.
Chapter 8, Solution 36.
For t = 0–, 3u(t) = 0. Thus, i(0) = 0, and v(0) = 20 V.
For t &gt; 0, we have the series RLC circuit shown below.
10 
i
10 
5H
+
30V
+

2
40 V
0.2 F
+ 
v

 = R/(2L) = (2 + 5 + 1)/(2x5) = 0.8
 o = 1/ LC = 1/ 5x 0.2 = 1
s 1,2 =     2  2o  0.8  j0.6
v(t) = V s + [(Acos(0.6t) + Bsin(0.6t))e-0.8t]
V s = 30 + 40 = 70 V and v(0) = 40 = 70 + A or A = –30
i(0) = Cdv(0)/dt = 0
But dv/dt = [–0.8(Acos(0.6t) + Bsin(0.6t))e–0.8t] + [0.6(–Asin(0.6t) + Bcos(0.6t))e–0.8t]
0
= dv(0)/dt = –0.8A + 0.6B which leads to B = 0.8x(–30)/0.6 = –40
v(t) = {70 – [(30cos(0.6t) + 40sin(0.6t))e–0.8t]} V
i = Cdv/dt
= 0.2{[0.8(30cos(0.6t) + 40sin(0.6)t)e-0.8t] + [0.6(30sin(0.6t) – 40cos(0.6t))e-0.8t]}
i(t) = 10sin(0.6t)e-0.8t A
Chapter 8, Solution 37.
For t = 0–, the equivalent circuit is shown below.
+
i2
6
6
6
v(0)
45V
+

i1
15V
+


18i 2 – 6i 1 = 0 or i 1 = 3i 2
–45 + 6(i 1 – i 2 ) + 15 = 0 or i 1 – i 2 = 30/6
(2)
From (1) and (2), (2/3)i 1 = 5 or i 1 = 7.5 and i 2 = i 1 – 5 = 2.5
i(0) = i 1 = 7.5A
–15 – 6i 2 + v(0) = 0
v(0) = 15 + 6x2.5 = 30
For t &gt; 0, we have a series RLC circuit.
R = 6||12 = 4
 o = 1/ LC = 1/ (1 / 2)(1 / 8) = 4
 = R/(2L) = (4)/(2x(1/2)) = 4
 =  o , therefore the circuit is critically damped
v(t) = V s +[(A + Bt)e-4t], and V s = 15
(1)
=5
v(0) = 30 = 15 + A, or A = 15
i C = Cdv/dt = C[–4(15 + Bt)e-4t] + C[(B)e-4t]
To find i C (0) we need to look at the circuit right after the switch is opened. At this time,
the current through the inductor forces that part of the circuit to act like a current source
and the capacitor acts like a voltage source. This produces the circuit shown below.
Clearly, i C (0+) must equal –i L (0) = –7.5A.
6
6
iC
6
30V
+

7.5A
i C (0) = –7.5 = C(–60 + B) which leads to –60 = –60 + B or B = 0
i C = Cdv/dt = (1/8)[–4(15 + 0t)e-4t] + (1/8)[(0)e-4t]
i C (t) = [–(1/2)(15)e-4t]
i(t) = –i C (t) = 7.5e-4t A
Chapter 8, Solution 38.
At t = 0—, the equivalent circuit is as shown.
2A
+
i
10 
v
i1
5
10 

i(0) = 2A, i 1 (0) = 10(2)/(10 + 15) = 0.8 A
v(0) = 5i 1 (0) = 4V
For t &gt; 0, we have a source-free series RLC circuit.
R = 5||(10 + 10) = 4 ohms
 o = 1/ LC = 1/ (1 / 3)(3 / 4) = 2
 = R/(2L) = (4)/(2x(3/4)) = 8/3
s 1,2 =     2   2o  -4.431, -0.903
i(t) = [Ae-4.431t + Be-0.903t]
i(0) = A + B = 2
(1)
di(0)/dt = (1/L)[-Ri(0) + v(0)] = (4/3)(-4x2 + 4) = -16/3 = -5.333
Hence, -5.333 = -4.431A – 0.903B
From (1) and (2), A = 1 and B = 1.
i(t) = [e-4.431t + e-0.903t] A
(2)
Chapter 8, Solution 39.
For t = 0–, the source voltages are equal to zero thus, the initial conditions are v(0) = 0 and
i L (0) = 0.
30 
0.5F 0.25H
60u(t)V
+ v 
20 
+

+

30u(t)V
For t &gt; 0, the circuit is shown above.
R = 20||30 = 12 ohms
 o = 1/ LC = 1/ (1 / 2)(1 / 4) =
8
 = R/(2L) = (12)/(0.5) = 24
Since  &gt;  o , we have an overdamped response.
s 1,2 =     2   2o  –47.83, –0.167
Thus,
v(t) = V s + [Ae–47.83t + Be–0.167t], where
V s = [60/(30+20)]20–30 = –6 volts.
v(0) = 0 = –6 + A + B or 6 = A + B
(1)
i(0) = Cdv(0)/dt = 0
But,
dv(0)/dt = –47.83A – 0.167B = 0 or
B = –286.4A
From (1) and (2),
A + (–286.4)A = 6 or A = 6/(–285.4) = –0.02102 and
B = –286.4x(–0.02102) = 6.02
v(t) = [–6 + (–0.021e-47.83t+6.02e-0.167t)] volts.
(2)
Chapter 8, Solution 40.
At t = 0-, v C (0) = 0 and i L (0) = i(0) = (6/(6 + 2))4 = 3A
For t &gt; 0, we have a series RLC circuit with a step input as shown below.
i
0.02 F
2H
+
6
v
14 

24V
12V
+

+ 
 o = 1/ LC = 1/ 2x 0.02 = 5
 = R/(2L) = (6 + 14)/(2x2) = 5
Since  =  o , we have a critically damped response.
v(t) = V s + [(A + Bt)e-5t], V s = 24 – 12 = 12V
v(0) = 0 = 12 + A or A = -12
i = Cdv/dt = C{[Be-5t] + [-5(A + Bt)e-5t]}
i(0) = 3 = C[-5A + B] = 0.02[60 + B] or B = 90
Thus, i(t) = 0.02{[90e-5t] + [-5(-12 + 90t)e-5t]}
i(t) = {(3 – 9t)e-5t} A
Chapter 8, Solution 41.
At t = 0–, the switch is open. i(0) = 0, and
v(0) = 5x100/(20 + 5 + 5) = 50/3
For t &gt; 0, we have a series RLC circuit shown in Figure (a). After source
transformation, it becomes that shown in Figure (b).
10 H
4
1H
i
20 
5A
5
10 F
+
20V
+

0.04F
v

(a)
(b)
 o = 1/ LC = 1/ 1x1 / 25 = 5
 = R/(2L) = (4)/(2x1) = 2
s 1,2 =     2   2o  -2  j4.583
Thus,
v(t) = V s + [(Acos( d t) + Bsin( d t))e-2t],
where  d = 4.583 and V s = 20
v(0) = 50/3 = 20 + A or A = -10/3
i(t) = Cdv/dt
= C(-2) [(Acos( d t) + Bsin( d t))e-2t] + C d [(-Asin( d t) + Bcos( d t))e-2t]
i(0) = 0 = -2A +  d B
B = 2A/ d = -20/(3x4.583) = -1.455
i(t) = C{[(0cos( d t) + (-2B -  d A)sin( d t))]e-2t}
= (1/25){[(2.91 + 15.2767) sin( d t))]e-2t}
i(t) = 727.5sin(4.583t)e-2t mA
Chapter 8, Solution 42.
For t = 0-, we have the equivalent circuit as shown in Figure (a).
i(0) = i(0) = 0, and v(0) = 4 – 12 = -8V
4V
5
 +
+ 
1
6
12V
i
1H
+
v(0)
+

+
v
12V


(a)
0.04F
(b)
For t &gt; 0, the circuit becomes that shown in Figure (b) after source transformation.
 o = 1/ LC = 1/ 1x1 / 25 = 5
 = R/(2L) = (6)/(2) = 3
s 1,2 =     2   2o  -3  j4
Thus,
v(t) = V s + [(Acos4t + Bsin4t)e-3t], V s = -12
v(0) = -8 = -12 + A or A = 4
i = Cdv/dt, or i/C = dv/dt = [-3(Acos4t + Bsin4t)e-3t] + [4(-Asin4t + Bcos4t)e-3t]
i(0) = -3A + 4B or B = 3
v(t) = {-12 + [(4cos4t + 3sin4t)e-3t]} A
Chapter 8, Solution 43.
For t&gt;0, we have a source-free series RLC circuit.

R
2L

R  2L  2x8x 0.5  8Ω
d  o 2   2  30
o 
1
LC




C
1
 L
2
o
o  900  64  964

1
 2.075 mF
964 x 0.5
Chapter 8, Solution 44.

R 1000

 500,
2L
2 x1
o  


o 
1
LC

underdamped.
1
100 x10
9
 10 4
Chapter 8, Solution 45.
 o = 1/ LC = 1/ 1x 0.5 =
2
 = 1/(2RC) = (1)/(2x2x0.5) = 0.5
Since  &lt;  o , we have an underdamped response.
s 1,2 =    o2   2  –0.5  j1.3229
i(t) = I s + [(Acos1.3229t + Bsin1.3229t)e-0.5t], I s = 4
Thus,
i(0) = 1 = 4 + A or A = -3
v = v C = v L = Ldi(0)/dt = 0
di/dt = [1.3229(-Asin1.3229t + Bcos1.3229t)e-0.5t] +
[-0.5(Acos1.3229t + Bsin1.3229t)e-0.5t]
di(0)/dt = 0 = 1.3229B – 0.5A or B = 0.5(–3)/1.3229 = –1.1339
Thus,
i(t) = {4 – [(3cos1.3229t + 1.1339sin1.3229t)e-t/2]} A
To find v(t) we use v(t) = v L (t) = Ldi(t)/dt.
From above,
di/dt = [1.3229(-Asin1.3229t + Bcos1.3229t)e-0.5t] +
[-0.5(Acos1.323t + Bsin1.323t)e-0.5t]
Thus,
v(t) = Ldi/dt = [1.323(-Asin1.323t + Bcos1.323t)e-0.5t] +
[-0.5(Acos1.323t + Bsin1.323t)e-0.5t]
= [1.3229(3sin1.3229t – 1.1339cos1.3229t)e-0.5t] +
[(1.5cos1.3229t + 0.5670sin1.3229t)e-0.5t]
v(t) = [(–0cos1.323t + 4.536sin1.323t)e-0.5t] V
= [(4.536sin1.3229t)e-t/2] V
Please note that the term in front of the cos calculates out to –3.631x10-5 which is zero for
all practical purposes when considering the rounding errors of the terms used to calculate
it.
Chapter 8, Solution 46.
Using Fig. 8.93, design a problem to help other students to better understand the step response of
a parallel RLC circuit.
Although there are many ways to work this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
Find i(t) for t &gt; 0 in the circuit in Fig. 8.93.
Figure 8.93
Solution
For t = 0-, u(t) = 0, so that v(0) = 0 and i(0) = 0.
For t &gt; 0, we have a parallel RLC circuit with a step input, as shown below.
+
i
8mH
5F
v

2 k
6mA
 = 1/(2RC) = (1)/(2x2x103 x5x10-6) = 50
 o = 1/ LC = 1/ 8x10  3 x 5x10  6 = 5,000
Since  &lt;  o , we have an underdamped response.
s 1,2 =     2  o2  -50  j5,000
Thus,
i(t) = I s + [(Acos5,000t + Bsin5,000t)e-50t], I s = 6mA
i(0) = 0 = 6 + A or A = -6mA
v(0) = 0 = Ldi(0)/dt
di/dt = [5,000(-Asin5,000t + Bcos5,000t)e-50t] + [-50(Acos5,000t + Bsin5,000t)e-50t]
di(0)/dt = 0 = 5,000B – 50A or B = 0.01(-6) = -0.06mA
Thus,
i(t) = {6 – [(6cos5,000t + 0.06sin5,000t)e-50t]} mA
Chapter 8, Solution 47.
At t = 0-, we obtain,
i L (0) = 3x5/(10 + 5) = 1A
and v o (0) = 0.
For t &gt; 0, the 10-ohm resistor is short-circuited and we have a parallel RLC circuit with
a step input.
 = 1/(2RC) = (1)/(2x5x0.01) = 10
 o = 1/ LC = 1/ 1x 0.01 = 10
Since  =  o , we have a critically damped response.
s 1,2 = -10
Thus,
i(t) = I s + [(A + Bt)e-10t], I s = 3
i(0) = 1 = 3 + A or A = -2
v o = Ldi/dt = [Be-10t] + [-10(A + Bt)e-10t]
v o (0) = 0 = B – 10A or B = -20
Thus, v o (t) = (200te-10t) V
Chapter 8, Solution 48.
For t = 0–, we obtain i(0) = –6/(1 + 2) = –2 and v(0) = 2x1 = 2.
For t &gt; 0, the voltage is short-circuited and we have a source-free parallel RLC circuit.
 = 1/(2RC) = (1)/(2x1x0.25) = 2
 o = 1/ LC = 1/ 1x 0.25 = 2
Since  =  o , we have a critically damped response.
s 1,2 =
Thus,
–2
i(t) = [(A + Bt)e-2t], i(0) = –2 = A
v = Ldi/dt = [Be-2t] + [–2(–2 + Bt)e–2t]
v o (0) = 2 = B + 4 or B = –2
Thus,
i(t) = [(–2 – 2t)e–2t] A
and v(t) = [(2 + 4t)e–2t] V
Chapter 8, Solution 49.
For t = 0-, i(0) = 3 + 12/4 = 6 and v(0) = 0.
For t &gt; 0, we have a parallel RLC circuit with a step input.
 = 1/(2RC) = (1)/(2x5x0.05) = 2
 o = 1/ LC = 1/ 5x 0.05 = 2
Since  =  o , we have a critically damped response.
s 1,2 =
Thus,
-2
i(t) = I s + [(A + Bt)e-2t], I s = 3
i(0) = 6 = 3 + A or A = 3
v = Ldi/dt or v/L = di/dt = [Be-2t] + [-2(A + Bt)e-2t]
v(0)/L = 0 = di(0)/dt = B – 2x3 or B = 6
Thus, i(t) = {3 + [(3 + 6t)e-2t]} A
Chapter 8, Solution 50.
For t = 0-, 4u(t) = 0, v(0) = 0, and i(0) = 30/10 = 3A.
For t &gt; 0, we have a parallel RLC circuit.
i
+
10 
3A
10 mF
6A
40 
v
10 H

I s = 3 + 6 = 9A and R = 10||40 = 8 ohms
 = 1/(2RC) = (1)/(2x8x0.01) = 25/4 = 6.25
 o = 1/ LC = 1/ 4x 0.01 = 5
Since  &gt;  o , we have a overdamped response.
s 1,2 =     2   o2  -10, -2.5
Thus,
i(t) = I s + [Ae-10t] + [Be-2.5t], I s = 9
i(0) = 3 = 9 + A + B or A + B = -6
di/dt = [-10Ae-10t] + [-2.5Be-2.5t],
v(0) = 0 = Ldi(0)/dt or di(0)/dt = 0 = -10A – 2.5B or B = -4A
Thus, A = 2 and B = -8
Clearly,
i(t) = { 9 + [2e-10t] + [-8e-2.5t]} A
Chapter 8, Solution 51.
Let i = inductor current and v = capacitor voltage.
At t = 0, v(0) = 0 and i(0) = i o .
For t &gt; 0, we have a parallel, source-free LC circuit (R = ).
 = 1/(2RC) = 0 and  o = 1/ LC which leads to s 1,2 =  j o
v = Acos o t + Bsin o t, v(0) = 0 A
i C = Cdv/dt = -i
dv/dt =  o Bsin o t = -i/C
dv(0)/dt =  o B = -i o /C therefore B = i o /( o C)
v(t) = -(i o /( o C))sin o t V where  o = 1 / LC
Chapter 8, Solution 52.
  300 
1
2 RC
d  o   2  400
2
(1)


o2  d2   2  160,000  90,000 
From (2),
C
1
 80 &micro;F
250,000x 50x10 3
From (1),
R
1
1

 20.83Ω.
2C 2x 300x80x10 6
1
LC
(2)
Chapter 8, Solution 53.
At t&lt;0, i(0  )  0, vc(0  )  120V
For t &gt;0, we have the circuit as shown below.
80 
i
120 V
+
_
120  V
dv
C
i
R
dt
10 mF


120  V  RC
di
dt
Substituting (2) into (1) yields
di
d2 i
120  L
 RCL 2  iR


dt
dt
or
But
dv
 iR
dt
vL  v  L
0.25 H
(1)
(2)
120 
1 di
1
d2 i
 80 x x10 x10 3 2  80i
4 dt
4
dt
(d2i/dt2) + 0.125(di/dt) + 400i = 600
Chapter 8, Solution 54.
Using Fig. 8.100, design a problem to help other students better understand general second-order
circuits.
Although there are many ways to work this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
For the circuit in Fig. 8.100, let I = 9A, R 1 = 40 Ω, R 2 = 20 Ω, C = 10 mF, R 3 = 50 Ω,and
L = 20 mH. Determine: (a) i(0+) and v(0+), (b) di(0+)/dt and dv(0+)/dt, (c) i(∞) and v(∞).
A
t=0
R3
i
B
I
+
R1
R2
C
v
L

Figure 8.100
For Prob. 8.54.
Solution
(a) When the switch is at A, the circuit has reached steady state. Under this condition, the circuit
is as shown below.
t=0
50Ω
A
i
B
9A
40Ω
+
20Ω
v

(a)
When the switch is at A, i(0–) = 9[(40x50)/(40+50)]/50 = 4 A and v(0–) = 50i(0–) = 200 V.
Since the current flowing through the inductor cannot change in zero time, i(0+) = i(0–) = 4 A.
Since the voltage across the capacitor cannot change in zero time, v(0–) = v(0–) = 200 V.
(b)
For the inductor, v L = L(di/dt) or di(0+)/dt = v L (0+)/0.02.
At t = 0+, the right hand loop becomes,
–200 + 50x4 + v L (0+) = 0 or v L (0+) = 0 and (di(0+)/dt) = 0.
For the capacitor, i C = C(dv/dt) or dv(0+)/dt = i C (0+)/0.01.
At t = 0+, and looking at the current flowing out of the node at the top of the circuit,
((200–0)/20) + i C + 4 = 0 or i C = –14 A.
Therefore,
dv(0+)/dt = –14/0.01 = –1.4 kV/s.
(c)
When the switch is in position B, the circuit reaches steady state. Since it is source-free, i
and v decay to zero with time.
Thus,
i(∞) = 0 A and v(∞) = 0 V.
Chapter 8, Solution 55.
At the top node, writing a KCL equation produces,
i/4 +i = C 1 dv/dt, C 1 = 0.1
5i/4 = C 1 dv/dt = 0.1dv/dt
i = 0.08dv/dt
But,
(1)
v =  (2i  (1 / C 2 )  idt ) , C 2 = 0.5
or
-dv/dt = 2di/dt + 2i
(2)
Substituting (1) into (2) gives,
-dv/dt = 0.16d2v/dt2 + 0.16dv/dt
0.16d2v/dt2 + 0.16dv/dt + dv/dt = 0, or d2v/dt2 + 7.25dv/dt = 0
Which leads to s2 + 7.25s = 0 = s(s + 7.25) or s 1,2 = 0, -7.25
From (1),
v(t) = A + Be-7.25t
(3)
v(0) = 4 = A + B
(4)
i(0) = 2 = 0.08dv(0+)/dt or dv(0+)/dt = 25
But,
dv/dt = -7.25Be-7.25t, which leads to,
dv(0)/dt = -7.25B = 25 or B = -3.448 and A = 4 – B = 4 + 3.448 = 7.448
Thus, v(t) = {7.448 – 3.448e-7.25t} V
Chapter 8, Solution 56.
For t &lt; 0, i(0) = 0 and v(0) = 0.
For t &gt; 0, the circuit is as shown below.
4
i
6
i
0.04F
20
+

io
0.25H
Applying KVL to the larger loop,
-20 +6i o +0.25di o /dt + 25  (i o  i)dt = 0
Taking the derivative,
6di o /dt + 0.25d2i o /dt2 + 25(i o + i) = 0
For the smaller loop,
4 + 25  (i  i o )dt = 0
Taking the derivative,
25(i + i o ) = 0 or i = -i o
From (1) and (2)
6di o /dt + 0.25d2i o /dt2 = 0
This leads to, 0.25s2 + 6s = 0 or s 1,2 = 0, -24
i o (t) = (A + Be-24t) and i o (0) = 0 = A + B or B = -A
As t approaches infinity, i o () = 20/10 = 2 = A, therefore B = -2
Thus, i o (t) = (2 - 2e-24t) = -i(t) or
i(t) = (-2 + 2e-24t) A
(1)
(2)
Chapter 8, Solution 57.
(a)
Let v = capacitor voltage and i = inductor current. At t = 0-, the switch is
closed and the circuit has reached steady-state.
v(0-) = 16V and i(0-) = 16/8 = 2A
At t = 0+, the switch is open but, v(0+) = 16 and i(0+) = 2.
We now have a source-free RLC circuit.
R 8 + 12 = 20 ohms, L = 1H, C = 4mF.
 = R/(2L) = (20)/(2x1) = 10
 o = 1/ LC = 1/ 1x (1 / 36) = 6
Since  &gt;  o , we have a overdamped response.
s 1,2 =     2  o2  -18, -2
Thus, the characteristic equation is (s + 2)(s + 18) = 0 or s2 + 20s +36 = 0.
i(t) = [Ae-2t + Be-18t] and i(0) = 2 = A + B
(b)
(1)
To get di(0)/dt, consider the circuit below at t = 0+.
i
12 
+
+
(1/36)F
v

8
vL

1H
-v(0) + 20i(0) + v L (0) = 0, which leads to,
-16 + 20x2 + v L (0) = 0 or v L (0) = -24
Ldi(0)/dt = v L (0) which gives di(0)/dt = v L (0)/L = -24/1 = -24 A/s
Hence -24 = -2A – 18B or 12 = A + 9B
From (1) and (2),
B = 1.25 and A = 0.75
(2)
i(t) = [0.75e-2t + 1.25e-18t] = -i x (t) or i x (t) = [-0.75e-2t - 1.25e-18t] A
v(t) = 8i(t) = [6e-2t + 10e-18t] A
Chapter 8, Solution 58.
(a) Let i =inductor current, v = capacitor voltage i(0) =0, v(0) = 4
dv(0)
[v(0)  Ri (0)]
(4  0)


  8 V/s
dt
RC
0.5
(b) For t  0 , the circuit is a source-free RLC parallel circuit.

1
1

 1,
2 RC 2 x0.5 x1
o 
1
LC

1
0.25 x1
2
 d   2 o   2  4  1  1.732
Thus,
v(t )  e t ( A1 cos1.732t  A2 sin 1.732t )
v(0) = 4 = A 1
dv
 e t A1 cos1.732t  1.732e t A1 sin 1.732t  e t A2 sin 1.732t  1.732e t A2 cos1.732t
dt
dv(0)
 8   A1  1.732 A2


A2  2.309
dt
v ( t )  e  t (4 cos 1.732t  2.309 sin 1.732t ) V
Chapter 8, Solution 59.
Let i = inductor current and v = capacitor voltage
v(0) = 0, i(0) = 40/(4+16) = 2A
For t&gt;0, the circuit becomes a source-free series RLC with
R
16
1
1

 2,  o 

 2,

    o  2
2L 2 x4
LC
4 x1 / 16
i (t )  Ae 2t  Bte 2t
i(0) = 2 = A
di
 2 Ae  2t  Be  2t  2 Bte  2t
dt
di(0)
1
1
 2A  B   [Ri(0)  v(0)]

  2A  B   (32  0),
dt
L
4

B  4
i (t )  2e 2t  4te 2t
t
t
t
t 64
t
1
 2t
v    idt  v(0)  32  e dt  64  te  2 t dt  16e  2 t  e  2 t (2 t  1)
C
0 4
0
0
0
0
v = –32te–2t V.
Checking,
v = Ldi/dt + Ri = 4(–4e–2t – 4e–2t + 8e–2t) + 16(2e–2t – 4te–2t) = –32te–2t V.
Chapter 8, Solution 60.
At t = 0-, 4u(t) = 0 so that i 1 (0) = 0 = i 2 (0)
(1)
Applying nodal analysis,
4 = 0.5di 1 /dt + i 1 + i 2
(2)
Also,
i 2 = [1di 1 /dt – 1di 2 /dt]/3 or 3i 2 = di 1 /dt – di 2 /dt
Taking the derivative of (2), 0 = d2i 1 /dt2 + 2di 1 /dt + 2di 2 /dt
From (2) and (3),
(3)
(4)
di 2 /dt = di 1 /dt – 3i 2 = di 1 /dt – 3(4 – i 1 – 0.5di 1 /dt)
= di 1 /dt – 12 + 3i 1 + 1.5di 1 /dt
Substituting this into (4),
d2i 1 /dt2 + 7di 1 /dt + 6i 1 = 24 which gives s2 + 7s + 6 = 0 = (s + 1)(s + 6)
Thus, i 1 (t) = I s + [Ae-t + Be-6t], 6I s = 24 or I s = 4
i 1 (t) = 4 + [Ae-t + Be-6t] and i 1 (0) = 4 + [A + B]
(5)
i 2 = 4 – i 1 – 0.5di 1 /dt = i 1 (t) = 4 + -4 - [Ae-t + Be-6t] – [-Ae-t - 6Be-6t]
= [-0.5Ae-t + 2Be-6t] and i 2 (0) = 0 = -0.5A + 2B
From (5) and (6),
A = -3.2 and B = -0.8
i 1 (t) = {4 + [-3.2e-t – 0.8e-6t]} A
i 2 (t) = [1.6e-t – 1.6e-6t] A
(6)
Chapter 8, Solution 61.
For t &gt; 0, we obtain the natural response by considering the circuit below.
a
1H
iL
+
4
vC
0.25F
6

At node a,
v C /4 + 0.25dv C /dt + i L = 0
(1)
But,
v C = 1di L /dt + 6i L
(2)
Combining (1) and (2),
(1/4)di L /dt + (6/4)i L + 0.25d2i L /dt2 + (6/4)di L /dt + i L = 0
d2i L /dt2 + 7di L /dt + 10i L = 0
s2 + 7s + 10 = 0 = (s + 2)(s + 5) or s 1,2 = -2, -5
Thus, i L (t) = i L () + [Ae-2t + Be-5t],
where i L () represents the final inductor current = 4(4)/(4 + 6) = 1.6
i L (t) = 1.6 + [Ae-2t + Be-5t] and i L (0) = 1.6 + [A+B] or -1.6 = A+B (3)
di L /dt = [-2Ae-2t - 5Be-5t]
and di L (0)/dt = 0 = -2A – 5B or A = -2.5B
(4)
From (3) and (4), A = -8/3 and B = 16/15
i L (t) = 1.6 + [-(8/3)e-2t + (16/15)e-5t]
v(t) = 6i L (t) = {9.6 + [-16e-2t + 6.4e-5t]} V
v C = 1di L /dt + 6i L = [ (16/3)e-2t - (16/3)e-5t] + {9.6 + [-16e-2t + 6.4e-5t]}
v C = {9.6 + [-(32/3)e-2t + 1.0667e-5t]}
i(t) = v C /4 = {2.4 + [-2.667e-2t + 0.2667e-5t]} A
Chapter 8, Solution 62.
This is a parallel RLC circuit as evident when the voltage source is turned off.
 = 1/(2RC) = (1)/(2x3x(1/18)) = 3
 o = 1/ LC = 1/ 2 x1 / 18 = 3
Since  =  o , we have a critically damped response.
s 1,2 =
-3
Let v(t) = capacitor voltage
Thus, v(t) = V s + [(A + Bt)e-3t] where V s = 0
But -10 + v R + v = 0 or v R = 10 – v
Therefore v R = 10 – [(A + Bt)e-3t] where A and B are determined from initial
conditions.
Chapter 8, Solution 63.
vs  0
d(0  vo )
vs
dv
C


 C o
dt
R
dt
R
2
dvo
v
di
di
vo  L


L 2  s
dt
dt
RC
dt
Thus,
v
d 2 i (t )
 s
2
RCL
dt
Chapter 8, Solution 64.
Using Fig. 8.109, design a problem to help other students to better understand second-order op
amp circuits.
Although there are many ways to work this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
Obtain the differential equation for v o (t) in the network of Fig. 8.109.
Figure 8.109
Solution
C2
R2
R1
vs
1
v1
C1
2

+
vo
At node 1,
(v s – v 1 )/R 1 = C 1 d(v 1 – 0)/dt or v s = v 1 + R 1 C 1 dv 1 /dt
At node 2,
C 1 dv 1 /dt = (0 – v o )/R 2 + C 2 d(0 – v o )/dt
or
From (1) and (2),
or
–R 2 C 1 dv 1 /dt = v o + R 2 C 2 dv o /dt
(1)
(2)
(v s – v 1 )/R 1 = C 1 dv 1 /dt = –(1/R 2 )(v o + R 2 C 2 dv o /dt)
v 1 = v s + (R 1 /R 2 )(v o + R 2 C 2 dv o /dt)
(3)
Substituting (3) into (1) produces,
v s = v s + (R 1 /R 2 )(v o + R 2 C 2 dv o /dt) + R 1 C 1 d{v s + (R 1 /R 2 )(v o + R 2 C 2 dv o /dt)}/dt
= v s + (R 1 /R 2 )(v o )+ (R 1 C 2 )dv o /dt + R 1 C 1 dv s /dt + (R 1 R 1 C 1 /R 2 )dv o /dt
+ ((R 1 )2 C 1 C 2 )[d2v o /dt2]
((R 1 )2 C 1 C 2 )[d2v o /dt2] + [(R 1 C 2 ) + (R 1 R 1 C 1 /R 2 )]dv o /dt + (R 1 /R 2 )(v o ) = –R 1 C 1 dv s /dt
Simplifying we get,
[d2v o /dt2] + {[(R 1 C 2 ) + (R 1 R 1 C 1 /R 2 )]/((R 1 )2 C 1 C 2 )}dv o /dt + {(R 1 /R 2 )(v o )/ ((R 1 )2
C 1 C 2 )} = –{R 1 C 1 /((R 1 )2 C 1 C 2 )}dv s /dt
d2v o /dt2 + [(1/ R 1 C 1 ) + (1/(R 2 C 2 ))]dv o /dt + [1/(R 1 R 2 C 1 C 2 )](v o ) = –[1/(R 1 C 2 )]dv s /dt
Another way to successfully work this problem is to give actual values of the resistors and
capacitors and determine the actual differential equation. Alternatively, one could give a
differential equations and ask the other students to choose actual value of the differential
equation.
Chapter 8, Solution 65.
At the input of the first op amp,
(v o – 0)/R = Cd(v 1 – 0)
(1)
At the input of the second op amp,
(-v 1 – 0)/R = Cdv 2 /dt
(2)
Let us now examine our constraints. Since the input terminals are essentially at ground,
then we have the following,
v o = -v 2 or v 2 = -v o
(3)
Combining (1), (2), and (3), eliminating v 1 and v 2 we get,
d 2 vo  1 
d 2vo

v

 100 v o  0


o
dt 2  R 2 C 2 
dt 2
Which leads to s2 – 100 = 0
Clearly this produces roots of –10 and +10.
And, we obtain,
v o (t) = (Ae+10t + Be-10t)V
At t = 0, v o (0+) = – v 2 (0+) = 0 = A + B, thus B = –A
This leads to v o (t) = (Ae+10t – Ae-10t)V. Now we can use v 1 (0+) = 2V.
From (2), v 1 = –RCdv 2 /dt = 0.1dv o /dt = 0.1(10Ae+10t + 10Ae-10t)
v 1 (0+) = 2 = 0.1(20A) = 2A or A = 1
Thus, v o (t) = (e+10t – e-10t) V
It should be noted that this circuit is unstable (clearly one of the poles lies in the righthalf-plane).
Chapter 8, Solution 66.
We apply nodal analysis to the circuit as shown below.
v2
10pF
60 k
60 k
v1
–
v2
+
vs
20 pF
+
_
+
vo
–
At node 1,
vs  v1 v1  v2
d

 10 pF (v1  vo )
dt
60k
60k
But v2  vo
d( v 1  v o )
v s  2 v1  v o  6x10  7
dt
At node 2,
v1  v2
d
 20 pF (v2  0), v2  vo
60k
dt
dv
v1  v o  1.2 x10  6 o
dt
Substituting (2) into (1) gives
(1)
(2)

dv 
d 2 v o 

v s  2 v o  1.2 x10  6 o   v o  6 x10  7 1.2 x10  6

dt 

dt 2 

v s = v o + 2.4x10–6(dv o /dt) + 7.2x10–13(d2v o /dt2).
Chapter 8, Solution 67.
At node 1,
d ( v1  v o )
v in  v1
d( v1  0)
 C2
 C1
R1
dt
dt
At node 2,
C2
(1)
 vo
d ( v 1  0) 0  v o
dv1
, or


dt
R2
dt
C2R 2
(2)
From (1) and (2),
v in  v1  
v1  v in 
dv
v
R 1C1 dv o
 R 1 C1 o  R 1 o
C 2 R 2 dt
dt
R2
dv
v
R 1C1 dv o
 R 1C1 o  R 1 o
C 2 R 2 dt
dt
R2
(3)
C1
R2
R1
v in
C2
1
v1
2
0V

+
vo
From (2) and (3),

vo
d 2 v o R 1 dv o
dv
dv
R C dv o
 1  in  1 1
 R 1 C1

C2R 2
dt
dt
C 2 R 2 dt
R 2 dt
dt 2
d 2 vo
vo
1  1
1  dv o
1 dv in






2
R 2  C1 C 2  dt
C1C 2 R 2 R 1
R 1C1 dt
dt
But C 1 C 2 R 1 R 2 = 10-4 x10-4 x104 x104 = 1
1
R2
 1
1 
2
2
 

 4
2

4
 C1 C 2  R 2 C1 10 x10
d 2 vo
dv
dv
 2 o  v o   in
2
dt
dt
dt
Which leads to s2 + 2s + 1 = 0 or (s + 1)2 = 0 and s = –1, –1
Therefore,
v o (t) = [(A + Bt)e-t] + V f
As t approaches infinity, the capacitor acts like an open circuit so that
V f = v o () = 0
v in = 10u(t) mV and the fact that the initial voltages across each capacitor is 0
means that v o (0) = 0 which leads to A = 0.
v o (t) = [Bte-t]
dv o
= [(B – Bt)e-t]
dt
dv o (0 )
v (0 )
 o
0
dt
C2R 2
From (2),
From (1) at t = 0+,
dv (0)
dv o (0 )
1 0
1
 C 1 o

 1
R1
dt
dt
C1 R 1
Substituting this into (4) gives B = –1
Thus,
v(t) = –te-tu(t) V
(4)
Chapter 8, Solution 68.
The schematic is as shown below. The unit step is modeled by VPWL as shown. We
insert a voltage marker to display V after simulation. We set Print Step = 25 ms and
final step = 6s in the transient box. The output plot is shown below.
Chapter 8, Solution 69.
The schematic is shown below. The initial values are set as attributes of L1 and C1. We
set Print Step to 25 ms and the Final Time to 20s in the transient box. A current marker
is inserted at the terminal of L1 to automatically display i(t) after simulation. The result
is shown below.
Chapter 8, Solution 70.
The schematic is shown below.
After the circuit is saved and simulated, we obtain the capacitor voltage v(t) as shown
below.
Chapter 8, Solution 71.
The schematic is shown below. We use VPWL and IPWL to model the 39 u(t) V and 13
u(t) A respectively. We set Print Step to 25 ms and Final Step to 4s in the Transient
box. A voltage marker is inserted at the terminal of R2 to automatically produce the plot
of v(t) after simulation. The result is shown below.
Chapter 8, Solution 72.
When the switch is in position 1, we obtain IC=10 for the capacitor and IC=0 for the
inductor. When the switch is in position 2, the schematic of the circuit is shown below.
When the circuit is simulated, we obtain i(t) as shown below.
Chapter 8, Solution 73.
Design a problem, using PSpice, to help other students to better understand source-free RLC
circuits.
Although there are many ways to work this problem, this is an example based on a somewhat
similar problem worked in the third edition.
Problem
The step response of an RLC circuit is given by
Given that i L (0) = 3 A and v C (0) = 24 V, solve for v C (t) and I C (t).
Solution
(a)
For t &lt; 0, we have the schematic below. When this is saved and simulated, we obtain
the initial inductor current and capacitor voltage as
i L (0) = 3 A and v c (0) = 24 V.
(b)
For t &gt; 0, we have the schematic shown below. To display i(t) and v(t), we insert
current and voltage markers as shown. The initial inductor current and capacitor voltage are also
incorporated. In the Transient box, we set Print Step = 25 ms and the Final Time to 4s. After
simulation, we automatically have i o (t) and v o (t) displayed as shown below.
Chapter 8, Solution 74.
The dual is constructed as shown below.
0.5 
0.25 
2
9V
+
_
4
1/6 
6
1
3A
1
9A
–
+
The dual is redrawn as shown below.
1/6 
9A
1/2 
1
1/4 
–
+
3V
3V
Chapter 8, Solution 75.
The dual circuit is connected as shown in Figure (a). It is redrawn in Figure (b).
0.1 
12V
+

10 
12A
24A
0.5 F
24V
0.25 
4
10 H
10 H
10 F
(a)
0.1 
2F
0.5 H
24A
12A
0.25 
(b)
+

Chapter 8, Solution 77.
The dual is obtained from the original circuit as shown in Figure (a). It is redrawn in
Figure (b).
0.1 
0.05 
1/3 
10 
20 
60 A
30 
120 A
+ 
– +
60 V
2V
120 V
+ 
4H
1F
1H
2A
4F
(a)
0.05 
60 A
120 A
1H
0.1 
1/30 
1/4 F 2V
(b)
+

Chapter 8, Solution 77.
The dual is constructed in Figure (a) and redrawn in Figure (b).
– +
5A
5V
2
1/3 
1/2 
1F
1
1/4 H
1H
3
1
1/4 F
12V
+

12 A
(a)
1
1/2 
1/3 
1/4 F
12 A
1H
5V
+

(b)
Chapter 8, Solution 78.
The voltage across the igniter is v R = v C since the circuit is a parallel RLC type.
v C (0) = 12, and i L (0) = 0.
 = 1/(2RC) = 1/(2x3x1/30) = 5
 o  1 / LC  1 / 60 x10 3 x1 / 30 = 22.36
 &lt;  o produces an underdamped response.
s1, 2     2   o2 = –5  j21.794
v C (t) = e-5t(Acos21.794t + Bsin21.794t)
(1)
v C (0) = 12 = A
dv C /dt = –5[(Acos21.794t + Bsin21.794t)e-5t]
+ 21.794[(–Asin21.794t + Bcos21.794t)e-5t]
(2)
dv C (0)/dt = –5A + 21.794B
But,
dv C (0)/dt = –[v C (0) + Ri L (0)]/(RC) = –(12 + 0)/(1/10) = –120
Hence,
–120 = –5A + 21.794B, leads to B (5x12 – 120)/21.794 = –2.753
At the peak value, dv C (t o )/dt = 0, i.e.,
0
= A + Btan21.794t o + (A21.794/5)tan21.794t o – 21.794B/5
(B + A21.794/5)tan21.794t o = (21.794B/5) – A
tan21.794t o = [(21.794B/5) – A]/(B + A21.794/5) = –24/49.55 = –0.484
Therefore,
21.7945t o = |–0.451|
t o = |–0.451|/21.794 = 20.68 ms
Chapter 8, Solution 79.
For critical damping of a parallel RLC circuit,
  o


1

2 RC
Hence,
C
L
0.25

 434 F
2
4 x144
4R
1
LC
Chapter 8, Solution 80.
t 1 = 1/|s 1 | = 0.1x10-3 leads to s 1 = –1000/0.1 = –10,000
t 2 = 1/|s 2 | = 0.5x10-3 leads to s 1 = –2,000
s1     2  o2
s 2     2   o2
s 1 + s 2 = –2 = –12,000, therefore  = 6,000 = R/(2L)
L = R/12,000 = 50,000/12,000 = 4.167H
s 2     2  o2 = –2,000
   2   o2 = 2,000
6,000   2   o2 = 2,000
 2   o2 = 4,000
2 – o2 = 16x106
o2 = 2 – 16x106 = 36x106 – 16x106
 o = 103 20  1 / LC
C = 1/(20x106x4.167) = 12 nF
Chapter 8, Solution 81.
t = 1/ = 0.25 leads to  = 4
But,
 1/(2RC) or,
C = 1/(2R) = 1/(2x4x200) = 625 F
d  o2   2
o2  d2   2  (24 x10 3 ) 2  16  (24 x10 3 0 2 = 1/(LC)
This results in L = 1/(642x106x625x10-6) = 2.533 H
Chapter 8, Solution 82.
For t = 0-, v(0) = 0.
For t &gt; 0, the circuit is as shown below.
R1
a
+
+
C1
vo
R2
C2
v


At node a,
(v o – v/R 1 = (v/R 2 ) + C 2 dv/dt
v o = v(1 + R 1 /R 2 ) + R 1 C 2 dv/dt
60 = (1 + 5/2.5) + (5x106 x5x10-6)dv/dt
60 = 3v + 25dv/dt
v(t) = V s + [Ae-3t/25]
where
3V s = 60 yields V s = 20
v(0) = 0 = 20 + A or A = –20
v(t) = 20(1 – e-3t/25)V
Chapter 8, Solution 83.
i = i D + Cdv/dt
(1)
–v s + iR + Ldi/dt + v = 0
(2)
Substituting (1) into (2),
v s = Ri D + RCdv/dt + Ldi D /dt + LCd2v/dt2 + v = 0
LCd2v/dt2 + RCdv/dt + Ri D + Ldi D /dt = v s
d2v/dt2 + (R/L)dv/dt + (R/LC)i D + (1/C)di D /dt = v s /LC
Chapter 9, Solution 1.
(a) V m = 50 V.
2
 0.2094s = 209.4ms
 30
(c ) Frequency f = ω/(2π) = 30/(2π) = 4.775 H z .
(d) At t=1ms, v(0.01) = 50cos(30x0.01rad + 10˚)
= 50cos(1.72˚ + 10˚) = 44.48 V and ωt = 0.3 rad.
(b) Period T 
2

Chapter 9, Solution 2.
(a)
amplitude = 15 A
(b)
 = 25 = 78.54 rad/s
(c)
f =
(d)
I s = 1525 A
I s (2 ms) = 15 cos((500 )(2  10 -3 )  25)
= 15 cos( + 25) = 15 cos(205)
= –13.595 A

= 12.5Hz
2
Chapter 9, Solution 3.
(a)
10 sin(t + 30) = 10 cos(t + 30 – 90) = 10cos(t – 60)
(b)
–9 sin(8t) = 9cos(8t + 90)
(c)
–20 sin(t + 45) = 20 cos(t + 45 + 90) = 20cos(t + 135)
(a) 10cos(t – 60), (b) 9cos(8t + 90), (c) 20cos(t + 135)
Chapter 9, Solution 4.
Design a problem to help other students to better understand sinusoids.
Although there are many ways to work this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
(a) Express v = 8 cos(7t + 15) in sine form.
(b) Convert i = –10sin(3t - 85) to cosine form.
Solution
(a)
v = 8 cos(7t + 15) = 8 sin(7t + 15 + 90) = 8 sin(7t + 105)
(b)
i = –10 sin(3t – 85) = 10 cos(3t – 85 + 90) = 10 cos(3t + 5)
Chapter 9, Solution 5.
v 1 = 45 sin(t + 30) V = 45 cos(t + 30  90) = 45 cos(t  60) V
v 2 = 50 cos(t – 30) V
This indicates that the phase angle between the two signals is 30 and that v 1 lags
v2.
Chapter 9, Solution 6.
(a)
v(t) = 10 cos(4t – 60)
i(t) = 4 sin(4t + 50) = 4 cos(4t + 50 – 90) = 4 cos(4t – 40)
Thus, i(t) leads v(t) by 20.
(b)
v 1 (t) = 4 cos(377t + 10)
v 2 (t) = -20 cos(377t) = 20 cos(377t + 180)
Thus, v 2 (t) leads v 1 (t) by 170.
(c)
x(t) = 13 cos(2t) + 5 sin(2t) = 13 cos(2t) + 5 cos(2t – 90)
X = 130 + 5-90 = 13 – j5 = 13.928-21.04
x(t) = 13.928 cos(2t – 21.04)
y(t) = 15 cos(2t – 11.8)
phase difference = -11.8 + 21.04 = 9.24
Thus, y(t) leads x(t) by 9.24.
Chapter 9, Solution 7.
If f() = cos + j sin,
df
 -sin  j cos   j (cos   j sin )  j f ( )
d
df
 j d
f
Integrating both sides
ln f = j + ln A
f = Aej = cos + j sin
f(0) = A = 1
i.e. f() = ej = cos + j sin
Chapter 9, Solution 8.
(a)
(b)
6045
6045
+ j2 =
+ j2
7.5  j10
12.5 - 53.13
= 4.898.13 + j2 = –0.6788+j4.752+j2
= –0.6788 + j6.752
(6–j8)(4+j2) = 24–j32+j12+16 = 40–j20 = 44.72–26.57˚
20
32  20
32  20
20
+
+
=
(6  j8)(4  j2) - 10  j24
44.72  26.57 26112.62
= 0.71566.57˚+0.7692–112.62˚ = 0.7109+j0.08188–0.2958–j0.71
= 0.4151–j0.6281
(c)
20 + (16–50)(1367.38) = 20+20817.38 = 20 + 198.5+j62.13
= 218.5+j62.13
Chapter 9, Solution 9.
(a)
(530)(6  j8  1.1197  j0.7392)  (530)(7.13  j7.261)
 (530)(10.176  45.52) 
50.88–15.52˚.
(b)
(1060)(35  50)
 60.02–110.96˚.
(3  j5)  (5.83120.96)
Chapter 9, Solution 10.
Design a problem to help other students to better understand phasors.
Although there are many ways to work this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
Given that z 1 = 6 – j8, z 2 = 10–30, and z 3 = 8e–j120, find:
(a) z 1 + z 2 + z 3
(b) z 1 z 2 / z 3
Solution
(a) z1  6  j8, z 2  8.66  j 5, and z 3  4  j 6.9282
z1  z 2  z 3 = (10.66 – j19.928)Ω
(b)
z1 z 2
= [(10–53.13˚)(10–30˚)/(8–120˚)] = 12.536.87˚Ω = (10 + j7.5) Ω
z3
Chapter 9, Solution 11.
(a)
V  21  15o V
(b) i(t )  8sin(10t  70o  180o )  8cos(10t  70o  180o  90o )  8cos(10t  160o )
I = 8 160&deg; mA
(c ) v(t )  120sin(103 t  50o )  120 cos(103 t  50o  90o )
V = 120 –140&deg; V
(d)
i(t )  60 cos(30t  10o )  60 cos(30t  10o  180o )
I = 60 –170&deg; mA
Chapter 9, Solution 12.
Let X = 440 and Y = 20–30. Evaluate the following quantities and express
(X + Y)/X*
(X - Y)*
(X + Y)/X
X = 3.064+j2.571; Y = 17.321–j10
(20.38  j 7.429)(4  40)
 (21.69  20.03)(4  40)  86.76  60.03
= 86.76–60.03˚
(a)
(X + Y)X* =
(b)
(X – Y)* = (–14.257+j12.571)* = 19.41–139.63˚
(c)
(X + Y)/X = (21.69–20.03˚)/(440˚) = 5.422–60.03˚
Chapter 9, Solution 13.
(a) ( 0.4324  j 0.4054) ( 0.8425  j 0.2534)   1.2749  j 0.1520
(b)
50  30 o
  2.0833 = –2.083
24150 o
(c) (2+j3)(8-j5) –(-4) = 35 +j14
Chapter 9, Solution 14.
(a)
3  j14
14.318  77.91

 0.7788169.71   0.7663  j0.13912
 7  j17 18.385112.38
(b)
(62.116  j 231.82  138.56  j80)(60  j80)
24186  6944.9

  1.922  j11.55
(67  j84)(16.96  j10.5983)
246.06  j 2134.7
(c)
= [(22.3663.43˚)/(553.13˚)]2[(11.1826.57˚)(25.61–51.34˚)]0.5
= [4.47210.3˚]2[286.3–24.77˚]0.5 = (19.99920.6˚)(16.921–12.38˚) = 338.48.22˚
or 334.9+j48.38
Chapter 9, Solution 15.
(a)
(b)
(c)
10  j6 2  j3
= -10 – j6 + j10 – 6 + 10 – j15
-5
-1  j
= –6 – j11
20  30 - 4 - 10
= 6015 + 64-10
160
345
= 57.96 + j15.529 + 63.03 – j11.114
= 120.99 + j4.415
1 j  j 0
j
1 j
1
j 1 j
1 j  j
j
1
= 1  1  0  1  0  j2 (1  j)  j2 (1  j)
0
j
= 1  1 (1  j  1  j)
= 1 – 2 = –1
Chapter 9, Solution 16.
(a)
–20 cos(4t + 135) = 20 cos(4t + 135  180)
= 20 cos(4t  45)
The phasor form is 20–45
(b)
8 sin(20t + 30) = 8 cos(20t + 30 – 90)
= 8 cos(20t – 60)
The phasor form is 8–60
(c)
20 cos(2t) + 15 sin(2t) = 20 cos(2t) + 15 cos(2t – 90)
The phasor form is 200 + 15–90 = 20 – j15 = 25-36.87
Chapter 9, Solution 17.
V  V1  V2  10  60o  12  30o  5  j8.66  10.392  j 6  15.62  9.805o
v(t) = 15.62cos(50t–9.8˚) V
Chapter 9, Solution 18.
(a)
v1 ( t ) = 60 cos(t + 15)
(b)
V2 = 6 + j8 = 1053.13
v 2 ( t ) = 10 cos(40t + 53.13)
(c)
i1 ( t ) = 2.8 cos(377t – /3)
(d)
I 2 = -0.5 – j1.2 = 1.3247.4
i 2 ( t ) = 1.3 cos(103t + 247.4)
Chapter 9, Solution 19.
(a)
310  5-30 = 2.954 + j0.5209 – 4.33 + j2.5
= -1.376 + j3.021
= 3.32114.49
Therefore,
3 cos(20t + 10) – 5 cos(20t – 30)
= 3.32 cos(20t + 114.49)
(b)
40-90 + 30-45 = -j40 + 21.21 – j21.21
= 21.21 – j61.21
= 64.78-70.89
Therefore,
40 sin(50t) + 30 cos(50t – 45) = 64.78 cos(50t – 70.89)
(c)
Using sin = cos(  90),
20-90 + 1060  5-110 = -j20 + 5 + j8.66 + 1.7101 + j4.699
= 6.7101 – j6.641
= 9.44-44.7
Therefore,
20 sin(400t) + 10 cos(400t + 60) – 5 sin(400t – 20)
= 9.44 cos(400t – 44.7)
Chapter 9, Solution 20.
7.5cos(10t+30˚) A can be represented by 7.530˚ and 120cos(10t+75˚) V can be
represented by 12075˚. Thus,
Z = V/I = (12075˚)/(7.530˚) = 1645˚ or (11.314+j11.314) Ω.
Chapter 9, Solution 21.
(a) F  515 o  4 30 o  90 o  6.8296  j 4.758  8.323634.86 o
f ( t )  8.324 cos( 30t  34.86 o )
(b) G  8  90 o  450 o  2.571  j 4.9358  5.565  62.49 o
g ( t )  5.565 cos( t  62.49 o )
(c) H 


1
100 o  50  90 o ,
j
  40
i.e. H  0.25  90 o  1.25  180 o   j0.25  1.25  1.2748  168.69 o
h(t) = 1.2748cos(40t – 168.69&deg;)
Chapter 9, Solution 22.
t
dv
Let f(t) = 10v(t )  4  2  v(t )dt
dt

2V
F  10V  j 4V 
,   5, V  5545o
j
F  10V  j 20V  j 0.4V  (10  j 20.4)V  22.7263.89(5545)  1249.6108.89 o
f(t) = 1249.6cos(5t+108.89˚)
Chapter 9, Solution 23.
(a) v = [110sin(20t+30˚) + 220cos(20t–90˚)] V leads to V = 110(30˚–90˚) +
220–90˚ = 55–j95.26 – j220 = 55–j315.3 = 320.1–80.11˚ or
v = 320.1cos(20t–80.11˚) A.
(b) i = [30cos(5t+60˚)–20sin(5t+60˚)] A leads to I = 3060˚ – 20(60˚–90˚) =
15+j25.98 – (17.321–j10) = –2.321+j35.98 = 36.0593.69˚ or
i = 36.05cos(5t+93.69˚) A.
(a) 320.1cos(20t–80.11˚) A, (b) 36.05cos(5t+93.69˚) A
Chapter 9, Solution 24.
(a)
V
 10 0,   1
j
V (1  j)  10
10
V
 5  j5  7.07145
1 j
V
Therefore,
v(t) = 7.071cos(t + 45) V
(b)
4V
 20(10  90),   4
j

4
V  j4  5    20  - 80
j4 

20 - 80
 3.43 - 110.96
V
5  j3
jV  5V 
Therefore,
v(t) = 3.43cos(4t – 110.96) V
Chapter 9, Solution 25.
(a)
2jI  3I  445,   2
I (3  j 4)  445
445
445
I

 0.8 - 8.13
3  j4 553.13
Therefore,
i(t) = 800cos(2t – 8.13) mA
(b)
I
 jI  6I  522,   5
j
(- j2  j5  6) I  522
522
522

 0.745 - 4.56
I
6  j3 6.70826.56
Therefore,
i(t) = 745 cos(5t – 4.56) mA
10
Chapter 9, Solution 26.
I
 10,   2
j

1
I  j2  2    1
j2 

1
 0.4  - 36.87
I
2  j1.5
Therefore,
i(t) = 0.4 cos(2t – 36.87)
jI  2I 
Chapter 9, Solution 27.
V
 110  - 10,   377
j

j100 
  110 - 10
V  j377  50 

377 
V (380.682.45)  110 - 10
V  0.289  - 92.45
jV  50V  100
Therefore, v(t) = 289 cos(377t – 92.45) mV.
Chapter 9, Solution 28.
i (t ) 
vs (t ) 156 cos(377t  45)

 10.4cos(377t+45˚) A.
15
R
Chapter 9, Solution 29.
Z
1
1

 - j 0.5
6
jC j (10 )(2  10 -6 )
V  IZ  (4 25)(0.5 - 90)  2  - 65
Therefore
v(t) = 2 sin(106t – 65) V.
Chapter 9, Solution 30.
Since R and C are in parallel, they have the same voltage across them. For the resistor,
100  20o
V  IR R

 IR  V / R 
 2.5  20o mA
40k
iR  2.5cos(60t  20o ) mA
For the capacitor,
iC  C
dv
 50 x106 (60) x100sin(60t  20o )  300sin(60t  20o ) mA
dt
Chapter 9, Solution 31.
j L  j 2 x 240 x103  j 0.48
1
1
C  5mF



  j100
jC j 2 x5 x103
Z  80  j 0.48  j100  80  j 99.52 =
L  240mH


V
10  00
 0.0783  51.206o
I 
Z 80  j 99.52
i(t) = 78.3cos(2t+51.21˚) mA
Chapter 9, Solution 32.
Using Fig. 9.40, design a problem to help other students to better understand phasor relationships
for circuit elements.
Although there are many ways to work this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
Two elements are connected in series as shown in Fig. 9.40.
If i = 12 cos (2t - 30) A, find the element values.
Figure 9.40
Solution
V = 18010,
Z
I = 12-30,
 = 2
V 18010
 1540  11.49  j 9.642 

I 12 - 30
One element is a resistor with R = 11.49 .
The other element is an inductor with L = 9.642 or
L = 4.821 H.
Chapter 9, Solution 33.
110  v 2R  v 2L
v L  110 2  v 2R
v L  110 2  85 2  69.82 V
Chapter 9, Solution 34.
v o  0 when jX L –jX C = 0 so X L = X C or L 

1
C
1
(5  10 3 )(20  10 3 )

  
1
LC
.
Chapter 9, Solution 35.
vs (t )  50 cos 200t


Vs  50  0o ,   200
1
1

j
jC j 200 x5 x103

 j L  j 20 x103 x 200  j 4


5mF
20mH
Z in  10  j  j 4  10  j 3
I
Vs 50  0o

 4.789  16.7o
Z in 10  j 3
i(t) = 4.789cos(200t–16.7&deg;) A
Chapter 9, Solution 36.
Using Fig. 9.43, design a problem to help other students to better understand impedance.
Although there are many ways to work this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
In the circuit in Fig. 9.43, determine i. Let v s = 60 cos(200t - 10) V.
Figure 9.43
Solution
Let Z be the input impedance at the source.
100 mH
10 F




jL  j 200 x100 x10 3  j 20
1
1

  j 500
jC j10 x10 6 x 200
1000//-j500 = 200 –j400
1000//(j20 + 200 –j400) = 242.62 –j239.84
Z  2242.62  j 239.84  2255  6.104 o
I
60  10 o
 26.61  3.896 o mA
o
2255  6.104
i  266.1 cos( 200t  3.896 o ) mA
Chapter 9, Solution 37.
Y = (1/4) + (1/(j8)) + (1/(–j10)) = 0.25 – j0.025
= (250–j25) mS
Chapter 9, Solution 38.
Using Fig. 9.45, design a problem to help other students to better understand admittance.
Although there are many ways to work this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
Find i(t) and v(t) in each of the circuits of Fig. 9.45.
Figure 9.45
Solution
(a)
1
F 

6
1
1

 - j2
jC j (3)(1 / 6)
- j2
(10 45)  4.472  - 18.43
4  j2
Hence, i(t) = 4.472 cos(3t – 18.43) A
I
V  4I  (4)( 4.472  - 18.43)  17.89  - 18.43
Hence, v(t) = 17.89 cos(3t – 18.43) V
(b)
1
F 

12
1
1

 - j3
jC j ( 4)(1 / 12)
3H


jL  j ( 4)(3)  j12
V 50 0

 10 36.87
Z 4  j3
Hence, i(t) = 10 cos(4t + 36.87) A
I
j12
(50 0)  41.6 33.69
8  j12
Hence, v(t) = 41.6 cos(4t + 33.69) V
V
Chapter 9, Solution 39.
Z eq  4  j 20  10 //(  j14  j 25)  9.135  j 27.47 
= (9.135+j27.47) Ω
I
V
12

 0.4145  71.605o
Z eq 9.135  j 27.47
i(t) = 414.5cos(10t–71.6˚) mA
Chapter 9, Solution 40.
(a)
For   1 ,
1H 

jL  j (1)(1)  j
1
1
0.05 F 


 - j20
jC j (1)(0.05)
- j40
Z  j  2 || (- j20)  j 
 1.98  j0.802
2  j20
4 0
40
V


 1.872  - 22.05
Z 1.98  j0.802 2.13622.05
Hence,
i o ( t )  1.872 cos(t – 22.05) A
Io 
(b)
For   5 ,
1H 

jL  j (5)(1)  j5
1
1
0.05 F 


 - j4
jC j (5)(0.05)
- j4
Z  j5  2 || (- j4)  j5 
 1.6  j4.2
1  j2
40
40
V


 0.89 - 69.14
Z 1.6  j4 4.49469.14
Hence,
i o ( t )  890cos(5t – 69.14) mA
Io 
(c)
For   10 ,
1H 
 jL  j (10)(1)  j10
1
1
0.05 F 


 - j2
jC j (10)(0.05)
- j4
Z  j10  2 || (- j2)  j10 
 1  j9
2  j2
V 4 0
4 0

 0.4417  - 83.66

Z 1  j9 9.05583.66
Hence,
i o ( t )  441.7cos(10t – 83.66) mA
Io 
Chapter 9, Solution 41.
  1,
1H 

jL  j (1)(1)  j
1
1
1F 


 -j
jC j (1)(1)
- j1
Z  1  (1  j) || (- j)  1 
 2 j
1
I
Vs
10

,
Z 2 j
I c  (1  j) I
V  (- j)(1  j) I  (1  j) I 
(1  j)(10)
 6.325 - 18.43
2 j
Thus,
v(t) = 6.325cos(t – 18.43) V
Chapter 9, Solution 42.
  200
50 F 

1
1

 - j100
jC j (200)(50  10 -6 )
0.1 H 
 jL  j (200)(0.1)  j20
(50)(-j100) - j100
50 || -j100 

 40  j20
50  j100
1 - j2
Vo 
j20
j20
(600) 
(600)  17.14 90
j20  30  40  j20
70
Thus,
v o ( t )  17.14 sin(200t + 90) V
or
v o ( t )  17.14 cos(200t) V
Chapter 9, Solution 43.
Z in  50  j80 //(100  j 40)  50 
Io 
j80(100  j 40)
 105.71  j 57.93
100  j 40
60  0o
 0.4377  0.2411  0.4997  28.85o A = 499.7–28.85˚ mA
Z in
Chapter 9, Solution 44.
  200
10 mH 

jL  j (200)(10  10 -3 )  j2
1
1


 -j
5 mF 
jC j (200)(5  10 -3 )
3 j
1 1
1
Y  
 0.25  j0.5 
 0.55  j0.4
10
4 j2 3  j
1
1
Z 
 1.1892  j0.865
Y 0.55  j0.4
60
60
I

 0.96  - 7.956
5  Z 6.1892  j0.865
Thus,
i(t) = 960cos(200t – 7.956) mA
Chapter 9, Solution 45.
We obtain I o by applying the principle of current division twice.
I
I2
Z1
I2
-j2 
Z2
(a)
Z 1  - j2 ,
Io
2
(b)
Z 2  j4  (-j2) || 2  j4 
I2 
Z1
- j2
- j10
I
(50) 
Z1  Z 2
- j 2  1  j3
1 j
Io 
 - j   - j10  - 10
- j2
 
 
I 2  
 –5 A
2 - j2
1- j 1 j  11
- j4
 1  j3
2 - j2
Chapter 9, Solution 46.
i s  5 cos(10 t  40) 
 I s  540
1
1
0.1 F 


 -j
jC j (10)(0.1)
0.2 H 

Z1  4 || j2 
Let
jL  j (10)(0.2)  j2
j8
 0.8  j1.6 ,
4  j2
Z2  3  j
Z1
0.8  j1.6
Is 
(540)
Z1  Z 2
3.8  j0.6
(1.78963.43)(540)
Io 
 2.32594.46
3.847 8.97
Io 
Thus,
i o ( t )  2.325cos(10t + 94.46) A
Chapter 9, Solution 47.
First, we convert the circuit into the frequency domain.
Ix
50˚
Ix 
+

2
j4
-j10
20 
5
5
5


 0.460752.63
 j10(20  j4) 2  4.588  j8.626 10.854  52.63
2
 j10  20  j4
i s (t) = 460.7cos(2000t +52.63˚) mA
Chapter 9, Solution 48.
Converting the circuit to the frequency domain, we get:
10 
V 1 30 
Ix
20-40˚
+

j20
We can solve this using nodal analysis.
V1  20  40 V1  0
V 0

 1
0
10
j 20
30  j 20
V1 (0.1  j 0.05  0.02307  j 0.01538)  2  40
240
 15.643  24.29
0.12307  j 0.03462
15.643  24.29
Ix 
 0.43389.4
30  j 20
i x  0.4338 sin(100t  9.4 ) A
V1 
-j20
Chapter 9, Solution 49.
Z T  2  j2 || (1  j)  2 
I
( j2)(1  j)
4
1 j
Ix
1
j2 
j2
j2
I
I,
j2  1  j
1 j
1 j
1 j
I
Ix 
j2
j4
Ix 
Vs  I Z T 
-j 
where I x  0.50 
1 j
1 j
 1  j  1.414 - 45
(4) 
j
j4
v s ( t )  1.4142 sin(200t – 45) V
1
2
Chapter 9, Solution 50.
Since ω = 100, the inductor = j100x0.1 = j10 Ω and the capacitor = 1/(j100x10-3)
= -j10Ω.
j10
Ix
+
540˚
-j10
20 
vx

Using the current dividing rule:
 j10
540   j 2.540  2.5  50
 j10  20  j10
V x  20 I x  50  50
Ix 
v x (t) = 50cos(100t–50&deg;) V
Chapter 9, Solution 51.
0.1 F 

1
1

 - j5
jC j (2)(0.1)
0.5 H 

jL  j (2)(0.5)  j
The current I through the 2- resistor is
Is
1
,
I
Is 
1  j5  j  2
3  j4
I s  (5)(3  j4)  25 - 53.13
where I 
Therefore,
i s ( t )  25cos(2t – 53.13) A
10
0  5
2
Chapter 9, Solution 52.
We begin by simplifying the circuit. First we replace the parallel inductor and resistor
with their series equivalent.
5 || j5 
j25
j5

 2.5  j2.5
5  j5 1  j
Next let Z1  10 , and Z 2  - j 5  2.5  j 2.5  2.5  j 2.5 .
I2
Z1
IS
By current division I 2 
Z2
Z1
10
4
Is 
Is 
I .
12.5  j2.5
5 j s
Z1  Z 2
Since Vo  I 2 (2.5  j2.5) we can now find I s .
 4 
10 (1  j)
 I s (2.5)(1  j) 
830  
I
5 j s
 5  j
Is 
(830)(5  j)
 2.884–26.31 A.
10 (1  j)
Chapter 9, Solution 53.
Convert the delta to wye subnetwork as shown below.
Z1
Io
Z2
2
Z3
+
10 
8
60  30 V
o
-
Z
 j2x 4
8  90
j6x 4

 1  j1,
Z2 
 3  j3,
4  j4 5.656945
4  j4
12
Z3 
 1.5  j1.5
4  j4
( Z 3  8) //( Z 2  10)  (9.5  j1.5) //(13  j3)  5.6910.21  5.691  j0.02086
Z1 
Z  2  Z1  5.691  j0.02086  6.691  j0.9791
Io 
60  30 o
60  30 o

 8.873  21.67 o A
Z
6.7623  8.33 o
Chapter 9, Solution 54.
Since the left portion of the circuit is twice as large as the right portion, the
equivalent circuit is shown below.
+ 
–
2Z
V2
+
Vs
+
V1
Z
–
V1  I o (1  j)  2 (1  j)
V2  2V1  4 (1  j)
V 2 + V s + V 1 = 0 or
Vs  V1  V2  6 (1  j) = (6180)(1.4142–45)
Vs  8.485135 V
Chapter 9, Solution 55.
12 
I
I1
I2
-j20 V
+

-j4 
Z
+
Vo
j8 

Vo
4
  -j0.5
j 8 j8
I (Z  j8) (-j0.5)(Z  j8) Z
I2  1

 j
- j4
- j4
8
Z
Z
I  I 1  I 2  -j0.5   j   j0.5
8
8
- j20  12 I  I 1 (Z  j8)
Z j - j
- j20  12     (Z  j8)
 8 2 2
3
1
- 4 - j26  Z   j 
2
2
- 4 - j26 26.31261.25
 16.64279.68

Z
1 1.5811 - 18.43
3
j
2
2
I1 
Z = (2.798 – j16.403) 
Chapter 9, Solution 56.
1
1

  j 53.05
jC j 377 x50 x106
60mH

 j L  j 377 x60 x103  j 22.62
Z in  12  j 53.05  j 22.62 // 40  21.692  j 35.91 
50  F


Chapter 9, Solution 57.
2H


jL  j 2
1
j
j C
j2(2  j)
Z  1  j2 //( 2  j)  1 
 2.6  j1.2
j2  2  j
1F


Y 1
Z
 0.3171  j 0.1463 S
Chapter 9, Solution 58.
Using Fig. 9.65, design a problem to help other students to better understand impedance
combinations.
Although there are many ways to work this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
At  = 50 rad/s, determine Z in for each of the circuits in Fig. 9.65.
Figure 9.65
Solution
(a)

10 mF 
1
1

 - j2
jC j (50)(10  10 -3 )
10 mH 

jL  j (50)(10  10 -3 )  j0.5
Z in  j0.5  1 || (1  j2)
1  j2
Z in  j0.5 
2  j2
Z in  j0.5  0.25 (3  j)
Z in  0.75 + j0.25 
(b)
0.4 H 

jL  j (50)(0.4)  j20
0.2 H 

jL  j (50)(0.2)  j10
1
1

 - j20
jC j (50)(1  10 -3 )
1 mF 

For the parallel elements,
1
1
1
1



Z p 20 j10 - j20
Z p  10  j10
Then,
Z in  10 + j20 + Z p = 20 + j30 
1
Chapter 9, Solution 59.
0.25 F
0.5 H




1
1

  j 0.4
jC j10 x0.25
j L  j10 x0.5  j 5
Z in  j5 (5  j0.4) 
(590)(5.016  4.57)
 3.69142.82
6.79442.61
= (2.707+j2.509) Ω.
Chapter 9, Solution 60.
Z  (25  j15)  (20  j 50) //(30  j10)  25  j15  26.097  j 5.122
Z = (51.1+j9.878) Ω
Chapter 9, Solution 61.
All of the impedances are in parallel.
1
1
1
1
1


 
Z eq 1  j 1  j2 j5 1  j3
1
 (0.5  j0.5)  (0.2  j0.4)  (- j0.2)  (0.1  j0.3)  0.8  j0.4
Z eq
Z eq 
1
 (1 + j0.5) 
0.8  j0.4
Chapter 9, Solution 62.
2 mH 

jL  j (10  10 3 )(2  10 -3 )  j20
1
1
1 F 


 - j100
3
jC j (10  10 )(1  10 -6 )
50 
+
10 A
+
V
j20 

+

V in

-j100 
V  (10)(50)  50
Vin  (10)(50  j20  j100)  (2)(50)
Vin  50  j80  100  150  j80
Z in 
Vin
 150 – j80 
10
2V
Chapter 9, Solution 63.
First, replace the wye composed of the 20-ohm, 10-ohm, and j15-ohm impedances with
the corresponding delta.
200  j150  j300
 20  j45
10
200  j450
200  j450
z2 
 30  j13.333, z3 
 10  j22.5
j15
20
z1 
8
–j12 
–j16 
z2
10 
z1
ZT
z3
–j16 
10 
Now all we need to do is to combine impedances.
z 2 (10  j16) 
(30  j13.333)(10  j16)
 8.721  j8.938
40  j29.33
z3 (10  j16)  21.70  j3.821
ZT  8  j12  z1 (8.721  j8.938  21.7  j3.821)  34.69  j6.93
Chapter 9, Solution 64.
 j10(6  j8)
 19  j5
6  j2
3090
I
 0.3866  j1.4767  1.527104.7 A
ZT
ZT  4 
Z T = (19–j5) Ω
I = 1.527104.7&deg; A
Chapter 9, Solution 65.
Z T  2  (4  j6) || (3  j4)
(4  j6)(3  j4)
ZT  2 
7  j2
Z T  6.83 + j1.094  = 6.9179.1 
I
120 10
V

 17.350.9 A
Z T 6.917 9.1
Chapter 9, Solution 66.
(20  j5)(40  j10) 170

(12  j)
60  j5
145
Z T  14.069 – j1.172  = 14.118-4.76
Z T  (20  j5) || (40  j10) 
I
V
6090
 4.2594.76

Z T 14.118 - 4.76
I
I1
I2
20 
j10 
+
V ab
40  j10
8  j2
I
I
60  j5
12  j
20  j5
4 j
I2 
I
I
60  j5
12  j
I1 
Vab  -20 I 1  j10 I 2
- (160  j40)
10  j40
Vab 
I
I
12  j
12  j
- 150
(-12  j)(150)
Vab 
I
I
12  j
145
Vab  (12.457 175.24)(4.2597.76)
Vab  52.94273 V

Chapter 9, Solution 67.
(a)
20 mH 

jL  j (10 3 )(20  10 -3 )  j20
1
1
12.5 F 


 - j80
3
jC j (10 )(12.5  10 -6 )
Z in  60  j20 || (60  j80)
( j20)(60  j80)
Z in  60 
60  j60
Z in  63.33  j23.33  67.494 20.22
Yin 
(b)
1
 14.8-20.22 mS
Z in
10 mH 

20 F 

jL  j (10 3 )(10  10 -3 )  j10
1
1

 - j50
3
jC j (10 )(20  10 -6 )
30 || 60  20
Z in  - j50  20 || (40  j10)
Z in  - j50 
(20)(40  j10)
= –j50 + 20(41.231 14.036&deg;)/(60.828 9.462&deg;)
60  j10
= –j50 + (13.5566 4.574&deg; = –j50 + 13.51342 + j1.08109
= 13.51342 – j48.9189 = 50.751 –74.56&deg;
Z in  13.5  j48.92  50.75 - 74.56
Yin 
1
 19.70474.56 mS = 5.246 + j18.993 mS
Z in
Chapter 9, Solution 68.
Yeq 
1
1
1


5  j2 3  j - j4
Yeq  (0.1724  j0.069)  (0.3  j0.1)  ( j0.25)
Yeq  (472.4 + j219) mS
Chapter 9, Solution 69.
1
1
1
1
 
 (1  j2)
Yo 4 - j2 4
Yo 
4
(4)(1  j2)

 0.8  j1.6
1  j2
5
Yo  j  0.8  j0.6
1
1 1
1
 

 (1)  ( j0.333)  (0.8  j0.6)

1 - j3 0.8  j0.6
Yo
1
Yo 
 1.8  j0.933  2.02827.41
Yo   0.4932 - 27.41  0.4378  j0.2271
Yo   j5  0.4378  j4.773
0.4378  j4.773
1
1
1
 
 0.5 
22.97
Yeq 2 0.4378  j4.773
1
 0.5191  j0.2078
Yeq
Yeq 
0.5191  j0.2078
 (1.661 + j0.6647) S
0.3126
Chapter 9, Solution 70.
Make a delta-to-wye transformation as shown in the figure below.
a
Z an
Z bn
Z eq
n
Z cn
b
c
8
2
-j5 
(- j10)(10  j15)
(10)(15  j10)

 7  j9
5  j10  10  j15
15  j5
(5)(10  j15)

 4.5  j3.5
15  j5
(5)(- j10)

 -1  j3
15  j5
Z an 
Z bn
Z cn
Z eq  Z an  (Z bn  2) || (Z cn  8  j5)
Z eq  7  j9  (6.5  j3.5) || (7  j8)
(6.5  j3.5)(7  j8)
13.5  j4.5
 7  j9  5.511  j0.2
Z eq  7  j9 
Z eq
Z eq  12.51  j9.2  15.53-36.33 
Chapter 9, Solution 71.
We apply a wye-to-delta transformation.
j4 
Z ab
b
a
Z bc
Z ac
Z eq
-j2 
1
c
2  j2  j4 2  j2

 1 j
j2
j2
2  j2
Z ac 
 1 j
2
2  j2
Z bc 
 -2  j2
-j
Z ab 
( j4)(1  j)
 1.6  j0.8
1  j3
(1)(1  j)
 0.6  j0.2
1 || Z ac  1 || (1  j) 
2 j
j4 || Z ab  1 || Z ac  2.2  j0.6
j4 || Z ab  j4 || (1  j) 
1
1
1
1



Z eq - j2 - 2  j2 2.2  j0.6
 j0.5  0.25  j0.25  0.4231  j0.1154
 0.173  j0.3654  0.404364.66
Z eq  2.473-64.66  = (1.058 – j2.235) 
Chapter 9, Solution 72.
Transform the delta connections to wye connections as shown below.
a
j2 
j2 
-j18 
-j9 
j2 
R1
R2
R3
b
- j9 || - j18  - j6 ,
R1 
(20)(20)
 8 ,
20  20  10
(20)(10)
R3 
 4
50
Z ab  j2  ( j2  8) || (j2  j6  4)  4
Z ab  4  j2  (8  j2) || (4  j4)
(8  j2)(4  j4)
Z ab  4  j2 
12 - j2
Z ab  4  j2  3.567  j1.4054
Z ab  (7.567 + j0.5946) 
R2 
(20)(10)
 4,
50
Chapter 9, Solution 73.
Transform the delta connection to a wye connection as in Fig. (a) and then
transform the wye connection to a delta connection as in Fig. (b).
a
j2 
j2 
-j18 
-j9 
j2 
R1
R2
R3
b
( j8)(- j6)
48

 - j4.8
j8  j8  j6 j10
Z 2  Z1  -j4.8
( j8)( j8) - 64
Z3 

 j6.4
j10
j10
Z1 
(2  Z1 )(4  Z 2 )  (4  Z 2 )(Z 3 )  (2  Z1 )(Z 3 ) 
(2  j4.8)(4  j4.8)  (4  j4.8)( j6.4)  (2  j4.8)( j6.4)  46.4  j9.6
46.4  j9.6
 1.5  j7.25
j6.4
46.4  j9.6
Zb 
 3.574  j6.688
4  j4.8
46.4  j9.6
Zc 
 1.727  j8.945
2  j4.8
Za 
(690)(7.58361.88)
 07407  j3.3716
3.574  j12.688
(-j4)(1.5  j7.25)
- j4 || Z a 
 0.186  j2.602
1.5  j11.25
j6 || Z b 
j12 || Z c 
(1290)(9.1179.07)
 0.5634  j5.1693
1.727  j20.945
Z eq  ( j6 || Z b ) || (- j4 || Z a  j12 || Z c )
Z eq  (0.7407  j3.3716) || (0.7494  j2.5673)
Z eq  1.50875.42  = (0.3796 + j1.46) 
Chapter 9, Solution 74.
One such RL circuit is shown below.
20 
V
20 
+
+
j20 
Vi =
j20 
Vo

10
Z
We now want to show that this circuit will produce a 90 phase shift.
Z  j20 || (20  j20) 
V
( j20)(20  j20) - 20  j20

 4 (1  j3)
20  j40
1  j2
Z
4  j12
1  j3 1
Vi 
(10) 
 (1  j)
Z  20
6  j3 3
24  j12
Vo 
 j  1
 j
j20
 (1  j)    0.333390
V 
 3
20  j20
1  j  3
This shows that the output leads the input by 90.
Chapter 9, Solution 75.
Since cos(t )  sin(t  90) , we need a phase shift circuit that will cause the
output to lead the input by 90. This is achieved by the RL circuit shown
below, as explained in the previous problem.
10 
10 
+
Vi

+
j10 
j10 
Vo

This can also be obtained by an RC circuit.
Chapter 9, Solution 76.
(a) v2  8sin 5t  8cos(5t  90o )
v 1 leads v 2 by 70o.
(b) v2  6sin 2t  6 cos(2t  90o )
v 1 leads v 2 by 180o.
(c ) v1  4 cos10t  4 cos(10t  180o )
v2  15sin10t  15cos(10t  90o )
v 1 leads v 2 by 270o.
Chapter 9, Solution 77.
(a)
- jX c
V
R  jX c i
1
1
where X c 

 3.979
C (2)(2  10 6 )(20  10 -9 )
Vo 
Vo
- j3.979


Vi 5 - j3.979
Vo

Vi
3.979
25  15.83
3.979
5  3.979
2
2
(-90  tan -1 (3.979 5))
(-90  38.51)
Vo
 0.6227  - 51.49
Vi
Therefore, the phase shift is 51.49 lagging
(b)
  -45  -90  tan -1 (X c R )
 R  X c 
45  tan -1 (X c R ) 
  2f 
f
1
C
1
RC
1
1

 1.5915 MHz
2RC (2)(5)(20  10 -9 )
Chapter 9, Solution 78.
8+j6
R
Z
-jX
Z  R //8  j (6  X ) 
R[8  j (6  X )]
5
R  8  j (6  X )
i.e 8R + j6R – jXR = 5R + 40 + j30 –j5X
Equating real and imaginary parts:
8R = 5R + 40 which leads to R=13.333Ω
X= 6 Ω.
Chapter 9, Solution 79.
(a)
Consider the circuit as shown.
20 
V2
40 
V1
30 
+
Vi
+
j10 
j30 
j60 

Vo

Z2
Z1
( j30)(30  j60)
 3  j21
30  j90
( j10)(43  j21)
Z 2  j10 || (40  Z1 ) 
 1.535  j8.896  9.02880.21
43  j31
Z1  j30 || (30  j60) 
Let Vi  10 .
Z2
(9.02880.21)(10)
Vi 
21.535  j8.896
Z 2  20
V2  0.387557.77
V2 
Z1
3  j21
(21.21381.87)(0.387557.77)
V2 
V2 
43  j21
47.8526.03
Z1  40
V1  0.1718113.61
V1 
j60
j2
2
V1 
V1  (2  j)V1
30  j60
1  j2
5
Vo  (0.8944 26.56)(0.1718113.6)
Vo  0.1536 140.2
Vo 
Therefore, the phase shift is 140.2
(b)
(c)
If Vi  120 V , then
Vo  (120)(0.1536 140.2)  18.43140.2 V
and the magnitude is 18.43 V.
Chapter 9, Solution 80.
200 mH 

Vo 
jL  j (2)(60)(200  10 -3 )  j75.4 
j75.4
j75.4
Vi 
(1200)
R  50  j75.4
R  50  j75.4
(a)
When R  100  ,
j75.4
(75.490)(1200)
Vo 
(120 0) 
150  j75.4
167.8826.69
Vo  53.8963.31 V
(b)
When R  0  ,
j75.4
(75.490)(120 0)
Vo 
(1200) 
50  j75.4
90.47 56.45
Vo  10033.55 V
(c)
To produce a phase shift of 45, the phase of Vo = 90 + 0   = 45.
Hence,  = phase of (R + 50 + j75.4) = 45.
For  to be 45,
R + 50 = 75.4
Therefore,
R = 25.4 
Chapter 9, Solution 81.
Let
Z1  R 1 ,
Z2  R 2 
1
,
jC 2
Zx 
Z3
Z
Z1 2
Rx 
R3 
1
1 
R 2 


jC x R 1 
jC 2 
Rx 
R3
1200
(600)  1.8 k
R2 
400
R1
Z 3  R 3 , and Z x  R x 
R1
 400 
1 R3   1 
(0.3  10 -6 )  0.1 F
   
 C x 
C2  

1200 
R3
Cx  R1   C2 
1
.
jC x
Chapter 9, Solution 82.
Cx 
R1
 100 
(40  10 -6 )  2 F
Cs  
 2000 
R2
Chapter 9, Solution 83.
Lx 
R2
 500 
(250  10 -3 )  104.17 mH
Ls  
1200 
R1
Chapter 9, Solution 84.
Let
1
Z2  R 2 ,
,
jC s
R1
jC s
R1
Z1 

1
jR 1C s  1
R1 
jC s
Z1  R 1 ||
Since Z x 
Z 3  R 3 , and Z x  R x  jL x .
Z3
Z ,
Z1 2
R x  jL x  R 2 R 3
jR 1C s  1 R 2 R 3

(1  jR 1C s )
R1
R1
Equating the real and imaginary components,
R 2R 3
Rx 
R1
L x 
R 2R 3
(R 1C s ) implies that
R1
L x  R 2 R 3Cs
Given that R 1  40 k , R 2  1.6 k , R 3  4 k , and C s  0.45 F
R 2 R 3 (1.6)(4)

k  0.16 k  160 
R1
40
L x  R 2 R 3 C s  (1.6)(4)(0.45)  2.88 H
Rx 
Chapter 9, Solution 85.
Let
1
,
jC 2
R4
- jR 4
Z4 

jR 4 C 4  1 R 4 C 4  j
Z1  R 1 ,
Since Z 4 
Z2  R 2 
Z3
Z
Z1 2
Z 3  R 3 , and Z 4  R 4 ||

 Z1 Z 4  Z 2 Z 3 ,

- jR 4 R 1
j 

 R 3 R 2 
R 4 C 4  j
C 2 

jR 3
- jR 4 R 1 (R 4 C 4  j)
 R 3R 2 
2
2 2
 R 4C4  1
C 2
Equating the real and imaginary components,
R 1R 4
 R 2R 3
2
 R 24 C 24  1
(1)
R3
R 1 R 24 C 4

2 R 24 C 24  1 C 2
(2)
Dividing (1) by (2),
1
 R 2 C 2
R 4 C 4
1
2 
R 2C2R 4C4
1
  2f 
R 2C2 R 4C4
f
1
2 R 2 R 4 C 2 C 4
1
.
jC 4
Chapter 9, Solution 86.
1
1
1


240 j95 - j84
Y  4.1667  10 -3  j0.01053  j0.0119
Y
Z
1
1000
1000


Y 4.1667  j1.37 4.386118.2
Z = 228-18.2 
Chapter 9, Solution 87.
1
-j
 50 
jC
(2)(2  10 3 )(2  10 -6 )
Z1  50  j39.79
Z1  50 
Z 2  80  jL  80  j (2)(2  10 3 )(10  10 -3 )
Z 2  80  j125.66
Z 3  100
1
1
1
1



Z Z1 Z 2 Z 3
1
1
1
1



Z 100 50  j39.79 80  j125.66
1
 10 -3 (10  12.24  j9.745  3.605  j5.663)
Z
 (25.85  j4.082)  10 -3
 26.17  10 -3 8.97
Z = 38.21–8.97 
Chapter 9, Solution 88.
(a)
Z  - j20  j30  120  j20
Z = (120 – j10) 
(b)
If the frequency were halved,
1
1

would cause the capacitive
C 2f C
impedance to double, while L  2f L would cause the inductive
impedance to halve. Thus,
Z  - j40  j15  120  j40
Z = (120 – j65) 
Chapter 9, Solution 89.
An industrial load is modeled as a series combination of an inductor and a
resistance as shown in Fig. 9.89. Calculate the value of a capacitor C across the
series combination so that the net impedance is resistive at a frequency of 2 kHz.
10 
C
5 mH
Figure 9.89
For Prob. 9.89.
Solution
Step 1.
There are different ways to solve this problem but perhaps the easiest way is to
convert the series R L elements into their parallel equivalents. Then all you need
to do is to make the inductance and capacitance cancel each other out to result in a
purely resistive circuit.
X L = 2x103x5x10–3 = 10 which leads to Y = 1/(10+j10) = 0.05–j0.05 or a 20Ω
resistor in parallel with a j20Ω inductor. X c = 1/(2x103C) and the parallel
combination of the capacitor and inductor is equal to,
[(–jX C )(j20)/(–jX C +j20)].
Step 2.
Now we just need to set X C = 20 = 1/(2x103C) which will create an open circuit.
C = 1/(20x2x103) = 25 &micro;F.
Chapter 9, Solution 90.
Let
Vs  1450 ,
X  L  (2)(60) L  377 L
Vs
1450
I

80  R  jX 80  R  jX
V1  80 I 
50 
(80)(145)
80  R  jX
(80)(145)
80  R  jX
(1)
Vo  (R  jX) I 
110 
(R  jX)(1450)
80  R  jX
(R  jX)(145)
80  R  jX
(2)
From (1) and (2),
50
80

110
R  jX
11 
R  jX  (80)  
5
R 2  X 2  30976
(3)
From (1),
(80)(145)
 232
50
6400  160R  R 2  X 2  53824
160R  R 2  X 2  47424
80  R  jX 
(4)
Subtracting (3) from (4),
160R  16448 
 R  102.8 
From (3),
X 2  30976  10568  20408
X  142.86  377L 
 L  378.9 mH
Chapter 9, Solution 91.
1
 R || jL
jC
-j
jLR
Z in 

C R  jL
Z in 

- j  2 L2 R  jLR 2

C
R 2   2 L2
To have a resistive impedance, Im(Z in )  0 .
Hence,
-1
LR 2

0
C R 2  2 L2
1
LR 2

C R 2  2 L2
R 2   2 L2
C
2 LR 2
where   2 f  2  10 7
9  10 4  (4 2  1014 )(400  10 12 )
(4 2  1014 )(20  10  6 )(9  10 4 )
9  16 2
C
nF
72 2
C
C = 235 pF
Chapter 9, Solution 92.
(a) Z o 
(b)  
Z

Y
10075 o
 471.413.5 o 
45048 o x10  6
ZY  10075 o x 450 48 o x10 6  212.1 61.5 o mS
Chapter 9, Solution 93.
Z  Zs  2 Z  ZL
Z  (1  0.8  23.2)  j(0.5  0.6  18.9)
Z  25  j20
IL 
VS
1150

Z 32.02 38.66
I L  3.592–38.66 A
Chapter 10, Solution 1.
We first determine the input impedance.
1H


j L  j1x10  j10
1F


1
j C

1
  j 0.1
j10 x1
1
 1
1
1
Zin  1 

   1.0101 j 0.1  1.015  5.653o
 j10  j 0.1 1
I
2  0o
 1.9704  5.653o
o
1.015  5.653
i(t) = 1.9704cos(10t+5.65˚) A
Chapter 10, Solution 2.
Using Fig. 10.51, design a problem to help other students better understand nodal
analysis.
Although there are many ways to work this problem, this is an example based on the
same kind of problem asked in the third edition.
Problem
Solve for V o in Fig. 10.51, using nodal analysis.
2
+
o
40 V
j4 
+
_
–j5 
Vo
–
Figure 10.51 For Prob. 10.2.
Solution
Consider the circuit shown below.
2
Vo
–j5
+
40 V- _
j4
o
At the main node,
4  Vo
V
V
 o  o
2
 j5 j 4

 40  Vo (10  j )
V o = 40/(10–j) = (40/10.05)5.71˚ = 3.985.71˚ V
Chapter 10, Solution 3.
 4
2 cos(4 t ) 
 20
16 sin(4 t ) 
 16 - 90  -j16
2H 

jL  j8
1
1
1 12 F 


 - j3
jC j (4)(1 12)
The circuit is shown below.
4
-j16 V
-j3 
+

Vo
1
j8 
20 A
Applying nodal analysis,
- j16  Vo
Vo
Vo
2

4  j3
1 6  j8

1
1 
- j16
V
 2  1 

4  j3
 4  j3 6  j8  o
Vo 
3.92  j2.56 4.682  - 33.15

 3.835 - 35.02
1.22  j0.04
1.2207 1.88
Therefore,
v o ( t )  3.835cos(4t – 35.02) V
6
Chapter 10, Solution 4.
Step 1.
Convert the circuit into the frequency domain and solve for the node
voltage, V 1 , using analysis. The find the current I C = V 1 /[1+(1/(j4x0.25)] which
then produces V o = 1xI C . Finally, convert the capacitor voltage back into the time
domain.
Ix
j1
–j1 Ω
V1
160&ordm; V
0.5I x
Vo
Note that we represented 16sin(4t–10&ordm;) volts by 160&ordm; V. That will make our
calculations easier and all we have to do is to offset our answer by a –10&ordm;.
Our node equation is [(V 1 –16)/j] – (0.5I x ) + [(V 1 –0)/(1–j)] = 0. We have two
unknowns, therefore we need a constraint equation. I x = [(16–V 1 )/j] = j(V 1 –16).
Once we have V 1 , we can find I o = V 1 /(1–j) and V o = 1xI o .
Step 2.
Now all we need to do is to solve our equations.
[(V 1 –16)/j] – [0.5j(V 1 –16] + [(V 1 –0)/(1–j)] = [–j–j0.5+0.5+j0.5]V 1 +j16+j8 = 0
or
[0.5–j]V 1 = –j24 or V 1 = j24/(–0.5+j) = (2490&ordm;)/(1.118116.57&ordm;)
= 21.47–26.57&ordm; V.
Finally, I x = V 1 /(1–j) = (21.47–26.57&ordm;) (0.707145&ordm;) = 15.18118.43&ordm; A and
V o = 1xI o = 15.18118.43&ordm; V or
v o (t) = 15.181sin(4t–10&ordm;+18.43&ordm;) = 15.181sin(4t–8.43&ordm;) volts.
Chapter 10, Solution 5.
0.25H
2 F
j L  j 0.25 x4 x103  j1000




1
j C

1
  j125
j 4 x10 x2 x106
3
Consider the circuit as shown below.
2000
Io
Vo
250o V +
_
-j125
j1000
+
–
At node V o ,
Vo  25 Vo  0 Vo  10I o


0
2000
j1000
 j125
Vo  25  j2Vo  j16Vo  j160I o  0
(1  j14)Vo  j160I o  25
But I o = (25–V o )/2000
(1  j14)Vo  j2  j0.08Vo  25
Vo 
25  j2
25.084.57

1.7768  81.37
1  j14.08 14.11558.94
Now to solve for i o ,
25  Vo 25  0.2666  j1.7567

 12.367  j0.8784 mA
2000
2000
 12.3984.06
Io 
i o = 12.398cos(4x103t + 4.06˚) mA.
10I o
Chapter 10, Solution 6.
Let V o be the voltage across the current source. Using nodal analysis we get:
Vo  4Vx
Vo
20
3
 0 where Vx 
Vo
20  j10
20  j10
20
Combining these we get:
Vo
4Vo
Vo

3
 0  (1  j0.5  3)Vo  60  j30
20 20  j10
20  j10
Vo 
60  j30
20(3)
or Vx 

 2  j0.5
 2  j0.5
29.11–166˚ V.
Chapter 10, Solution 7.
At the main node,
120  15 o  V
V
V
 630 o 

40  j20
 j30 50


115.91  j31.058
 5.196  j3 
40  j20

1
j
1
V

 
 40  j20 30 50 
V
 3.1885  j4.7805
 124.08  154 o V
0.04  j0.0233
Chapter 10, Solution 8.
  200,
100mH
50F




jL  j200x 0.1  j20
1
1

  j100
jC j200x 50x10  6
The frequency-domain version of the circuit is shown below.
0.1 V o
40 
V1
20 
615 o
+
Vo
-
Io
V2
-j100 
j20 
At node 1,
or
V
V1
V  V2
 1
615 o  0.1V1  1 
20  j100
40
5.7955  j1.5529  (0.025  j 0.01)V1  0.025V2
(1)
At node 2,
V1  V2
V
 0.1V1  2
40
j20
From (1) and (2),

0  3V1  (1  j2)V2
(0.025  j0.01)  0.025 V1   (5.7955  j1.5529) 
 


3
(1  j2)  V2  
0


Using MATLAB,
or
AV  B
(2)
V = inv(A)*B
leads to V1  70.63  j127.23,
V2  110.3  j161.09
V  V2
Io  1
 7.276  82.17 o
40
Thus,
i o ( t )  7.276 cos( 200t  82.17 o ) A
Chapter 10, Solution 9.
10 cos(10 3 t ) 
 10 0,   10 3
10 mH 


50 F 
jL  j10
1
1

 - j20
3
jC j (10 )(50  10 -6 )
Consider the circuit shown below.
20 
V1
-j20 
V2
j10 
Io
100 V
+

20 
+
4 Io
30 
Vo

At node 1,
10  V1 V1 V1  V2


20
20
- j20
10  (2  j) V1  jV2
(1)
At node 2,
V1  V2
V
V2
V1
 (4) 1 
, where I o 
has been substituted.
20
- j20
20 30  j10
(-4  j) V1  (0.6  j0.8) V2
0.6  j0.8
(2)
V1 
V2
-4 j
Substituting (2) into (1)
(2  j)(0.6  j0.8)
10 
V2  jV2
-4 j
170
or
V2 
0.6  j26.2
Vo 
Therefore,
30
3
170
V2 

 6.154 70.26
30  j10
3  j 0.6  j26.2
v o ( t )  6.154 cos(103 t + 70.26) V
Chapter 10, Solution 10.


50 mH
2F


jL  j2000x50 x10  3  j100,
1
1

  j250
jC j2000 x 2x10  6
  2000
Consider the frequency-domain equivalent circuit below.
V1
360o
2k 
-j250
j100
V2
0.1V 1
4k 
At node 1,
36 
V1
V
V  V2
 1  1
2000 j100  j250


36  (0.0005  j0.006)V1  j0.004V2
(1)
0  (0.1  j0.004)V1  (0.00025  j0.004)V2
(2)
At node 2,
V1  V2
V
 0.1V1  2
 j250
4000

Solving (1) and (2) gives
Vo  V2  535.6  j893.5  8951.193.43o
v o (t) = 8.951 sin(2000t +93.43o) kV
Chapter 10, Solution 11.
Consider the circuit as shown below.
–j5 
Io
2
2
V1
40o V
+
_
V2
j8 
2I o
At node 1,
V1  4
V  V2
 2I o  1
0
2
2
V1  0.5V2  2I o  2
But, I o = (4–V 2 )/(–j5) = –j0.2V 2 + j0.8
Now the first node equation becomes,
V 1 – 0.5V 2 + j0.4V 2 – j1.6 = 2 or
V 1 + (–0.5+j0.4)V 2 = 2 + j1.6
At node 2,
V2  V1 V2  4 V2  0


0
2
 j5
j8
–0.5V 1 + (0.5 + j0.075)V 2 = j0.8
Using MATLAB to solve this, we get,
&gt;&gt; Y=[1,(-0.5+0.4i);-0.5,(0.5+0.075i)]
Y=
1.0000
-0.5000
-0.5000 + 0.4000i
0.5000 + 0.0750i
&gt;&gt; I=[(2+1.6i);0.8i]
I=
2.0000 + 1.6000i
0 + 0.8000i
&gt;&gt; V=inv(Y)*I
V=
4.8597 + 0.0543i
4.9955 + 0.9050i
I o = –j0.2V 2 + j0.8 = –j0.9992 + 0.01086 + j0.8 = 0.01086 – j0.1992
= 199.586.89˚ mA.
Chapter 10, Solution 12.
Using Fig. 10.61, design a problem to help other students to better understand Nodal analysis.
Although there are many ways to work this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
By nodal analysis, find i o in the circuit in Fig. 10.61.
Figure 10.61
Solution
20 sin(1000t ) 
 20 0,   1000
10 mH 


50 F 
jL  j10
1
1

 - j20
3
jC j (10 )(50  10 -6 )
The frequency-domain equivalent circuit is shown below.
2 Io
V1
10 
V2
Io
200 A
At node 1,
20 
-j20 
j10 
20  2 I o 
V1 V1  V2

,
20
10
V2
j10
2V2 V1 V1  V2
20 


j10 20
10
400  3V1  (2  j4) V2
(1)
Io 
At node 2,
or
2V2 V1  V2
V
V

 2  2
j10
10
- j20 j10
j2 V1  (-3  j2) V2
V1  (1  j1.5) V2
(2)
Substituting (2) into (1),
400  (3  j4.5) V2  (2  j4) V2  (1  j0.5) V2
Therefore,
V2 
400
1  j0.5
Io 
V2
40

 35.74 - 116.6
j10 j (1  j0.5)
i o ( t )  35.74 sin(1000t – 116.6) A
where
Chapter 10, Solution 13.
Nodal analysis is the best approach to use on this problem. We can make our work easier
by doing a source transformation on the right hand side of the circuit.
–j2 
18 
j6 
+
4030&ordm; V
+

Vx
3

Vx  4030 Vx Vx  50


0
 j2
3
18  j6
which leads to V x = 29.3662.88˚ A.
500&ordm; V
+

Chapter 10, Solution 14.
At node 1,
0  V1 0  V1 V2  V1


 2030
- j2
10
j4
- (1  j2.5) V1  j2.5 V2  173.2  j100
(1)
V2 V2 V2  V1


 20 30
j2 - j5
j4
- j5.5 V2  j2.5 V1  173.2  j100
(2)
At node 2,
Equations (1) and (2) can be cast into matrix form as
1  j2.5 j2.5  V1   - 200 30

 j2.5
- j5.5 V2   200 30 


1  j2.5
j2.5
j2.5
- j5.5
 20  j5.5  20.74 - 15.38
1 
- 200 30 j2.5
 j3 (20030)  600120
200 30 - j5.5
2 
1  j2.5 - 20030
 (200 30)(1  j5)  1020108.7
j2.5
20030
1
 28.93135.38 V

2
V2 
 49.18124.08 V

V1 
Chapter 10, Solution 15.
We apply nodal analysis to the circuit shown below.
5A
2
j
V1
V2
I
-j20 V
+

-j2 
4
2I
At node 1,
- j20  V1
V
V  V2
 5 1  1
2
- j2
j
- 5  j10  (0.5  j0.5) V1  j V2
(1)
At node 2,
V1  V2 V2
,

j
4
V
where I  1
- j2
5
V2 
V1
0.25  j
5  2I 
(2)
Substituting (2) into (1),
j5
 0.5 (1  j) V1
0.25  j
j40
(1  j) V1  -10  j20 
1  j4
160 j40

( 2  - 45) V1  -10  j20 
17 17
V1  15.81313.5
- 5  j10 
I
V1
 (0.590)(15.81313.5)
- j2
I  7.90643.49 A
Chapter 10, Solution 16.
Consider the circuit as shown in the figure below.
j4 
V1
V2
+ Vx –
–j3 
20o A
345o A
5
At node 1,
V  0 V1  V2
2 1

0
5
j4
(0.2  j0.25)V1  j0.25V2  2
(1)
At node 2,
V2  V1 V2  0

 345  0
j4
 j3
j0.25V1  j0.08333V2  2.121  j2.121
In matrix form, (1) and (2) become
0.2  j0.25
 j0.25

j0.25   V1  
2





j0.08333 V2  2.121  j2.121
Solving this using MATLAB, we get,
&gt;&gt; Y=[(0.2-0.25i),0.25i;0.25i,0.08333i]
Y=
0.2000 - 0.2500i
0 + 0.2500i
0 + 0.2500i
0 + 0.0833i
&gt;&gt; I=[2;(2.121+2.121i)]
I=
(2)
2.0000
2.1210 + 2.1210i
&gt;&gt; V=inv(Y)*I
V=
5.2793 - 5.4190i
9.6145 - 9.1955i
V s = V 1 – V 2 = –4.335 + j3.776 = 5.749138.94˚ V.
Chapter 10, Solution 17.
Consider the circuit below.
j4 
10020 V
+

1
Io
2
V1
V2
3
-j2 
At node 1,
10020  V1 V1 V1  V2


j4
3
2
V1
100 20 
(3  j10)  j2 V2
3
(1)
At node 2,
10020  V2 V1  V2 V2


1
2
- j2
100 20  -0.5 V1  (1.5  j0.5) V2
(2)
From (1) and (2),
10020  - 0.5
0.5 (3  j)  V1 
10020  1  j10 3
- j2  V2 

 

- 0.5
1.5  j0.5
1  j10 3
- j2
 0.1667  j4.5
1 
10020 1.5  j0.5
 -55.45  j286.2
10020
- j2
2 
- 0.5
10020
 -26.95  j364.5
1  j10 3 10020
1
 64.74  - 13.08

2
V2 
 81.17  - 6.35

V  V2  1   2 - 28.5  j78.31

Io  1

2
2
0.3333  j 9
V1 
I o  9.25-162.12 A
Chapter 10, Solution 18.
Consider the circuit shown below.
8
V1
j6  V
2
4
j5 
+
2
445 A
Vx
+
2 Vx
-j 
-j2 

Vo

At node 1,
V1 V1  V2

2
8  j6
200 45  (29  j3) V1  (4  j3) V2
(1)
445 
At node 2,
V1  V2
V
V2
 2Vx  2 
,
8  j6
- j 4  j5  j2
(104  j3) V1  (12  j41) V2
12  j41
V1 
V
104  j3 2
(2)
Substituting (2) into (1),
(12  j41)
V  (4  j3) V2
104  j3 2
200 45  (14.2189.17) V2
20045
V2 
14.2189.17
200 45  (29  j3)
- j2
- j2
- 6  j8
V2 
V2 
V2
4  j5  j2
4  j3
25
10233.13
200 45
Vo 

25
14.2189.17
Vo 
Vo  5.63189 V
where Vx  V1
Chapter 10, Solution 19.
We have a supernode as shown in the circuit below.
j2 
V1
V2
4
V3
+
2
-j4 
Vo
0.2 V o

Notice that
Vo  V1 .
At the supernode,
V3  V2 V2 V1 V1  V3



4
- j4 2
j2
0  (2  j2) V1  (1  j) V2  (-1  j2) V3
At node 3,
V1  V3 V3  V2
0.2V1 

j2
4
(0.8  j2) V1  V2  (-1  j2) V3  0
(2)
Subtracting (2) from (1),
0  1.2V1  j V2
(3)
But at the supernode,
V1  12 0  V2
or
V2  V1  12
(4)
Substituting (4) into (3),
0  1.2V1  j (V1  12)
j12
V1 
 Vo
1.2  j
Vo 
1290
1.56239.81
Vo  7.68250.19 V
(1)
Chapter 10, Solution 20.
The circuit is converted to its frequency-domain equivalent circuit as shown below.
R
+
V m 0
+

jL
Vo
1
jC

Let
Z  jL ||
1

jC
L
C
jL 
1
jC
If
Vo  A , then
A
and
jL
1  2 LC
jL
jL
1  2 LC
Vm 
V
jL
R (1  2 LC)  jL m
R
1  2 LC


L Vm
L
90  tan -1


R (1  2 LC) 
R 2 (1  2 LC) 2  2 L2 
Z
Vo 
V 
RZ m
Vo 

L Vm
R 2 (1   2 LC) 2   2 L2
  90  tan -1
L
R (1   2 LC)
Chapter 10, Solution 21.
(a)
Vo

Vi
1
jC
R  jL 
1
jC
As    ,
(b)
Vo

Vi
1
LC
Vo

Vi
,
jL
R  jL 
As    ,
1
LC
1
jC

1
jRC 
1

-j L
R C
LC
 2 LC
1  2 LC  jRC
Vo
 0
Vi
Vo 1
  1
Vi 1
At   0 ,
At  
1
1  2 LC  jRC
Vo 1
  1
Vi 1
Vo
 0
Vi
At   0 ,
At  

,
Vo

Vi
1
jRC 
1
LC

j L
R C
Chapter 10, Solution 22.
Consider the circuit in the frequency domain as shown below.
R1
R2
Vs
1
jC
+

jL
Let
Z  (R 2  jL) ||
+
Vo

1
jC
1
(R  jL)
R 2  jL
jC 2
Z

1
1  jR 2  2 LC
R 2  jL 
jC
R 2  jL
Vo
1  2 LC  jR 2 C
Z


R 2  jL
Vs Z  R 1
R1 
1  2 LC  jR 2 C
Vo
R 2  jL

2
Vs R 1  R 2   LCR 1  j (L  R 1 R 2 C)
Chapter 10, Solution 23.
V  Vs
V

 jCV  0
1
R
jL 
j C
V
jRCV
 2LC  1
 jRCV  Vs
 1  2LC  jRC  jRC  j3RLC2 

 V  Vs
2


1   LC


V 
(1   2 LC )V s
1   2 LC  jRC ( 2   2 LC )
Chapter 10, Solution 24.
Design a problem to help other students to better understand mesh analysis.
Although there are many ways to work this problem, this is an example based on the
same kind of problem asked in the third edition.
Problem
Use mesh analysis to find V o in the circuit in Prob. 10.2.
Solution
Consider the circuit as shown below.
2
+
o
40 V
I1
+
_
j4 
–j5 
Vo
I2
–
For mesh 1,
4  (2  j 5) I1  j 5 I1
For mesh 2,
(1)
0  j 5I 1  ( j 4  j 5) I 2

 I1 
1
I2
5
(2)
Substituting (2) into (1),
1
4  (2  j 5) I 2  j 5I 2
5

 I2 
1
0.1  j
V o = j4I 2 = j4/(0.1+j) = j4/(1.00499 84.29&deg;) = 3.98 5.71&deg; V
Chapter 10, Solution 25.
 2
10 cos(2 t ) 
 100
6 sin(2t ) 
 6  - 90  -j6
2H 

jL  j4
1
1
0.25 F 


 - j2
jC j (2)(1 4)
The circuit is shown below.
4
j4 
Io
100 V
+

-j2 
I1
I2
+

6-90 V
For loop 1,
- 10  (4  j2) I 1  j2 I 2  0
5  (2  j) I 1  j I 2
(1)
For loop 2,
j2 I 1  ( j4  j2) I 2  (- j6)  0
I1  I 2  3
(2)
In matrix form (1) and (2) become
 2  j j   I 1   5
 1 1   I    3

 2   
  2 (1  j) ,
I o  I1  I 2 
 1  5  j3 ,
 2  1  j3
1   2
4

 1  j  1.414245

2 (1  j )
Therefore,
i o ( t )  1.4142cos(2t + 45) A
Chapter 10, Solution 26.
0.4 H
1 F




j L  j10 3 x 0.4  j 400
1
j C

1
  j1000
j10 x10 6
3
The circuit becomes that shown below.
2 k
–j1000
Io
100o
+
_
+
_
I1
j400
–j20
I2
For loop 1,
10  (12000  j 400) I1  j 400 I 2  0

 1  (200  j 40) I1  j 40 I 2 (1)
For loop 2,
 j 20  ( j 400  j1000) I 2  j 400 I1  0 
 12  40 I1  60 I 2
(2)
In matrix form, (1) and (2) become
 1   200  j 40  j 40   I1 
 12    40
60   I 2 

 
I 1 =0.0025-j0.0075, I 2 = -0.035+j0.005
I o = I 1 – I 2 = 0.0375 – j0.0125 = 39.5 –18.43&deg; mA
i o (t) = 39.5cos(103t–18.43&deg;) mA
Chapter 10, Solution 27.
For mesh 1,
- 40 30  ( j10  j20) I 1  j20 I 2  0
4 30  - j I 1  j2 I 2
(1)
50 0  (40  j20) I 2  j20 I 1  0
5  - j2 I 1  (4  j2) I 2
(2)
For mesh 2,
From (1) and (2),
 430  - j
j2  I 1 
 5    - j2 - (4  j2)  I 

 
 2 
  -2  4 j  4.472116.56
 1  -(4 30)(4  j2)  j10  21.01211.8
 2  - j5  8120  4.44 154.27
I1 
1
 4.69895.24 A

I2 
2
 992.837.71 mA

Chapter 10, Solution 28.
1
1

  j0.25
jC j1x 4
The frequency-domain version of the circuit is shown below, where
1H


V1  100 o ,
jL  j4,


1F
V2  20  30 o .
1
j4
j4
1
-j0.25
+
+
V1
I1
1
I2
V2
-
-
V1  100 o ,
V2  20  30 o
Applying mesh analysis,
10  (2  j3.75)I1  (1  j0.25)I 2
(1)
 20  30 o  (1  j0.25)I1  (2  j3.75)I 2
(2)
From (1) and (2), we obtain
10

  2  j3.75  1  j0.25  I1 

  
 
  17.32  j10    1  j0.25 2  j3.75  I 2 
I1  2.741  41.07 o ,
I 2  4.11492 o
Hence,
i 1 (t) = 2.741cos(4t–41.07˚)A, i 2 (t) = 4.114cos(4t+92˚)A.
Chapter 10, Solution 29.
Using Fig. 10.77, design a problem to help other students better understand mesh analysis.
Although there are many ways to work this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
By using mesh analysis, find I 1 and I 2 in the circuit depicted in Fig. 10.77.
Figure 10.77
Solution
For mesh 1,
(5  j5) I 1  (2  j) I 2  30 20  0
30 20  (5  j5) I 1  (2  j) I 2
(1)
For mesh 2,
(5  j3  j6) I 2  (2  j) I 1  0
0  - (2  j) I 1  (5  j3) I 2
(2)
From (1) and (2),
3020  5  j5 - (2  j)  I 1 
 0    - (2  j) 5 - j3  I 
 2 

 
  37  j6  37.489.21
 1  (30 20)(5.831 - 30.96)  175 - 10.96
 2  (30 20)(2.356 26.56)  67.0846.56
I1 
1
 4.67–20.17 A

I2 
2
 1.7937.35 A

Chapter 10, Solution 30.
300mH


j L  j100 x300 x10 3  j30
200mH


j L  j100 x200 x10 3  j20
400mH


j L  j100 x400 x10 3  j40
1
  j200
j100 x50 x10 6
The circuit becomes that shown below.
50  F
1


j C

j40
j20
20 
+
12090
10 
o
+
_
I1
j30
–j200
I2
vo
I 3`
-
+
_
800o
For mesh 1,
120  90o  (20  j30) I1  j 30 I 2  0

 j120  (20  j 30) I1  j 30 I 2
For mesh 2,
 j 30 I1  ( j 30  j 40  j 200) I 2  j 200 I 3  0

 0  3I1  13I 2  20 I 3
For mesh 3,
80  j200I 2  (10  j180)I 3  0  8  j20I 2  (1  j18)I 3
(3)
We put (1) to (3) in matrix form.
0   I1   j12
2  j3  j3
  3  13
20  I 2    0 

 0
j20 1  j18  I 3    8
This is an excellent candidate for MATLAB.
&gt;&gt; Z=[(2+3i),-3i,0;-3,-13,20;0,20i,(1-18i)]
Z=
2.0000 + 3.0000i
0 - 3.0000i
0
(1)
(2)
-3.0000
0
-13.0000
20.0000
0 +20.0000i 1.0000 -18.0000i
&gt;&gt; V=[12i;0;-8]
V=
0 +12.0000i
0
-8.0000
&gt;&gt; I=inv(Z)*V
I=
2.0557 + 3.5651i
0.4324 + 2.1946i
0.5894 + 1.9612i
V o = –j200(I 2 – I 3 ) = –j200(–0.157+j0.2334) = 46.68 + j31.4 = 56.2633.93˚
v o = 56.26cos(100t + 33.93˚) V.
Chapter 10, Solution 31.
Consider the network shown below.
80 
100120 V
+

I1
Io
-j40 
j60 
I2
-j40 
20 
I3
+

60-30 V
For loop 1,
- 100120  (80  j40) I1  j40 I 2  0
10 20  4 (2  j) I 1  j4 I 2
(1)
j40 I 1  ( j60  j80) I 2  j40 I 3  0
0  2 I1  I 2  2 I 3
(2)
60  - 30  (20  j40) I 3  j40 I 2  0
- 6 - 30  j4 I 2  2 (1  j2) I 3
(3)
For loop 2,
For loop 3,
From (2),
2 I 3  I 2  2 I1
Substituting this equation into (3),
- 6  - 30  -2 (1  j2) I 1  (1  j2) I 2
(4)
From (1) and (4),
 10120   4 (2  j)
j4  I 1 
 - 6  - 30   - 2 (1  j2) 1  j2 I 

 
 2 

2 
8  j4
- j4
 32  j20  37.7432
- 2  j4 1  j2
8  j4 10120
 -4.928  j82.11  82.2593.44
- 2  j4 - 6 - 30
Io  I2 
2
 2.17961.44 A

Chapter 10, Solution 32.
Consider the circuit below.
j4 
Io
+
2
4-30 V
Vo
I1
+

3 Vo
I2

For mesh 1,
where
(2  j4) I 1  2 (4  - 30)  3 Vo  0
Vo  2 (4  - 30  I 1 )
Hence,
(2  j4) I 1  8 - 30  6 (4  - 30  I 1 )  0
4  - 30  (1  j) I 1
I 1  2 2 15
or
Io 
3 Vo
3

(2)(4 - 30  I 1 )
- j2 - j2
I o  j3 (4 - 30  2 2 15)
I o  8.48515 A
Vo 
- j2 I o
 5.657-75 V
3
-j2 
5A
Chapter 10, Solution 33.
Consider the circuit shown below.
I4
2
j
I
-j20 V
+

I1
-j2 
I2
2I
I3
4
For mesh 1,
j20  (2  j2) I 1  j2 I 2  0
(1  j) I 1  j I 2  - j10
(1)
For the supermesh,
( j  j2) I 2  j2 I 1  4 I 3  j I 4  0
(2)
Also,
I 3  I 2  2 I  2 (I 1  I 2 )
I 3  2 I1  I 2
(3)
I4  5
(4)
For mesh 4,
Substituting (3) and (4) into (2),
(8  j2) I 1  (- 4  j) I 2  j5
(5)
Putting (1) and (5) in matrix form,
 1 j
j  I 1   - j10 
8  j2 4  j I    j5 


 2  
  -3  j5 ,
I  I1  I 2 
 1  -5  j40 ,
 2  -15  j85
 1   2 10  j45


- 3  j5

7.90643.49 A
Chapter 10, Solution 34.
The circuit is shown below.
Io
I2
5
3A
20 
8
4090 V
+

-j2 
I3
10 
I1
j15 
j4 
For mesh 1,
- j40  (18  j2) I 1  (8  j2) I 2  (10  j4) I 3  0
For the supermesh,
(13  j2) I 2  (30  j19) I 3  (18  j2) I 1  0
(1)
(2)
Also,
I2  I3  3
Adding (1) and (2) and incorporating (3),
- j40  5 (I 3  3)  (20  j15) I 3  0
3  j8
I3 
 1.46538.48
5  j3
I o  I 3  1.46538.48 A
(3)
Chapter 10, Solution 35.
4
j2 
Consider the circuit shown below.
I3
8
1
-j3 
10 
20 V
+

I1
-j4 A
I2
-j5 
For the supermesh,
- 20  8 I 1  (11  j8) I 2  (9  j3) I 3  0
(1)
Also,
I 1  I 2  j4
(2)
(13  j) I 3  8 I 1  (1  j3) I 2  0
(3)
For mesh 3,
Substituting (2) into (1),
(19  j8) I 2  (9  j3) I 3  20  j32
(4)
Substituting (2) into (3),
- (9  j3) I 2  (13  j) I 3  j32
(5)
From (4) and (5),
 19  j8 - (9  j3)  I 2   20  j32 
 - (9  j3) 13  j  I    j32 

 3  

  167  j69 ,
I2 
 2  324  j148
 2 324  j148 356.2  - 24.55



167  j69 180.69  - 22.45
I 2  1.971–2.1 A
Chapter 10, Solution 36.
Consider the circuit below.
j4 
-j3 
+
490 A
I1
2
Vo
I2
+


2
2
I3
20 A
Clearly,
I 1  4 90  j4
and
I 3  -2
For mesh 2,
(4  j3) I 2  2 I 1  2 I 3  12  0
(4  j3) I 2  j8  4  12  0
- 16  j8
I2 
 -3.52  j0.64
4  j3
Thus,
Vo  2 (I 1  I 2 )  (2)(3.52  j4.64)  7.04  j9.28
Vo  11.64852.82 V
120 V
Chapter 10, Solution 37.
I1
+
120  90 o V
-
Ix
Z
Z=80-j35 
I2
Iz
o
120  30 V
+
Iy
Z
I3
For mesh x,
ZI x  ZI z   j120
(1)
ZI y  ZI z  12030 o  103.92  j60
(2)
 ZI x  ZI y  3ZI z  0
(3)
For mesh y,
For mesh z,
Putting (1) to (3) together leads to the following matrix equation:
0
(80  j35)  I x  
 j120

 (80  j35)

  

0
(80  j35) (80  j35)  I y     103.92  j60 


 (80  j35) (80  j35) (240  j105)  I  
0

 z  

Using MATLAB, we obtain
 - 0.2641  j2.366 


I  inv(A) * B   - 2.181 - j0.954 
 - 0.815  j1.1066 


I 1  I x  0.2641  j 2.366  2.38  96.37 o A
I 2  I y  I x  1.9167  j1.4116  2.38143.63 o A
I 3   I y  2.181  j 0.954  2.38 23.63 o A


AI  B
Chapter 10, Solution 38.
Consider the circuit below.
Io
I1
j2 
1
I2
2
20 A
+

-j4 
40 A
I3
I4
1090 V
1
A
Clearly,
I1  2
(1)
(2  j4) I 2  2 I 1  j4 I 4  10 90  0
(2)
For mesh 2,
Substitute (1) into (2) to get
(1  j2) I 2  j2 I 4  2  j5
For the supermesh,
(1  j2) I 3  j2 I 1  (1  j4) I 4  j4 I 2  0
j4 I 2  (1  j2) I 3  (1  j4) I 4  j4
At node A,
I3  I4  4
Substituting (4) into (3) gives
j2 I 2  (1  j) I 4  2 (1  j3)
From (2) and (5),
1  j2 j2  I 2   2  j5
 j2 1  j I    2  j6


 4  
  3  j3 ,
 1  9  j11
-  1 - (9  j11) 1
 (-10  j)

3  j3
3

I o  3.35174.3 A
Io  -I2 
(3)
(4)
(5)
Chapter 10, Solution 39.
For mesh 1,
(28  j15)I1  8I 2  j15I 3  1264 o
(1)
 8I1  (8  j9)I 2  j16I 3  0
(2)
j15I1  j16I 2  (10  j)I 3  0
(3)
For mesh 2,
For mesh 3,
In matrix form, (1) to (3) can be cast as
8
j15  I1  1264 o 
 (28  j15)

 
(8  j9)  j16  I 2    0 
 8



j15
 j16 (10  j)  I 3   0 



or
AI  B
Using MATLAB,
I = inv(A)*B
I 1  0.128  j 0.3593  381.4109.6&deg; mA
I 2  0.1946  j 0.2841  344.3124.4&deg; mA
I 3  0.0718  j 0.1265  145.5–60.42&deg; mA
I x  I 1  I 2  0.0666  j 0.0752  100.548.5&deg; mA
381.4109.6&deg; mA, 344.3124.4&deg; mA, 145.5–60.42&deg; mA, 100.548.5&deg; mA
Chapter 10, Solution 40.
Let I o = I o1 + I o2 , where I o1 is due to the dc source and I o2 is due to the ac source. For I o1 ,
consider the circuit in Fig. (a).
Clearly,
4
2
I o1
+

8V
(a)
I o1 = 8/2 = 4 A
For I o2 , consider the circuit in Fig. (b).
4
2
I o2
100 V
+

j4 
(b)
If we transform the voltage source, we have the circuit in Fig. (c), where 4 || 2  4 3  .
I o2
2.50 A
4
2
j4 
(c)
By the current division principle,
43
Io2 
(2.50)
4 3  j4
I o 2  0.25  j 0.75  0.79 - 71.56
Thus,
I o 2  0.79 cos(4t  71.56) A
Therefore,
I o = I o1 + I o2 = [4 + 0.79cos(4t–71.56)] A
Chapter 10, Solution 41.
We apply superposition principle. We let
vo = v1 + v2
where v 1 and v 2 are due to the sources 6cos2t and 4sin4t respectively. To find v 1 ,
consider the circuit below.
-j2
+
+
_
60o
2
V1
–
1/ 4F


1
j C

1
  j2
j2 x1/ 4
(6) = 3+j3 = 4.243 45&deg;
Thus,
v 1 (t) = 4.243cos(2t+45&deg;) volts.
To get v 2 (t), consider the circuit below,
–j
+
40o
+
_
2
V2
–
1/ 4 F


1
j C

1
  j1
j 4 x1/ 4
or
v 2 (t) = 3.578sin(4t+25.56&deg;) volts.
Hence,
v o = [4.243cos(2t+45˚) + 3.578sin(4t+25.56˚)] volts.
Chapter 10, Solution 42.
Using Fig. 10.87, design a problem to help other students to better understand the
superposition theorem.
Although there are many ways to work this problem, this is an example based on the
same kind of problem asked in the third edition.
Problem
Solve for I o in the circuit of Fig. 10.87.
j10 
o
200 V
+
_
50 
Io
60 
–j40 
+
_
3045o V
Figure 10.87 For Prob. 10.42.
Solution
Let I o  I1  I 2
where I 1 and I 2 are due to 20&lt;0o and 30&lt;45o sources respectively. To get I 1 , we use the
circuit below.
I1
j10 
60 
o
200 V
+
_
50 
–j40 
Let Z 1 = -j40//60 = 18.4615 –j27.6927, Z 2 = j10//50=1.9231 + j9.615
Transforming the voltage source to a current source leads to the circuit below.
I1
Z2
Z1
–j2
Using current division,
Z2
I1 
( j 2)  0.6217  j 0.3626
Z1  Z 2
To get I 2 , we use the circuit below.
j10 
I2
50 
60 
–j40 
+
_
After transforming the voltage source, we obtain the circuit below.
I2
Z2
Z1
0.545o
Using current division,
 Z1
I2 
(0.5  45o )  0.5275  j 0.3077
Z1  Z 2
3045o V
Hence, I o = I 1 + I 2 = 0.0942+j0.0509 = 109 30&deg; mA.
Chapter 10, Solution 43.
Let I x  I 1  I 2 , where I 1 is due to the voltage source and
I 2 is due to the current source.
 2
5 cos(2t  10) 
 510
10 cos(2t  60) 
 10  - 60
4H 

1

F 
8
jL  j8
1
1

 -j4
jC j (2)(1 / 8)
For I 1 , consider the circuit in Fig. (a).
-j4 
3
I1
+

j8 
10-60 V
(a)
I1 
10 - 60 10  - 60

3  j8  j4
3  j4
For I 2 , consider the circuit in Fig. (b).
-j4 
510 A
3
j8 
(b)
I2 
- j8
- j40 10
(510) 
3  j8  j4
3  j4
1
(10 - 60  j4010)
3  j4
49.51 - 76.04
Ix 
 9.902 - 129.17
553.13
I x  I1  I 2 
Therefore,
i x  9.902 cos(2t – 129.17) A
I2
Chapter 10, Solution 44.
Let v x  v1  v 2 , where v 1 and v 2
respectively.
For v 1 ,   6 , 5 H


are due to the current source and voltage source
jL  j30
The frequency-domain circuit is shown below.
20 
16 
Is
Let Z  16 //(20  j30) 
+
V1
-
16(20  j30)
 11.8  j3.497  12.3116.5 o
36  j30
V1  I s Z  (1210 o )(12.3116.5 o )  147.726.5 o
For v 2 ,   2 , 5 H
j30




v1  147.7 cos(6 t  26.5 o ) V
jL  j10
The frequency-domain circuit is shown below.
20 
16 
j10
+
V2
+
Vs
-
-
Using voltage division,
-
V2 
16(500 o )
16
 21.41  15.52 o
Vs 
36  j10
16  20  j10


v 2  21.41sin(2t  15.52 o ) V
Thus,
v x  [147.7cos(6t+26.5&deg;)+21.41sin(2t–15.52&deg;)] V
Chapter 10, Solution 45.
Let i  i1  i2 , where i 1 and i 2 are due to 16cos(10t +30o) and 6sin4t sources respectively.
To find i 1 , consider the circuit below.
I1
20 
+
_
16 30o V
jX
X   L  10 x300 x103  3
Type equation here.
= 0.7913 21.47&deg;
i 1 (t) = 791.1cos(10t+21.47&deg;) mA.
To find i 2 (t), consider the circuit below,
I2
20 
+
_
60o V
jX
X   L  4 x300 x103  1.2
= 0.2995 176.57&deg; or
i 2 (t) = 299.5sin(4t+176.57&deg;) mA.
Thus,
i(t) = i 1 (t) + i 2 (t) = [791.1cos(10t+21.47&deg;) + 299.5sin(4t+176.57&deg;)] mA.
Chapter 10, Solution 46.
Let v o  v1  v 2  v 3 , where v1 , v 2 , and v 3 are respectively due to the 10-V dc source, the ac
current source, and the ac voltage source. For v1 consider the circuit in Fig. (a).
6
2H
+
1/12 F
+

v1
10 V

(a)
The capacitor is open to dc, while the inductor is a short circuit. Hence,
v1  10 V
For v 2 , consider the circuit in Fig. (b).
 2
2H 
 jL  j4
1
1
1
F 


 - j6
jC j (2)(1 / 12)
12
+
6
-j6 
40 A
V2

(b)
Applying nodal analysis,
V
V
V 1 j j 
4  2  2  2      V2
6 - j6 j4  6 6 4 
V2 
Hence,
24
 21.4526.56
1  j0.5
v 2  21.45 sin( 2 t  26.56) V
For v 3 , consider the circuit in Fig. (c).
3
j4 
2H 
 jL  j6
1
1
1


 - j4
F 
12
jC j (3)(1 / 12)
6
j6 
+
120 V
+

-j4 
V3

(c)
At the non-reference node,
12  V3 V3 V3


6
- j4 j6
12
V3 
 10.73 - 26.56
1  j0.5
Hence,
v 3  10.73 cos(3t  26.56) V
Therefore,
v o  [10+21.45sin(2t+26.56)+10.73cos(3t–26.56)] V
Chapter 10, Solution 47.
Let i o  i1  i 2  i 3 , where i1 , i 2 , and i 3 are respectively due to the 24-V dc source, the
ac voltage source, and the ac current source. For i1 , consider the circuit in Fig. (a).
1
24 V
1/6 F
2H
 +
i1
2
4
(a)
Since the capacitor is an open circuit to dc,
24
4A
i1 
42
For i 2 , consider the circuit in Fig. (b).
1
2H 
 jL  j2
1
1

 - j6
F 
6
jC
1
j2 
-j6 
I2
10-30 V
+

2
I1
I2
4
(b)
For mesh 1,
- 10  - 30  (3  j6) I 1  2 I 2  0
10  - 30  3 (1  2 j) I 1  2 I 2
(1)
0  -2 I 1  (6  j2) I 2
I 1  (3  j) I 2
(2)
For mesh 2,
Substituting (2) into (1)
Hence,
10  - 30  13  j15 I 2
I 2  0.504 19.1
i 2  0.504 sin( t  19.1) A
For i 3 , consider the circuit in Fig. (c).
3
2H 
 jL  j6
1
1
1
F 


 - j2
jC j (3)(1 / 6)
6
1
j6 
-j2 
I3
2
20 A
4
(c)
2 || (1  j2) 
2 (1  j2)
3  j2
Using current division,
2 (1  j2)
 (20)
2 (1  j2)
3  j2
I3 

2 (1  j2)
13  j3
4  j6 
3  j2
I 3  0.3352  - 76.43
Hence
i 3  0.3352 cos(3t  76.43) A
Therefore,
i o  [4 + 0.504 sin(t + 19.1) + 0.3352 cos(3t – 76.43)] A
Chapter 10, Solution 48.
Let i O  i O1  i O 2  i O 3 , where i O1 is due to the ac voltage source, i O 2 is due to the dc
voltage source, and i O3 is due to the ac current source. For i O1 , consider the circuit in
Fig. (a).
  2000
50 cos(2000t ) 
 500
40 mH 

jL  j (2000)(40  10 -3 )  j80
1
1

 - j25
jC j (2000)(20  10 -6 )

20 F 
I
500 V
-j25 
I O1
+

80 
j80 
100 
60 
(a)
80 || (60  100)  160 3
50
30
I

160 3  j80  j25 32  j33
Using current division,
- 80 I
-1
10180
 I
80  160 3
4645.9
 0.217 134.1
i O1  0.217 cos(2000 t  134.1) A
I O1 
I O1
Hence,
For i O 2 , consider the circuit in Fig. (b).
i O2
80 
100 
60 
+

(b)
24 V
i O2 
24
 0.1 A
80  60  100
For i O3 , consider the circuit in Fig. (c).
  4000
2 cos(4000t ) 
 20
40 mH 

20 F 

jL  j (4000)(40  10 -3 )  j160
1
1

 - j12.5
jC j (4000)(20  10 -6 )
-j12.5 
I2
I O3
80 
j160 
I3
20 A
60 
I1
(c)
`
For mesh 1,
I1  2
(1)
For mesh 2,
(80  j160  j12.5) I 2  j160 I 1  80 I 3  0
Simplifying and substituting (1) into this equation yields
(8  j14.75) I 2  8 I 3  j32
(2)
For mesh 3,
240 I 3  60 I 1  80 I 2  0
Simplifying and substituting (1) into this equation yields
I 2  3 I 3  1.5
(3)
100 
Substituting (3) into (2) yields
(16  j44.25) I 3  12  j54.125
12  j54.125
I3 
 1.17827.38
16  j44.25
Hence,
I O 3  - I 3  -1.17827.38
i O 3  -1.1782 sin( 4000t  7.38) A
Therefore,
i O  {0.1 + 0.217 cos(2000t + 134.1) – 1.1782 sin(4000t + 7.38)} A
Chapter 10, Solution 49.
8 sin( 200t  30) 
 830,   200
5 mH 


1 mF 
jL  j (200)(5  10 -3 )  j
1
1

 - j5
jC j (200)(1  10 -3 )
After transforming the current source, the circuit becomes that shown in the figure below.
5
4030 V
I
3
I
j
+

-j5 
40 30
40 30

 4.47256.56
5  3  j  j5
8  j4
i  [4.472sin(200t+56.56)] A
Chapter 10, Solution 50.
Using Fig. 10.95, design a problem to help other students to better understand source
transformation.
Although there are many ways to work this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
Use source transformation to find v o in the circuit in Fig. 10.95.
Figure 10.95
Solution
5 cos(10 5 t )
 50,   10 5
0.4 mH 


0.2 F 
jL  j (10 5 )(0.4  10 -3 )  j40
1
1

 - j50
5
jC j (10 )(0.2  10 -6 )
After transforming the voltage source, we get the circuit in Fig. (a).
j40 
+
20 
0.250
-j50 
80 
Vo

(a)
Let
and
Z  20 || - j 50 
- j100
2  j5
Vs  (0.250) Z 
- j25
2  j5
With these, the current source is transformed to obtain the circuit in Fig.(b).
j40 
Z
+
Vs
+

80 
Vo

(b)
By voltage division,
80
80
- j25

Vs 
- j100
2  j5
Z  80  j40
 80  j40
2  j5
8 (- j25)
Vo 
 3.615 - 40.6
36  j42
v o  3.615 cos(105 t – 40.6) V
Vo 
Therefore,
Chapter 10, Solution 51.
There are many ways to create this problem, here is one possible solution. Let V 1 =
40 30&deg; V, X L = 10 Ω, X C = 20 Ω, R 1 = R 2 = 80 Ω, and V 2 = 50 V.
If we let the voltage across the capacitor be equal to V x , then
I o = [V x /(–j20)] + [(V x –50)/80] = (0.0125+j0.05)V x – 0.625 = (0.051539 75.96&deg;)V x –
0.625.
The following circuit is obtained by transforming the voltage sources.
Vx
4-60 V
j10 
-j20 
40 
V x = (4-60+1.25)/(–j0.1+j0.05+0.025) = (2–j3.4641+1.25)/(0.025–j0.05)
= (3.25–j3.4641)/( 0.025–j0.05) = (4.75 –46.826&deg;)/(0.055902 –63.435&deg;)
= 84.97 16.609&deg; V.
Therefore,
1.250 A
I o = (0.051539 75.96&deg;)(84.97 16.609&deg;) – 0.625 = 4.3793 92.569&deg; – 0.625
= –0.196291+j4.3749 – 0.625 = –0.821291+j4.3749 = 4.451 100.63&deg; A.
Chapter 10, Solution 52.
We transform the voltage source to a current source.
60 0
Is 
 6  j12
2  j4
The new circuit is shown in Fig. (a).
-j2 
Ix
2
4
6
I s = 6 – j12
590 A
j4 
-j3 
(a)
Let
6 (2  j4)
 2.4  j1.8
8  j4
Vs  I s Z s  (6  j12)(2.4  j1.8)  36  j18  18 (2  j)
Z s  6 || (2  j4) 
With these, we transform the current source on the left hand side of the circuit to a voltage
source. We obtain the circuit in Fig. (b).
Zs
-j2 
Ix
Vs
4
+

j5 A
-j3 
(b)
Let
Z o  Z s  j2  2.4  j0.2  0.2 (12  j)
Vs
18 (2  j)
Io 

 15.517  j6.207
Z o 0.2 (12  j)
With these, we transform the voltage source in Fig. (b) to a current source. We obtain the circuit
in Fig. (c).
Ix
4
Io
j5 A
Zo
-j3 
(c)
Using current division,
Zo
2.4  j0.2
Ix 
(I o  j5) 
(15.517  j1.207)
Z o  4  j3
6.4  j3.2
I x  5  j1.5625  5.23817.35 A
Chapter 10, Solution 53.
We transform the voltage source to a current source to obtain the circuit in Fig. (a).
-j3 
j4 
+
4
50 A
j2 
2
Vo
-j2 

(a)
Let
j8
 0.8  j1.6
4  j2
Vs  (50) Z s  (5)(0.8  j1.6)  4  j8
Z s  4 || j2 
With these, the current source is transformed so that the circuit becomes that shown in
Fig. (b).
Zs
-j3 
j4 
+
Vs
+

2
-j2 
Vo

(b)
Let
Z x  Z s  j3  0.8  j1.4
V
4  j8
Ix  s 
 3.0769  j4.6154
Z s 0.8  j1.4
With these, we transform the voltage source in Fig. (b) to obtain the circuit in Fig. (c).
j4 
+
Ix
Zx
2
-j2 
Vo

(c)
Let
Z y  2 || Z x 
1.6  j2.8
 0.8571  j0.5714
2.8  j1.4
Vy  I x Z y  ( 3.0769  j4.6154)  (0.8571  j0.5714)  j5.7143
With these, we transform the current source to obtain the circuit in Fig. (d).
j4 
Zy
+
Vy
+

-j2 
Vo

(d)
Using current division,
Vo 
- j2 ( j5.7143)
- j2
Vy 
 (3.529 – j5.883) V
Z y  j4  j2
0.8571  j0.5714  j4  j2
Chapter 10, Solution 54.
50 x( j 30)
 13.24  j 22.059
50  j 30
We convert the current source to voltage source and obtain the circuit below.
50 //(  j 30) 
40 
+
V s =115.91 –j31.06V
13.24 – j22.059 
j20 
+
-
I
134.95-j74.912 V
V
-
+
-
Applying KVL gives
-115.91 + j31.058 + (53.24-j2.059)I -134.95 + j74.912 = 0
or I 
 250.86  j105.97
 4.7817  j1.8055
53.24  j 2.059
But  Vs  (40  j20)I  V  0


V  Vs  (40  j20)I
V  115.91  j 31.05  (40  j 20)( 4.7817  j1.8055)  124.06  154 o V
which agrees with the result in Prob. 10.7.
Chapter 10, Solution 55.
(a)
To find Z th , consider the circuit in Fig. (a).
j20 
10 
Z th
-j10 
(a)
( j20)(- j10)
j20  j10
 10  j20  22.36-63.43 
Z N  Z th  10  j20 || (- j10)  10 
To find Vth , consider the circuit in Fig. (b).
j20 
10 
+
5030 V
+

-j10 
V th

(b)
Vth 
IN 
(b)
- j10
(50 30)  -5030 V
j20  j10
Vth
- 50 30
 2.236273.4 A

Z th 22.36  - 63.43
To find Z th , consider the circuit in Fig. (c).
-j5 
8
j10 
(c)
Z th
Z N  Z th  j10 || (8  j5) 
( j10)(8  j5)
 1026 
j10  8  j5
To obtain Vth , consider the circuit in Fig. (d).
-j5 
Io
40 A
8
j10 
+
V th

(d)
By current division,
8
32
Io 
(4 0) 
8  j10  j5
8  j5
Vth  j10 I o 
IN 
j320
 33.9258 V
8  j5
Vth 33.92 58

 3.39232 A
10 26
Z th
Chapter 10, Solution 56.
(a)
To find Z th , consider the circuit in Fig. (a).
j4 
6
-j2 
Z th
(a)
( j4)(- j2)
 6  j4
j4  j2
= 7.211-33.69 
Z N  Z th  6  j4 || (- j2)  6 
By placing short circuit at terminals a-b, we obtain,
I N  20 A
Vth  Z th I th  (7.211 - 33.69) (2 0)  14.422-33.69 V
(b)
To find Z th , consider the circuit in Fig. (b).
j10 
30 
60 
-j5 
(b)
30 || 60  20
(- j5)(20  j10)
20  j5
= 5.423-77.47 
Z N  Z th  - j5 || (20  j10) 
Z th
To find Vth and I N , we transform the voltage source and combine the 30 
and 60  resistors. The result is shown in Fig. (c).
j10 
445 A
20 
a
IN
-j5 
(c)
20
2
(4 45)  (2  j)(445)
20  j10
5
= 3.57818.43 A
IN 
Vth  Z th I N  (5.423 - 77.47) (3.57818.43)
= 19.4-59 V
b
Chapter 10, Solution 57.
Using Fig. 10.100, design a problem to help other students to better understand Thevenin and
Norton equivalent circuits.
Although there are many ways to work this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
Find the Thevenin and Norton equivalent circuits for the circuit shown in
Fig. 10.100.
Figure 10.100
Solution
To find Z th , consider the circuit in Fig. (a).
5
-j10 
2
Z th
j20 
(a)
( j20)(5  j10)
5  j10
 18  j12  21.633-33.7 
Z N  Z th  2  j20 || (5  j10)  2 
To find Vth , consider the circuit in Fig. (b).
5
-j10 
2
+
60120 V
+

j20 
V th

(b)
j20
j4
(60 120) 
(60120)
5  j10  j20
1  j2
= 107.3146.56 V
Vth 
IN 
Vth 107.3146.56

 4.961-179.7 A
Z th 21.633 - 33.7
Chapter 10, Solution 58.
Consider the circuit in Fig. (a) to find Z eq .
8
Z eq
j10 
-j6 
(a)
Z eq  j10 || (8  j 6) 
( j10)(8  j 6)
 5 (2  j )
8  j4
= 11.1826.56 
Consider the circuit in Fig. (b) to find V Thev .
Io
+
8
j10 
545 A
-j6 
(b)
Io 
4  j3
8  j6
(545)
(545) 
4  j2
8  j6  j10
VThev  j10 I o 
( j10)(4  j 3)(545)
(2)(2  j )
= 55.971.56 V
V Thev
Chapter 10, Solution 59.
Calculate the output impedance of the circuit shown in Fig. 10.102.
–j2 Ω
+
0.2V o
Vo
10 

j40 Ω
Figure 10.102
For Prob. 10.59.
Solution
Since there are no independent sources, we need to inject a current, best value is
to make it 1 amp, into the terminals on the right and then to determine the voltage
at the terminals.
–j2 Ω
V1
+
0.2V o
Vo
10 

a
1A
j40 Ω
b
Clearly V o = –(–j2) = j2 and V 1 = (0.2V o + 1)j40 = (1+j0.4)j40 = –16+j40 V.
Next, V ab = 10 – j2 – 16 + j40 = –6+j38 = 38.47 98.97&deg; V or
Z eq = (–6+j38) Ω.
Chapter 10, Solution 60.
(a)
To find Z eq , consider the circuit in Fig. (a).
10 
-j4 
a
j5 
Z eq
4
b
(a)
Z eq  4 || (- j 4  10 || j 5)  4 || (- j 4  2  j 4)
Z eq  4 || 2
= 1.333 
To find VThev , consider the circuit in Fig. (b).
10 
-j4 
V1
V2
+
200 V
+

j5 
40 A
4
V Thev

(b)
At node 1,
20  V1 V1 V1  V2


10
j5
- j4
(1  j0.5) V1  j2.5 V2  20
(1)
At node 2,
V1  V2 V2

- j4
4
V1  (1  j) V2  j16
4
Substituting (2) into (1) leads to
28  j16  (1.5  j3) V2
28  j16
 8  j5.333
V2 
1.5  j3
(2)
Therefore,
VThev  V2  9.61533.69 V
(b)
To find Z eq , consider the circuit in Fig. (c).
Z eq
c
d
10 
-j4 
j5 
4
(c)

j10 

Z eq  - j 4 || (4  10 || j 5)  - j 4 ||  4 
2  j 

Z eq  - j 4 || (6  j 4) 
- j4
(6  j 4)  (2.667 – j4) 
6
To find VThev ,we will make use of the result in part (a).
V2  8  j5.333  (8 3 ) (3  j2)
V1  (1  j) V2  j16  j16  (8 3) (5  j)
VThev  V1  V2  16 3  j8  9.61456.31 V
Chapter 10, Solution 61.
Find the Thevenin equivalent at terminals a-b of the circuit in Fig. 10.104.
V 1 and V 2
4
a
Ix
-j3 
o
20 A
+
1.5I x
V oc
I sc

b
Figure 10.104
For Prob. 10.61.
Solution
Step 1.
First we solve for the open circuit voltage using the above circuit and
writing two node equations. Then we solve for the short circuit current which
only need one node equation. For being able to solve for V oc , we need to solve
these three equations,
–2 + [(V 1 –0)/(–j3)] + [(V 1 –V oc )/4] = 0 and
[(V oc –V 1 )/4] – 1.5I x = 0 where I x = [(V 1 –0)/(–j3)].
To solve for I sc , all we need to do is to solve these three equations,
–2 + [(V 2 –0)/(–j3)] + [(V 2 –0)/4] = 0, I sc = [V 2 /4] + 1.5I x , and
I x = [V 2 /–j3].
Finally, V Thev = V oc and Z eq = V oc /I sc .
Step 2.
Now all we need to do is to solve for the unknowns. For V oc ,
I x = j0.33333V 1 and (0.25+(1.5)(j0.33333))V 1 = 0.25V oc or
(0.25+j0.5)V 1 = (0.55902 63.43&deg;)V 1 = 0.25V oc or
V 1 = (0.44721 –63.43&deg;)V oc which leads to,
(0.25+j0.33333)V 1 – 0.25V oc = 2
= (0.41666
53.13&deg;)(0.44721 –63.43&deg;)V oc – 0.25V oc
= (0.186335 –10.3&deg;)V oc – 0.25V oc = (0.183333–0.25–j0.033333)V oc
= (–0.066667–j0.033333)V oc = (0.074536 –153.435&deg;)V oc = 2 or
V oc = V Thev = 26.83 153.44&deg; V = (–24+j12) V.
Now for I sc ,
I sc = [V 2 /4] + 1.5I x = (0.25+(1.5)(j0.33333))V 2 = (0.25+j0.5)V 2 .
[(V 2 –0)/(–j3)] + [(V 2 –0)/4] = 2 = (0.25+j0.3333)V 2
= (0.41667 53.13&deg;)V 2 = 2 or V 2 = 4.8 –53.13&deg;
I sc = (0.25+j0.5)V 2 = (0.55901 63.435&deg;)(4.8 –53.13&deg;)
= 2.6832 10.305&deg; A
Finally,
Z eq = V oc /I sc = 26.833 153.435&deg;/2.6832 10.305&deg;
= 10 143.13&deg; Ω or = (–8+j6) Ω.
Chapter 10, Solution 62.
First, we transform the circuit to the frequency domain.
12 cos( t ) 
 120,   1
2H 

1
F 

4
1
F 

8
jL  j2
1
 - j4
jC
1
 - j8
jC
To find Z eq , consider the circuit in Fig. (a).
3 Io
Io
4
Vx
j2 
1
Ix
2
-j4 
-j8 
+

1V
(a)
At node 1,
Vx Vx
1  Vx

 3Io 
,
j2
4 - j4
Thus,
where I o 
Vx 2 Vx 1  Vx


- j4
4
j2
Vx  0.4  j0.8
At node 2,
I x  3Io 
1 1  Vx

- j8
j2
I x  (0.75  j0.5) Vx  j
3
8
I x  -0.1  j0.425
Z eq 
1
 -0.5246  j 2.229  2.29 - 103.24 
Ix
- Vx
4
To find VThev , consider the circuit in Fig. (b).
3 Io
Io
4
j2 
V1
1
120 V
+

V2
2
-j4 
-j8 
+
V Thev

(b)
At node 1,
12  V1
V
V  V2
12  V1
 3Io  1  1
, where I o 
4
4
- j4
j2
24  (2  j) V1  j2 V2
(1)
V1  V2
V2
 3Io 
j2
- j8
72  (6  j4) V1  j3 V2
(2)
At node 2,
From (1) and (2),
 24  2  j - j2  V1 
 72    6  j4 - j3  V 
 2
  
  -5  j6 ,
Vth  V2 
Thus,
 2  - j24
2
 3.073 - 219.8

2
(2)(3.073 - 219.8)
Vth 
2  Z th
1.4754  j2.229
6.146 - 219.8
Vo 
 2.3 - 163.3
2.673 - 56.5
Vo 
Therefore,
v o  2.3cos(t–163.3) V
Chapter 10, Solution 63.
Transform the circuit to the frequency domain.
4 cos(200t  30) 
 4 30,   200
10 H 


5 F 
jL  j (200)(10)  j2 k
1
1

 - j k
jC j (200)(5  10 -6 )
Z N is found using the circuit in Fig. (a).
-j k
j2 k
ZN
2 k
(a)
Z N  - j  2 || j2  - j  1  j  1 k
We find I N using the circuit in Fig. (b).
-j k
430 A
j2 k
2 k
(b)
j2 || 2  1  j
By the current division principle,
1 j
IN 
(4 30)  5.657 75
1 j  j
Therefore,
i N (t) = 5.657 cos(200t + 75) A
Z N  1 k
IN
Chapter 10, Solution 64.
Z N is obtained from the circuit in Fig. (a).
60 
40 
ZN
-j30 
j80 
(a)
Z N  (60  40) || ( j80  j30)  100 || j50 
(100)( j50)
100  j50
Z N  20  j40  44.7263.43 
To find I N , consider the circuit in Fig. (b).
60 
I1
40 
I2
-j30 
IN
360 A
Is
j80 
(b)
I s  360
For mesh 1,
100 I 1  60 I s  0
I 1  1.860
For mesh 2,
( j80  j30) I 2  j80 I s  0
I 2  4.860
I N = I 2 – I 1 = 360 A
Chapter 10, Solution 65.
Using Fig. 10.108, design a problem to help other students to better understand Norton’s
theorem.
Although there are many ways to work this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
Compute i o in Fig. 10.108 using Norton's theorem.
Figure 10.108
Solution
5 cos(2 t ) 
 50,   2
4H 

1
F 

4
1
F 

2
jL  j (2)(4)  j8
1
1

 - j2
jC j (2)(1 / 4)
1
1

 -j
jC j (2)(1 / 2)
To find Z N , consider the circuit in Fig. (a).
2
ZN
-j2 
-j 
(a)
Z N  - j || (2  j2) 
- j (2  j2) 1
 (2  j10)
2  j3
13
To find I N , consider the circuit in Fig. (b).
50 V
2
+ 
-j2 
-j 
IN
(b)
IN 
50
 j5
-j
The Norton equivalent of the circuit is shown in Fig. (c).
Io
ZN
IN
j8 
(c)
Using current division,
ZN
(1 13)(2  j10)( j5) 50  j10

IN 
Io 
(1 13)(2  j10)  j8 2  j94
Z N  j8
I o  0.1176  j0.5294  0542 - 77.47
Therefore, i o  542 cos(2t – 77.47) mA
Chapter 10, Solution 66.
  10
0.5 H 

jL  j (10)(0.5)  j5
1
1


 - j10
10 mF 
jC j (10)(10  10 -3 )
To find Z th , consider the circuit in Fig. (a).
-j10 
Vx
+
10 
j5 
Vo
2 Vo
1A

(a)
Vx
Vx

,
j5 10  j10
19 Vx
V
- 10  j10
1
 x 
 Vx 
10  j10
j5
21  j2
1  2 Vo 
Z N  Z th 
where Vo 
10Vx
10  j10
Vx
14.142 135

 670129.56 m
1
21.0955.44
To find Vth and I N , consider the circuit in Fig. (b).
120 V
-j10 
 +
+
+
-j2 A
10 
Vo
j5 
I
V th


(b)
where
2 Vo
(10  j10  j5) I  (10)(- j2)  j5 (2 Vo )  12  0
Vo  (10)(- j2  I )
Thus,
(10  j105) I  -188  j20
188  j20
I
- 10  j105
Vth  j5 (I  2 Vo )  j5 (19I  j40)   j95 I  200
 j 95 (188  j 20)
(95  90)(189.066.07)
Vth 
 200 
 200
- 10  j105
105.4895.44
 170.28  179.37  200  170.27  j1.8723  200  29.73  j1.8723
Vth  29.79–3.6 V
IN 
Vth 29.79  3.6

 44.46–133.16 A
Z th 0.67129.56
Chapter 10, Solution 67.
Z N  Z Th  10 //(13  j 5)  12 //( 8  j 6) 
Va 
10
(6045 o )  13.78  j21.44,
23  j5
10(13  j 5) 12(8  j 6)

 11.243  j1.079
23  j 5
20  j 6
Vb 
(8  j6)
(6045 o )  12.069  j26.08
20  j6
VTh  Va  Vb  1.711  j 4.64  4.945  69.76 o V,
IN 
VTh 4.945  69.76

 437.8  75.24 o mA
11.295 5.48
Z Th
Chapter 10, Solution 68.
1H


jL  j10x1  j10
1
1
1



  j2
F
1
20
j C
j10x
20
We obtain V Th using the circuit below.
Io
4
a
+
+
6&lt;0o
-
V o /3
+
-
-j2 j10
b
j10( j2)
  j2.5
j10  j2
Vo  4I o x ( j2.5)   j10I o
1
 6  4I o  Vo  0
3
j10 //( j2) 
(1)
(2)
Combining (1) and (2) gives
Io 
6
,
4  j10 / 3
Vo
4I o
VTh  Vo   j10I o 
 j60
 11.52  50.19 o
4  j10 / 3
v Th  11.52 sin(10t  50.19 o )
To find R Th, we insert a 1-A source at terminals a-b, as shown below.
Io
4
a
+
+
-
V o /3
-j2 j10
Vo
4I o
-
1
4I o  Vo  0
3

V
Io   o
12
Vo Vo

 j2 j10
Combining the two equations leads to
1  4I o 
Vo 
1
 1.2293  j1.4766
0.333  j0.4
V
Z Th  o  1.2293  1.477
1
1&lt;0o
Chapter 10, Solution 69.
This is an inverting op amp so that
Vo - Z f
-R


 –jRC
1 jC
Vs
Zi
When Vs  Vm and   1 RC ,
1
Vo  - j 
 RC  Vm  - j Vm  Vm  - 90
RC
Therefore,
v o ( t )  Vm sin(t  90)  - V m cos(t)
Chapter 10, Solution 70.
Using Fig. 10.113, design a problem to help other students to better understand op amps in AC
circuits.
Although there are many ways to work this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
The circuit in Fig. 10.113 is an integrator with a feedback resistor. Calculate v o (t) if v s =
2 cos 4  104t V.
Figure 10.113
Solution
This may also be regarded as an inverting amplifier.
2 cos(4  10 4 t ) 
 2 0,   4  10 4
1
1
10 nF 


 - j2.5 k
4
jC j (4  10 )(10  10 -9 )
Vo - Z f

Vs
Zi
where Z i  50 k and Z f  100k || (- j2.5k ) 
Thus,
If Vs  2 0 ,
Vo - (- j2)

Vs 40  j
- j100
k .
40  j
Vo 
j4
490

 0.191.43
40  j 40.01 - 1.43
Therefore,
v o ( t )  100 cos(4x104 t + 91.43) mV
Chapter 10, Solution 71.
8 cos(2t  30 o )
 830 o
1
1
0. 5F



  j1M
jC j2x 0.5x10  6
At the inverting terminal,
Vo  830 o Vo  830 o 830 o


 j1000k
10k
2k


Vo (1  j100)  830  800  60  4000   60
Vo 
6.928  j4  2400  j4157 4800  59.9

 4829.53o
1  j100
100  89.43
v o (t) = 48cos(2t+29.53o) V
Chapter 10, Solution 72.
4 cos(10 4 t ) 
 4 0,   10 4
1
1
1 nF 


 - j100 k
4
jC j (10 )(10 -9 )
Consider the circuit as shown below.
50 k
Vo
+

40 V
+

-j100 k
At the noninverting node,
4  Vo
Vo

50
- j100
Io 

 Vo 
Vo
Io
100 k
4
1  j0.5
Vo
4

mA  35.78 - 26.56 A
100k (100)(1  j0.5)
Therefore,
i o ( t )  35.78cos(104t–26.56) A
Chapter 10, Solution 73.
As a voltage follower, V2  Vo
1
1

 -j20 k
3
jC1 j (5  10 )(10  10 -9 )
1
1
C 2  20 nF 


 -j10 k
3
jC 2 j (5  10 )(20  10 -9 )

C1  10 nF 
Consider the circuit in the frequency domain as shown below.
-j20 k
I s 10 k
20 k
+

V1
VS
+

-j10 k
Z in
At node 1,
Vs  V1 V1  Vo V1  Vo


10
- j20
20
2 Vs  (3  j)V1  (1  j)Vo
(1)
At node 2,
V1  Vo Vo  0

20
- j10
V1  (1  j2)Vo
(2)
Substituting (2) into (1) gives
2 Vs  j6Vo
or
V2
1
Vo  -j Vs
3
2
1
V1  (1  j2)Vo    j  Vs
3
3
Io
Vo
Vs  V1 (1 3)(1  j)

Vs
10k
10k
1 j

30k
Is 
Is
Vs
Vs 30k

 15 (1  j) k
Is 1  j
Z in  21.21–45 k
Z in 
Chapter 10, Solution 74.
Zi  R1 
1
,
jC1
Zf  R 2 
1
jC 2
1
 C   1  j R 2 C 2 
V
- Zf
jC 2


   1  
Av  o 
1
Vs
Zi
C 2   1  jR 1C1 

R1 
jC1
R2 
Av  –
At   0 ,
As    ,
Av  –
C1
C2
R2
R1
At  
1
,
R 1 C1
 C   1  j R 2 C 2 R 1C1 

A v  – 1  
1 j

 C2  
At  
1
,
R 2C2
C 

1 j

A v  – 1  
 C 2   1  j R 1C1 R 2 C 2 
Chapter 10, Solution 75.
  2  10 3
C1  C 2  1 nF 

1
1

 -j500 k
3
jC1 j (2  10 )(1  10 -9 )
Consider the circuit shown below.
100 k
Let V s = 10V.
-j500 k
-j500 k
V2
+

V1
40 k
VS
+

100 k
Vb
+
Vo
20 k

At node 1,
[(V 1 –10)/(–j500k)] + [(V 1 –V o )/105] + [(V 1 –V 2 )/(–j500k)] = 0
or (1+j0.4)V 1 – j0.2V 2 – V o = j2
(1)
[(V 2 –V 1 )/(–j500k)] + [(V 2 –0)/100k] + 0 = 0 or
–j0.2V 1 + (1+j0.2)V 2 = 0 or V 1 = [–(1+j0.2)/(–j0.2)]V 2
= (1–j5)V 2
(2)
At node 2,
At node b,
Vb =
R3
V
Vo  o  V 2
3
R3  R4
From (2) and (3),
V 1 = (0.3333–j1.6667)V o
Substituting (3) and (4) into (1),
(1+j0.4)(0.3333–j1.6667)V o – j0.06667V o – V o = j2
(1+j0.4)(0.3333–j1.6667) = (1.07721.8˚)(1.6997–78.69˚)
= 1.8306–56.89˚ = 1–j1.5334
(3)
(4)
(1–1+j(–1.5334–0.06667))V o = (–j1.6001)V o = 1.6001–90˚
Therefore,
V o = 290˚/(1.6001–90˚) = 1.2499180˚
Since V s = 10,
V o /V s = 0.12499180˚.
Chapter 10, Solution 76.
Let the voltage between the -jk  capacitor and the 10k  resistor be V 1.
230 o  V1 V1  Vo V1  Vo


 j4k
10k
20k


(1)
230 o  (1  j0.6)V1  j0.6Vo
= 1.7321+j1
Also,
V1  Vo
V
 o
 j2k
10k

V1  (1  j5)Vo
(2)
Solving (2) into (1) yields
230  (1  j0.6)(1  j5)Vo  j0.6Vo  (1  3  j0.6  j5  j6)Vo
= (4+j5)V o
230
Vo 
 0.3124  21.34 o V
6.40351.34
= 312.4–21.34˚ mV
I o = (V 1 –V o )/20k = V o /(–j4k) = (0.3124/4k)(–21.43+90)˚
= 78.168.57˚ &micro;A
We can easily check this answer using MATLAM. Using equations (1) and (2) we can
identify the following matrix equations:
YV = I where
&gt;&gt; Y=[1-0.6i,0.6i;1,-1-0.5i]
Y=
1.0000 - 0.6000i
0 + 0.6000i
1.0000
-1.0000 - 5.0000i
&gt;&gt; I=[1.7321+1i;0]
I=
1.7321 + 1.0000i
0
&gt;&gt; V=inv(Y)*I
V=
0.8593 + 1.3410i
0.2909 - 0.1137i = V o = 312.3–21.35˚ mV. The answer checks.
Chapter 10, Solution 77.
Consider the circuit below.
R3
2
R1
1
+

VS
At node 1,
V1
C2
R2

+
V1
C1
+
Vo

Vs  V1
 jC V1
R1
Vs  (1  jR 1C1 ) V1
(1)
At node 2,
0  V1 V1  Vo

 jC 2 (V1  Vo )
R3
R2
 R3

 jC 2 R 3 
V1  (Vo  V1 ) 
R2



1
 V1
Vo  1 
 (R 3 R 2 )  jC 2 R 3 
From (1) and (2),
Vo 


Vs
R2

1 
1  jR 1C1  R 3  jC 2 R 2 R 3 
Vo
R 2  R 3  jC 2 R 2 R 3

Vs (1  jR 1C 1 ) ( R 3  jC 2 R 2 R 3 )
(2)
Chapter 10, Solution 78.
2 sin(400t ) 
 20,   400
1
1
0.5 F 


 - j5 k
jC j (400)(0.5  10 -6 )
1
1


 - j10 k
0.25 F 
jC j (400)(0.25  10 -6 )
Consider the circuit as shown below.
20 k
10 k V
1
-j5 k
V2
+

Vo
40 k
20 V
+

-j10 k
10 k
20 k
At node 1,
V
V  V2 V1  Vo
2  V1
 1  1

10
- j10
- j5
20
4  (3  j6) V1  j4 V2  Vo
(1)
V1  V2 V2

10
 j5
V1  (1  j0.5) V2
(2)
At node 2,
But
20
1
Vo  Vo
20  40
3
From (2) and (3),
1
V1   (1  j0.5) Vo
3
Substituting (3) and (4) into (1) gives
1
4
1

4  (3  j6)   (1  j0.5) Vo  j Vo  Vo  1  j  Vo
3
3
6

24
Vo 
 3.945  9.46
6 j
Therefore,
v o ( t )  3.945sin(400t–9.46) V
V2 
(3)
(4)
Chapter 10, Solution 79.
0.5 cos(1000t ) 
 0.50,   1000
1
1


 - j10 k
0.1 F 
jC j (1000)(0.1  10 -6 )
1
1
0.2 F 


 - j5 k
jC j (1000)(0.2  10 -6 )
Consider the circuit shown below.
20 k
-j10 k
40 k
10 k
V s = 0.50

+
+

V1

+
-j5 k
Since each stage is an inverter, we apply Vo 
- Zf
V to each stage.
Zi i
Vo 
- 40
V1
- j5
(1)
V1 
- 20 || (- j10)
Vs
10
(2)
and
From (1) and (2),
 - j8  - ( 20)(-j10) 
Vo  
 0.50

 10  20  j10 
Vo  1.6 ( 2  j)  35.7826.56
Therefore,
v o ( t )  3.578cos(1000t + 26.56) V
+
Vo

Chapter 10, Solution 80.
4 cos(1000t  60) 
 4  - 60,   1000
1
1
0.1 F 


 - j10 k
jC j (1000)(0.1  10 -6 )
1
1


 - j5 k
0.2 F 
jC j (1000)(0.2  10 -6 )
The two stages are inverters so that
 20
20  - j5 

Vo  
V 
 (4 - 60) 
50 o  10 
 - j10
-j
-j 2
  ( j2)  (4 - 60)   Vo
2
2 5
(1  j 5) Vo  4 - 60
4 - 60
Vo 
 3.922  - 71.31
1 j 5
Therefore,
v o ( t )  3.922 cos(1000t – 71.31) V
1
Chapter 10, Solution 81.
We need to get the capacitance and inductance corresponding to –j2  and j4 .
1
1
 j2

 C

 0.5 F
 X c 1x 2
X
j4

 L  L  4H

The schematic is shown below.
When the circuit is simulated, we obtain the following from the output file.
FREQ
VM(5)
VP(5)
1.592E-01 1.127E+01 -1.281E+02
From this, we obtain
V o = 11.27128.1o V.
Chapter 10, Solution 82.
The schematic is shown below. We insert PRINT to print V o in the output file. For AC
Sweep, we set Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. After
simulation, we print out the output file which includes:
FREQ
1.592 E-01
which means that
VM(\$N_0001)
7.684 E+00
V o = 7.68450.19o V
VP(\$N_0001)
5.019 E+01
Chapter 10, Solution 83.
The schematic is shown below. The frequency is f   / 2 
1000
 159.15
2
When the circuit is saved and simulated, we obtain from the output file
FREQ
1.592E+02
VM(1)
6.611E+00
VP(1)
-1.592E+02
Thus,
v o = 6.611cos(1000t – 159.2o) V
Chapter 10, Solution 84.
The schematic is shown below. We set PRINT to print V o in the output file. In AC
Sweep box, we set Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. After
simulation, we obtain the output file which includes:
FREQ
VM(\$N_0003)
1.592 E-01
1.664 E+00
VP(\$N_0003)
E+02
Namely,
V o = 1.664–146.4o V
–1.646
Chapter 10, Solution 85.
Using Fig. 10.127, design a problem to help other students to better understand performing AC
analysis with PSpice.
Although there are many ways to solve this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
Use PSpice to find V o in the circuit of Fig. 10.127. Let R 1 = 2 Ω, R 2 = 1 Ω, R 3 = 1 Ω, R 4 = 2 Ω,
I s = 20˚ A, X L = 1 Ω, and X C = 1 Ω.
Solution
The schematic is shown below. We let   1 rad/s so that L=1H and C=1F.
When the circuit is saved and simulated, we obtain from the output file
FREQ
1.591E-01
VM(1)
2.228E+00
VP(1)
-1.675E+02
From this, we conclude that
V o = 2.228–167.5&deg; V.
Chapter 10, Solution 86.
The schematic is shown below. We insert three pseudocomponent PRINTs at nodes 1, 2,
and 3 to print V 1 , V 2 , and V 3 , into the output file. Assume that w = 1, we set Total Pts =
1, Start Freq = 0.1592, and End Freq = 0.1592. After saving and simulating the circuit,
we obtain the output file which includes:
FREQ
VM(\$N_0002)
1.592 E-01
6.000 E+01
FREQ
VM(\$N_0003)
1.592 E-01
2.367 E+02
FREQ
VM(\$N_0001)
1.592 E-01
1.082 E+02
VP(\$N_0002)
3.000
E+01
VP(\$N_0003)
-8.483
E+01
VP(\$N_0001)
E+02
Therefore,
V 1 = 6030o V V 2 = 236.7-84.83o V V 3 = 108.2125.4o V
1.254
Chapter 10, Solution 87.
The schematic is shown below. We insert three PRINTs at nodes 1, 2, and 3. We set
Total Pts = 1, Start Freq = 0.1592, End Freq = 0.1592 in the AC Sweep box. After
simulation, the output file includes:
FREQ
VM(\$N_0004)
1.592 E-01
1.591 E+01
FREQ
VM(\$N_0001)
1.592 E-01
5.172 E+00
FREQ
VM(\$N_0003)
1.592 E-01
2.270 E+00
VP(\$N_0004)
1.696
E+02
VP(\$N_0001)
-1.386
E+02
VP(\$N_0003)
E+02
Therefore,
V 1 = 15.91169.6o V V 2 = 5.172-138.6o V V 3 = 2.27-152.4o V
-1.524
Chapter 10, Solution 88.
The schematic is shown below. We insert IPRINT and PRINT to print I o and V o in the
output file. Since w = 4, f = w/2 = 0.6366, we set Total Pts = 1, Start Freq = 0.6366,
and End Freq = 0.6366 in the AC Sweep box. After simulation, the output file includes:
FREQ
VM(\$N_0002)
6.366 E-01
3.496 E+01
1.261
FREQ
IM(V_PRINT2)
IP
6.366 E-01
8.912 E-01
VP(\$N_0002)
E+01
(V_PRINT2)
-8.870 E+01
Therefore,
V o = 34.9612.6o V,
v o = 34.96 cos(4t + 12.6o)V,
I o = 0.8912-88.7o A
i o = 0.8912cos(4t – 88.7o )A
Chapter 10, Solution 89.
Consider the circuit below.
R1
R2
V in
2
R3
1
V in
4
R4
3

+
0  Vin Vin  V2

R1
R2
R
- Vin  V2  2 Vin
R1
(1)
At node 3,
V2  Vin Vin  V4

R3
1 jC
Vin  V2
- Vin  V4 
jCR 3
(2)
From (1) and (2),
- Vin  V4 
- R2
V
jCR 3 R 1 in
Thus,
I in 
Vin  V4
R2

V
R4
jCR 3 R 1 R 4 in
Z in 
Vin jCR 1R 3 R 4

 jL eq
R2
I in
L eq 
I in

+
At node 1,
where
C
R 1R 3 R 4C
R2
+

V in
Chapter 10, Solution 90.
Let
Z 4  R ||
1
R

jC 1  jRC
1
1  jRC
Z3  R 

jC
jC
Consider the circuit shown below.
Z3
Vi
+

+
Z4
Vo 
R1
Vo
R2
R2
Z4
Vi 
V
R1  R 2 i
Z3  Z 4
R
Vo
R2
1  jC


R
1  jRC R 1  R 2
Vi

1  jC
jC

jRC
R2

2
jRC  (1  jRC)
R1  R 2
Vo
R2
jRC


2
2 2
Vi 1   R C  j3RC R 1  R 2
For Vo and Vi to be in phase,
Vo
must be purely real. This happens when
Vi
1  2 R 2 C 2  0
1

 2f
RC
or
f
1
2RC
At this frequency,
Vo 1
R2
Av 
 
Vi 3 R 1  R 2
Chapter 10, Solution 91.
(a)
Let
V2  voltage at the noninverting terminal of the op amp
Vo  output voltage of the op amp
Z p  10 k  R o
1
jC
Z s  R  jL 
As in Section 10.9,
Zp
V2


Vo Z s  Z p
Ro
R  R o  jL 
j
C
CR o
V2

Vo C (R  R o )  j ( 2 LC  1)
For this to be purely real,
o2 LC  1  0 
 o 
fo 
1
2 LC

1
LC
1
2 (0.4  10 -3 )(2  10 -9 )
f o  180 kHz
(b)
At oscillation,
o CR o
Ro
V2


Vo o C (R  R o ) R  R o
This must be compensated for by
Vo
80
Av 
 1
5
V2
20
Ro
1

R  Ro 5

 R  4R o  40 k
Chapter 10, Solution 92.
Let
V2  voltage at the noninverting terminal of the op amp
Vo  output voltage of the op amp
Zs  R o
RL
1
1

Z p  jL ||
|| R 
1
1
L  jR ( 2 LC  1)
jC
 jC 
R
jL
As in Section 10.9,
RL
V2
L  jR (2 LC  1)


RL
Vo Z s  Z p
Ro 
L  jR (2 LC  1)
V2
RL

Vo RL  R o L  jR o R (2 LC  1)
Zp
For this to be purely real,
o2 LC  1 
 f o 
(a)
1
2 LC
At   o ,
o RL
V2
R


Vo o RL  o R o L R  R o
This must be compensated for by
Vo
Rf
1000k
Av 
 1
 1
 11
Ro
V2
100k
Hence,
R
1


 R o  10R  100 k
R  R o 11
(b)
fo 
1
2 (10  10 -6 )(2  10 -9 )
f o  1.125 MHz
Chapter 10, Solution 93.
As shown below, the impedance of the feedback is
jL
1
jC2
ZT 
1
jC1
ZT

1
1 

||  jL 
jC1 
jC 2 
-j 
-j 
1
 jL 

 LC 2
C1 
C 2 

ZT 

-j
-j
j (C1  C 2   2 LC1C 2 )
 jL 
C1
C 2
In order for Z T to be real, the imaginary term must be zero; i.e.
C1  C 2  o2 LC1 C 2  0
C  C2
1
o2  1

LC1C 2
LC T
1
fo 
2 LC T
Chapter 10, Solution 94.
If we select C1  C 2  20 nF
C1 C 2
C1
CT 

 10 nF
C1  C 2
2
Since f o 
1
2 LC T
L
,
1
1

 10.13 mH
2
2
(2f ) C T (4 )(2500  10 6 )(10  10 -9 )
Xc 
1
1

 159 
C 2 ( 2 )(50  10 3 )(20  10 -9 )
We may select R i  20 k and R f  R i , say R f  20 k .
Thus,
C1  C 2  20 nF,
L  10.13 mH
R f  R i  20 k
Chapter 10, Solution 95.
First, we find the feedback impedance.
C
ZT
L2
L1

1 

Z T  jL1 ||  jL 2 
jC 


j 

jL1  jL 2 

2 L1C (1  L 2 )
C 
ZT 

j
j (2 C (L1  L 2 )  1)
jL1  jL 2 
C
In order for Z T to be real, the imaginary term must be zero; i.e.
o2 C (L 1  L 2 )  1  0
1
 o  2 f o 
C (L1  L 2 )
1
fo 
2 C ( L1  L2 )
Chapter 10, Solution 96.
(a)
Consider the feedback portion of the circuit, as shown below.
jL
Vo
V2 
+

V1
R
V2
R
jL
V
R  jL 1
jL

 V1 
R  jL
V2
jL
Applying KCL at node 1,
Vo  V1 V1
V1


jL
R R  jL

1
1

Vo  V1  jL V1  
 R R  jL 
 j2RL  2 L2 

Vo  V1 1 
R (R  jL) 

(2)
From (1) and (2),
 R  jL  j2RL  2 L2 
V
1 
Vo  
R (R  jL)  2
 jL 
Vo R 2  jRL  j2RL  2 L2

V2
jRL
V2

Vo
1
R  2 L2
3
jRL
2
V2
1

Vo 3  j (L R  R L )
(1)
(b)
V2
must be real,
Vo
Since the ratio
o L
R

0
R
o L
o L 
R2
o L
 o  2 f o 
R
L
fo 
(c)
R
2 L
When   o
V2 1

Vo 3
This must be compensated for by A v  3 . But
Av  1
R2
3
R1
R 2  2 R1
Chapter 11, Solution 1.
v( t )  160 cos(50t )
i(t) = –33sin(50t–30˚) = 33cos(50t–30˚+180˚–90˚) = 33cos(50t+60˚)
p(t) = v(t)i(t) = 160x33cos(50t)cos(50t+60˚)
= 5280(1/2)[cos(100t+60˚)+cos(60˚)] = [1.320+2.640cos(100t+60˚)] kW.
P = [V m I m /2]cos(0–60˚) = 0.5x160x33x0.5 = 1.320 kW.
Chapter 11, Solution 2.
Using current division,
j1 Ω
I1
I2
Vo
-j4 Ω
I1 
j1  j 4
 j6
(2) 
5  j1  j 4
5  j3
I2 
5
10
(2) 
5  j1  j 4
5  j3
20o A
5Ω
.
For the inductor and capacitor, the average power is zero. For the resistor,
1
1
P  | I1 |2 R  (1.029) 2 (5)  2.647 W
2
2
Vo  5I1  2.6471  j 4.4118
1
1
S  Vo I *  (2.6471  j 4.4118) x 2  2.6471  j 4.4118
2
2
Hence the average power supplied by the current source is 2.647 W.
Chapter 11, Solution 3.
I
+
–
90  F
C
1600˚


R
1
1

  j 5.5556
6
j C j 90 x10 x 2 x103
I = 160/60 = 2.667A
The average power delivered to the load is the same as the average power absorbed by
the resistor which is
P avg = 0.5|I|260 = 213.4 W.
Chapter 11, Solution 4.
Using Fig. 11.36, design a problem to help other students better understand instantaneous
and average power.
Although there are many ways to work this problem, this is an example based on the
same kind of problem asked in the third edition.
Problem
Find the average power dissipated by the resistances in the circuit of Fig. 11.36.
Additionally, verify the conservation of power. Note, we do not talk about rms values of
voltages and currents until Section 11.4, all voltages and currents are peak values.
5Ω
2030o V
j4 Ω
+
–
8Ω
–j6 Ω
Figure 11.36 For Prob. 11.4.
Solution
We apply nodal analysis. At the main node,
I1
2030o V
+
–
5 ΩI 2
Vo
j4 Ω
8Ω
–j6 Ω
20  30o  Vo Vo
V

 o

 Vo  5.152  j10.639 = 11.82164.16˚
5
j 4 8  j6
For the 5-Ω resistor,
20  30o  Vo
I1 
 2.438  3.0661o A
5
The average power dissipated by the resistor is
1
1
P1  | I1 |2 R1  x 2.4382 x5  14.86 W
2
2
For the 8-Ω resistor,
I 2 = V o /(8–j6) = (11.812/10)(64.16+36.87)˚ = 1.1812101.03˚ A
The average power dissipated by the resistor is
P 2 = 0.5|I 2 |2R 2 = 0.5(1.1812)28 = 5.581 W
The complex power supplied is
S = 0.5(V s )(I 1 )* = 0.5(2030˚)(2.4383.07˚) = 24.3833.07˚
= (20.43+13.303) VA
Adding P 1 and P 2 gives the real part of S, showing the conservation of power.
P = 14.86+5.581 = 20.44 W which checks nicely.
Chapter 11, Solution 5.
Converting the circuit into the frequency domain, we get:
2
1
8–40˚
I1 
+

j6
8  40
 1.6828  25.38
j6(2  j2)
1
j6  2  j2
1.6828 2
1  1.4159 W
P1 
2
P 1Ω = 1.4159 W
P 3H = P 0.25F = 0 W
I 2 
j6
1.6828  25.38  2.258
j6  2  j2
2.258 2
P2 
2  5.097 W
2
P 2Ω = 5.097 W
–j2
Chapter 11, Solution 6.

 j L  j103 x 20 x103  j 20
1
1
40F 

  j25
jC j10 3 x 40x10  6
20 mH
We apply nodal analysis to the circuit below.
Vo
+
20I x
–
Ix
j20
50
60o
–j25
10
V  20I x
V 0
6 o
 o
0
10  j20
50  j25
Vo
But I x 
. Substituting this and solving for V o leads
50  j25


1
20
1
1

Vo  6


 10  j20 (10  j20) (50  j25) 50  j25 


1
20
1

Vo  6


 22.3663.43 (22.3663.43)(55.9  26.57) 55.9  26.57 
0.02  j0.04  0.012802  j0.009598  0.016  j0.008Vo  6
(0.0232 – j0.0224)V o = 6 or V o = 6/(0.03225–43.99˚) = 186.0543.99˚ volts.
|I x | = 186.05/55.9 = 3.328
We can now calculate the average power absorbed by the 50-Ω resistor.
P avg = [(3.328)2/2]x50 = 276.8 W.
Chapter 11, Solution 7.
Applying KVL to the left-hand side of the circuit,
820  4 I o  0.1Vo
(1)
Applying KCL to the right side of the circuit,
V
V1
8Io  1 
0
j5 10  j5
10
10  j5
But,
 V1 
Vo 
V1 
Vo
10  j5
10
Vo
10  j5
Hence,
8Io 
Vo 
0
j50
10
I o  j0.025 Vo
(2)
Substituting (2) into (1),
820  0.1 Vo (1  j)
8020
Vo 
1 j
I1 
Vo
8

 - 25
10
2
P
1
 1  64 
2
I1 R    (10)  160W
2
 2  2 
Chapter 11, Solution 8.
We apply nodal analysis to the following circuit.
V 1 I o -j20 
V2
I2
60 A
j10 
0.5 I o
40 
At node 1,
6
V1 V1  V2
V1  j120  V2

j10
- j20
(1)
At node 2,
0.5 I o  I o 
But,
Hence,
V2
40
V1  V2
- j20
1.5 (V1  V2 ) V2

- j20
40
3V1  (3  j) V2
Io 
(2)
Substituting (1) into (2),
j360  3V2  3V2  j V2  0
j360 360
V2 

(-1  j6)
6  j 37
I2 
V2
9

(-1  j6)
40 37
1
1 9 
2
 (40)  43.78 W
P  I2 R  
2
2  37 
2
Chapter 11, Solution 9.
This is a non-inverting op amp circuit. At the output of the op amp,
 Z 
 (10  j 6) x103 
Vo  1  2  Vs   1 
 (8.66  j 5)  20.712  j 28.124
(2  j 4) x103 

 Z1 
The current through the 20-k resistor is
Io 
Vo
 0.1411  j1.491 mA or |I o | = 1.4975 A
20k  j12k
P = [|I o |2/2]R = [1.48752/2]10–6x20x103
= 22.42 mW
Chapter 11, Solution 10.
No current flows through each of the resistors. Hence, for each resistor,
P  0 W . It should be noted that the input voltage will appear at the output of
each of the op amps.
Chapter 11, Solution 11.
  377 ,
R  10 4 ,
C  200  10 -9
RC  (377)(10 4 )(200  10 -9 )  0.754
tan -1 (RC)  37.02
Z ab 
10k
1  (0.754) 2
 - 37.02  7.985 - 37.02 k
i (t )  33 sin(377t  22)  33 cos(377t  68) mA
I = 33–68˚ mA
2
I 2 Z ab 33 x10 3 (7.985 - 37.02)  103
S

2
2
S = 4.348–37.02˚ VA


P  S cos(37.02)  3.472 W
Chapter 11, Solution 12.
We find the Thevenin impedance using the circuit below.
j2 Ω
4Ω
-j3 Ω
5Ω
We note that the inductor is in parallel with the 5-Ω resistor and the combination
is in series with the capacitor. That whole combination is in parallel with the 4-Ω
resistor. Thus,

5xj2 

4  j3 
5  j2  4(0.6896  j1.2758) 4(1.4502  61.61)

Z Thev 


5xj2
4.69  j1.2758
4.86  15.22
4  j3 
5  j2
 1.1936  46.39
Z Thev = 0.8233 – j0.8642 or Z L = [823.3 + j864.2] mΩ.
We obtain V Th using the circuit below. We apply nodal analysis.
j2 Ω
I
4Ω
–j3 Ω
V2
+
o
1650 V
+
–
V Th
–
5Ω
V2  165 V2  165 V2  0
0


4  j3
5
j2
(0.16  j 0.12  j 0.5  0.2)V2  (0.16  j 0.12  j 0.5)165 4.125
(0.5235  46.55)V2  (0.4123  67.17)165
Thus, V 2 = 129.94–20.62˚V = 121.62–j45.76
I = (165 – V 2 )/(4 – j3) = (165 – 121.62 + j45.76)/(4 – j3)
= (63.0646.52˚)/(5–36.87˚) = 12.61383.39˚ = 1.4519+j12.529
V Thev = 165 – 4I = 165 – 5.808 – j50.12 = [159.19 – j50.12] V
= 166.89–17.48˚V
We can check our value of V Thev by letting V 1 = V Thev . Now we can use nodal
analysis to solve for V 1 .
At node 1,
V1  165 V1  V2 V2  0


 0  (0.25  j 0.3333)V1  (0.2  j 0.3333)V2  41.25
4
 j3
5
At node 2,
V2  V1 V2  165

 0   j 0.3333V1  ( j 0.1667)V2   j82.5
j2
 j3
&gt;&gt; Y=[(0.25+0.3333i),-0.3333i;-0.3333i,(0.2-0.1667i)]
Y=
0.2500 + 0.3333i
0 - 0.3333i
0 - 0.3333i 0.2000 - 0.1667i
&gt;&gt; I=[41.25;–82.5i]
I=
41.2500
0 -20.0000i
&gt;&gt; V=inv(Y)*I
V=
159.2221 – 50.1018i
121.6421–45.7677i
Please note, these values check with the ones obtained above.
To calculate the maximum power to the load,
|I L | = (166.89/(2x0.8233)) = 101.34 A
P avg = [(|I L | rms )20.8233]/2 = 4.228 mW.
Chapter 11, Solution 13.
For maximum power transfer to the load, Z L = [120 – j60] Ω.
I L = 165/(240) = 0.6875 A
P avg = [|I L |2120]/2 = 28.36 W.
Chapter 11, Solution 14.
Using Fig. 11.45, design a problem to help other students better understand maximum
average power transfer.
Although there are many ways to work this problem, this is an example based on the
same kind of problem asked in the third edition.
Problem
It is desired to transfer maximum power to the load Z in the circuit of Fig. 11.45. Find Z
and the maximum power. Let is  5cos 40t A.
40 mF
is
7.5 mH
8Ω
12 Ω
Z
Figure 11.45 For Prob. 11.14.
Solution
We find the Thevenin equivalent at the terminals of Z.
40 mF
7.5 mH
1
1

 j 0.625
jC j 40 x 40 x103

 j L  j 40 x7.5 x103  j 0.3


To find Z Th , consider the circuit below.
j0.3
-j0.625
12 Ω
8Ω
Z Th
ZTh  8  j 0.625  12 // j 0.3  8  j 0.625 
12 x0.3
 8.0075  j 0.3252
12  0.3
Z L = (Z Thev )* = [8.008 + j0.3252] Ω.
To find V Th , consider the circuit below.
-j0.625
8Ω
I1
50o
j0.3
12 Ω
+
V Th
–
By current division,
I 1 = 5(j0.3)/(12+j0.3) = 1.590˚/12.0041.43˚ = 0.1249688.57˚
= 0.003118 + j0.12492A
V Thev rms = 12I 1 / 2 = 1.060388.57˚V
I Lrms = 1.060388.57˚/2(8.008) = 66.288.57˚mA
P avg = |I Lrms |28.008 = 35.09 mW.
Chapter 11, Solution 15.
To find Z eq , insert a 1-A current source at the load terminals as shown in Fig. (a).
1
-j 
1
2
+
j
Vo
2 Vo
1A

(a)
At node 1,
Vo Vo V2  Vo


1
j
-j

 Vo  j V2
(1)
At node 2,
1  2 Vo 
V2  Vo
-j

 1  j V2  (2  j) Vo
(2)
Substituting (1) into (2),
1  j V2  (2  j)( j) V2  (1  j) V2
1
V2 
1 j
V
1 j
Z eq  2 
 0.5  j 0.5
1
2
Z L  Z *eq  [0.5  j 0.5] 
We now obtain VThev from Fig. (b).
1
-j 
+
+
120 V
+

Vo
j
2 Vo


(b)
 2 Vo 
Vo  12 Vo

0
1
j
V Thev
Vo 
- 12
1 j
– Vo  (- j  2 Vo )  VTh  0
(12)(1  j 2)
VThev  (1 - j2)Vo 
1 j
2
Pmax


VThev
 0.5  j 0.5  0.5  j 0.5 



0.5 
2
2
 12 5 


 2 

 0.5
2
22 x0.5
= 90 W
Chapter 11, Solution 16.
1
1

  j5
jC j 4 x1 / 20
We find the Thevenin equivalent at the terminals of Z L . To find V Thev , we use the circuit shown
below.
0.5V o
  4,
1H


jL  j 4,


1 / 20F
2
4
V1
V2
+
+
10&lt;0o
-
+
Vo
-
-j5
j4
V Thev
-
At node 1,
10  V1
V
V  V2
 1  0.5V1  1
 j5
2
4

5  V1 (1.25  j0.2)  0.25V2
(1)
At node 2,
V1  V2
V
 0.25V1  2
4
j4


0  0.5V1  V2 (0.25  j 0.25)
Solving (1) and (2) leads to
VThev  V2  6.1947  j 7.0796  9.407248.81o
(2)
To obtain R eq , consider the circuit shown below. We replace Z L by a 1-A current source.
0.5V 1
2
4
V1
-j5
V2
j4
1A
At node 1,
V1
V
V  V2
 1  0.25V1  1
0 

2  j5
4
At node 2,
0  V1 (1  j 0.2)  0.25V2
(3)
V1  V2
V
 0.25V1  2

  1  0.5V1  V2 (0.25  j 0.25)
(4)
4
j4
Solving (1) and (2) gives
V
Z eq  2  1.9115  j 3.3274  3.83760.12 o and Z L = 3.837–60.12&deg; Ω
1
1
Pmax 
| VTh | 2
2 Z eq  Z L
1.9115 
2
9.4072 2
 5.787 W
2 x 4 x1.9115
Chapter 11, Solution 17.
We find Z eq at terminals a-b following Fig. (a).
-j10 
30 
a
b
40 
j20 
(a)
Z eq   j10  30 ||  j 20  40  
(30  j10)(40  j 20)
= 20 Ω = Z L
70  j10
We obtain VThev from Fig. (b).
I1
I2
-j10 
30 
j5 A
+ V Thev 
j20 
40 
(b)
Using current division,
30  j20
I1 
( j5)  -1.1  j2.3
70  j10
40  j10
I2 
( j5)  1.1  j2.7
70  j10
VTh  30 I 2  j10 I 1  10  j70
Pmax 
VTh
2
2 Z eq  Z L 
2
ZL 
5000
20  31.25 W
(2)(2 x 20) 2
Chapter 11, Solution 18.
We find Z Th at terminals a-b as shown in the figure below.
40 
40 
j30 
-j10 
80 
a
Z th
b
Z Th  j 30  40 || 40  80 || (-j10)  j30  20 
(80)(-j10)
80  j10
Z Th  21.23  j 20.154
Z L  Z *Th  [21.23–j20.15] Ω
Chapter 11, Solution 19.
Z Th  - j2  6 || (3  j)  -j2 
(6)(3  j)
9 j
Z Th  2.049  j1.561
R L  Z Th  2.576 Ω
To get VTh , let Z  6 || (3  j)  2.049  j0.439 .
By transforming the current sources, we obtain
VTh  (330) Z  67.62  j14.487 = 69.1612.09˚
2
P max
69.16
2.576
=
= 258.5 W.
2.049  j1.561  2.576
2
Chapter 11, Solution 20.
Combine j20  and -j10  to get j20 || -j10  -j20 .
To find Z Th , insert a 1-A current source at the terminals of R L , as shown in Fig. (a).
Io
4 Io
40 
V1
V2
+ 
-j20 
-j10 
1A
(a)
At the supernode,
V1
V
V
 1  2
40 - j20 - j10
40  (1  j2) V1  j4 V2
1
Also,
V1  V2  4 I o ,
1.1 V1  V2

 V1 
Substituting (2) into (1),
V 
40  (1  j2)  2   j4 V2
 1.1 
44
V2 
1  j6.4
V2
 1.05  j6.71 
1
R L  Z Th  6.792 
Z Th 
(1)
where I o 
V2
1.1
- V1
40
(2)
To find VTh , consider the circuit in Fig. (b).
Io
4 Io
40 
V1
+ 
V2
+
1650 V
+

-j20 
-j10 
V th

(b)
At the supernode,
165  V1
V
V
 1  2
40
- j20 - j10
165  (1  j 2) V1  j 4 V2
Also,
V1  V2  4 I o , where I o 
V1 
V2  16.5
1.1
(3)
165  V1
40
(4)
Substituting (4) into (3),
150–j30  (0.9091  j 5.818) V2
VTh  V2 
150  j 30
152.97  11.31

 25.98  92.43
0.9091  j 5.818
5.88981.12
2
P max
25.98
6.792
=
= 21.51 W
1.05  j 6.71  6.792
2
Chapter 11, Solution 21.
We find Z Th at terminals a-b, as shown in the figure below.
100 
-j10 
a
40 
50 
Z th
j30 
b
Z Th  50 || [ - j10  100 || (40  j30) ]
where 100 || (40  j30) 
(100)(40  j30)
 31.707  j14.634
140  j30
Z Th  50 || (31.707  j4.634) 
Z Th  19.5  j1.73
R L  Z Th  19.58 
(50)(31.707  j4.634)
81.707  j4.634
Chapter 11, Solution 22.
i(t) = [2–2cos(2t)] amps
Chapter 11, Solution 23.
Using Fig. 11.54, design a problem to help other students to better understand how to find
the rms value of a waveshape.
Although there are many ways to work this problem, this is an example based on the
same kind of problem asked in the third edition.
Problem
Determine the rms value of the voltage shown in Fig. 11.54.
v(t) (V)
10
0
1
2
3
4
t (s)
Figure 11.54 For Prob. 11.23.
Solution
T
2
rms
V
1
1
1
100
  v 2 (t )dt   102 dt 
T 0
30
3
V rms = 5.7735 V
Chapter 11, Solution 24.
 5, 0  t  1
v( t )  
- 5, 1  t  2
T  2,
2

Vrms
1
2
 5
1
0
2

dt  1 (-5) 2 dt 
2
25
[1  1]  25
2
Vrms  5 V
Chapter 11, Solution 25.

1 T 2
1 1
2
3
f ( t )dt   0 (4) 2 dt   1 0dt  2 4 2 dt

0
T
3
1
32
 [16  0  16] 
3
3
2
f rms

f rms 
32
 3.266
3
f rms = 3.266

Chapter 11, Solution 26.
5 0t 2
v(t )  
20 2  t  4
T  4,
2
Vrms

4
1 2 2
1
10
dt
(20) 2 dt   [200  800 ]  250





0
2
 4
4
Vrms  15.811 V.
Chapter 11, Solution 27.
i( t )  t , 0  t  5
T  5,
1 5 2
1 t3
t
dt



5 0
5 3
 2.887 A
I 2rms 
I rms
5
0

125
 8.333
15
Chapter 11, Solution 28.
2
Vrms


5
1 2
2
(
4
t
)
dt

0 2 dt


2
5 0
1 16 t 3 2 16
 
 (8)  8.533
5 3 0 15
2

Vrms
Vrms  2.92 V
2
Vrms
8.533
P

 4.267 W
R
2
Chapter 11, Solution 29.
5  t  15
 60  6t
i (t )  
- 120  6t 15  t  25
T  20 ,
25
1  15
2
(
60
6
t
)
dt
(-120  6t) 2 dt 






5
15
20 
25
1 15
   (900  180t  9t 2 ) dt   (9t 2  360t  3600) dt 

15
5  5
1
3
2
25

900t  90t 2  3t 3 15
5  3t  180t  3600t 15
5
1
 [750  750]  300
5
I eff2 
I eff2
I eff2
I eff2



 
I eff  17.321 A
2
P  I eff
R  (17.321)2x12 = 3.6 kW.
Chapter 11, Solution 30.
t 0t2
v( t )  
- 1 2  t  4
2
Vrms

1
4
 t
2
0
2

4

1 8
dt  2 (-1) 2 dt    2   1.1667

43
Vrms  1.08 V
Chapter 11, Solution 31.
V
2
rms
2
1
2
 1 4
1
1

2
  v(t )dt    (2t ) dt   (4) 2 dt     16  8.6667
2 0
20

1
 2 3
Vrms  2.944 V
Chapter 11, Solution 32.
2
1 1
(10t 2 ) 2 dt   0 dt 


1
2  0
5
1
t
 50 0 t 4 dt  50  10  10
5
I 2rms 
I 2rms
I rms  3.162 A
Chapter 11, Solution 33.
I
2
rms
T
3
4

1 2
1 1 2
  i (t )dt    25t dt   25dt   (5t  20) 2 dt 
6 0
T 0
1
3

2
I rms

4
1  t3 1
t3
25
25(3
1)
(25



 100t 2  400t )   11.1056

3
6 3 0
3
I rms = 3.332 A
Chapter 11, Solution 34.
1
T

1  9t 3
 
3 3

2
2

f rms
T
0
0
f 2 (t )dt 
3
1 2
2

(
3
t
)
dt
6 2 dt 




0
2
3

 36  20

f rms  20  4.472
f rms = 4.472
Chapter 11, Solution 35.
2
Vrms

6
5
4
2
1 1 2
2
2
2
10
dt

20
dt

30
dt

20
dt

10 2 dt





5
4
2
1
6 0
1
 [100  400  1800  400  100 ]  466.67
6
2

Vrms
Vrms  21.6 V

Chapter 11, Solution 36.
(a) I rms = 10 A
2
 3 

(b) V rms  4  


 2
36
 9.055 A
(c)
I rms  64 
2
2
(d)
2
V rms 
25 16

 4.528 V
2
2
Vrms  16 
9
 4.528 V (checked)
2
Chapter 11, Solution 37.
Design a problem to help other students to better understand how to determine the rms value of
the sum of multiple currents.
Although there are many ways to work this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
Calculate the rms value of the sum of these three currents:
i 1 = 8, i 2 = 4 sin(t + 10),
i 3 = 6 cos(2t + 30) A
Solution
i  i1  i2  i3  8  4 sin(t  10 o )  6 cos(2t  30 o )
I rms 
I 2 1rms  I 2 2 rms  I 2 3 rms  64 
16 36

 90  9.487 A
2
2
Chapter 11, Solution 38.
V 2 2202
S1  * 
 390.32
Z1
124
S2 
V2
2202

 944.4  j1180.5
Z 2* 20  j 25
S3 
V2
2202

 300  j 267.03
Z 3* 90  j80
S  S1  S2  S3  1634.7  j 913.47  1872.6  29.196o VA
(a) P = Re(S) = 1634.7 W
(b) Q = Im (S) = 913.47 VA (leading)
(c ) pf = cos (29.196o) = 0.8732
Chapter 11, Solution 39.
(a) Z L = 4.2 + j3.6 = 5.5317 40.6o
pf = cos 40.6 = 0.7592
2
Vrms
2202
S * 
 6.643  j 5.694 kVA
Z
5.5317  40.6o
P = 6.643 kW
Q = 5.695 kVAR
(b) C 
P(tan 1  tan  2 ) 6.643 x103 (tan 40.6o  tan 0o )

 312  F ,
2
Vrms
2 x60 x 2202
{It is important to note that this capacitor will see a peak voltage of 220 2 =
311.08V, this means that the specifications on the capacitor must be at least this or
greater!}
Chapter 11, Solution 40.
Design a problem to help other students to better understand apparent power and power
factor.
Although there are many ways to work this problem, this is an example based on the
same kind of problem asked in the third edition.
Problem
A load consisting of induction motors is drawing 80 kW from a 220-V, 60 Hz power line
at a pf of 0.72 lagging. Find the capacitance of a capacitor required to raise the pf to
0.92.
Solution
pf 1  0.72  cos  1

 1  43.940
pf 2  0.92  cos  2

  2  23.07 0
C
P(tan 1  tan  2 ) 80 x103 (0.9637  0.4259)

 2.4 mF ,
2
2 x60 x(220) 2
Vrms
{Again, we need to note that this capacitor will be exposed to a peak voltage of
311.08V and must be rated to at least this level, preferably higher!}
Chapter 11, Solution 41.
(a)
- j2 || ( j5  j2)  -j2 || -j3 
(-j2)(-j3)
 -j6
j
Z T  4  j6  7.211 - 56.31
pf  cos(-56.31)  0.5547 (leading)
(b)
j2 || (4  j) 
( j2)(4  j)
 0.64  j1.52
4  j3
Z  1 || (0.64  j1.52  j) 
0.64  j0.44
 0.479321.5
1.64  j0.44
pf  cos(21.5)  0.9304 (lagging)
Chapter 11, Solution 42.
pf  0.707  cos 

   45o
(a) S=120,
S  S cos   jS sin   84.84  j84.84 VA
(b) S  Vrms I rms
2
(c) S  I rms
Z

 I rms 

 Z
(d) If Z = R + jL, then
 L  2 fL  71.278
S
2
I rms
S
120

 1.091 A rms
Vrms 110
 71.278  j 71.278 
R = 71.278 Ω

 L
71.278
 0.1891 H = 189.1 mH.
2 x60
Chapter 11, Solution 43.
Design a problem to help other students to better understand complex power.
Although there are many ways to work this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
The voltage applied to a 10-ohm resistor is
v(t) = 5 + 3 cos(t + 10) + cos(2t + 30) V
(a) Calculate the rms value of the voltage.
(b) Determine the average power dissipated in the resistor.
Solution
(a) Vrms  V 2 1rms  V 2 2 rms  V 2 3 rms  25 
(b) P 
V 2 rms
 30 / 10  3 W
R
9 1
  30  5.477 V
2 2
Chapter 11, Solution 44.
40  F


1
1

  j12.5
jC j 2000 x 40 x106
60mH


j L  j 2000 x60 x103  j120
We apply nodal analysis to the circuit shown below.
But
100  Vo
4I  V
V
 x o  o
30  j12.5
20
j120
V
I x  o . Solving for V o leads to
j120
Vo  2.9563  j1.126
Io
1000o
Io 
30 Ω
-j12.5
20 Ω
Vo
Ix
j120
+
–
100  Vo
 2.7696  j1.1165
30  j12.5
1
1
S  Vs I o*  (100)(2.7696  j.1165)  138.48  j 55.825 VA
2
2
S = (138.48 – j55.82) VA
+
–
4I x
Chapter 11, Solution 45.
(a) V 2 rms  20 2 
60 2
 2200
2
I rms  1 2 


Vrms  46.9 V
0.5 2
 1.125  1.061 A
2
(b) p(t) = v(t)i(t) = 20 + 60cos100t – 10sin100t – 30(sin100t)(cos100t); clearly
the average power = 20W.
Chapter 11, Solution 46.
(a)
S  V I *  (220 30)(0.5 - 60)  110  - 30
S  [95.26  j 55] VA
Apparent power = 110 VA
Real power = 95.26 W
Reactive power = 55 VAR
(b)
S  V I *  (250 - 10)(6.2 25)  155015
S  [497.2  j 401.2] VA
Apparent power = 1550 VA
Real power = 1497.2 W
Reactive power = 401.2 VAR
pf is lagging because current lags voltage
(c)
S  V I *  (1200)(2.415)  28815
S  [278.2  j 74.54] VA
Apparent power = 288 VA
Real power = 278.2 W
Reactive power = 74.54 VAR
pf is lagging because current lags voltage
(d)
S  V I *  (16045)(8.5 - 90)  1360 - 45
S  [961.7 – j961.7] VA
Apparent power = 1360 VA
Real power = 961.7 W
Reactive power = - 961.7 VAR
Chapter 11, Solution 47.
(a)
(b)
V  112 10 ,
I  4 - 50
1
S  V I *  22460  [112  j194] VA
2
Average power = 112 W
Reactive power = 194 VAR
V  160 0 ,
I  445
1
S  V I *  320 - 45  226.3 – j226.3
2
Average power = 226.3 W
Reactive power = –226.3 VAR
(c)
S
2
V
Z*

(80) 2
 12830  110.85 + j64
50 - 30
Average power = 110.85 W
Reactive power = 64 VAR
(d)
2
S  I Z  (100)(10045)  [7.071  j 7.071] kVA
Average power = 7.071 kW
Reactive power = 7.071 kVAR
Chapter 11, Solution 48.
(a)
S  P  jQ  [269  j150] VA
(b)
pf  cos   0.9 
   25.84
Q  S sin  
 S 
Q
2000

 4588.31
sin  sin(25.84)
P  S cos   4129.48
S  [4.129  j 2] kVA
(c)
Q 450

 0.75
S 600
pf  0.6614
Q  S sin  
 sin  
  48.59 ,
P  S cos   (600)(0.6614)  396.86
S  [396.9  j 450] VA
(d)
2
(220) 2

 1210
S
40
Z
V
P  S cos  
 cos  
  34.26
Q  S sin   681.25
S  [1  j 0.6812] kVA
P 1000

 0.8264
S 1210
Chapter 11, Solution 49.
(a)
(b)
4
sin(cos -1 (0.86)) kVA
0.86
S  [4  j 2.373] kVA
S  4 j
pf 
P 1.6
0.8  cos  

 sin   0.6
S
2
S  1.6  j2 sin   [1.6  j1.2] kVA
(c)
S  Vrms I *rms  (20820)(6.550) VA
S  1.352 70  [0.4624  j1.2705] kVA
2
(d)
V
(120) 2
14400

S * 
Z
40  j60 72.11 - 56.31
S  199.7 56.31  [110.77  j166.16] VA
Chapter 11, Solution 50.
(a)
S  P  jQ  1000  j
S  1000  j750
But,
Vrms
S
Vrms
1000
sin(cos -1 (0.8))
0.8
2
Z*
2
(220) 2
Z 

 30.98  j23.23
S
1000  j750
Z  [30.98  j 23.23] 
*
(b)
2
S  I rms Z
Z
S
I rms
(c)
2
Vrms

1500  j2000
 [10.42  j13.89] 
(12) 2
2
2
(120) 2
 1.6  - 60
2S
(2)(4500 60)
S
Z  1.660  [0.8  j1.386] 
Z 
*

V

Chapter 11, Solution 51.
(a)
Z T  2  (10  j5) || (8  j6)
(10  j5)(8  j6)
110  j20
ZT  2 
 2
18  j
18  j
Z T  8.152  j0.768  8.1885.382
pf  cos(5.382)  0.9956 (lagging)
(b)
S  VI 
*
V
2

Z*
S  31.265.382
(16) 2
(8.188 - 5.382)
P  S cos   31.12 W
(c)
Q  S sin   2.932 VAR
(d)
S  S  31.26 VA
(e)
S  31.265.382  (31.12+j2.932) VA
(a) 0.9956 (lagging, (b) 31.12 W, (c) 2.932 VAR, (d) 31.26 VA, (e) [31.12+j2.932]
VA
Chapter 11, Solution 52.
2000
0.6  2000  j1500
0.8
S B  3000 x 0.4  j3000 x 0.9165  1200  j2749
S A  2000  j
SC  1000  j500
S  S A  S B  SC  4200  j749
4200
(a)
pf 
(b)
S  Vrms I rms 
 I rms 
4200 2  749 2
I rms = 35.5555.11˚ A.
4200  j749
 35.55  55.11
12045
Chapter 11, Solution 53.
S = S A + S B + S C = 4000(0.8–j0.6) + 2400(0.6+j0.8) + 1000 + j500
= 5640 + j20 = 56400.2˚
(a)

I rms

S B S A  SC
S
56400.2
 47  29.8



Vrms
Vrms
Vrms
120 30
I  47 29.8  47 29.8 A
(b)
pf = cos(0.2˚) ≈ 1.0 lagging.
Chapter 11, Solution 54.
Consider the circuit shown below.
8 - 20
 1.6 16.87
4  j3
8 - 20
I2 
 1.6  - 110
j5
I1 
I  I 1  I 2  (-0.5472  j1.504)  (1.531  j0.4643)
I  0.9839  j1.04  1.432 - 46.58
For the source,
S  V I *  (8 - 20)(1.43246.58)
S  11.45626.58  (10.24+j3.12) VA
For the capacitor,
S  I1
2
Z c  (1.6) 2 (-j3)  –j7.68 VA
For the resistor,
S  I1
2
Z R  (1.6) 2 (4)  10.24 VA
For the inductor,
S  I2
2
Z L  (1.6) 2 ( j 5)  j12.8 VA
Chapter 11, Solution 55.
Using Fig. 11.74, design a problem to help other students to better understand the conservation
of AC power.
Although there are many ways to work this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
Find the complex power absorbed by each of the five elements in the circuit of Fig.
11.74.
Figure 11.74
Solution
We apply mesh analysis to the following circuit.
-j20 
j10 
I3
400 V rms
+

I1
20 
I2
+

5090 V rms
For mesh 1,
40  (20  j20) I1  20 I 2
2  (1  j) I1  I 2
(1)
For mesh 2,
- j50  (20  j10) I 2  20 I1
- j5  -2 I1  (2  j) I 2
Putting (1) and (2) in matrix form,
 2  1  j - 1  I1 
 - j5   - 2 2  j I 
  
 2 
(2)
  1 j ,
 1  4  j3 ,
 2  -1  j5

4  j3 1
I1  1 
 (7  j)  3.5358.13

1 j
2
 2 - 1  j5
 2  j3  3.605 - 56.31

I2 

1 j
I 3  I1  I 2  (3.5  j0.5)  (2  j3)  1.5  j3.5  3.80866.8
For the 40-V source,
1

S  -V I 1*  -(40)   (7  j)   [-140  j20] VA
2

For the capacitor,
S  I1
For the resistor,
2
Z c  - j250 VA
S  I3
2
R  290 VA
For the inductor,
2
S  I 2 Z L  j130 VA
For the j50-V source,
S  V I *2  ( j50)(2  j3)  [-150  j100] VA
Chapter 11, Solution 56.
(6)(- j 2)
12  90

 1.897365  71.565  0.6  j1.8
6  j2
6.32456  18.435
3  j 4  [(-j2) || 6]  3.6  j2.2
- j 2 || 6 
The circuit is reduced to that shown below.
Io
+
230 A
Vo
5
3.6 + j2.2 

Io 
3.6  j 2.2
4.21931.4296
(230) 
(230)  0.9505547.08
8.6  j 2.2
8.8769414.3493
Vo  5 I o  4.7527547.08
S  Vo I*s  (4.7527547.08)(2  30)
S  9.505517.08  (9.086+j2.792) VA
Chapter 11, Solution 57.
Consider the circuit as shown below.
4
Vo
-j1 
V1
2
+
240 V
+

1
j2 
V2
2 Vo

At node o,
24  Vo Vo Vo  V1


4
1
-j
24  (5  j4) Vo  j4 V1
(1)
Vo  V1
V1
 2 Vo 
-j
j2
V1  (2  j4) Vo
(2)
At node 1,
Substituting (2) into (1),
24  (5  j4  j8  16) Vo
- 24
(-24)(2 - j4)
Vo 
,
V1 
11  j4
11  j4
The voltage across the dependent source is
V2  V1  (2)(2 Vo )  V1  4 Vo
(-24)(6  j4)
- 24
V2 
 (2  j4  4) 
11  j4
11  j4
S  V2 I *  V2 ( 2 Vo* )
S
(-24)(6  j4) - 48
 1152 


(6  j4)
11  j4
11 - j4  137 
S  (50.45–j33.64) VA
Chapter 11, Solution 58.
From the left portion of the circuit,
0.2
Io 
 0.4 mA
500
20 I o  8 mA which then leads to the following circuit,
I x -j3 k
j1 k
4 k
8 mA
10 k
From the right portion of the circuit,
16
4
Ix 
mA
(8 mA) 
7 j
4  10  j  j3
S  Ix
2
R
(16  10 -3 ) 2
 (10  10 3 )
50
S  51.2 mVA
It should be noted that the complex power delivered to a resistor is always watts.
Chapter 11, Solution 59.
Let V o represent the voltage across the current source and then apply nodal
analysis to the circuit and we get:
Vo
Vo
240  Vo


50
- j20 40  j30
88  (0.36  j0.38) Vo
88
Vo 
 168.13 - 46.55
0.36  j0.38
4
Vo
 8.4143.45
- j20
Vo
I2 
 3.363 - 83.42
40  j30
I1 
Reactive power in the inductor is
2
S  I 2 Z L  (3.363) 2 ( j30)  j339.3 VAR
Reactive power in the capacitor is
2
S  I1 Z c  (8.41) 2 (- j20)  –j1.4146 kVAR
Chapter 11, Solution 60.
20
sin(cos -1 (0.8))  20  j15
0.8
16
S 2  16  j
sin(cos -1 (0.9))  16  j7.749
0.9
S1  20  j
S  S1  S 2  36  j22.749  42.58532.29
But
S  Vo I *  6 Vo
Vo 
S
 7.098  32.29
6
pf  cos(32.29)  0.8454 (lagging)
Chapter 11, Solution 61.
Consider the network shown below.
Io
I2
+
I1
S2
Vo
So
S1
S3

S 2  1.2  j0.8 kVA
S3  4  j
4
sin(cos -1 (0.9))  4  j1.937 kVA
0.9
Let
S 4  S 2  S 3  5.2  j1.137 kVA
But
S 4  Vo I *2
S 4 (5.2  j1.137)  10 3

 11.37  j 52
10090
Vo
I 2  11.37  j 52
I *2 
Similarly,
S1  2  j
But
S1  Vo I 1*
2
sin(cos -1 (0.707))  2 (1  j) kVA
0.707
S1 (1.4142  j1.4142)  10 3

 -14.142  j14.142
Vo
j100
I1  – 14.142 + j14.142
I 1* 
I o  I 1  I 2  - 2.772  j66.14  66.292.4&deg; A
S o  Vo I o*
S o  (10090)(66.2 - 92.4) VA
S o  6.62–2.4&deg; kVA
66.292.4&deg; A, 6.62–2.4&deg; kVA
Chapter 11, Solution 62.
Consider the circuit below.
0.2 + j0.04 
I
I2
0.3 + j0.15 
I1
Vs
+

S 2  15  j
But
+
+
V1
V2


15
sin(cos -1 (0.8))  15  j11.25
0.8
S 2  V2 I *2
S 2 15  j11.25

V2
120
I 2  0.125  j0.09375
I *2 
V1  V2  I 2 (0.3  j0.15)
V1  120  (0.125  j0.09375)(0.3  j0.15)
V1  120.02  j0.0469
S1  10  j
But
10
sin(cos -1 (0.9))  10  j4.843
0.9
S1  V1 I 1*
S 1 11.11125.84

V1 120.02 0.02
I 1  0.093 - 25.82  0.0837  j0.0405
I 1* 
I  I 1  I 2  0.2087  j0.053
Vs  V1  I (0.2  j0.04)
Vs  (120.02  j0.0469)  (0.2087  j0.053)(0.2  j0.04)
Vs  120.06  j0.0658
Vs  120.060.03 V
Chapter 11, Solution 63.
Let
S  S1  S 2  S 3 .
S1  12  j
12
sin(cos -1 (0.866))  12  j6.929
0.866
S 2  16  j
16
sin(cos -1 (0.85))  16  j9.916
0.85
S3 
(20)(0.6)
 j 20  15  j 20
sin(cos -1 (0.6))
S  43  j 22.987  V I *o
I *o 
S (43  j 22.99) x10 3

 195.45  j104.5  221.628.13
V
220
I o  221.6–28.13˚ A
Chapter 11, Solution 64.
I2
I1
8
+

Is
1200&ordm; V
j12
I s + I 2 = I 1 or I s = I 1 – I 2
I1 
But,
120
8  j12
 4.615  j6.923
2500  j400
S
 I 2 

 20.83  j3.333
S  VI 2 
V
120
or I 2  20.83  j3.333
I s = I 1 – I 2 = –16.22 – j10.256 = 19.19–147.69˚ A.
Chapter 11, Solution 65.
C  1 nF 

1
-j
 4
 -j100 k
jC 10  10 -9
At the noninverting terminal,
Vo
40  Vo
4


 Vo 
100
- j100
1 j
4
Vo 
 - 45
2
4
v o (t) 
cos(10 4 t  45)
2
2
 4 1   1 
Vrms
 
W


P
R
 2 2   50  10 3 
2
P  80 W
Chapter 11, Solution 66.
As an inverter,
- Zf
- (2  j4)
Vo 
Vs 
 (4 45)
Zi
4  j3
Io 
Vo
- (2  j4)(445)
mA
mA 
(6 - j2)(4  j3)
6  j2
The power absorbed by the 6-k resistor is
2
P  Io
2
 20  4 
  10 -6  6  10 3
R  

 40  5 
P  1.92 mW
Chapter 11, Solution 67.
  2,
3H


jL  j 6,
0.1F


1
1

  j5
jC j 2 x0.1
 j 50
 2  j4
10  j 5
The frequency-domain version of the circuit is shown below.
Z 2 =2-j4 
10 //(  j 5) 
Z 1 =8+j6 
I1
+
Io
+
+
0.620 o V
Z 3  12
Vo
-
(a) I 1 
0.620 o  0 0.5638  j 0.2052

 0.06  16.87 o
8  j6
8  j6
1
S  Vs I *1  (0.320 o )(0.06  16.87 o )  14.4  j10.8 mVA  1836.86 o mVA
2
S = (14.4+j10.8) mVA = 1836.86&deg; mVA
V
Z2
(2  j 4)
(0.620 o )  0.022499.7 o
Vs ,
Io  o  
12(8  j 6)
Z1
Z3
1
P  | I o | 2 R  0.5(0.0224) 2 (12)  2.904 mW
2
(b) Vo  
P = 2.904 mW
(a) 1836.86&deg; mVA, (b) 2.904 mW
Chapter 11, Solution 68.
Let
where
S  SR  SL  Sc
1 2
I R  j0
2 o
1
S L  PL  jQ L  0  j I o2 L
2
1 2 1
S c  Pc  jQ c  0  j I o 
2
C
S R  PR  jQ R 
Hence,
S

1 
1 2

I o  R  jL 

C 
2 
Chapter 11, Solution 69.
(a)
(b)
(c)
Given that Z  10  j12
12

   50.19
tan  
10
pf  cos   0.6402
2
(120) 2
 295.12  j354.09
2 Z * (2)(10  j12)
The average power absorbed  P  Re(S)  295.1 W
S
V

For unity power factor, 1  0 , which implies that the reactive power due
to the capacitor is Q c  354.09
But
C
Qc 
V2
1
 C V 2
2 Xc 2
2 Qc
(2)(354.09)
 130.4 F
2 
V
(2 )(60)(120) 2
Chapter 11, Solution 70.
Design a problem to help other students to better understand power factor correction.
Although there are many ways to work this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
An 880-VA, 220-V, 50-Hz load has a power factor of 0.8 lagging. What value of parallel
capacitance will correct the load power factor to unity?
Solution
pf  cos   0.8 

sin   0.6
Q  S sin   (880)(0.6)  528
If the power factor is to be unity, the reactive power due to the capacitor is
Q c  Q  528 VAR
2
Vrms
Qc
 C V 2 
 C 
Xc
V2
(528)
 34.72 &micro;F
C
(2)(50)(220) 2
But
Q
Chapter 11, Solution 71.
Q 1 = 60 kVAR, pf = 0.85 or θ 1 = 31.79˚
Q 1 = S 1 sinθ 1 = 60k or S 1 = 113.89k and P 1 = 113.89cos(31.79) = 96.8kW
S 1 = 96.8 + j60 kVA
For load 2, S 2 = 90 – j50 kVA
For load 3, S 3 =100 kVA
Hence,
S = S 1 + S 2 + S 3 = 286.8 + j10kVA = 2872˚kVA
But
Thus,
S = (V rms )2/Z* or Z* = 1202/2872˚k = 0.05017–2˚
Z = 0.050172˚Ω or [50.14 + j1.7509] mΩ.
(b) From above, pf = cos2˚ = 0.9994.
(c) I rms = V rms /Z = 120/0.050172˚ = 2.392–2˚ kA or [2.391 – j0.08348] kA.
Chapter 11, Solution 72.
(a) P  S cos 1

 S
P
2.4

 3.0 kVA
cos 1 0.8
pf  0.8  cos 1

 1  36.87 o
Q  S sin 1  3.0sin 36.87 o  1.8 kVAR
Hence, S = 2.4 + j1.8 kVA
P
1.5
S1  1 
 2.122 kVA
cos  0.707
pf  0.707  cos 

   45o
Q1  P1  1.5 kVAR

 S1  1.5  j1.5 kVA
Since, S  S1  S 2

 S2  S  S1  (2.4  j1.8)  (1.5  j1.5)  0.9  j 0.3 kVA
S 2  0.9497  18.43o
pf = cos 18.43o = 0.9487

  2  25.84o
(b) pf  0.9  cos  2
P(tan 1  tan  2 ) 2400(tan 36.87  tan 25.84)
C

 117.5  F
2
2 x60 x(120) 2
Vrms
Chapter 11, Solution 73.
(a)
S  10  j15  j22  10  j7 kVA
S  S  10 2  7 2  12.21 kVA
(b)
S  V I*

 I * 
S 10,000  j7,000

240
V
I  41.667  j29.167  50.86 - 35 A
(c)
7
1  tan -1    35 ,
10 
 2  cos -1 (0.96)  16.26
Q c  P1 [ tan 1  tan  2 ]  10 [ tan(35) - tan(16.26) ]
Q c  4.083 kVAR
C
(d)
Qc
4083

 188.03 F
2
 Vrms (2 )(60)(240) 2
S 2  P2  jQ 2 ,
P2  P1  10 kW
Q 2  Q1  Q c  7  4.083  2.917 kVAR
S 2  10  j2.917 kVA
But
S 2  V I *2
S 2 10,000  j2917

240
V
I 2  41.667  j12.154  43.4 - 16.26 A
I *2 
Chapter 11, Solution 74.
(a)
1  cos -1 (0.8)  36.87
P1
24
S1 

 30 kVA
cos 1 0.8
Q1  S1 sin 1  (30)(0.6)  18 kVAR
S1  24  j18 kVA
 2  cos -1 (0.95)  18.19
P2
40
S2 

 42.105 kVA
cos  2 0.95
Q 2  S 2 sin  2  13.144 kVAR
S 2  40  j13.144 kVA
S  S1  S 2  64  j31.144 kVA
 31.144 
  25.95
  tan -1 
 64 
pf  cos   0.8992
(b)
 2  25.95 ,
1  0
Q c  P [ tan  2  tan 1 ]  64 [ tan(25.95)  0 ]  31.144 kVAR
C
Qc
31,144

 5.74 mF
2
 Vrms (2 )(60)(120) 2
Chapter 11, Solution 75.
(a)
S1 
V
Z1*
2

(240) 2
5760

 517.75  j323.59 VA
80  j50 8  j5
S2 
(240) 2
5760

 358.13  j208.91 VA
120  j70 12  j7
S3 
(240) 2
 960 VA
60
S  S1  S 2  S 3  [1.8359  j 0.11468] kVA
(b)
(c)
 114.68 
  3.574
  tan -1 
1835.88 
pf  cos   0.998 {leading}
Since the circuit already has a leading power factor, near unity, no
compensation is necessary.
Chapter 11, Solution 76.
The wattmeter reads the real power supplied by the current source. Consider the
circuit below.
4
120 V
-j3 
+

Vo
j2 
8
12  Vo Vo Vo


4  j3
j2
8
36.14  j23.52
Vo 
 0.7547  j11.322  11.347 86.19
2.28  j3.04
330 
S  Vo I *o  (11.34786.19)(3 - 30)
S  34.0456.19 VA
P  Re(S)  18.942 W
330 A
Chapter 11, Solution 77.
The wattmeter measures the power absorbed by the parallel combination of 0.1 F and 150
.
120 cos(2 t ) 
 1200 ,
 2
4H 

jL  j8
1
0.1 F 

 -j5
jC
Consider the following circuit.
6
1200 V
Z  15 || (-j5) 
I
j8 
I
+

(15)(-j5)
 1.5  j4.5
15  j5
120
 14.5 - 25.02
(6  j8)  (1.5  j4.5)
1
1 2
1
V I *  I Z   (14.5) 2 (1.5  j4.5)
2
2
2
S  157.69  j473.06 VA
S
P  Re(S)  157.69 W
Z
Chapter 11, Solution 78.
The wattmeter reads the power absorbed by the element to its right side.
2 cos(4 t ) 
 20 ,
 4
1H 
 jL  j4
1
1
F 

 -j3
12
jC
Consider the following circuit.
10 
200 V
I
+

Z  5  j4  4 || - j3  5  j4 
Z
(4)(- j3)
4  j3
Z  6.44  j2.08
I
20
 1.207  - 7.21
16.44  j2.08
S
1 2
1
I Z   (1.207) 2 (6.44  j2.08)
2
2
P  Re(S)  4.691 W
Chapter 11, Solution 79.
The wattmeter reads the power supplied by the source and partly absorbed by the 40- 
resistor.
  100,
10 mH
j100 x10x10  3  j,


500F


1
1

  j20
jC j100 x 500 x10  6
The frequency-domain circuit is shown below.
20
I
40
Io
j
V1
V2
+1
2 Io
10&lt;0o
-j20
-
At node 1,
V  V2 V1  V2 3(V1  V2 ) V1  V2
10  V1
 2I o  1



j
20
20
j
40
10  (7  j40)V1  (6  j40)V2

(1)
At node 2,
V1  V 2 V1  V 2
V

 2
j
20
 j 20


0  (20  j )V1  (19  j )V 2
Solving (1) and (2) yields V 1 = 1.5568 –j4.1405
I
10  V1
 0.2111  j0.1035,
40
S
1
V1I   0.04993  j0.5176
2
P = Re(S) = 50 mW.
(2)
Chapter 11, Solution 80.
(a)
(b)
I
V 110

 17.19 A
Z 6.4
S
V
Z
2

(110) 2
 1890.6
6.4
cos   pf  0.825 
   34.41
P  S cos   1559.8  1.6 kW
Chapter 11, Solution 81.
Design a problem to help other students to better understand how to correct power factor
to values other than unity.
Although there are many ways to work this problem, this is an example based on the
same kind of problem asked in the third edition.
Problem
A 120-V rms, 60-Hz electric hair dryer consumes 600 W at a lagging pf of 0.92.
Calculate the rms-valued current drawn by the dryer.
How would you power factor correct this to a value of 0.95?
Solution

   23.074o
P
P  S cos 

 S
 652.17 VA
0.92
S = P+jQ = 600 + j652.17sin23.09o = 600 +j255.6
*
S  Vrms I rms
But
.
S
600  j 255.6
*
I rms


Vrms
120
P = 600 W,
pf  0.92
I rms = 5 – j2.13 = 5.435–23.07˚A.
To correct this to a pf = 0.95, I would add a capacitor in parallel with the hair dryer
(remember, series compensation will increase the power delivered to the load and
probably burn out the hair dryer.
pf = 0.95 = 600/S or S = 631.6 VA and θ = 18.19˚ and VARs = 197.17
Thus,
VARs cap = 255.6 – 197.17 = 58.43 = 120xI C or I C = 58.43/120 = 0.4869A
Next,
X C = 120/0.4869 = 246.46 = 1/(377xC) or C = 10.762 &micro;F
Chapter 11, Solution 82.
(a) P1  5,000,
Q1  0
P2  30,000 x0.82  24,600,
Q2  30,000 sin(cos 1 0.82)  17,171
S  S1  S 2  (P1  P2 )  j(Q1  Q 2 )  29,600  j17,171
S | S | 34.22 kVA
Q = 17.171 kVAR
(b)
(c ) pf 
P 29,600

 0.865
S 34,220
Qc  P (tan  1  tan  2 )


 29,600 tan(cos 1 0.865)  tan(cos 1 0.9)  2833 VAR
(c)
C
Qc
2833

 130.46  F
2
V rms 2x 60 x 240 2
Chapter 11, Solution 83.
1
1
(a) S  VI   (21060 o )(8  25 o )  84035 o
2
2
P  S cos   840 cos 35 o  688.1 W
(b) S = 840 VA
(c) Q  S sin   840 sin 35 o  481.8 VAR
(d) pf  P / S  cos 35 o  0.8191 (lagging)
Chapter 11, Solution 84.
(a)
Maximum demand charge  2,400  30  \$72,000
Energy cost  \$0.04  1,200  10 3  \$48,000
Total charge = \$120,000
(b)
To obtain \$120,000 from 1,200 MWh will require a flat rate of
\$120,000
per kWh  \$0.10 per kWh
1,200  10 3
Chapter 11, Solution 85.


j 2x60 x15 x10 3  j 5.655
(a) 15 mH
We apply mesh analysis as shown below.
I1
+
Ix
o
120&lt;0 V
-
10 
In
30 
Iz
10 
+
120&lt;0o V
Iy
j5.655 
I2
For mesh x,
(1)
120 = 10 I x - 10 I z
For mesh y,
120 = (10+j5.655) I y - (10+j5.655) I z
(2)
For mesh z,
0 = -10 I x –(10+j5.655) I y + (50+j5.655) I z
(3)
Solving (1) to (3) gives
I x =20, I y =17.09-j5.142, I z =8
Thus,
I 1 =I x =20 A
I 2 =-I y =-17.09+j5.142 = 17.85163.26 o A
I n =I y - I x = –2.91 –j5.142 = 5.907  119.5 o A
(b) S1  (120)I  x  120x 20  2400,
S 2  (120)I  y  2051  j617
S  S 1  S 2  [4.451  j 0.617] kVA
(c ) pf = P/S = 4451/4494 = 0.9904 (lagging)
Chapter 11, Solution 86.
For maximum power transfer
Z L  Z *Th 
 Z i  Z Th  Z *L
Z L  R  jL  75  j (2)(4.12  10 6 )(4  10 -6 )
Z L  75  j103.55 
Z i  [75  j103.55] 
Chapter 11, Solution 87.
Z  R  jX
VR  I R
Z
2

 R 
 R 2  X2
VR
80

 1.6 k
50  10 -3
I

 X 2  Z
2
 R 2  (3) 2  (1.6) 2
X  2.5377 k
X
 2.5377 
  57.77
  tan -1    tan -1 
R 
 1.6 
pf  cos   0.5333
Chapter 11, Solution 88.
(a)
S  (110)(2 55)  22055
P  S cos   220 cos(55)  126.2 W
(b)
S  S  220 VA
Chapter 11, Solution 89.
(a)
Apparent power  S  12 kVA
P  S cos   (12)(0.78)  9.36 kW
Q  S sin   12 sin(cos -1 (0.78))  7.51 kVAR
S  P  jQ  [9.36  j 7.51] kVA
(b)
S
V
Z*
2

 Z 
Z  [2.866 + j2.3] Ω
*
V
S
2
(210) 2

= 2.866 – j2.3
(9.36  j7.51)  10 3
Chapter 11, Solution 90
P1  2000 kW ,
cos 1  0.85 
 1  31.79
P1
S1 
 2352.94 kVA
cos 1
Q1  S1 sin 1  1239.5 kVAR
P2  300 kW ,
cos  2  0.8 
  2  36.87
P2
S2 
 375 kVA
cos  2
Q 2  S 2 sin  2  225 kVAR
S  S1  S 2  (P1  P2 )  j (Q1  Q 2 )  P  jQ
P  2000  300  2300 kW
Q  1239.5  225  1464.5 kVAR
The minimum operating pf for a 2300 kW load and not exceeding the kVA rating of the
generator is
P
2300

 0.9775
cos  
S1 2352.94
  12.177
or
The maximum load kVAR for this condition is
Q m  S1 sin   2352.94 sin(12.177)
Q m  496.313 kVAR
The capacitor must supply the difference between the total load kVAR ( i.e. Q ) and the
permissible generator kVAR ( i.e. Q m ). Thus,
Q c  Q  Q m  968.2 kVAR
Chapter 11, Solution 91
The nameplate of an electric motor has the following information:
Line voltage: 220 V rms
Line current: 15 A rms
Line frequency: 60 Hz
Power: 2700 W
Determine the power factor (lagging) of the motor. Find the value of the
capacitance C that must be connected across the motor to raise the pf to unity.
Solution
I = V/Z which leads to Z = [220/15] θ = 14.6667 θ, S = (220)(15) θ = 3.3 θ
kVA, where cos–1(2700/3300) = cos–1(0.818182) = 35.097&deg;, and X L =
3300sin(35.097&deg;) = 1897.38 = X C . This leads to C = 1/[377(1897.38)] = 1.398
&micro;F.
pf = 0.8182 (lagging)
C = 1.398 &micro;F
0.8182 (lagging), 1.398 &micro;F
Chapter 11, Solution 92
(a)
Apparent power drawn by the motor is
P
60

 80 kVA
Sm 
cos  0.75
Q m  S 2  P 2  (80) 2  (60) 2  52.915 kVAR
Total real power
P  Pm  Pc  PL  60  0  20  80 kW
Total reactive power
Q  Q m  Q c  Q L  52.915  20  0  32.91 kVAR
Total apparent power
S  P 2  Q 2  86.51 kVA
(b)
pf 
(c)
I
P
80

 0.9248
S 86.51
S 86510

 157.3 A
V
550
Chapter 11, Solution 93
(a)
P1  (5)(0.7457)  3.7285 kW
P1 3.7285

 4.661 kVA
S1 
pf
0.8
Q1  S1 sin(cos -1 (0.8))  2.796 kVAR
S1  3.7285  j2.796 kVA
P2  1.2 kW ,
S 2  1.2  j0 kVA
Q 2  0 VAR
P3  (10)(120)  1.2 kW ,
S 3  1.2  j0 kVA
Q 3  0 VAR
Q 4  1.6 kVAR ,
cos  4  0.6 
 sin  4  0.8
Q4
S4 
 2 kVA
sin  4
P4  S 4 cos  4  (2)(0.6)  1.2 kW
S 4  1.2  j1.6 kVA
S  S1  S 2  S 3  S 4
S  7.3285  j1.196 kVA
Total real power = 7.3285 kW
Total reactive power = 1.196 kVAR
(b)
 1.196 
  9.27
  tan -1 
 7.3285 
pf  cos   0.987
Chapter 11, Solution 94
cos 1  0.7 
 1  45.57
S1  1 MVA  1000 kVA
P1  S1 cos 1  700 kW
Q1  S1 sin 1  714.14 kVAR
For improved pf,
cos  2  0.95 
  2  18.19
P2  P1  700 kW
P2
700
S2 

 736.84 kVA
cos  2 0.95
Q 2  S 2 sin  2  230.08 kVAR
P 1 = P 2 = 700 kW
1
2
Q2
S2
S1
Q1
(a)
Qc
Reactive power across the capacitor
Q c  Q1  Q 2  714.14  230.08  484.06 kVAR
Cost of installing capacitors  \$30  484.06  \$14,521.80
(b)
Substation capacity released  S1  S 2
 1000  736.84  263.16 kVA
Saving in cost of substation and distribution facilities
 \$120  263.16  \$31,579.20
(c)
Yes, because (a) is greater than (b). Additional system capacity obtained
by using capacitors costs only 46% as much as new substation and
distribution facilities.
Chapter 11, Solution 95
(a)
Source impedance
Zs  R s  jXc
ZL  R L  jX 2
Z L  Z *s 
 R s  R L , X c  X L
1
Xc  XL 

 L
C
1
or

 2 f
LC
f
1
2 LC

1
2 (80  10 -3 )(40  10 -9 )
2
(b)
 2.814 kHz
2
 V

s
 4   4.6  4  431.8 mW (since V is in rms)
P
s
 (10  4) 
 14 


Chapter 11, Solution 96
Z Th
+

V Th
(a)
VTh  146 V, 300 Hz
Z Th  40  j8 
Z L  Z *Th  [40  j 8] 
(b)
P
VTh
2
8 R Th
(146) 2

 66.61 W
(8)(40)
ZL
Chapter 11, Solution 97
Z T  (2)(0.1  j)  (100  j20)  100.2  j22 
Vs
240

I
Z T 100.2  j22
2
P  I R L  100 I
2

(100)(240) 2
 547.3 W
(100.2) 2  (22) 2
Chapter 12, Solution 1.
(a)
If Vab  400 , then
400
Van 
 - 30  231 - 30 V
3
Vbn  231 - 150 V
Vcn  231 - 270 V
(b)
For the acb sequence,
Vab  Van  Vbn  Vp 0  Vp 120
 1
3
Vab  Vp 1   j   Vp 3 - 30
2 
 2
i.e. in the acb sequence, Vab lags Van by 30.
Hence, if Vab  400 , then
400
Van 
30  23130 V
3
Vbn  231150 V
Vcn  231 - 90 V
Chapter 12, Solution 2.
Since phase c lags phase a by 120, this is an acb sequence.
Vbn  120(30  120)  120150 V
Chapter 12, Solution 3.
Since Vbn leads Vcn by 120, this is an abc sequence.
Van  440(130  120)  440–110˚ V.
Chapter 12, Solution 4.
Knowing the line-to-line voltages we can calculate the wye voltages and can let the value
of V a be a reference with a phase shift of zero degrees.
V L = 440 =
V p or V p = 440/1.7321 = 254 V or V an = 254 0&deg; V which
determines, using abc roataion, both V bn = 254 –120&deg; and V cn = 254 120&deg;.
I a = V an /Z Y = 254/(40 30&deg;) = 6.35–30˚ A
I b = I a –120˚ = 6.35–150˚ A
I c = I a +120˚ = 6.3590˚ A
Chapter 12, Solution 5.
V AB = 1.7321xV AN +30˚ = 207.8(32˚+30˚) = 207.862˚ V or
v AB = 207.8cos(ωt+62˚) V
v BC = 207.8cos(ωt–58˚) V
and
v CA = 207.8cos(ωt+182˚) V
207.8cos(ωt+62˚) V, 207.8cos(ωt–58˚) V, 207.8cos(ωt+182˚) V
Chapter 12, Solution 6.
Using Fig. 12.41, design a problem to help other students to better understand balanced wye-wye
connected circuits.
Although there are many ways to work this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
For the Y-Y circuit of Fig. 12.41, find the line currents, the line\ voltages, and the load
voltages.
Figure 12.41
Solution
Z Y  10  j5  11.1826.56
The line currents are
Van
220 0

Ia 
 19.68 - 26.56 A
Z Y 11.1826.56
I b  I a  - 120  19.68 - 146.56 A
I c  I a 120  19.6893.44 A
The line voltages are
Vab  220 3 30  38130 V
Vbc  381 - 90 V
Vca  381 - 210 V
VAN  I a Z Y  Van  2200 V
VBN  Vbn  220 - 120 V
VCN  Vcn  220120 V
Chapter 12, Solution 7.
This is a balanced Y-Y system.
4400 V
+

Z Y = 6  j8 
Using the per-phase circuit shown above,
4400
Ia 
 4453.13 A
6  j8
I b  I a  - 120  44 - 66.87 A
I c  I a 120  44173.13 A
Chapter 12, Solution 8.
Consider the per phase equivalent circuit shown below.
Zl
V an
+
_
ZL
5.396–35.1˚ A
I a = V an /( Z l + Z L ) = (100 20&deg;)/(10.6+j15.2) = (100 20&deg;)/(18.531 55.11&deg;)
= 5.396 –35.11&deg; amps.
I b = I a –120&deg; = 5.396 –155.11&deg; amps.
Ic = Ia
120&deg; = 5.396 84.89&deg; amps.
V La = I a Z L = (4.414–j3.103)(10+j14) = (5.396 –35.11&deg; 17.205 54.46&deg;)
= 92.84 19.35&deg; volts.
V Lb = V La –120&deg; = 94.84 –100.65&deg; volts.
V Lc = V La
120&deg; = 94.84 139.35&deg; volts.
Chapter 12, Solution 9.
Ia 
Van
1200

 4.8 - 36.87 A
Z L  Z Y 20  j15
I b  I a  - 120  4.8 - 156.87 A
I c  I a 120  4.883.13 A
As a balanced system, I n  0 A
Chapter 12, Solution 10.
Since the neutral line is present, we can solve this problem on a per-phase basis.
For phase a,
Ia 
Van
4400
440


 15.28320.32
Z A  2 27  j10 28.79  20.32
Ib 
Vbn
440 - 120

 20 - 120
ZB  2
22
Ic 
Vcn
440120 440120


 33.8597.38
12  j 5
1322.62
ZC  2
For phase b,
For phase c,
The current in the neutral line is
I n  -(I a  I b  I c ) or - I n  I a  I b  I c
- I n  (14.332  j 5.308)  (10  j17.321)  (4.346  j33.57)
I n  0.014  j 21.56  21.56–89.96&deg;A
Chapter 12, Solution 11.
Given that V p = 240 and that the system is balanced, V L = 1.7321V p = 415.7 V.
I p = V L /|2–j3| = 415.7/3.606 = 115.29 A and
I L = 1.7321x115.29 = 199.69 A.
Chapter 12, Solution 12.
Using Fig. 12.45, design a problem to help other students to better understand wye-delta
connected circuits.
Although there are many ways to solve this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
Solve for the line currents in the Y- circuit of Fig. 12.45. Take Z  = 6045.
Figure 12.45
Solution
Ia
1100 V
ZY 
+

Z
 20 45 
3
110 0
 5.5 - 45 A
2045
I b  I a  - 120  5.5 - 165 A
I c  I a 120  5.575 A
Ia 
ZY
Chapter 12, Solution 13.
Convert the delta load to wye as shown below.
1100o V rms
2
–+
110–120o V rms
ZY
2
ZY
–+
110120o V rms
ZY
2
–+
1
ZY  Z   3  j 2 
3
We consider the single phase equivalent shown below.
2
1100˚ V rms
+
_
3 – j2 
I a = 110/(2 + 3 – j2) = 20.4321.8&deg; A
I L = |I a | = 20.43 A
S = 3|I a |2Z Y = 3(20.43)2(3–j2) = 4514–33.96˚ = 3744 – j2522
P = Re(S) = 3.744 kW.
Chapter 12, Solution 14.
We apply mesh analysis with Z L = (12+j12) Ω.
a
Ia
1+j2 Ω
+

1000
I1
A
n
100120
+
c


100–120
+
I2
ZL
I3
ZL
Ib
b
1+j2 Ω
B
C
ZL
1+j2 Ω
Ic
For mesh 1,
 100  100  120o  I1 (14  j16)  (1  j 2) I 2  (12  j12) I 3  0 or
(14  j16) I1  (1  j 2) I 2  (12  j12) I 3  100  50  j86.6  150  j86.6
(1)
For mesh 2,
100120 o  100  120 o  I 1 (1  j 2)  (12  j12) I 3  (14  j16) I 2  0 or
 (1  j 2) I 1  (14  j16) I 2  (12  j12) I 3  50  j86.6  50  j86.6   j173.2
(2)
For mesh 3,
 (12  j12) I 1  (12  j12) I 2  (36  j36) I 3  0 or I 3 = I 1 + I 2
(3)
Solving for I 1 and I 2 using (1) to (3) gives
I 1 = 12.804–50.19&deg; A = (8.198 – j9.836) A and
I 2 = 12.804–110.19&deg; A = (–4.419 – j12.018) A
I a = I 1 = 12.804–50.19&deg; A
I b = I 2 – I 1 = 12.804–170.19&deg; A
I c = –I 2 = 12.80469.81&deg; A
As a check we can convert the delta into a wye circuit. Thus,
Z Y = (12+j12)/3 = 4+j4 and I a = 100/(1+j2+4+j4) = 100/(5+j6)
= 100/(7.8102 50.19&deg;) =
12.804 –50.19&deg; A.
Chapter 12, Solution 15.
Convert the delta load, Z  , to its equivalent wye load.
Z
 8  j10
Z Ye 
3
(12  j5)(8  j10)
 8.076  - 14.68
20  j5
Z p  7.812  j2.047
Z p  Z Y || Z Ye 
Z T  Z p  Z L  8.812  j1.047
Z T  8.874  - 6.78
We now use the per-phase equivalent circuit.
Vp
210
Ia 
,
where Vp 
Zp  ZL
3
Ia 
210
3 (8.874  - 6.78)
I L  I a  13.66 A
 13.66 6.78
Chapter 12, Solution 16.
(a)
I CA  - I AC  5(-30  180)  5150
This implies that
I AB  530
I BC  5  90
I a  I AB 3  - 30  8.660˚ A
I b  8.66–120˚ A
I c  8.66120˚ A
(b)
Z 
VAB 1100

 22–30˚ Ω.
I AB
530
Chapter 12, Solution 17.
I a = 1.7321xI AB –30˚ or
I AB = I a /(1.7321–30˚) = 2.887(–25˚+30˚) = 2.8875˚ A
I BC = I AB –120˚ = 2.887–115˚ A
I CA = I AB +120˚ = 2.887125˚ A
2.8875˚ A, 2.887–115˚ A, 2.887125˚ A
Chapter 12, Solution 18.
VAB  Van 3 30  (22060)( 3 30)  381.190
Z   12  j9  1536.87
I AB 
VAB 381.190

 25.453.13˚ A
Z  1536.87
I BC  I AB  - 120  25.4–66.87˚ A
I CA  I AB 120  25.4173.13˚ A
Chapter 12, Solution 19.
Z   30  j10  31.62 18.43
The phase currents are
Vab
1730
 5.47  - 18.43 A
I AB 

Z  31.62 18.43
I BC  I AB  - 120  5.47  - 138.43 A
I CA  I AB 120  5.47 101.57 A
The line currents are
I a  I AB  I CA  I AB 3  - 30
I a  5.47 3  - 48.43  9.474 - 48.43 A
I b  I a  - 120  9.474 - 168.43 A
I c  I a 120  9.47471.57 A
Chapter 12, Solution 20.
Using Fig. 12.51, design a problem to help other students to better understand balanced deltadelta connected circuits.
Although there are many ways to solve this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
Refer to the - circuit in Fig. 12.51. Find the line and phase currents. Assume that the
load impedance is 12 + j9 per phase.
Figure 12.51
Solution
Z   12  j9  1536.87
The phase currents are
2100
 14 - 36.87 A
1536.87
 I AB  - 120  14 - 156.87 A
 I AB 120  1483.13 A
I AB 
I BC
I CA
The line currents are
I a  I AB 3  - 30  24.25 - 66.87 A
I b  I a  - 120  24.25 - 186.87 A
I c  I a 120  24.2553.13 A
Chapter 12, Solution 21.
(a)
I AC 
 230120
 230120

 17.96  98.66 A
10  j8
12.80638.66
I AC = 17.96–98.66˚ A
230  120 2300

10  j8
10  j8
 17.96  158.66  17.96  38.66
 16.729  j 6.536  14.024  j11.220  30.75  j 4.684
I bB  I BC  I BA  I BC  I AB 
(b)
I bB = 31.1171.34˚ A.
Chapter 12, Solution 22.
Convert the -connected source to a Y-connected source.
Vp
440
Van 
 - 30 
 - 30  254 - 30
3
3
Z
(4  j6)(4  j5)
Z  Z Y ||
 (4  j6) || (4  j5) 
8 j
3
Z  5.723  j0.2153
ZL
V an
Ia
+

Z
Van
254  30

 32.88–28.4˚ A
Z L  Z 7.723  j 0.2153
I b  I a  - 120  32.88–148.4˚ A
Ia 
I c  I a 120  32.8891.6˚ A
Chapter 12, Solution 23.
(a)
I AB 
I a  I AB
V AB
202

Z
2560 o
202 3  30 o
3  30 
 13.995  90 o
o
25 60
o
I L | I a | 13.995 A
(b)
 202 3 
 cos 60 o
P  P1  P2  3V L I L cos   3 ( 202)

25


= 2.448 kW
Chapter 12, Solution 24.
Convert both the source and the load to their wye equivalents.
Z
 20 30  17.32  j10
ZY 
3
Vab
Van 
 - 30  240.20
3
We now use per-phase analysis.
1+j
V an
Ia 
+

Ia
2030 
Van
240.2

 11.24 - 31 A
(1  j)  (17.32  j10) 21.37 31
I b  I a  - 120  11.24 - 151 A
I c  I a 120  11.2489 A
But
I AB 
I a  I AB 3  - 30
11.24  - 31
3  - 30
 6.489 - 1 A
I BC  I AB  - 120  6.489 - 121 A
I CA  I AB 120  6.489119 A
Chapter 12, Solution 25.
Convert the delta-connected source to an equivalent wye-connected source and consider
the single-phase equivalent.
Ia 
where
440 (10  30)
3 ZY
ZY  3  j 2  10  j8  13  j 6  14.318  24.78
Ia 
440  20
\= 17.742 4.78&deg; amps.
3 (14.318  24.78)
I b = I a –120&deg; = 17.742 –115.22&deg; amps.
I c = I a +120&deg; = 17.742 124.78&deg; amps.
Chapter 12, Solution 26.
Using Fig. 12.55, design a problem to help other students to better understand balanced delta
connected sources delivering power to balanced wye connected loads.
Although there are many ways to solve this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
For the balanced circuit in Fig. 12.55, V ab = 1250 V. Find the line currents I aA , I bB ,
and I cC .
Figure 12.55
Solution
Transform the source to its wye equivalent.
Vp
Van 
 - 30  72.17  - 30
3
Now, use the per-phase equivalent circuit.
Van
I aA 
,
Z  24  j15  28.3 - 32
Z
I aA 
72.17  - 30
 2.55 2 A
28.3 - 32
I bB  I aA  - 120  2.55 - 118 A
I cC  I aA 120  2.55122 A
Chapter 12, Solution 27.
Since Z L and Z  are in series, we can lump them together so that
ZY  2  j  6  j 4  8  j 5
VP
 30o
208  30o

Ia  3
ZY
3(8  j 5)
208(0.866  j 0.5)(6  j 4)
VL  (6  j 4) I a 
 80.81  j 43.54
3(8  j 5)
|V L | = 91.79 V
Chapter 12, Solution 28.
VL  Vab  440  3VP or V P = 440/1.7321 = 254
For reference, let V AN = 2540˚ V which leads to
V BN = 254–120˚ V and V CN = 254120˚ V.
The line currents are found as follows,
I a = V AN /Z Y = 254/2530˚ = 10.16–30˚ A.
This leads to, I b = 10.16–150˚ A and I c = 10.1690˚ A.
Chapter 12, Solution 29.
We can replace the delta load with a wye load, Z Y = Z Δ /3 = 17+j15Ω.
The per-phase equivalent circuit is shown below.
Zl
V an
+
_
ZY
I a = V an /|Z Y + Z l | = 240/|17+j15+0.4+j1.2| = 240/|17.4+j16.2| = 240/23.77 =
10.095
S = 3[(I a )2(17+j15)] = 3x101.91(17+j15)
= [5.197+j4.586] kVA.
Chapter 12, Solution 30.
Since this a balanced system, we can replace it by a per-phase equivalent, as
shown below.
+
ZL
Vp
3V 2 p
S  3S p  * ,
Z p
Vp 
VL
3
V 2L
(208) 2
S * 
 1.442145 o kVA
o
Z p 30  45
P  S cos   1.02 kW
Chapter 12, Solution 31.
(a)
Pp  6,000,
cos  0.8,
Q p  S P sin   4.5 kVAR
Sp 
PP
 6 / 0.8  7.5 kVA
cos 
S  3S p  3(6  j 4.5)  18  j13.5 kVA
For delta-connected load, V p = V L = 240 (rms). But
S
(b)
3V 2 p
Z*p


Z*p 
Pp  3V L I L cos 
3V 2 p
3( 240) 2

,
S
(18  j13.5) x10 3


IL 
6000
3 x 240 x 0.8
Z P  [6.144  j 4.608] 
 18.04 A
(c ) We find C to bring the power factor to unity
Qc  Q p  4.5 kVA


C
Qc
4500

 207.2 F
2
V rms 2x 60 x 240 2
Chapter 12, Solution 32.
Design a problem to help other students to better understand power in a balanced threephase system.
Although there are many ways to solve this problem, this is an example based on the
same kind of problem asked in the third edition.
Problem
A balanced wye load is connected to a 60-Hz three-phase source with V ab = 2400˚V.
The load has lagging pf =0.5 and each phase draws 5 kW. (a) Determine the load
impedance Z Y . (b) Find I a , I b , and I c .
Solution
(a) | Vab | 3V p  240

 Vp 
240
 138.56
3
Van  V p  30o
pf  0.5  cos 

   60o
P
5

 S

 10 kVA
P  S cos 
cos  0.5
Q  S sin   10sin 60  8.66
S p  5  j8.66 kVA
But
SP 
V p2
Z *p
V p2
138.562

 Z 

 0.96  j1.663
S p (5  j8.66) x103
*
p
Z p = [0.96 + j1.663] 
(b)
Van 138.56  30o
Ia 

 72.17  90o A = 72.17 –90&deg; A
ZY 0.96  j1.6627
I b  I a  120o  72.17  210o A = 72.17 150&deg; A
I c  I a  120o  72.17  30o A = 72.17 30&deg; A
Chapter 12, Solution 33.
S  3 VL I L 
S  S  3 VL I L
IL  Ip ,
VL  3 Vp
S  3 Vp I p
IL  Ip 
S
4800

 7.69 A
3 Vp (3)(208)
VL  3 Vp  3  208  360.3 V
Chapter 12, Solution 34.
Vp 
Ia 
VL
3
Vp
ZY


220
3
220
3 (10  j16)

127.02
 6.73258
18.868  58
I L  I p  6.732A
S  3 VL I L   3  220  6.732 - 58  2565  58
S = [1.3592–j2.175] kVA
Chapter 12, Solution 35.
(a) This is a balanced three-phase system and we can use per phase equivalent circuit.
The delta-connected load is converted to its wye-connected equivalent
Z '' y 
1
Z   (60  j 30) / 3  20  j10
3
IL
+
230 V
-
Z y  Z ' y // Z '' y  (40  j10) //( 20  j10)  13.5  j 5.5
IL 
230
 [14.61  j5.953] A
13.5  j5.5
(b) S  3Vs I * L  [10.081  j4.108] kVA
(c ) pf = P/S = 0.9261
Z’y
Z’’y
Chapter 12, Solution 36.
(a)
(b) S  3V p I * p
S = 1 [0.75 + sin(cos-10.75) ] = 0.75 + j0.6614 MVA


I*p 
S
(0.75  j 0.6614) x10 6

 59.52  j 52.49
3V p
3x 4200
PL | I p | 2 Rl  (79.36) 2 (4)  25.19 kW
(c) V s  V L  I p (4  j )  4.4381  j 0.21 kV  4.443 - 2.709 o kV
Chapter 12, Solution 37.
S
P
12

 20
pf 0.6
S  S  20  12  j16 kVA
But
S  3 VL I L 
IL 
S  3 Ip
20  10 3
3  208
 55.51 A
2
Zp
For a Y-connected load, I L  I p .
Zp 
S
3 IL
2

(12  j16)  10 3
(3)(55.51) 2
Z p  [1.298  j1.731] 
Chapter 12, Solution 38.
As a balanced three-phase system, we can use the per-phase equivalent shown below.
Ia 
110 0
110 0

(1  j2)  (9  j12) 10  j14
S p  Ia
2
ZY 
(110) 2
 (9  j12)
(10 2  14 2 )
The complex power is
(110) 2
S  3S p  3
 (9  j12)
296
S  (1.1037+j1.4716) kVA
Chapter 12, Solution 39.
Consider the system shown below.
5
a
A
4
-j6 
100120
c
+

+

 +
100-120
1000
I1
5
b
8
B
I2
j3 
I3
C
10 
5
For mesh 1,
100  (18  j6) I 1  5 I 2  (8  j6) I 3
(1)
100  - 120  20 I 2  5 I 1  10 I 3
20 - 120  - I 1  4 I 2  2 I 3
(2)
0  - (8  j6) I 1  10 I 2  (22  j3) I 3
(3)
For mesh 2,
For mesh 3,
To eliminate I 2 , start by multiplying (1) by 2,
200  (36  j12) I 1  10 I 2  (16  j12) I 3
(4)
Subtracting (3) from (4),
200  (44  j18) I 1  (38  j15) I 3
(5)
Multiplying (2) by 5 4 ,
25 - 120  -1.25 I 1  5 I 2  2.5 I 3
(6)
87.5  j21.65  (16.75  j6) I 1  (10.5  j6) I 3
(7)
In matrix form, (5) and (7) become

  44  j18 - 38  j15  I 1 
200
87.5  j12.65  16.75  j6 - 10.5  j6  I 

 
 3 
  192.5  j26.25 ,
 1  900.25  j935.2 ,
 3  110.3  j1327.6
I1 
 1 1298.1 - 46.09

 6.682  - 38.33  5.242  j4.144

194.28 - 7.76
I3 
 3 1332.2 - 85.25

 6.857 - 77.49  1.485  j6.694

194.28 - 7.76
We obtain I 2 from (6),
1
1
I 2  5 - 120  I 1  I 3
4
2
I 2  (-2.5  j4.33)  (1.3104  j1.0359)  (0.7425  j3.347)
I 2  -0.4471  j8.713
The average power absorbed by the 8- resistor is
2
2
P1  I 1  I 3 (8)  3.756  j2.551 (8)  164.89 W
The average power absorbed by the 4- resistor is
2
P2  I 3 (4)  (6.8571) 2 (4)  188.1 W
The average power absorbed by the 10- resistor is
2
2
P3  I 2  I 3 (10)  - 1.9321  j2.019 (10)  78.12 W
Thus, the total real power absorbed by the load is
P  P1  P2  P3  431.1 W
Chapter 12, Solution 40.
Transform the delta-connected load to its wye equivalent.
Z
 7  j8
ZY 
3
Using the per-phase equivalent circuit above,
100 0
Ia 
 8.567  - 46.75
(1  j0.5)  (7  j8)
I p  I a  I a  8.567
S  3 Ip
2
Z p  (3)(8.567) 2 (7  j8)
P  Re(S)  (3)(8.567) 2 (7)  1.541 kW
Chapter 12, Solution 41.
S
But
P 5 kW

 6.25 kVA
pf
0.8
S  3 VL I L
IL 
S
3 VL

6.25  10 3
3  400
 9.021 A
Chapter 12, Solution 42.
The load determines the power factor.
40
tan  
 1.333 
   53.13
30
pf  cos   0.6 (leading)
 7.2 
S  7.2  j (0.8)  7.2  j9.6 kVA
 0.6 
But
S  3 Ip
2
Ip

2
Zp
S
(7.2  j9.6)  10 3

 80
3Zp
(3)(30  j40)
I p  8.944 A
I L  I p  8.944 A
VL 
S
3 IL

12  10 3
3 (8.944)
 774.6 V
Chapter 12, Solution 43.
S  3 Ip
2
Zp ,
I p  I L for Y-connected loads
S  (3)(13.66) 2 (7.812  j2.047)
S  [4.373  j1.145] kVA
Chapter 12, Solution 44.
Vp  VL ,
IL  3 Ip
S  3 VL I L
IL 
S
3 VL

(12 2  5 2 )  10 3
3 (240)
 31.273
At the source,
VL'  VL  I L Z l + I L Z l
VL'  2400  2(31.273)(1  j 3) = 240+62.546+j187.638
VL'  302.546+j187.638 = 356 31.81&deg;
VL'  356 V
Also, at the source,
S’ = 3(31.273)2(1+j3) + (12,000+j5,000) = 2,934+12,000+j(8,802+5,000)
= 14,934+j13,802 = 20,335 42.744&deg; thus, θ = 42.744&deg;.
pf = cos(42.744&deg;) = 0.7344
Checking, V Y = 240/1.73205 = 138.564, S = 3(138.564)2/(Z Y )* = 12,000+l5,000, and Z Y
= 57,600/(12,000–j5,000) = 57.6/(13 –22.62&deg;) = 4.4308 22.62&deg; = 4.09+j1.70416. The
total load seen by the source is 1+j3+4.09+j1.70416 = 5.09+j4.70416 = 6.9309 42.74&deg;
per phase. This leads to θ = Tan-1(4.70416/5.09) = Tan-1(0.9242) = 42.744&deg;. Clearly, the
answer checks. I l = 138.564/4.4308 = 31.273 A. Again the answer checks. Finally,
3(31.273)2(5.09+j4.70416) = 2,934(6.9309 42.74&deg;) = 20,335 42.74&deg;, the same as we
calculated above.
Chapter 12, Solution 45.
S  3 VL I L 
IL 
IL 
S -
3 VL
,
(635.6)  - 
3  440
P 450  10 3
S 

 635.6 kVA
pf
0.708
 834  - 45 A
At the source,
VL  440 0  I L (0.5  j2)
VL  440  (834  - 45)(2.062 76)
VL  440  1719.7 31
VL  1914.1  j885.7
VL  2.109 24.83 V
Chapter 12, Solution 46.
IL  Ip ,
VL  3 Vp
I p  Vp Z
2
S  3V I 
*
p p
S
VL
2
Z*
3 Vp

Z*
3 VL
(110) 2

 121 W
100
2
S  3V I 
S
2
Z*
Vp  VL ,
IL  3 Ip ,
*
p p
3
3 Vp
Z*

3 VL
I p  Vp Z
2
Z*
(3)(110) 2
 363 W
100
This shows that the delta-connected load will absorb three times more average
power than the wye-connected load using the same elements.. This is also evident
Z
from Z Y   .
3
Chapter 12, Solution 47.
pf  0.8 (lagging) 
   cos -1 (0.8)  36.87
S1  250 36.87  200  j150 kVA
   cos -1 (0.95)  -18.19
S 2  300  - 18.19  285  j93.65 kVA
pf  1.0 
   cos -1 (1)  0
S 3  450 kVA
S T  S1  S 2  S 3  935  j56.35  936.7 3.45 kVA
S T  3 VL I L
IL 
936.7  10 3
3 (13.8  10 3 )
 39.19 A rms
pf  cos   cos(3.45)  0.9982 (lagging)
Chapter 12, Solution 48.
(a) We first convert the delta load to its equivalent wye load, as shown below.
A
A
18-j12 
ZA
40+j15 
ZB
ZC
C
B
C
B
60 
(40  j15)(18  j12)
 7.577  j1.923
118  j 3
60(40  j15).
ZB 
 20.52  j7.105
118  j3
60(18  j12)
ZC 
 8.992  j 6.3303
118  j 3
The system becomes that shown below.
ZA 
`
a
2+j3
A
+
240&lt;0o
-
ZA
I1
240&lt;120o
+
c
-
I2
240&lt;-120o
+
b
ZB
ZC
2+j3
B
2+j3
We apply KVL to the loops. For mesh 1,
C
 240  240  120 o  I 1 (2Z l  Z A  Z B )  I 2 ( Z B  Z l )  0
or
(32.097  j11.13) I 1  (22.52  j10.105) I 2  360  j 207.85
For mesh 2,
240120 o  240  120 o  I 1 ( Z B  Z l )  I 2 (2Z l  Z B  Z C )  0
or
(1)
 (22.52  j10.105) I 1  (33.51  j 6.775) I 2   j 415.69
Solving (1) and (2) gives
I 1  23.75  j 5.328,
I 2  15.165  j11.89
(2)
I aA  I 1  24.34  12.64o A,
I bB  I 2  I 1  10.81  142.6o A
I cC   I 2  19.27141.9o A
(b) S a  ( 2400o )( 24.3412.64o )  5841.612.64o
S b  ( 240  120o )(10.81142.6o )  2594.4 22.6o
S c  ( 240120o )(19.27  141.9o )  4624.8  21.9o
S  S a  S b  S c  12.386  j0.55 kVA  12.4 2.54o kVA
Chapter 12, Solution 49.
(a) For the delta-connected load, Z p  20  j10,
S
V p  VL  220 (rms) ,
3V 2 p
3 x 220 2

 5808  j 2904  6.94326.56 o kVA
*
(20  j10)
Z p
P = 5.808 kW
(b) For the wye-connected load, Z p  20  j10,
S
3V 2 p
3 x 220 2

 2.16426.56 o kVA
3(20  j10)
Z*p
P = 1.9356 kW
V p  VL / 3 ,
Chapter 12, Solution 50.
S  S 1  S 2  8(0.6  j 0.8)  4.8  j 6.4 kVA,
Hence,
S 1  3 kVA
S 2  S  S 1  1.8  j 6.4 kVA
But S 2 
3V 2 p
,
Z*p
Z
*
p

Vp 
V *L
S2
VL
3


240 2

(1.8  j 6.4) x10 3
S2 
.V 2 L
Z*p


Z p  2.346  j 8.34
Chapter 12, Solution 51.
This is an unbalanced system.
240  0o 240  0o
I AB 

 19.2-j14.4 A
Z1
8  j6
I BC 
240120
240120

 50.62147.65˚ = [–42.76+j27.09] A
Z2
4.7413  27.65
I CA 
240  120 240  120

 [–12–j20.78] A
Z3
10
At node A,
I aA  I AB  I CA  (19.2  j14.4)  (12  j 20.78)  31.2  j 6.38 A
I bBI b  I BC  I AB  (42.76  j 27.08)  (19.2  j14.4)  61.96  j 41.48 A
I cCI c  I CA  I
BC
 (12  j 20.78)  (42.76  j 27.08)  30.76  j 47.86 A
Chapter 12, Solution 52.
Since the neutral line is present, we can solve this problem on a per-phase basis.
Van 120 120

Ia 
 6 60
20 60
Z AN
Vbn 120 0
 4 0
Ib 

Z BN
30 0
Vcn 120  - 120
 3 - 150
Ic 

Z CN
40 30
Thus,
- In
- In
- In
- In
 Ia  Ib  Ic
 6 60  4 0  3 - 150
 (3  j5.196)  (4)  (-2.598  j1.5)
 4.405  j3.696  5.7540
I n  5.75220 A
Chapter 12, Solution 53.
Using Fig. 12.61, design a problem that will help other students to better understand
unbalanced three-phase systems.
Although there are many ways to solve this problem, this is an example based on the
same kind of problem asked in the third edition.
Problem
In the wye-wye system shown in Fig. 12.61, loads connected to the source are
unbalanced. (a) Calculate I a , I b , and I c . (b) Find the total power delivered to the load.
Take V P = 240 V rms.
Ia
+
_
V P 0˚
V P 120˚
+ –
100
V P –120˚
–+
80
Ib
60 
Ic
Figure 12.61
For Prob. 12.53.
Solution
Applying mesh analysis as shown below, we get.
Ia
+
_
V P 0˚
100
I1
V P 120˚
+ –
V P –120˚
– +
80
Ib
60 
I2
Ic
240–120˚ – 240 + 160I 1 – 60I 2 = 0 or 160I 1 – 60I 2 = 360+j207.84
(1)
240120˚ – 240–120˚ – 60I 1 + 140I 2 = 0 or – 60I 1 + 140I 2 = –j415.7
(2)
In matrix form, (1) and (2) become
160 60   I1  360  j 207.84 
 60 140   I     j 415.7 

 2 

Using MATLAB, we get,
&gt;&gt; Z=[160,-60;-60,140]
Z=
160 -60
-60 140
&gt;&gt; V=[(360+207.8i);-415.7i]
V=
1.0e+002 *
3.6000 + 2.0780i
0 - 4.1570i
&gt;&gt; I=inv(Z)*V
I=
2.6809 + 0.2207i
1.1489 - 2.8747i
I 1 = 2.681+j0.2207 and I 2 = 1.1489–j2.875
I a = I 1 = 2.694.71˚ A
I b = I 2 – I 1 = –1.5321–j3.096 = 3.454–116.33˚ A
I c = –I 2 = 3.096111.78˚ A
S a | I a |2 Z a  (2.69) 2 x100  723.61
Sb | I b |2 Z b  (3.454)2 x60  715.81
Sc | I c |2 Z c  (3.0957) 2 x80  766.67
S  S a  Sb  Sc  2.205 kVA
Chapter 12, Solution 54.
Consider the load as shown below.
Ia
A
N
Ib
B
Ic
Ia 
210  0o
 2.625 A
80
Ib 
2100
210

 1.9414–56.31˚ A
60  j90 108.1756.31
210  0o
 2.625  90o A
j80
*
S a  VI a  210 x 2.625  551.25
Ic 
Sb  VI b* 
| V |2
2102

 226.15  j 339.2
Z b*
60  j 90
| V |2 2102

 j 551.25
Z c*
 j80
S  S a  Sb  Sc  777.4  j890.45 VA
Sc 
C
1
Chapter 12, Solution 55.
The phase currents are:
I AB = 240/j25 = 9.6–90˚ A
I CA = 240120˚/40 = 6120˚ A
I BC = 240–120˚/3030˚ = 8–150˚ A
The complex power in each phase is:
S AB | I AB |2 Z AB  (9.6) 2 j 25  j 2304
S AC | I AC |2 Z AC  (6) 2 40  0o  1440
S BC | I BC |2 Z BC  (8) 2 30  30o  1662.77  j960
The total complex power is
S  S AB  S AC  S BC  3102.77  j 3264 VA
= [3.103+j3.264] kVA
Chapter 12, Solution 56.
Using Fig. 12.63, design a problem to help other students to better understand unbalanced threephase systems.
Although there are many ways to solve this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
Refer to the unbalanced circuit of Fig. 12.63. Calculate:
(a) the line currents
(b) the real power absorbed by the load
(c) the total complex power supplied by the source
Figure 12.63
Solution
(a)
Consider the circuit below.
a
A
4400 + 
b
440120
+

j10 
I1
B
 +
440-120
I2
c
For mesh 1,
440 - 120  440 0  j10 (I 1  I 3 )  0
I3
-j5 
20 
C
I1  I 3 
(440)(1.5  j0.866)
 76.21 - 60
j10
For mesh 2,
440120  440 - 120  20 (I 2  I 3 )  0
(440)( j1.732)
I3  I2 
 j38.1
20
For mesh 3,
j10 (I 3  I 1 )  20 (I 3  I 2 )  j5 I 3  0
Substituting (1) and (2) into the equation for mesh 3 gives,
(440)(-1.5  j0.866)
I3 
 152.4260
j5
From (1),
I 1  I 3  76.21 - 60  114.315  j66  13230
From (2),
I 2  I 3  j38.1  76.21  j93.9  120.9350.94
I a  I 1  13230 A
I b  I 2  I 1  -38.105  j27.9  47.23143.8 A
I c  - I 2  120.9230.9 A
(b)
2
S AB  I 1  I 3 ( j10)  j58.08 kVA
2
S BC  I 2  I 3 (20)  29.04 kVA
2
S CA  I 3 (-j5)  (152.42) 2 (-j5)  -j116.16 kVA
S  S AB  S BC  S CA  29.04  j58.08 kVA
Real power absorbed = 29.04 kW
(c)
Total complex supplied by the source is
S  29.04  j58.08 kVA
(1)
(2)
(3)
Chapter 12, Solution 57.
We apply mesh analysis to the circuit shown below.
Ia
+
Va
–
80  j 50
I1
–
20  j 30
–
Vc
+
60  j 40
Vb
+
Ib
I2
Ic
(100  j80) I 1  (20  j 30) I 2  Va  Vb  165  j 95.263
 (20  j 30) I 1  (80  j10) I 2  Vb  Vc   j190.53
Solving (1) and (2) gives I 1  1.8616  j 0.6084,
I 2  0.9088  j1.722 .
I a  I 1  1.9585  18.1o A,
(1)
(2)
I b  I 2  I 1  0.528  j1.1136  1.4656  130.55 o A
I c   I 2  1.947117.8 o A
Chapter 12, Solution 58.
The schematic is shown below. IPRINT is inserted in the neutral line to measure the
current through the line. In the AC Sweep box, we select Total Ptss = 1, Start Freq. =
0.1592, and End Freq. = 0.1592. After simulation, the output file includes
FREQ
IM(V_PRINT4)
IP(V_PRINT4)
1.592 E–01
2.156 E+01
–8.997 E+01
i.e.
I n = 21.56–89.97 A
ACMAG=440V
ACMAG=440V
ACMAG=440V
Chapter 12, Solution 59.
The schematic is shown below. In the AC Sweep box, we set Total Pts = 1, Start Freq
= 60, and End Freq = 60. After simulation, we obtain an output file which includes
i.e.
FREQ
VM(1)
VP(1)
6.000 E+01
2.206 E+02
–3.456 E+01
FREQ
VM(2)
VP(2)
6.000 E+01
2.141 E+02
–8.149 E+01
FREQ
VM(3)
VP(3)
6.000 E+01
4.991 E+01
–5.059 E+01
V AN = 220.6–34.56, V BN = 214.1–81.49, V CN = 49.91–50.59 V
Chapter 12, Solution 60.
The schematic is shown below. IPRINT is inserted to give I o . We select Total Pts = 1,
Start Freq = 0.1592, and End Freq = 0.1592 in the AC Sweep box. Upon simulation,
the output file includes
from which,
FREQ
IM(V_PRINT4)
IP(V_PRINT4)
1.592 E–01
1.953 E+01
–1.517 E+01
I o = 19.53–15.17 A
+
–
Chapter 12, Solution 61.
The schematic is shown below. Pseudocomponents IPRINT and PRINT are inserted to
measure I aA and V BN . In the AC Sweep box, we set Total Pts = 1, Start Freq = 0.1592,
and End Freq = 0.1592. Once the circuit is simulated, we get an output file which
includes
FREQ
VM(2)
VP(2)
1.592 E–01
2.308 E+02
–1.334 E+02
FREQ
IM(V_PRINT2)
IP(V_PRINT2)
1.592 E–01
1.115 E+01
3.699 E+01
from which
I aA = 11.1537 A, V BN = 230.8–133.4 V
Chapter 12, Solution 62.
Using Fig. 12.68, design a problem to help other students to better understand how to use PSpice
to analyze three-phase circuits.
Although there are many ways to solve this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
The circuit in Fig. 12.68 operates at 60 Hz. Use PSpice to find the source current I ab and
the line current I bB .
Figure 12.68
Solution
Because of the delta-connected source involved, we follow Example 12.12. In the AC Sweep
box, we type Total Pts = 1, Start Freq = 60, and End Freq = 60. After simulation, the output
file includes
FREQ
IM(V_PRINT2)
IP(V_PRINT2)
6.000 E+01
5.960 E+00
–9.141 E+01
FREQ
IM(V_PRINT1)
IP(V_PRINT1)
6.000 E+01
7.333 E+07
1.200 E+02
From which
I ab = 3.432-46.31 A, I bB = 10.39–78.4 A
Chapter 12, Solution 63.
Let   1 so that L  X/  20 H, and C 
1
 0.0333 F
X
The schematic is shown below..
.
When the file is saved and run, we obtain an output file which includes the following:
FREQ
IM(V_PRINT1)IP(V_PRINT1)
1.592E-01 1.867E+01 1.589E+02
FREQ
IM(V_PRINT2)IP(V_PRINT2)
1.592E-01 1.238E+01 1.441E+02
From the output file, the required currents are:
I aA  18.67158.9 o A, I AC  12.38144.1 o A
Chapter 12, Solution 64.
We follow Example 12.12. In the AC Sweep box we type Total Pts = 1, Start Freq =
0.1592, and End Freq = 0.1592. After simulation the output file includes
FREQ
IM(V_PRINT1)
IP(V_PRINT1)
1.592 E–01
4.710 E+00
7.138 E+01
FREQ
IM(V_PRINT2)
IP(V_PRINT2)
1.592 E–01
6.781 E+07
–1.426 E+02
FREQ
IM(V_PRINT3)
IP(V_PRINT3)
1.592 E–01
3.898 E+00
–5.076 E+00
FREQ
IM(V_PRINT4)
IP(V_PRINT4)
1.592 E–01
3.547 E+00
6.157 E+01
FREQ
IM(V_PRINT5)
IP(V_PRINT5)
1.592 E–01
1.357 E+00
9.781 E+01
FREQ
IM(V_PRINT6)
IP(V_PRINT6)
1.592 E–01
3.831 E+00
–1.649 E+02
from this we obtain
I aA = 4.7171.38 A, I bB = 6.781–142.6 A, I cC = 3.898–5.08 A
I AB = 3.54761.57 A, I AC = 1.35797.81 A, I BC = 3.831–164.9 A
Chapter 12, Solution 65.
Due to the delta-connected source, we follow Example 12.12. We type Total Pts = 1,
Start Freq = 0.1592, and End Freq = 0.1592. The schematic is shown below. After it
is saved and simulated, we obtain an output file which includes
Thus,
FREQ
IM(V_PRINT1)
IP(V_PRINT1)
1.592E-01
1.140E+01
8.664E+00
FREQ
IM(V_PRINT2)
IP(V_PRINT2)
1.592E-01
1.140E+01
-1.113E+02
FREQ
IM(V_PRINT3)
IP(V_PRINT3)
1.592E-01
1.140E+01
1.287E+02
I aA = 11.0212 A, I bB = 11.02–108 A, I cC = 11.02132 A
Since this is a balanced circuit, we can perform a quick check. The load resistance is
large compared to the line and source impedances so we will ignore them (although it
would not be difficult to include them).
Converting the sources to a Y configuration we get:
V an = 138.56 –20˚ Vrms
and
Z Y = 10 – j6.667 = 12.019–33.69˚
Now we can calculate,
I aA = (138.56 –20˚)/(12.019–33.69˚) = 11.52813.69˚
Clearly, we have a good approximation which is very close to what we really have.
Chapter 12, Solution 66.
VL
Vp 
(b)
Because the load is unbalanced, we have an unbalanced three-phase
system. Assuming an abc sequence,
120 0
I1 
 2.50 A
48
120 - 120
I2 
 3 - 120 A
40
120120
I3 
 2120 A
60
3

208
(a)
3
 120 V


3
3
- I N  I 1  I 2  I 3  2.5  (3)  - 0.5  j   (2)  - 0.5  j 
2 
2 


IN  j
3
 j0.866  0.86690 A
2
Hence,
I1  2.5 A ,
I2  3 A ,
(c)
P1  I12 R 1  (2.5) 2 (48)  300 W
P2  I 22 R 2  (3) 2 (40)  360 W
P3  I 32 R 3  (2) 2 (60)  240 W
(d)
PT  P1  P2  P3  900 W
I3  2 A ,
I N  0.866 A
Chapter 12, Solution 67.
(a)
The power to the motor is
PT  S cos   (260)(0.85)  221 kW
The motor power per phase is
1
Pp  PT  73.67 kW
3
Hence, the wattmeter readings are as follows:
Wa  73.67  24  97.67 kW
Wb  73.67  15  88.67 kW
Wc  73.67  9  82.67 kW
(b)
The motor load is balanced so that I N  0 .
24,000
 200 A
Ia 
120
15,000
Ib 
 125 A
120
9,000
Ic 
 75 A
120
If we let
I a  I a 0  2000 A
I b  125 - 120 A
I c  75120 A
Then,
- I N  Ia  Ib  Ic


3
3
- I N  200  (125) - 0.5  j   (75) - 0.5  j 
2 
2 


- I N  100  j43.3 A
I N  108.97 A
Chapter 12, Solution 68.
(a)
S  3 VL I L  3 (330)(8.4)  4801 VA
(b)
P  S cos  
 pf  cos  
pf 
4500
 0.9372
4801.24
(c)
I p  I L  8.4 A
(d)
Vp 
VL
3

330
3
 190.53 V
P
S
Chapter 12, Solution 69.
S 1  S1 cos  1  jS1 sin 1
pf  0.85  cos 1

 1  31.79o
S 1  13.6  j8.43 kVA
S 2  12 x0.6  j12 x0.8  7.2  j 9.6 kVA
S 3  8  j 0 kVA
Therefore,
S = S 1 + S 2 + S 3 = [28.8+j18.03] kVA
Although we can solve this using a delta load, it will be easier to assume our load is
wye connected. We also need the wye voltages and will assume that the phase angle
on V an = 208/1.73205 = 120.089 is –30 degrees.
Since S = 3VI* or I* = S/(3V) = (33,978 32.048&deg;)/[3(120.089) –30&deg;] =
94.31 62.05&deg; A.
I a = 94.31 –62.05&deg; A, I b = 94.31 177.95&deg; A, I c = 94.31 57.95&deg; A
I = 138.46 – j86.68 = 163.35-32˚ A.
Chapter 12, Solution 70.
PT  P1  P2  1200  400  800
Q T  P2  P1  -400  1200  -1600
tan  
Q T - 1600

 -2 
   -63.43
PT
800
pf  cos   0.4472 (leading)
Zp 
VL 240

 40
IL
6
Z p  40 - 63.43 
Chapter 12, Solution 71.
(a)
If Vab  2080 , Vbc  208 - 120 , Vca  208120 ,
Vab 2080
I AB 

 10.4 0
20
Z Ab
Vbc
208 - 120
I BC 

 14.708 - 75
Z BC 10 2  - 45
I CA 
Vca
208120
 16 97.38

Z CA 1322.62
I aA  I AB  I CA  10.40  16 97.38
I aA  10.4  2.055  j15.867
I aA  20.171 - 51.87
I cC  I CA  I BC  16 97.83  14.708 - 75
I cC  30.64 101.03
P1  Vab I aA cos( Vab  IaA )
P1  (208)(20.171) cos(0  51.87)  2.590 kW
P2  Vcb I cC cos( Vcb   IcC )
But
Vcb  -Vbc  20860
P2  (208)(30.64) cos(60  101.03)  4.808 kW
(b)
PT  P1  P2  7398.17 W
Q T  3 (P2  P1 )  3840.25 VAR
S T  PT  jQ T  7398.17  j3840.25 VA
S T  S T  8.335 kVA
Chapter 12, Solution 72.
From Problem 12.11,
VAB  220 130 V
and
I aA  30 180 A
P1  (220)(30) cos(130  180)  4.242 kW
VCB  -VBC  220190
I cC  30 - 60
P2  (220)(30) cos(190  60)  - 2.257 kW
Chapter 12, Solution 73.
Consider the circuit as shown below.
I1
Ia
240-60 V
+

Z
Z
240-120 V

+
Z
I2
Ib
Ic
Z  10  j30  31.6271.57
240 - 60
 7.59 - 131.57
31.6271.57
240  - 120
Ib 
 7.59 - 191.57
31.62 71.57
Ia 
I c Z  240 - 60  240  - 120  0
- 240
Ic 
 7.59108.43
31.6271.57
I 1  I a  I c  13.146 - 101.57
I 2  I b  I c  13.146138.43
P1  Re  V1 I 1*   Re  (240 - 60)(13.146 101.57)   2.360 kW
P2  Re  V2 I *2   Re  (240  - 120)(13.146 - 138.43)   - 632.8 W
Chapter 12, Solution 74.
Consider the circuit shown below.
Z = 60  j30 
2080 V
+

I1
Z
208-60 V

+
I2
Z
For mesh 1,
208  2 Z I 1  Z I 2
For mesh 2,
- 208 - 60  - Z I 1  2 Z I 2
In matrix form,

  2 Z - Z  I 1 
208
 - 208 - 60   - Z 2 Z  I 
 2 

 
  3Z 2 ,
 1  (208)(1.5  j0.866) Z ,
 2  (208)( j1.732) Z
I1 
 1 (208)(1.5  j0.866)
 1.78956.56


(3)(60  j30)
I2 
 2 (208)( j1.732)

 1.79116.56
(3)(60  j30)

P1  Re  V1 I 1*   Re  (208)(1.789 - 56.56)   208.98 W
P2  Re  V2 (- I 2 ) *   Re  (208 - 60))(1.7963.44)   371.65 W
Chapter 12, Solution 75.
(a)
I
V 12

 20 mA
R 600
(b)
I
V 120

 200 mA
R 600
Chapter 12, Solution 76.
If both appliances have the same power rating, P,
P
I
Vs
For the 120-V appliance,
For the 240-V appliance,
 P2 R

2
Power loss = I 2 R   120
2
P R
 240 2
Since
P
.
120
P
.
I2 
240
I1 
for the 120-V appliance
for the 240-V appliance
1
1
, the losses in the 120-V appliance are higher.
2 
120
240 2
Chapter 12, Solution 77.
Pg  PT  Pload  Pline ,
But
pf  0.85
PT  3600 cos   3600  pf  3060
Pg  3060  2500  (3)(80)  320 W
Chapter 12, Solution 78.
51
 0.85 
 1  31.79
60
Q1  S1 sin 1  (60)(0.5268)  31.61 kVAR
cos 1 
P2  P1  51 kW
cos  2  0.95 
  2  18.19
P2
S2 
 53.68 kVA
cos  2
Q 2  S 2 sin  2  16.759 kVAR
Q c  Q1  Q 2  3.61  16.759  14.851 kVAR
Qc
 4.95 kVAR
3
Q c1
4950
 67.82 F
C
2 
(2 )(60)(440) 2
V
Q c1 
Chapter 12, Solution 79.
Consider the per-phase equivalent circuit below.
Ia
a
A
+

V an
2
n
Ia 
Z Y = 12 + j5 
N
Van
2550

 17.15 - 19.65 A
Z Y  2 14  j5
Thus,
I b  I a  - 120  17.15 - 139.65 A
I c  I a 120  17.15 100.35 A
VAN  I a Z Y  (17.15 - 19.65)(1322.62)  223 2.97 V
Thus,
VBN  VAN  - 120  223 - 117.63 V
VCN  VAN 120  223122.97 V
Chapter 12, Solution 80.
S  S1  S 2  S 3  6[0.83  j sin(cos 1 0.83)]  S 2  8(0.7071  j 0.7071)
S  10.6368  j 2.31  S 2 kVA
(1)
But
S  3VL I L   3 (208)(84.6)(0.8  j 0.6) VA  24.383  j18.287 kVA
(2)
From (1) and (2),
S 2  13.746  j 20.6  24.7656.28 kVA
Thus, the unknown load is 24.76 kVA at 0.5551 pf lagging.
Chapter 12, Solution 81.
 1  -36.87
S1  150  - 36.87 kVA
pf  1.0 
  2  0
S 2  100 0 kVA
pf  0.6 (lagging) 
 3  53.13
S 3  20053.13 kVA
S 4  80  j95 kVA
S  S1  S 2  S 3  S 4
S  420  j165  451.221.45 kVA
S  3 VL I L
IL 
S
3 VL

451.2  10 3
3  480
 542.7 A
For the line,
S L  3 I 2L Z L  (3)(542.7) 2 (0.02  j0.05)
S L  17.67  j44.18 kVA
At the source,
S T  S  S L  437.7  j209.2
S T  485.125.55 kVA
VT 
ST
3 IL

485.1  10 3
3  542.7
 516 V
Chapter 12, Solution 82.
S 1  400(0.8  j 0.6)  320  j 240 kVA,
S2  3
V 2p
Z*p
For the delta-connected load, V L  V p
S 2  3x
(2400) 2
 1053.7  j842.93 kVA
10  j8
S  S 1  S 2  1.3737  j1.0829 MVA
Let I = I 1 + I 2 be the total line current. For I 1 ,
S1  3V p I *1 ,
I *1 
S1
Vp 
VL
3
(320  j 240) x10 3

,
3VL
3 (2400)
For I 2 , convert the load to wye.
I 1  76.98  j 57.735
2400
3  30 o  273.1  j 289.76
10  j8
I  I 1  I 2  350  j 347.5
I 2  I p 3  30 o 
V s  V L  Vline  2400  I ( 3  j 6)  5.185  j1.405 kV


| V s | 5.372 kV
Chapter 12, Solution 83.
S1  120 x746 x0.95(0.707  j 0.707)  60.135  j 60.135 kVA,
S 2  80 kVA
S  S1  S 2  140.135  j 60.135 kVA
But | S | 3V L I L


IL 
|S|
3V L

152.49 x10 3
3 x 480
 183.42 A
Chapter 12, Solution 84.
We first find the magnitude of the various currents.
For the motor,
S
IL 
3 VL

4000
440 3
 5.248 A
For the capacitor,
IC 
Q c 1800

 4.091 A
VL
440
For the lighting,
Vp 
440
I Li 
PLi 800

 3.15 A
Vp 254
3
 254 V
Consider the figure below.
Ia
a
IC
+
-jX C
V ab
b

Ib
I2
Ic
I3
c
I Li
In
n
If Van  Vp 0 ,
Vab  3 Vp 30
Vcn  Vp 120
IC 
I1
Vab
 4.091120
-j X C
R
I1 
Vab
 4.091(  30)
Z
where   cos -1 (0.72)  43.95
I 1  5.249 73.95
I 2  5.249  - 46.05
I 3  5.249193.95
I Li 
Vcn
 3.15120
R
Thus,
I a  I 1  I C  5.24973.95  4.091120
I a  8.60893.96 A
I b  I 2  I C  5.249 - 46.05  4.091120
I b  9.271 - 52.16 A
I c  I 3  I Li  5.249193.95  3.15120
I c  6.827 167.6 A
I n  - I Li  3.15 - 60 A
Chapter 12, Solution 85.
Let
ZY  R
Vp 
VL
3

240
3
 138.56 V
Vp2
27
 9 kW 
P  Vp I p 
2
R
Vp2
(138.56) 2
R

 2.133 
P
9000
Thus,
Z Y  2.133 
Chapter 12, Solution 86.
Consider the circuit shown below.
1
a
A
+

1200 V rms
24 – j2 
I1
1
n
N
I2
+

1200 V rms
15 + j4 
1
b
B
For the two meshes,
120  (26  j2) I 1  I 2
120  (17  j4) I 2  I 1
(1)
(2)
In matrix form,
120  26  j2
- 1  I 1 
120   - 1
17  j4 I 2 
  
  449  j70 ,
 1  (120)(18  j4) ,
 2  (120)(27  j2)
 1 120  18.44 12.53

 4.87 3.67

454.42 8.86
 2 120  27.07  - 4.24
I2 

 7.15 - 13.1
454.42 8.86

I1 
I aA  I 1  4.87  3.67 A
I bB  - I 2  7.15166.9 A
I nN  I 2  I 1 
 2  1

I nN 
(120)(9  j6)
 2.856 - 42.55 A
449  j70
Chapter 12, Solution 87.
 jL  j (2)(60)(5010 -3 )  j18.85
L  50 mH 
Consider the circuit below.
1
115 V
+

I1
20 
2
15 + j18.85 
115 V
+

I2
30 
1
Applying KVl to the three meshes, we obtain
23 I 1  2 I 2  20 I 3  115
- 2 I 1  33 I 2  30 I 3  115
- 20 I 1  30 I 2  (65  j18.85) I 3  0
In matrix form,
- 20
  I 1  115
 23 - 2
 I   115
 - 2 33
- 30
 2  

- 20 - 30 65  j18.85 I 3   0 
  12,775  j14,232 ,
 2  (115)(1825  j471.3) ,
(1)
(2)
(3)
 1  (115)(1975  j659.8)
 3  (115)(1450)
 1 115  208218.47

 12.52 - 29.62

1921448.09
 2 115  1884.9 14.48
I2 

 11.33 - 33.61
19124 48.09

   1 (115)(-150  j188.5)
I n  I 2  I1  2

 1.448 - 176.6 A

12,775  j14,231.75
I1 
S 1  V1 I *1  (115)(12.52 29.62)  [1.252  j 0.7116] kVA
S 2  V2 I *2  (115)(1.3333.61)  [1.085  j 0.7212] kVA
Chapter 13, Solution 1.
For coil 1, L 1 – M 12 + M 13 = 12 – 8 + 4 = 8
For coil 2, L 2 – M 21 – M 23 = 16 – 8 – 10 = – 2
For coil 3, L 3 + M 31 – M 32 = 20 + 4 – 10 = 14
L T = 8 – 2 + 14 = 20H
or
L T = L 1 + L 2 + L 3 – 2M 12 – 2M 23 + 2M 13
L T = 12 + 16 + 20 – 2x8 – 2x10 + 2x4 = 48 – 16 – 20 + 8
= 20H
Chapter 13, Solution 2.
Using Fig. 13.73, design a problem to help other students to better understand mutual inductance.
Although there are many ways to solve this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
Determine the inductance of the three series-connected inductors of Fig. 13.73.
Figure 13.73
Solution
L = L 1 + L 2 + L 3 + 2M 12 – 2M 23 –2M 31
= 10 + 12 +8 + 2x6 – 2x6 –2x4
= 22H
Chapter 13, Solution 3.
L 1 + L 2 + 2M = 500 mH
(1)
L 1 + L 2 – 2M = 300 mH
(2)
2L 1 + 2L 2 = 800 mH
But,
L 1 = 3L 2, , or 8L 2 + 400,
and L 2 = 100 mH
L 1 = 3L 2 = 300 mH
From (2),
150 + 50 – 2M = 150 leads to M = 50 mH
k = M/ L1 L2  50 / 100 x300 = 0.2887
300 mH, 100 mH, 50 mH, 0.2887
Chapter 13, Solution 4.
(a)
For the series connection shown in Figure (a), the current I enters each coil from
its dotted terminal. Therefore, the mutually induced voltages have the same sign as the
self-induced voltages. Thus,
L eq = L 1 + L 2 + 2M
M
Is
L1
I
I1
Vs
M
I2
+
–
L2
L2
L1
L eq
(a)
(b)
(b)
For the parallel coil, consider Figure (b).
Is = I1 + I2
and
Z eq = V s /I s
Applying KVL to each branch gives,
V s = jL 1 I 1 + jMI 2
(1)
V s = jMI 1 + j L 2 I 2
(2)
 Vs   jL1
 V    jM
 s 
or
jM   I1 
jL 2  I 2 
 = –2L 1 L 2 + 2M2,  1 = jV s (L 2 – M),  2 = jV s (L 1 – M)
I 1 =  1 /, and I 2 =  2 /
I s = I 1 + I 2 = ( 1 +  2 )/ = j(L 1 + L 2 – 2M)V s /( –2(L 1 L 2 – M2))
= (L 1 + L 2 – 2M)V s /( j(L 1 L 2 – M2))
Z eq = V s /I s = j(L 1 L 2 – M2)/(L 1 + L 2 – 2M) = jL eq
i.e.,
L eq = (L 1 L 2 – M2)/(L 1 + L 2 – 2M)
Chapter 13, Solution 5.
(a) If the coils are connected in series,
L  L1  L2  2 M  50  120  2(0.5) 50 x120  247.4 mH
(b) If they are connected in parallel,
L
50 x120  38.72 2
L1 L2  M 2

mH  48.62 mH
L1  L2  2M 50  120  2 x38.72
(a) 247.4 mH, (b) 48.62 mH
Chapter 13, Solution 6.
M  k L1 L2  0.6 40 x5  8.4853 mH
40mH
5mH


3
j L  j 2000 x 40 x10  j80
3
j L  j 2000 x5 x10  j10


3
8.4853mH


j M  j 2000 x8.4853 x10  j16.97
We analyze the circuit below.
16.77 
I1
+
I2

j80 
+
j10 
V1
V2

_
_
V1  j80 I1  j16.97 I 2
V2  16.97 I1  j10 I 2
(1)
(2)
But, V 1 = 200˚ V and I 2 = 4–90˚ A. Substituting these into (1) produces
I 1 = [(V 1 +j16.97I 2 )/j80] = [(20+j16.97(–j4))/j80] = 1.0986–90˚ A or
i 1 = 1.0986sin(ωt) A
From (2), V 2 = –16.97x(–j1.0986) + j10(–j4) = 40 + j18.643 = 44.1325˚ or
v 2 = 44.13cos(ωt+25˚) V.
Chapter 13, Solution 7.
We apply mesh analysis to the circuit as shown below.
j1 
2
1
–j1 

12
+
_
j6 
I1
+
j4 
I2
1

For mesh 1,
(2+j6)I 1 + jI 2 = 24
For mesh 2,
jI 1 + (2–j+j4)I 2 = jI 1 + (2+j3)I 2 = 0 or I 1 = (–3+j2)I 2
Substituting into the first equation results in I 2 = (–0.8762+j0.6328) A.
V o = I 2 x1 = 1.081144.16˚ V.
Vo
_
Chapter 13, Solution 8.
2H
1H




j L  j 4 x 2  j8
j L  j 4 x1  j 4
Consider the circuit below.
j4
4
2 0o
+
_
I1


j8
j4
+
I2
2  (4  j8) I1  j 4 I 2
0   j 4 I1  (2  j 4) I 2
2
v(t)
_
(1)
(2)
In matrix form, these equations become
 2   4  j8  j 4   I1 
0    j 4 2  j 4  I 
  
 2
I 2 = 0.2353 – j0.0588
V = 2I 2 = 0.4851 –14.04o
Thus,
v(t) = 485.1cos(4t–14.04&deg;) mV
Chapter 13, Solution 9.
Consider the circuit below.
2
830o
+
–
I1
2
j4
j4
I2
-j1
-j2V
+
–
For loop 1,
830 = (2 + j4)I 1 – jI 2
For loop 2,
(j4 + 2 – j)I 2 – jI 1 + (–j2) = 0
or
Substituting (2) into (1),
(1)
I 1 = (3 – j2)i 2 – 2
830 + (2 + j4)2 = (14 + j7)I 2
I 2 = (10.928 + j12)/(14 + j7) = 1.03721.12
V x = 2I 2 = 2.07421.12 V
(2)
Chapter 13, Solution 10.


2H
j L  j 2 x 2  j 4
0.5 H


j L  j 2 x0.5  j
1
1
1
F



j
jC j 2 x1/ 2
2
Consider the circuit below.
j


+
24  0
+
_
j4
I1
j4
I2
Vo
–j
_
24  j 4 I1  jI 2
0   jI1  ( j 4  j ) I 2
In matrix form,

 0   I1  3I 2
 24   j 4  j   I1 
 0    1 3   I 
  
 2
Solving this,
I 2   j 2.1818,
Vo   jI 2  2.1818
v o (t) = –2.1818cos2t V
(1)
(2)
Chapter 13, Solution 11.
3
800mH


j L  j 600 x800 x10  j 480
600mH


j L  j 600 x600 x10  j 360
3
3
j L  j 600 x1200 x10  j 720
j
1

= –j138.89
12F 
jC 600x12x10  6


1200mH
After transforming the current source to a voltage source, we get the circuit shown below.
200

j480
-j138.89
150
Ix
800  0
+
_
I1
j360
j720
I2
+
_
110  30

For mesh 1,
800  (200  j 480  j 720) I1  j 360 I 2  j 720 I 2 or
800  (200  j1200) I1  j 360 I 2
(1)
For mesh 2,
11030˚ + 150–j138.89+j720)I 2 + j360I 1 = 0 or
95.2628  j 55   j 360 I1  (150  j 581.1) I 2
(2)
In matrix form,
 j 360   I1 
800

  200  j1200

 95.2628  j 55   j 360
150  j 581.1  I 2 

 
Solving this using MATLAB leads to:
&gt;&gt; Z = [(200+1200i),-360i;-360i,(150+581.1i)]
Z=
1.0e+003 *
0.2000 + 1.2000i
0 - 0.3600i
0 - 0.3600i 0.1500 + 0.5811i
&gt;&gt; V = [800;(-95.26-55i)]
V=
1.0e+002 *
8.0000
-0.9526 - 0.5500i
&gt;&gt; I = inv(Z)*V
I=
0.1390 - 0.7242i
0.0609 - 0.2690i
I x = I 1 – I 2 = 0.0781 – j0.4552 = 0.4619–80.26˚.
Hence,
i x (t) = 461.9cos(600t–80.26˚) mA.
Chapter 13, Solution 12.
Let   1.
j8
j4

+
j12
1V
-
j16
I1
j20
I2

Applying KVL to the loops,
1  j16 I1  j8 I 2
0  j8 I1  j 36 I 2
(1)
(2)
Solving (1) and (2) gives I 1 = –j0.0703. Thus
Z
1
 jLeq
I1


Leq 
1
 14.225 H.
jI1
We can also use the equivalent T-section for the transform to find the equivalent
inductance.
Chapter 13, Solution 13.
4
4
I1
80  0
j5
–j Ω
j5
I2
+
_
j2I 2
+
–
–
+
j2I 1
j2 
–80 + (4+j5)I 1 + j2I 2 = 0 or (4+j5)I 1 + j2I 2 = 80
j2I 1 +(4+j6)I 2 = 0 or I 2 = [–j2/(7.2111 56.31&deg;)]I 1 = (0.27735 –146.31&deg;)J 1
[4+j5 + j2(–0.230769–j0.153846)]I 1 = [4+j5+0.307692–j0.461538]I 1 = 80
[4.307692+j4.538462]I 1 = 80 or I 1 = 80/(6.2573 46.494&deg;)
= 12.78507 –46.494&deg; A.
Z in = 80/I 1 = 6.2573 46.494&deg; Ω = (4.308 + j4.538) Ω
An alternate approach would be to use the equation,
Z in  4  j(5) 
4
4
 4  j5 
j5  4  j  j2
7.211156.31
= 4+j5+0.5547 –56.31&deg; = 4+0.30769+j(5–0.46154)
= [4.308+j4.538] Ω.
Chapter 13, Solution 14.
To obtain V Th , convert the current source to a voltage source as shown below.
j2
5
j6 
j8 
-j3 
2
a
j50 V
+
+
–
I
V Th
40 V
+
–
–
b
Note that the two coils are connected series aiding.
L = L 1 + L 2 – 2M
jL = j6 + j8 – j4 = j10
Thus,
–j50 + (5 + j10 – j3 + 2)I + 40 = 0
I = (– 40 + j50)/ (7 + j7)
But,
–j50 + (5 + j6)I – j2I + V Th = 0
V Th = j50 – (5 + j4)I = j50 – (5 + j4)(–40 + j50)/(7 + j7)
V Th = 26.7434.11 V
To obtain Z Th , we set all the sources to zero and insert a 1-A current source at the terminals
a–b as shown below.
j2
5
j6 
I1
a
j8 
+
Vo
1A
–
b
-j3 
2
I2
Clearly, we now have only a super mesh to analyze.
(5 + j6)I 1 – j2I 2 + (2 + j8 – j3)I 2 – j2I 1 = 0
(5 + j4)I 1 + (2 + j3)I 2 = 0
(1)
But,
I 2 – I 1 = 1 or I 2 = I 1 – 1
(2)
Substituting (2) into (1),
(5 + j4)I 1 +(2 + j3)(1 + I 1 ) = 0
I 1 = –(2 + j3)/(7 + j7)
Now,
((5 + j6)I 1 – j2I 1 + V o = 0
V o = –(5 + j4)I 1 = (5 + j4)(2 + j3)/(7 + j7) = (–2 + j23)/(7 + j7) = 2.33250
Z Th = V o /1 = 2.33250 Ω.
Chapter 13, Solution 15.
The first step is to replace the mutually coupled circuits with the equivalent circuits using
dependent sources. To obtain I N , short-circuit a–b as shown in Figure (a) and solve for I sc .
j20 Ω
20 
a
+ 
+
–
j5(I 1 –I 2 )
I1
I sc
j10 Ω
I2
10030o
+
–
j5I 2
b
(a)
Now all we need to do is to write our two mesh equations.
Loop 1.
–10060˚ + 20I 1 + j10(I 1 –I 2 ) + j5I 2 = 0 or (20+j10)I 1 – j5I 2 = 10060˚
or (4+j2)I 1 – jI 2 = 2030˚
Loop 2.
–j5I 2 + j10(I 2 –I 1 ) + j20I 2 + j5(I 1 –I 2 ) = 0 or –j5I 1 + j20I 2 = 0 or I 1 = 4I 2
Substituting back into the first equation, we get, (4+j2)4I 2 – jI 2 = 2030˚ or (16+j7)I 2 = 2030˚.
Now to solve for I 2 = I sc = I N = (2030˚)/(16+j7) = (2030˚)/(17.46423.63˚) = 1.14526.37˚ A.
j20 Ω
20 
+ 
+
–
j5(I 1 –I 2 )
I1
j10 Ω
I2
10030o
+
–
j5I 2
a
+
V oc
–
b
(b)
To solve for Z N = Z eq = V oc /I sc , all we need to do is to solve for V oc . In circuit (b) we note that I 2 = 0 and
we get the mesh equation, –10030˚ + (20+j10)I 1 = 0 or I 1 = (10030˚)/(22.3626.57˚) = 4.4723.43˚ A.
V oc = j10I 1 – j5I 1 (induced voltage due to the mutual coupling) = j5I 1 = 22.3693.43˚ V.
Z eq = Z N = (22.3693.43˚)/(1.14526.37˚) = 19.52587.06˚ Ω.
or [1.0014+j19.498] Ω.
Chapter 13, Solution 16.
To find I N , we short-circuit a-b.
8
j
-j2 
a
 
j4 
+
800 V
-
j6 
I2
IN
I1
o
b
 80  (8  j2  j4)I1  jI 2  0


j6I 2  jI1  0

 I1  6I 2
(8  j2)I1  jI 2  80
(1)
(2)
Solving (1) and (2) leads to
I N  I2 
80
 1.584  j0.362  1.6246 –12.91&deg; A
48  j11
To find Z N , insert a 1-A current source at terminals a-b. Transforming the current source to
voltage source gives the circuit below.
j
8
-j2 
2
a
 
j4 
+
j6 
I1
I2
2V
-
b
0  (8  j2)I1  jI 2


I1 
jI 2
8  j2
= [j/(8.24621 14.036&deg;)]I 2 = 0.121268 75.964&deg;I 2
= (0.0294113+j0.117647)I 2
(3)
2  (2  j 6) I 2  jI 1  0
(4)
Solving (3) and (4) leads to (2+j6)I 2 – j(0.0294113+j0.117647)I 2 = –2 or
(2.117647+j5.882353)I 2 = –2 or I 2 = –2/(6.25192 70.201&deg;) = 0.319902 109.8&deg;.
V ab = 2(1+ I 2 ) = 2(1–0.1083629+j0.30099) = (1.78327+j0.601979) V = 1Z eq or
Z eq = (1.78327+j0.601979) = 1.8821 18.65&deg; Ω
An alternate approach would be to calculate the open circuit voltage.
–80 + (8+j2)I 1 – jI 2 = 0 or (8+j2)I 1 – jI 2 = 80
(2+j6)I 2 – jI 1 = 0 or I 1 = (2+j6)I 2 /j = (6–j2)I 2
(5)
(6)
Substituting (6) into (5) we get,
(8.24621 14.036&deg;)(6.32456 –18.435&deg;)I 2 – jI 2 = 80 or
[(52.1536 –4.399&deg;)–j]I 2 = [52–j5]I 2 = (52.2398 –5.492&deg;)I 2 = 80 or
I 2 = 1.5314 5.492&deg; A and V oc = 2I 2 = 3.0628 5.492&deg; V which leads to,
Z eq = V oc /I sc = (3.0628 5.492&deg;)/(1.6246 –12.91&deg;) = 1.8853 18.4&deg; Ω
This is in good agreement with what we determined before.
Chapter 13, Solution 17.
j L  j 40
40
40

 2666.67
L 15 x103

 
M  k L1 L2  0.6 12 x103 x30 x103  62.35
mHmH
11.384
If
Then
15 mH
40 Ω
12 mH
30 mH
11.384 mH
32 Ω
80 Ω
30.36 Ω
The circuit becomes that shown below.
j30.36 
22 
60 

j32 
j80 
Z L =j40

Z in  22  j 32 
 2M 2
 22  j 32 
(30.36) 2
60  j120
j80  60  j 40
921.7
 22  j 32 
 22  j 32  6.87  63.43  22  j 32  3.073  j 6.144
134.1663.43
= [25.07 + j25.86] Ω.
Chapter 13, Solution 18.
Replacing the mutually coupled circuit with the dependent source equivalent we get,
–j4 Ω
j2 
j5Ω
j5I 2
120V
j20Ω
–
+
+
–
j5I 1
+

I2
I1
4
j6 Ω
Now all we need to do is to find V oc and I sc . To calculate the open circuit voltage, we
note that I 2 is equal to zero. Thus,
–120 + (4 + j(–4+5+6))I 1 = 0 or I 1 = 120/(4+j7) = 120/(8.06226 60.255&deg;)
= 14.8842 –60.255&deg;.
V oc = V Thev = j5I 1 + (4+j6)I 1 = (4+j11)I 1
= (11.7047 70.017&deg;)(14.8842 –60.255&deg;) = 174.22 9.76&deg; V
To find the short circuit current (I sc = I 2 ), we need to solve the following mesh equations,
Mesh 1
–120 + (–j4+j5)I 1 – j5I 2 + (4+j6)(I 1 –I 2 ) = 0 or
(4+j7)I 1 – (4+j11)I 2 = 120
(1)
Mesh 2
(4+j6)(I 2 –I 1 ) – j5I 1 + j22I 2 = 0 or –(4+j11)I 1 + (4+j28)I 2 = 0 or
I 1 = (28.2843 81.87&deg;)I 2 /(11.7047 70.0169&deg;) = (2.4165 11.853&deg;)I 2
Substituting this into equation (1) we get,
(8.06226 60.255&deg;)(2.4165 11.853&deg;)I 2 – (4+j11)I 2 = 120 or
[(19.4825 72.108&deg;) – 4 –j11]I 2 = 120 and
[5.9855+j18.5403–4–j11]I 2 = (1.9855+j7.5403)I 2 = 120 or
I 2 = I sc = 120/(7.79733 75.248&deg;) = 15.3899 ─75.248&deg; A
Checking using MATLAB we get,
&gt;&gt; Z = [(4+7j) (-4-11j);(-4-11j) (4+28j)]
Z=
4.0000 + 7.0000i -4.0000 -11.0000i
-4.0000 -11.0000i 4.0000 +28.0000i
&gt;&gt; V = [120;0]
V=
120
0
&gt;&gt; I = inv(Z)*V
I=
16.6551 -33.2525i
(I 1 )
3.9188 -14.8829i
(I 2 = I sc ) = 15.3902 ─75.248&deg; (answer checks)
Finally,
Z eq = V Thev /I sc = (174.22 9.76&deg;)/(15.3899 ─75.248&deg;)
= (11.32 85.01&deg;) Ω
Chapter 13, Solution 19.
X La = X L1 – (–X M ) = 40 + 25 = 65 Ω and X Lb = X L2 – (–X M ) = 40 + 25 = 55 Ω.
Finally, X C = X M thus, the T-section is as shown below.
j65 
j55 
–j25 
Chapter 13, Solution 20.
Transform the current source to a voltage source as shown below.
k=0.5
4
j10
8
j10
I3
+
–
j12
-j5
I1
I2
200o
+
–
k = M/ L1 L 2 or M = k L1 L 2
M = k L1L 2 = 0.5(10) = 5
For mesh 1,
j12 = (4 + j10 – j5)I 1 + j5I 2 + j5I 2 = (4 + j5)I 1 + j10I 2
For mesh 2,
0 = 20 + (8 + j10 – j5)I 2 + j5I 1 + j5I 1
–20 = +j10I 1 + (8 + j5)I 2
From (1) and (2),
(1)
 j12 4  j5  j10   I1 
 20     j10 8  j5 I 
  
 2 
 = 107 + j60,  1 = –60 –j296,  2 = 40 – j100
I 1 =  1 / = 2.46272.18 A
I 2 =  2 / = 878–97.48 mA
I 3 = I 1 – I 2 = 3.32974.89 A
i 1 = 2.462 cos(1000t + 72.18) A
i 2 = 0.878 cos(1000t – 97.48) A
At t = 2 ms, 1000t = 2 rad = 114.6
i 1 (0.002) = 2.462cos(114.6 + 72.18) = –2.445A
(2)
–2.445
i 2 = 0.878cos(114.6 – 97.48) = –0.8391
The total energy stored in the coupled coils is
w = 0.5L 1 i 1 2 + 0.5L 2 i 2 2 + Mi 1 i 2
Since L 1 = 10 and  = 1000, L 1 = L 2 = 10 mH, M = 0.5L 1 = 5mH
w = 0.5(0.01)(–2.445)2 + 0.5(0.01)(–0.8391)2 + 0.05(–2.445)(–0.8391)
w = 43.67 mJ
Chapter 13, Solution 21.
Using Fig. 13.90, design a problem to help other students to better understand energy in a
coupled circuit.
Although there are many ways to solve this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
Find I 1 and I 2 in the circuit of Fig. 13.90. Calculate the power absorbed by the
4- resistor.
Figure 13.90
Solution
For mesh 1, 3630 = (7 + j6)I 1 – (2 + j)I 2
For mesh 2,
0 = (6 + j3 – j4)I 2 – 2I 1 – jI 1 = –(2 + j)I 1 + (6 – j)I 2
(1)
(2)
3630  7  j6  2  j  I1 
Placing (1) and (2) into matrix form, 

 
 0   2  j 6  j  I 2 
 = 45 + j25 = 51.4829.05,  1 = (6 – j)3630 = 21920.54
 2 = (2 + j)3630 = 80.556.57, I 1 =  1 / = 4.254–8.51 A , I 2 =  2 / = 1.563727.52 A
Power absorbed by the 4-ohm resistor,
= 0.5(I 2 )24 = 2(1.5637)2 = 4.89 watts
Chapter 13, Solution 22.
With more complex mutually coupled circuits, it may be easier to show the effects of the
coupling as sources in terms of currents that enter or leave the dot side of the coil. Figure
13.85 then becomes,
-j50
Io
I3
j20I c
j40
+ 
j10I b
j60
+ 
Ia
j30I c
 +

+
 +
Ix
j30I b
j20I a
500 V
+

j80
I1
I2
100 
Ib

+
j10I a
Note the following,
Ia = I1 – I3
Ib = I2 – I1
Ic = I3 – I2
and
Io = I3
Now all we need to do is to write the mesh equations and to solve for I o .
Loop # 1,
-50 + j20(I 3 – I 2 ) j 40(I 1 – I 3 ) + j10(I 2 – I 1 ) – j30(I 3 – I 2 ) + j80(I 1 – I 2 ) – j10(I 1 – I 3 )
=0
j100I 1 – j60I 2 – j40I 3 = 50
Multiplying everything by (1/j10) yields 10I 1 – 6I 2 – 4I 3 = - j5
(1)
Loop # 2,
j10(I 1 – I 3 ) + j80(I 2 –I 1 ) + j30(I 3 –I 2 ) – j30(I 2 – I 1 ) + j60(I 2 – I 3 ) – j20(I 1 – I 3 ) + 100I 2
=0
-j60I 1 + (100 + j80)I 2 – j20I 3 = 0
(2)
Loop # 3,
-j50I 3 +j20(I 1 –I 3 ) +j60(I 3 –I 2 ) +j30(I 2 –I 1 ) –j10(I 2 –I 1 ) +j40(I 3 –I 1 ) –j20(I 3 –I 2 ) = 0
-j40I 1 – j20I 2 + j10I 3 = 0
Multiplying by (1/j10) yields,
-4I 1 – 2I 2 + I 3 = 0
(3)
Multiplying (2) by (1/j20) yields -3I 1 + (4 – j5)I 2 – I 3 = 0
(4)
Multiplying (3) by (1/4) yields
(5)
-I 1 – 0.5I 2 – 0.25I 3 = 0
Multiplying (4) by (-1/3) yields I 1 – ((4/3) – j(5/3))I 2 + (1/3)I 3 = -j0.5
(7)
Multiplying [(6)+(5)] by 12 yields
(-22 + j20)I 2 + 7I 3 = 0
(8)
Multiplying [(5)+(7)] by 20 yields
-22I 2 – 3I 3 = -j10
(9)
(8) leads to I 2 = -7I 3 /(-22 + j20) = 0.235542.3o = (0.17418+j0.15849)I 3
(9) leads to I 3 = (j10 – 22I 2 )/3, substituting (1) into this equation produces,
I 3 = j3.333 + (-1.2273 – j1.1623)I 3
or
I 3 = I o = 1.304063o amp.
(10)
Chapter 13, Solution 23.
 = 10
0.5 H converts to jL 1 = j5 ohms
1 H converts to jL 2 = j10 ohms
0.2 H converts to jM = j2 ohms
25 mF converts to 1/(jC) = 1/(10x25x10-3) = –j4 ohms
The frequency-domain equivalent circuit is shown below.
j2I 2
j5
j10
+ 
120
For mesh 1,
For mesh 2,
From (1),
+

I1
j2I 1
+ 
–j4
I2
5
–12 + j5I 1 + j2I 2 + (–j4)(I 1 –I 2 ) = 0 or jI 1 + j6I 2 = 12 or
I 1 + 6I 2 = –j12
(1)
(–j4)(I 2 –I 1 ) + j10I 2 + j2I 1 + 5I 2 = 0 or
j6I 1 + (5+j6)I 2 = 0
(2)
I 1 = –j12 – 6I 2
Substituting this into (2) produces,
j6(–j12–6I 2 ) + (5+j6)I 2 = 0 = 72 + (5+j6–j36)I 2 or
(5–j30)I 2 = (30.414 –80.54&deg;)I 2 = –72 or I 2 = 2.367 –99.46&deg; A
I 1 = –j12 – 6( –0.38909–j2.3351) = 2.33454 + j(–12+14.0106)
= 2.33454 + j2.0106) = 3.081 40.74&deg; A
Checking using matrices,
I 1 = [(72–j60)/(5–j30)] = (93.723 –39.806&deg;)/(30.414 –80.538&deg;) = 3.082 40.73&deg; A and
I 2 = [–72/(30.414 –80.54)] = 2.367 –99.46&deg; A
Thus,
i 1 (t) = 3.081cos(10t + 40.74) A, i 2 (t) = 2.367cos(10t – 99.46) A.
At t = 15 ms,
10t = 10x15x10-3 = 0.15 rad = 8.59
i 1 = 3.081cos(49.33) = 2.00789 A
i 2 = 2.367cos(–90.87) = –0.03594 A
w = 0.5(5)(2.00789)2 + 0.5(1)(–0.03594)2 – (0.2)(2.00789)(–0.03594)
= 10.079056 + 0.0006458 + 0.0144327 = 10.094 J.
3.081cos(10t + 40.74) A, 2.367cos(10t – 99.46) A, 10.094 J.
Chapter 13, Solution 24.
(a)
k = M/ L1 L 2 = 1/ 4 x 2 = 0.3535
(b)
 = 4
1/4 F leads to 1/(jC) = –j/(4x0.25) = –j
1||(–j) = –j/(1 – j) = 0.5(1 – j)
1 H produces jM = j4
4 H produces j16
2 H becomes j8
j4
2
j8
120
+

I1
I2
0.5(1–j)
j16
12 = (2 + j16)I 1 + j4I 2
or
6 = (1 + j8)I 1 + j2I 2
0 = (j8 + 0.5 – j0.5)I 2 + j4I 1 or I 1 = (0.5 + j7.5)I 2 /(–j4)
(1)
(2)
Substituting (2) into (1),
24 = (–11.5 – j51.5)I 2 or I 2 = –24/(11.5 + j51.5) = –0.455–77.41
V o = I 2 (0.5)(1 – j) = 0.321757.59
v o = 321.7cos(4t + 57.6) mV
(c)
From (2),
I 1 = (0.5 + j7.5)I 2 /(–j4) = 0.855–81.21
i 1 = 0.885cos(4t – 81.21) A, i 2 = –0.455cos(4t – 77.41) A
At t = 2s,
4t = 8 rad = 98.37
i 1 = 0.885cos(98.37 – 81.21) = 0.8169
i 2 = –0.455cos(98.37 – 77.41) = –0.4249
w = 0.5L 1 i 1 2 + 0.5L 2 i 2 2 + Mi 1 i 2
= 0.5(4)(0.8169)2 + 0.5(2)(–.4249)2 + (1)(0.1869)(–0.4249) = 1.168 J
Chapter 13, Solution 25.
m = k L1 L 2 = 0.5 H
We transform the circuit to frequency domain as shown below.
12sin2t converts to 120,  = 2
0.5 F converts to 1/(jC) = –j
2 H becomes jL = j4
j1
Io 4 
1
a
3
–j1
120
+

j2
j2
j4
2
b
Applying the concept of reflected impedance,
Z ab = (2 – j)||(1 + j2 + (1)2/(j2 + 3 + j4))
= (2 – j)||(1 + j2 + (3/45) – j6/45)
= (2 – j)||(1 + j2 + (3/45) – j6/45)
= (2 – j)||(1.0667 + j1.8667)
=(2 – j)(1.0667 + j1.8667)/(3.0667 + j0.8667) = 1.508517.9 Ω
I o = 120/(Z ab + 4) = 12/(5.4355 + j0.4636) = 2.2–4.88
i o = 2.2sin(2t – 4.88) A
Chapter 13, Solution 26.
M = k L1L 2
M = k L1L 2 = 0.601 20x 40 = 17
The frequency-domain equivalent circuit is shown below.
j17
50 
20060
For mesh 1,
For mesh 2,
+

–j30
I1
Io
j20
j40
I2
10 
–200 60&deg; + (50–j30+j20)I 1 – j17I 2 = 0 or
(50–j10)I 1 – j17I 2 = 200 60&deg;
(1)
(10 + j40)I 2 – j17I 1 = 0 or –j17I 1 + (10+j40)I 2 = 0
(2)
In matrix form,
 j17   I1 
20060 50  j10



   or
0

   j17 10  j40 I 2 
j17 
10  j40

j17
50  j10
 I1 
20060


 I  500  400  289  j100  j2,000 

0


 2
I 1 = (10+j40)(200 60&deg;)/(1,189+j1,900)
= (41.231 75.964&deg;)(200 60&deg;)/(2,241.4 57.962&deg;) = 3.679 78&deg; A and
I 2 = j17(200 60&deg;)/(2,241.4 57.962&deg;) = 1.5169 92.04&deg; A
I o = I 2 = 1.5169 92.04&deg; A
It should be noted that switching the dot on the winding on the right only reverses the direction
of I o .
Chapter 13, Solution 27.


1H
j L  j 20
2H


j L  j 40
0.5 H


j L  j10
We apply mesh analysis to the circuit as shown below.
10 
j10 
8
I3

40  0
+
_
I1
j20 

j40 
50 
I2
To make the problem easier to solve, let us have I 3 flow around the outside loop as
shown.
For mesh 1,
–40 + 8I 1 + j20I 1 – j10I 2 = 0 or (8+j20)I 1 – j10I 2 = 40
(1)
For mesh 2,
j40I 2 – j10I 1 + 50(I 2 + I 3 ) = 0 or –j10I 1 + (50+j40)I 2 + 50I 3 = 0
(2)
For mesh 3,
–40 + 10I 3 + 50(I 3 + I 2 ) = 0 or 50I 2 + 60I 3 = 40
(3)
In matrix form, (1) to (3) become
 j10
0  40
8  j20
  j10 50  j40 50 I   0 

  
 0
50
60 40
&gt;&gt; Z=[(8+20i),-10i,0;-10i,(50+40i),50;0,50,60]
Z=
8.0000 +20.0000i
0 -10.0000i
0
0 -10.0000i 50.0000 +40.0000i 50.0000
0
50.0000
60.0000
&gt;&gt; V=[40;0;40]
V=
40
0
40
&gt;&gt; I=inv(Z)*V
I=
0.6354 - 1.5118i
0.0613 + 0.4682i
0.6156 - 0.3901i
Solving this leads to I 50 = I 2 + I 3 = 0.0613+0.6156 + j(0.4682–0.3901) =
0.6769+j0.0781 = 0.6814 6.58&deg; A or I 50rms |I 50rms | = 0.6814/1.4142 = 481.8 mA.
The power delivered to the 50- resistor is
P = (I 50rms )2R = (0.4818)250 = 11.608 W.
Chapter 13, Solution 28.
We find Z Th by replacing the 20-ohm load with a unit source as shown below.
j10 
8
-jX
 
j12 
j15 
+
1V
-
I2
I1
For mesh 1,
0  (8  jX  j12) I 1  j10 I 2
For mesh 2,
1  j15 I 2  j10 I 1  0


(1)
I 1  1.5 I 2  0.1 j
(2)
Substituting (2) into (1) leads to
 1.2  j 0.8  0.1X
I2 
12  j8  j1.5 X
Z Th 
| Z Th | 20 
1
12  j8  j1.5 X

 I 2 1.2  j 0.8  0.1X
12 2  (8  1.5 X ) 2
(1.2  0.1X )  0.8
2
2


Solving the quadratic equation yields X = 6.425 Ω
0  1.75 X 2  72 X  624
Chapter 13, Solution 29.
30 mH becomes jL = j30x10-3x103 = j30
50 mH becomes j50
Let X = M
Using the concept of reflected impedance,
Z in = 10 + j30 + X2/(20 + j50)
I 1 = V/Z in = 165/(10 + j30 + X2/(20 + j50))
p = 0.5|I 1 |2(10) = 320 leads to |I 1 |2 = 64 or |I 1 | = 8
8 = |165(20 + j50)/(X2 + (10 + j30)(20 + j50))|
= |165(20 + j50)/(X2 – 1300 + j1100)|
64 = 27225(400 + 2500)/((X2 – 1300)2 + 1,210,000)
or
(X2 – 1300)2 + 1,210,000 = 1,233,633
X = 33.86 or 38.13
If X = 38.127 = M
M = 38.127 mH
k = M/ L1 L 2 = 38.127/ 30x 50 = 0.984
j38.127
10 
1650
+

I1
j30
j50
I2
20 
165 = (10 + j30)I 1 – j38.127I 2
0 = (20 + j50)I 2 – j38.127I 1
In matrix form,
(1)
(2)
165  10  j30  j38.127  I1 
 0     j38.127 20  j50   I 
  
 2 
 = 154 + j1100 = 1110.7382.03,  1 = 888.568.2,  2 = j6291
I 1 =  1 / = 8–13.81, I 2 =  2 / = 5.6647.97
i 1 = 8cos(1000t – 13.83), i 2 = 5.664cos(1000t + 7.97)
At t = 1.5 ms, 1000t = 1.5 rad = 85.94
i 1 = 8cos(85.94 – 13.83) = 2.457
i 2 = 5.664cos(85.94 + 7.97) = –0.3862
w = 0.5L 1 i 1 2 + 0.5L 2 i 2 2 + Mi 1 i 2
= 0.5(30)(2.547)2 + 0.5(50)(–0.3862)2 – 38.127(2.547)(–0.3862)
= 130.51 mJ
Chapter 13, Solution 30.
Z in = j40 + 25 + j30 + (10)2/(8 + j20 – j6)
(a)
= 25 + j70 + 100/(8 + j14) = (28.08 + j64.62) ohms
jL a = j30 – j10 = j20, jL b = j20 – j10 = j10, jL c = j10
(b)
Thus the Thevenin Equivalent of the linear transformer is shown below.
j40
25 
j20
j10
j10
8
–j6
Z in
Z in = j40 + 25 + j20 + j10||(8 + j4) = 25 + j60 + j10(8 + j4)/(8 + j14)
= (28.08 + j64.62) ohms
Chapter 13, Solution 31.
Using Fig. 13.100, design a problem to help other students to better understand linear
transformers and how to find T-equivalent and –equivalent circuits.
Although there are many ways to solve this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
For the circuit in Fig. 13.99, find:
(a) the T-equivalent circuit,
(b) the -equivalent circuit.
Figure 13.99
Solution
(a)
L a = L 1 – M = 10 H
L b = L 2 – M = 15 H
Lc = M = 5 H
(b)
L 1 L 2 – M2 = 300 – 25 = 275
L A = (L 1 L 2 – M2)/(L 1 – M) = 275/15 = 18.33 H
L B = (L 1 L 2 – M2)/(L 1 – M) = 275/10 = 27.5 H
L C = (L 1 L 2 – M2)/M = 275/5 = 55 H
Chapter 13, Solution 32.
We first find Z in for the second stage using the concept of reflected impedance.
Lb
R
LB
Z in ’
Z in ’ = jL b + 2M b 2/(R + jL b ) = (jL b R - 2L b 2 + 2M b 2)/(R + jL b )
(1)
For the first stage, we have the circuit below.
La
LA
Z in ’
Z in
Z in = jL a + 2M a 2/(jL a + Z in )
= (–2L a 2 + 2M a 2 + jL a Z in )/( jL a + Z in )
(2)
Substituting (1) into (2) gives,
( jL b R   2 L2b   2 M 2b )
R  j L b
2 2
jL b R   L b   2 M 2b
jL a 
R  j L b
  2 L2a   2 M a2  jL a
=
–R2L a 2 + 2M a 2R – j3L b L a + j3L b M a 2 + jL a (jL b R – 2L b 2 + 2M b 2)
=
jRLa –2L a L b + jL b R – 2L a 2 + 2M b 2
2R(L a 2 + L a L b – M a 2) + j3(L a 2L b + L a L b 2 – L a M b 2 – L b M a 2)
Z in =
2(L a L b +L b 2 – M b 2) – jR(L a +L b )
Chapter 13, Solution 33.
Z in = 10 + j12 + (15)2/(20 + j 40 – j5) = 10 + j12 + 225/(20 + j35)
= 10 + j12 + 225(20 – j35)/(400 + 1225)
= (12.769 + j7.154) Ω
Chapter 13, Solution 34.
Using Fig. 13.103, design a problem to help other students to better understand how to find the
input impedance of circuits with transformers.
Although there are many ways to solve this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
Find the input impedance of the circuit in Fig. 13.102.
Figure 13.102
Solution
Insert a 1-V voltage source at the input as shown below.
j6 
1

+
j12 
o
1&lt;0 V
8

j10 
I1
j4 
I2
-j2 
For loop 1,
1  (1  j10) I 1  j 4 I 2
(1)
For loop 2,
0  (8  j 4  j10  j 2) I 2  j 2 I 1  j 6 I 1


0   jI 1  (2  j 3) I 2
(2)
Solving (1) and (2) leads to I 1 =0.019 –j0.1068
Z
1
 1.6154  j 9.077  9.219 79.91 o 
I1
Alternatively, an easier way to obtain Z is to replace the transformer with its equivalent T
circuit and use series/parallel impedance combinations. This leads to exactly the same result.
Chapter 13, Solution 35.
For mesh 1,
16  (10  j 4) I 1  j 2 I 2
(1)
For mesh 2,
0  j 2 I 1  (30  j 26) I 2  j12 I 3
(2)
For mesh 3,
0   j12 I 2  (5  j11) I 3
(3)
We may use MATLAB to solve (1) to (3) and obtain
I 1  1.3736  j0.5385  1.4754–21.41&deg; A
I 2  0.0547  j0.0549  77.5–134.85&deg; mA
I 3  0.0268  j0.0721  77–110.41&deg; mA
1.4754–21.41&deg; A, 77.5–134.85&deg; mA, 77–110.41&deg; mA
Chapter 13, Solution 36.
Following the two rules in section 13.5, we obtain the following:
(a)
V 2 /V 1 = –n,
I 2 /I 1 = –1/n
(b)
V 2 /V 1 = –n,
I 2 /I 1 = –1/n
(c)
V 2 /V 1 = n,
I 2 /I 1 = 1/n
(d)
V 2 /V 1 = n,
I 2 /I 1 = –1/n
(n = V 2 /V 1 )
Chapter 13, Solution 37.
(a) n 
V2 2400

5
V1
480
(b) S 1  I 1V1  S 2  I 2V 2  50,000
(c ) I 2 
50,000
 20.83 A
2400


I1 
50,000
 104.17 A
480
Chapter 13, Solution 38.
Design a problem to help other students to better understand ideal transformers.
Although there are many ways to solve this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
A 4-kVA, 2300/230-V rms transformer has an equivalent impedance of 210 on the
primary side. If the transformer is connected to a load with 0.6 power factor leading,
calculate the input impedance.
Solution
Z in = Z p + Z L /n2, n = v 2 /v 1 = 230/2300 = 0.1
v 2 = 230 V, s 2 = v 2 I 2 *
I 2 * = s 2 /v 2 = 17.391–53.13 or I 2 = 17.39153.13 A
Z L = v 2 /I 2 = 2300/17.39153.13 = 13.235–53.13
Z in = 210 + 1323.5–53.13
= 1.97 + j0.3473 + 794.1 – j1058.8
Z in = 1.324–53.05 kΩ
Chapter 13, Solution 39.
Referred to the high-voltage side,
Z L = (1200/240)2(0.810) = 2010
Z in = 60–30 + 2010 = 76.4122–20.31
I 1 = 1200/Z in = 1200/76.4122–20.31 = 15.720.31 A
Since S = I 1 v 1 = I 2 v 2 , I 2 = I 1 v 1 /v 2
= (1200/240)( 15.720.31) = 78.520.31 A
Chapter 13, Solution 40.
Consider the circuit as shown below.
R Th
I1
1:5
V Th
I2
+
_
200 
We reflect the 200- load to the primary side.
200
 108
52
I
I2  1  2
108
n
Z p  100 
I1 
10
,
108
P
1
1 2 2
| I 2 |2 RL  (
) (200)  34.3 mW
2
2 108
Chapter 13, Solution 41.
We reflect the 2-ohm resistor to the primary side.
Z in = 10 + 2/n2,
n = –1/3
Since both I 1 and I 2 enter the dotted terminals,
Z in = 10 + 18 = 28 ohms
I 1 = 140/28 = 500 mA and I 2 = I 1 /n = 0.5/(–1/3) = –1.5 A
Chapter 13, Solution 42.
We apply mesh analysis to the circuit as shown below.
50 
80
+
_
–j1 
I1
j20 
1:2
+
+
V1
V2
_
_
For mesh 1,
–80 + (50–j)I 1 + V 1 = 0
For mesh 2,
–V 2 + (2+j20)I 2 = 0
At the transformer terminals,
V 2 = 2V 1 or 2V 1 – V 2 = 0
I 1 = 2I 2 or I 1 – 2I 2 = 0
From (1) to (4),
Solving this with MATLAB,
&gt;&gt; A = [(50-j) 0 1 0;0 (2+20j) 0 -1;0 0 2 -1;1 -2 0 0]
A=
Columns 1 through 3
50.0000 - 1.0000i
0
1.0000
0
2.0000 +20.0000i
0
0
0
2.0000
1.0000
-2.0000
0
Column 4
0
-1.0000
-1.0000
I2
2
(1)
(2)
(3)
(4)
0
&gt;&gt; B = [80;0;0;0]
B=
80
0
0
0
&gt;&gt; C = inv(A)*B
C=
1.5743 - 0.1247i
0.7871 - 0.0623i
1.4106 + 7.8091i
2.8212 +15.6181i
(I 1 )
(I 2 )
(V 1 )
(V 2 )
I 2 = (787.1–j62.3) mA or 789.6 –4.53&deg; mA
The power absorbed by the 2- resistor is
P = |I 2 |2R = (0.7896)22 = 1.2469 W.
Chapter 13, Solution 43.
Transform the two current sources to voltage sources, as shown below.
10 
+
20 V
+
–
Using mesh analysis,
I1
12 
1:4
v1

+
v2

I2
12V
+
–
–20 + 10I 1 + v 1 = 0
20 = v 1 + 10I 1
12 + 12I 2 – v 2 = 0 or 12 = v 2 – 12I 2
At the transformer terminal, v 2 = nv 1 = 4v 1
I 1 = nI 2 = 4I 2
(1)
(2)
(3)
(4)
Substituting (3) and (4) into (1) and (2), we get,
Solving (5) and (6) gives
20 = v 1 + 40I 2
(5)
12 = 4v 1 – 12I 2
(6)
v 1 = 4.186 V and v 2 = 4v = 16.744 V
Chapter 13, Solution 44.
We can apply the superposition theorem. Let i 1 = i 1 ’ + i 1 ” and i 2 = i 2 ’ + i 2 ”
where the single prime is due to the DC source and the double prime is due to the
AC source. Since we are looking for the steady-state values of i 1 and i 2 ,
i 1 ’ = i 2 ’ = 0.
For the AC source, consider the circuit below.
R
1:n
+
i1”
v1

v 2 /v 1 = –n,
+
v2

i2
+
–
V n 0
I 2 ”/I 1 ” = –1/n
But v 2 = v m , v 1 = –v m /n or I 1 ” = v m /(Rn)
I 2 ” = –I 1 ”/n = –v m /(Rn2)
Hence,
i 1 (t) = (v m /Rn)cost A, and i 2 (t) = (–v m /(n2R))cost A
Chapter 13, Solution 45.
48 
4–90˚
ZL  8 
Z
+

Z
j
 8  j4 , n = 1/3
C
ZL
 9Z L  72  j36
n2
4  90
4  90
I

 0.03193  73.3
48  72  j36 125.28  16.7
We now have some choices, we can go ahead and calculate the current in the second loop and
calculate the power delivered to the 8-ohm resistor directly or we can merely say that the power
delivered to the equivalent resistor in the primary side must be the same as the power delivered
to the 8-ohm resistor. Therefore,
P8 
I2
72  0.5098x10  3 72  36.71 mW
2
The student is encouraged to calculate the current in the secondary and calculate the power
delivered to the 8-ohm resistor to verify that the above is correct.
Chapter 13, Solution 46.
(a)
Reflecting the secondary circuit to the primary, we have the circuit shown below.
Z in
1660
+

I1
+

1030/(–n) = –530
Z in = 10 + j16 + (1/4)(12 – j8) = 13 + j14
–1660 + Z in I 1 – 530 = 0 or I 1 = (1660 + 530)/(13 + j14)
Hence,
(b)
I 1 = 1.0725.88 A, and I 2 = –0.5I 1 = 0.536185.88 A
Switching a dot will not affect Z in but will affect I 1 and I 2 .
I 1 = (1660 – 530)/(13 + j14) = 0.625 25 A
and I 2 = 0.5I 1 = 0.312525 A
Chapter 13, Solution 47.

(1/3) F 1F 
1
1

  j1
jC j 3 x1/ 3
Consider the circuit shown below.
–j
1
2

+
4  0&ordm;
+
_
I1
I3
1:4

+
+
V1
V2
_
_
I2
5  v(t)
–
For mesh 1,
3I 1 – 2I 3 + V 1 = 4
For mesh 2,
5I 2 – V 2 = 0
For mesh 3,
–2I 1 (2–j)I 3 – V 1 + V 2 =0
At the terminals of the transformer,
V2  nV1  4V1
I 1 – I 3 = 4(I 2 – I 3 )
In matrix form,
(1)
(2)
(3)
(4)
(5)
2
0
1
0   I1  4
3
0
5
0
0  1  I 2  0

  2 0 2  j  1 1   I 3   0 

   
 4 1   V1  0
0
0
0
 1  4
3
0
0  V2  0
Solving this using MATLAB yields
&gt;&gt;A = [3,0,-2,1,0; 0,5,0,0,-1; -2,0,(2-j),-1,1 ;0,0,0,-4,1; 1,-4,3,0,0]
A=
3.0000
0
-2.0000
0
1.0000
0
5.0000
0
0
-4.0000
-2.0000
1.0000
0
0
0
-1.0000
2.0000 - 1.0000i -1.0000
1.0000
0
-4.0000
1.0000
3.0000
0
0
&gt;&gt;U = [4;0;0;0;0]
&gt;&gt;X = inv(A)*U
X=
1.2952 + 0.0196i
0.5287 + 0.0507i
0.2733 + 0.0611i
0.6609 + 0.0634i
2.6437 + 0.2537i
V = 5I 2 = V 2 = 2.6437+j0.2537 = 2.656 5.48&deg; V, therefore,
v(t) = 2.656cos(3t+5.48˚) V
Chapter 13, Solution 48.
Using Fig. 13.113, design a problem to help other students to better understand how ideal
transformers work.
Although there are many ways to solve this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
Find I x in the ideal transformer circuit of Fig. 13.112.
Figure 13.112
Solution
We apply mesh analysis.
8
10 
2:1
+
+

+

V1
V2
-
I1
1000o V
-
Ix
j6 
I2
-j4 
100  (8  j 4) I 1  j 4 I 2  V1
(1)
0  (10  j 2) I 2  j 4 I 1  V 2
(2)
But
V2
1
n
V1
2


V1  2V2
(3)
I2
1
   2
I1
n


I 1  0.5 I 2
(4)
Substituting (3) and (4) into (1) and (2), we obtain
100  (4  j 2) I 2  2V2
0  (10  j 4) I 2 V2
(1)a
(2)a
Solving (1)a and (2)a leads to I 2 = -3.5503 +j1.4793
I x  I 1  I 2  0.5 I 2  1.923157.4 o A
Chapter 13, Solution 49.
1
F
20
  2,


1
  j10
j C
Ix
2
-j10
I1
1:3
I2
1
+
2
+ 
V1
o
12&lt;0 V
-
+
V2
-
6

At node 1,
12  V1 V1  V2

 I1

 12  2 I 1  V1 (1  j 0.2)  j 0.2V2
 j10
2
At node 2,
V  V2 V2
I2  1


 0  6 I 2  j 0.6V1  (1  j 0.6)V2
 j10
6
1
At the terminals of the transformer, V2  3V1 ,
I 2   I1
3
Substituting these in (1) and (2),
12  6 I 2  V1 (1  j 0.8),
0  6 I 2  V1 (3  j 2.4)
Adding these gives V 1 =1.829 –j1.463 and
Ix 
V1  V2
4V1

 0.93751.34 o
 j10
 j10
i x (t) = 937cos(2t+51.34˚) mA.
(1)
(2)
Chapter 13, Solution 50.
The value of Z in is not effected by the location of the dots since n2 is involved.
Z in ’ = (6 – j10)/(n’)2, n’ = 1/4
Z in ’ = 16(6 – j10) = 96 – j160
Z in = 8 + j12 + (Z in ’ + 24)/n2,
n = 5
Z in = 8 + j12 + (120 – j160)/25 = 8 + j12 + 4.8 – j6.4
Z in = (12.8 + j5.6) Ω
Chapter 13, Solution 51.
Let Z 3 = 36 +j18, where Z 3 is reflected to the middle circuit.
Z R ’ = Z L /n2 = (12 + j2)/4 = 3 + j0.5
Z in = 5 – j2 + Z R ’ = [8 – j1.5] Ω
I 1 = 240/Z eq = 240/(8 – j1.5) = 240/8.14–10.62 = 8.9510.62 A
[8 – j1.5] Ω, 8.9510.62 A
Chapter 13, Solution 52.
For maximum power transfer,
40 = Z L /n2 = 10/n2 or n2 = 10/40 which yields n = 1/2 = 0.5
I = 120/(40 + 40) = 3/2
p = I2R = (9/4)x40 = 90 watts.
Chapter 13, Solution 53.
(a)
The Thevenin equivalent to the left of the transformer is shown below.
8
20 V
+

The reflected load impedance is Z L ’ = Z L /n2 = 200/n2.
For maximum power transfer,
(b)
8 = 200/n2 produces n = 5.
If n = 10, Z L ’ = 200/10 = 2 and I = 20/(8 + 2) = 2
p = I2Z L ’ = (2)2(2) = 8 watts.
Chapter 13, Solution 54.
(a)
Z Th
VS
I1
+

Z L /n2
For maximum power transfer,
Z Th = Z L /n2, or n2 = Z L /Z Th = 8/128
n = 0.25
(b)
I 1 = V Th /(Z Th + Z L /n2) = 10/(128 + 128) = 39.06 mA
(c)
v 2 = I 2 Z L = 156.24x8 mV = 1.25 V
But
v 2 = nv 1 therefore v 1 = v 2 /n = 4(1.25) = 5 V
Chapter 13, Solution 55.
We first reflect the 60- resistance to the middle circuit.
60
Z L'  20  2  26.67
3
We now reflect this to the primary side.
Z L' 26.67
ZL  2 
 1.667
1.6669
 Ω
4
16
Chapter 13, Solution 56.
We apply mesh analysis to the circuit as shown below.
2
1:2
+
+
v1
46V
+

v2

I1

I2
10 
5
For mesh 1,
46 = 7I 1 – 5I 2 + v 1
(1)
For mesh 2,
v 2 = 15I 2 – 5I 1
(2)
At the terminals of the transformer,
v 2 = nv 1 = 2v 1
(3)
I 1 = nI 2 = 2I 2
(4)
Substituting (3) and (4) into (1) and (2),
Combining (5) and (6),
46 = 9I 2 + v 1
(5)
v 1 = 2.5I 2
(6)
46 = 11.5I 2 or I 2 = 4
P 10 = 0.5|I 2 |2(10) = 80 watts.
Chapter 13, Solution 57.
(a)
Z L = j3||(12 – j6) = j3(12 – j6)/(12 – j3) = (12 + j54)/17
Reflecting this to the primary side gives
Z in = 2 + Z L /n2 = 2 + (3 + j13.5)/17 = 2.316820.04
I 1 = v s /Z in = 6090/2.316820.04 = 25.969.96 A(rms)
I 2 = I 1 /n = 12.9569.96 A(rms)
(b)
6090 = 2I 1 + v 1 or v 1 = j60 –2I 1 = j60 – 51.869.96
v 1 = 21.06147.44 V(rms)
v 2 = nv 1 = 42.12147.44 V(rms)
v o = v 2 = 42.12147.44 V(rms)
(c)
S = v s I 1 * = (6090)(25.9–69.96) = 1.55420.04 kVA
Chapter 13, Solution 58.
Consider the circuit below.
20 
I3
20 
1:5
+
800
+
–
V1
I1

For mesh1,
V2

+
I2
vo
100 

–80 + 20I 1 – 20I 3 + V 1 = 0 or
20I 1 – 20I 3 + V 1 = 80
(1)
V 2 = 100I 2 or 100I 2 – V 2 = 0
For mesh 2,
(2)
For mesh 3,
+
40I 3 – 20I 1 + V 2 – V 1 = 0 which leads to
–20 I 1 + 40I 3 – V 1 + V 2 = 0
(3)
At the transformer terminals, V 2 = –nV 1 = –5V 1 or 5V 1 + V 2
(4)
I 1 – I 3 = –n(I 2 – I 3 ) = –5(I 2 – I 3 ) or
I 1 + 5I 2 – 6I 3 = 0
(5)
Solving using MATLAB,
&gt;&gt;A = [ 20 0 -20 1 0 ; 0 100 0 0 -1; -20 0 40 -1 1; 0 0 0 5 1; 1 5 -6 0
0]
A=
20 0 -20 1 0
0 100 0 0 -1
-20 0 40 -1 1
0 0 0 5 1
1 5 -6 0 0
&gt;&gt; B = [ 80 0 0 0 0 ]'
B=
80
0
0
0
0
&gt;&gt; Y = inv(A)*B
Y=
5.9355
0.5161
1.4194
-10.3226
51.6129
P20,1 = 0.5*( I1 – I3 )^2*20 = 0.5*( 5.9355– 1.4194)^2*20 = 203.95
p 20 (the one between 1 and 3) = 0.5(20)(I 1 – I 3 )2 = 10(5.9355–1.4194)2
= 203.95 watts
p 20 (at the top of the circuit) = 0.5(20)I 3 2 = 20.15 watts
p 100 = 0.5(100)I 2 2 = 13.318 watts
Chapter 13, Solution 59.
We apply mesh analysis to the circuit as shown below.
10 

1:4

+
+
40  0
+
–
V_2
V1
_
I1
20 
I2
12
For mesh 1,
–40 + 22I 1 – 12I 2 + V 1 = 0
(1)
For mesh 2,
–12I 1 + 32I 2 – V 2 = 0
(2)
At the transformer terminals,
–4V 1 + V 2 = 0
I 1 – 4I 2 = 0
Putting (1), (2), (3), and (4) in matrix form, we obtain
0  40
 22  12 1
 12 32
0  1  0 

I
 0
0
4 1   0 

  
4
0
0 0
 1
&gt;&gt; A=[22,-12,1,0;-12,32,0,-1;0,0,-4,1;1,-4,0,0]
A=
22 -12 1 0
-12 32 0 -1
0 0 -4 1
1 -4 0 0
(3)
(4)
&gt;&gt; U=[40;0;0;0]
U=
40
0
0
0
&gt;&gt; X=inv(A)*U
X=
2.2222
0.5556
-2.2222
-8.8889
For 10- resistor,
P 10 = [(2.222)2/2]10 = 24.69 W
For 12- resistor,
P 12 = [(2.222–0.5556)2/2]12 = 16.661 W
For 20- resistor,
P 20 = [(0.5556)2/2]20 = 3.087 W.
24.69 W, 16.661 W, 3.087 W
Chapter 13, Solution 60.
(a)
Transferring the 40-ohm load to the middle circuit,
Z L ’ = 40/(n’)2 = 10 ohms where n’ = 2
10||(5 + 10) = 6 ohms
We transfer this to the primary side.
Z in = 4 + 6/n2 = 4 + 0.375 = 4.375 ohms, where n = 4
I 1 = 120/4.375 = 27.43 A and I 2 = I 1 /n = 6.857 A
4
1:4
I1
+
1200
+
–
v1
10 

Using current division, I 2 ’ = (10/25)I 2 = 2.7429 and
I 3 = I 2 ’/n’ = 1.3714 A
(b)
I2’
+
v2

5
I2
p = 0.5(I 3 )2(40) = 37.62 watts
10 
Chapter 13, Solution 61.
We reflect the 160-ohm load to the middle circuit.
Z R = Z L /n2 = 160/(4/3)2 = 90 ohms, where n = 4/3
2
1:5
I1
+
240
+
–
v1

Io
14 
Io’
+
vo

14 + 60||90 = 14 + 36 = 50 ohms
We reflect this to the primary side.
Z R ’ = Z L ’/(n’)2 = 50/52 = 2 ohms when n’ = 5
I 1 = 24/(2 + 2) = 6A
24 = 2I 1 + v 1 or v 1 = 24 – 2I 1 = 12 V
v o = –nv 1 = –60 V, I o = –I 1 /n 1 = –6/5 = –1.2
I o ‘ = [60/(60 + 90)]I o = –0.48A
I 2 = –I o ’/n = 0.48/(4/3) = 360 mA
60 
90 
Chapter 13, Solution 62.
(a)
Reflect the load to the middle circuit.
Z L ’ = 8 – j20 + (18 + j45)/32 = 10 – j15
We now reflect this to the primary circuit so that
Z in = 6 + j4 + (10 – j15)/n2 = 7.6 + j1.6 = 7.76711.89, where n = 5/2 =
2.5
I 1 = 40/Z in = 40/7.76711.89 = 5.15–11.89
S = v s I 1 * = (400)(5.1511.89) = 20611.89 VA
(b)
I 2 = –I 1 /n,
n = 2.5
I 3 = –I 2 /n’, n = 3
I 3 = I 1 /(nn’) = 5.15–11.89/(2.5x3) = 0.6867–11.89
p = |I 2 |2(18) = 18(0.6867)2 = 8.488 watts
Chapter 13, Solution 63.
Reflecting the (9 + j18)-ohm load to the middle circuit gives,
Z in ’ = 7 – j6 + (9 + j18)/(n’)2 = 7 – j6 + 1 + j2 = 8 – j4 when n’ = 3
Reflecting this to the primary side,
Z in = 1 + Z in ’/n2 = 1 + 2 – j = 3 – j, where n = 2
I 1 = 120/(3 – j) = 12/3.162–18.43 = 3.79518.43A
I 2 = I 1 /n = 1.897518.43 A
I 3 = –I 2 /n2 = 632.5161.57 mA
Chapter 13, Solution 64.
The Thevenin equivalent to the left of the transformer is shown below.
8 k
24  0 V
+
_
Z
30k
Z L'  2L  2
n
n
For maximum power transfer,
30k 
8k  

 n 2  30 / 8  3.75
n2
n =1.9365
Chapter 13, Solution 65.
40 
10 
I1
200 V
(rms)
-
50 
I2

1
+
1:2
I2
1:3
+
+
+
V1
-
V2
-
V3
-
I3

2
+
V4
-
20 

At node 1,
200  V1 V1  V4

 I1
10
40


200  1.25V1  0.25V4  10 I 1
(1)
At node 2,
V1  V4 V4

 I3
40
20


V1  3V4  40 I 3
(2)
At the terminals of the first transformer,
V2
 2

 V2  2V1
V1
I2
 1 / 2


I 1  2 I 2
I1
(3)
(4)
For the middle loop,
 V2  50 I 2  V3  0


V3  V2  50 I 2
(5)
At the terminals of the second transformer,
V4
3
V3


V4  3V3
(6)
I3
 1 / 3


I 2  3 I 3
I2
We have seven equations and seven unknowns. Combining (1) and (2) leads to
(7)
200  3.5V4  10 I 1  50 I 3
But from (4) and (7), I 1  2 I 2  2(3I 3 )  6 I 3 . Hence
200  3.5V4  110 I 3
(8)
From (5), (6), (3), and (7),
V4  3(V2  50 I 2 )  3V2  150 I 2  6V1  450 I 3
Substituting for V 1 in (2) gives
V4  6(3V4  40 I 3 )  450 I 3


I3 
19
V4
210
Substituting (9) into (8) yields
200  13.452V4


V4  14.87
P
V 24
 11.05 W
20
(9)
Chapter 13, Solution 66.
Design a problem to help other students to better understand how the ideal autotransformer
works.
Although there are many ways to solve this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
An ideal autotransformer with a 1:4 step-up turns ratio has its secondary connected to a
120- load and the primary to a 420-V source. Determine the primary current.
Solution
v 1 = 420 V
(1)
v 2 = 120I 2
(2)
v 1 /v 2 = 1/4 or v 2 = 4v 1
(3)
I 1 /I 2 = 4 or I 1 = 4 I 2
(4)
Combining (2) and (4),
v 2 = 120[(1/4)I 1 ] = 30 I 1
4v 1 = 30I 1
4(420) = 1680 = 30I 1 or I 1 = 56 A
Chapter 13, Solution 67.
(a)
V1 N 1  N 2
1


V2
N2
0.4
(b)
S 2  I 2V 2  5,000
(c )
S 2  S 1  I 1V1  5,000




V 2  0.4V1  0.4 x 400  160 V
I2 


5000
 31.25 A
160
I1 
5000
 12.5 A
400
Chapter 13, Solution 68.
This is a step-up transformer.
I2
+
N1
2 – j6
v2
I1
10 + j40
+
2030
N2

v1
+


For the primary circuit,
2030 = (2 – j6)I 1 + v 1
(1)
For the secondary circuit,
v 2 = (10 + j40)I 2
(2)
At the autotransformer terminals,
v 1 /v 2 = N 1 /(N 1 + N 2 ) = 200/280 = 5/7,
Also,
thus v 2 = 7v 1 /5
(3)
I 1 /I 2 = 7/5 or I 2 = 5I 1 /7
(4)
Substituting (3) and (4) into (2),
v 1 = (10 + j40)25I 1 /49
Substituting that into (1) gives
2030 = (7.102 + j14.408)I 1
I 1 = 2030/16.06363.76 = 1.245–33.76 A
I 2 = 5I 1 /7 = 889.3–33.76 mA
I o = I 1 – I 2 = [(5/7) – 1]I 1 = –2I 1 /7 = 355.7146.2 mA
p = |I 2 |2R = (0.8893)2(10) = 7.51 watts
Chapter 13, Solution 69.
We can find the Thevenin equivalent.
I2
+
+
N2
j125 
75 
v2
V Th
I1
1200
+
N1
v1
+




I1 = I2 = 0
As a step up transformer,
v 1 /v 2 = N 1 /(N 1 + N 2 ) = 600/800 = 3/4
v 2 = 4v 1 /3 = 4(120)/3 = 1600 rms = V Th .
To find Z Th , connect a 1-V source at the secondary terminals. We now have a
step-down transformer.
+
j125 
75 
I2
I1
v1
10 V
+
v2
+



v 1 = 1V, v 2 =I 2 (75 + j125)
But
v 1 /v 2 = (N 1 + N 2 )/N 1 = 800/200 which leads to v 1 = 4v 2 = 1
and v 2 = 0.25
I 1 /I 2 = 200/800 = 1/4 which leads to I 2 = 4I 1
Hence
0.25 = 4I 1 (75 + j125) or I 1 = 1/[16(75 + j125)
Z Th = 1/I 1 = 16(75 + j125)
Therefore, Z L = Z Th * = (1.2 – j2) k
Since V Th is rms, p = (|V Th |/2)2/R L = (80)2/1200 = 5.333 watts
Chapter 13, Solution 70.
This is a step-down transformer.
30 + j12
I1
+
I2
v1
1200
+

+
v2


20 – j40
I 1 /I 2 = N 2 /(N 1 + N 2 ) = 200/1200 = 1/6, or I 1 = I 2 /6
(1)
v 1 /v 2 = (N 2 + N 2 )/N 2 = 6, or v 1 = 6v 2
(2)
For the primary loop,
120 = (30 + j12)I 1 + v 1
(3)
For the secondary loop,
v 2 = (20 – j40)I 2
(4)
Substituting (1) and (2) into (3),
120 = (30 + j12)( I 2 /6) + 6v 2
and substituting (4) into this yields
120 = (49 – j38)I 2 or I 2 = 1.93537.79
p = |I 2 |2(20) = 74.9 watts.
Chapter 13, Solution 71.
Z in = V 1 /I 1
But
Hence
V 1 I 1 = V 2 I 2 , or V 2 = I 2 Z L and I 1 /I 2 = N 2 /(N 1 + N 2 )
V 1 = V 2 I 2 /I 1 = Z L (I 2 /I 1 )I 2 = Z L (I 2 /I 1 )2I 1
V 1 /I 1 = Z L [(N 1 + N 2 )/N 2 ] 2
Z in = [1 + (N 1 /N 2 )] 2Z L
Chapter 13, Solution 72.
(a)
Consider just one phase at a time.
1:n
A
a
B
b
C
c
n = V L / 3VLp  7200 /(12470 3 ) = 1/3
(b)
The load carried by each transformer is 60/3 = 20 MVA.
Hence
I Lp = 20 MVA/12.47 k = 1604 A
I Ls = 20 MVA/7.2 k = 2778 A
(c)
The current in incoming line a, b, c is
3I Lp  3x1603.85 = 2778 A
Current in each outgoing line A, B, C is
2778/(n 3 ) = 4812 A
20MVA
Chapter 13, Solution 73.
(a)
This is a three-phase -Y transformer.
(b)
V Ls = nv Lp / 3 = 450/(3 3 ) = 86.6 V, where n = 1/3
As a Y-Y system, we can use per phase equivalent circuit.
I a = V an /Z Y = 86.60/(8 – j6) = 8.6636.87
I c = I a 120 = 8.66156.87 A
I Lp = n 3 I Ls
I 1 = (1/3) 3 (8.6636.87) = 536.87
I 2 = I 1 –120 = 5–83.13 A
(c)
p = 3|I a |2(8) = 3(8.66)2(8) = 1.8 kw.
Chapter 13, Solution 74.
(a)
This is a - connection.
(b)
The easy way is to consider just one phase.
1:n = 4:1 or n = 1/4
n = V 2 /V 1 which leads to V 2 = nV 1 = 0.25(2400) = 600
i.e. V Lp = 2400 V and V Ls = 600 V
S = p/cos = 120/0.8 kVA = 150 kVA
p L = p/3 = 120/3 = 40 kw
4:1
IL
I Ls
V Lp
I ps
I pp
p Ls = V ps I ps
But
IL =
Hence,
I ps = 40,000/600 = 66.67 A
I Ls =
(c)
V Ls
3 I ps =
3 I p and V L = V p
3 x66.67 = 115.48 A
Similarly, for the primary side
p pp = V pp I pp = p ps or I pp = 40,000/2400 = 16.667 A
and
(d)
I Lp =
3 I p = 28.87 A
Since S = 150 kVA therefore S p = S/3 = 50 kVA
Chapter 13, Solution 75.
(a)
n = V Ls /( 3 V Lp ) = 900/(4500 3 ) = 0.11547
(b)
S =
3 V Ls I Ls or I Ls = 120,000/(900 3 ) = 76.98 A
I Ls = I Lp /(n 3 ) = 76.98/(2.887 3 ) = 15.395 A
Chapter 13, Solution 76.
Using Fig. 13.138, design a problem to help other students to better understand a wye-delta,
three-phase transformer and how they work.
Although there are many ways to solve this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
A Y- three-phase transformer is connected to a 60-kVA load with 0.85 power factor
(leading) through a feeder whose impedance is 0.05 + j0.1 per phase, as shown in Fig.
13.137 below. Find the magnitude of:
(a) the line current at the load,
(b) the line voltage at the secondary side of the transformer,
(c) the line current at the primary side of the transformer.
Figure 13.137
Solution
(a)
V L = 240 V = V AB
V AN = V L / 3 = 138.56 V
Since S =
3 V L I L then I L = 60,000/(240 3 ) = 144.34 A
1:n
0.05 
2640V
j0.1 
A
240V
j0.1 
0.05 
B
0.05 
(b)
j0.1 
C
Balanced
60kVA
0.85pf
Let V AN = |V AN |0 = 138.560
cos = pf = 0.85 or  = 31.79
I AA’ = I L  = 144.3431.79
V A’N’ = ZI AA’ + V AN
= 138.560 + (0.05 + j0.1)(144.3431.79)
= 138.036.69
V Ls = V A’N’
(c)
3 = 138.03
3 = 239.1 V
For Y- connections,
n =
3 V Ls /V ps =
3 x238.7/2640 = 0.1569
f Lp = nI Ls / 3 = 0.1569x144.34/ 3 = 13.05 A
Chapter 13, Solution 77.
(a)
This is a single phase transformer.
V 1 = 13.2 kV, V 2 = 120 V
n = V 2 /V 1 = 120/13,200 = 1/110, therefore n = 1/110
or 110 turns on the primary to every turn on the secondary.
(b)
P = VI or I = P/V = 100/120 = 0.8333 A
I 1 = nI 2 = 0.8333/110 = 7.576 mA
Chapter 13, Solution 78.
We convert the reactances to their inductive values.
X  L

 L
X

The schematic is as shown below.
AC = y es
MAG = y es
PHASE = y es
COUPLING = 0.5
L1_VALUE = 80H
L2_VALUE = 60H
IPRINT
IPRINT
R1
AC = y es
MAG = y es
PHASE = y es
TX1
20
100Vac
ACPHASE = -30 V1
R2
50
0Vdc
R3
40
0
FREQ
IM(V_PRINT1)IP(V_PRINT1)
1.592E-01 1.347E+00 -8.489E+01
FREQ
IM(V_PRINT2)IP(V_PRINT2)
1.592E-01 6.588E-01 -7.769E+01
Thus,
I 1 = 1.347–84.89˚ amps and I 2 = 658.8–77.69˚ mA
Chapter 13, Solution 79.
The schematic is shown below.
k 1 = 15 / 5000 = 0.2121, k 2 = 10 / 8000 = 0.1118
In the AC Sweep box, we type Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592.
After the circuit is saved and simulated, the output includes
FREQ
IM(V_PRINT1)
IP(V_PRINT1)
1.592 E–01
4.068 E–01
–7.786 E+01
FREQ
IM(V_PRINT2)
IP(V_PRINT2)
1.592 E–01
1.306 E+00
–6.801 E+01
FREQ
IM(V_PRINT3)
IP(V_PRINT3)
1.592 E–01
1.336 E+00
–5.492 E+01
Thus, I 1 = 1.306–68.01 A, I 2 = 406.8–77.86 mA, I 3 = 1.336–54.92 A
Chapter 13, Solution 80.
The schematic is shown below.
k 1 = 10 / 40x80 = 0.1768, k 2 = 20 / 40 x 60 = 0.4082
k 3 = 30 / 80 x 60 = 0.433
In the AC Sweep box, we set Total Pts = 1, Start Freq = 0.1592, and End Freq =
0.1592. After the simulation, we obtain the output file which includes
i.e.
FREQ
IM(V_PRINT1)
IP(V_PRINT1)
1.592 E–01
1.304 E+00
6.292 E+01
I o = 1.30462.92 A
Chapter 13, Solution 81.
The schematic is shown below.
k 1 = 2 / 4x8 = 0.3535, k 2 = 1 / 2 x8 = 0.25
In the AC Sweep box, we let Total Pts = 1, Start Freq = 100, and End Freq = 100.
After simulation, the output file includes
FREQ
1.000 E+02
IM(V_PRINT1)
1.0448 E–01
IP(V_PRINT1)
1.396 E+01
FREQ
1.000 E+02
IM(V_PRINT2)
2.954 E–02
IP(V_PRINT2)
–1.438 E+02
FREQ
1.000 E+02
IM(V_PRINT3)
2.088 E–01
IP(V_PRINT3)
2.440 E+01
i.e.
I 1 = 104.513.96 mA, I 2 = 29.54–143.8 mA,
I 3 = 208.824.4 mA.
Chapter 13, Solution 82.
The schematic is shown below. In the AC Sweep box, we type Total Pts = 1,
Start Freq = 0.1592, and End Freq = 0.1592. After simulation, we obtain the output
file which includes
FREQ
1.592 E–01
IM(V_PRINT1)
1.955 E+01
IP(V_PRINT1)
8.332 E+01
FREQ
1.592 E–01
IM(V_PRINT2)
6.847 E+01
IP(V_PRINT2)
4.640 E+01
FREQ
1.592 E–01
IM(V_PRINT3)
4.434 E–01
IP(V_PRINT3)
–9.260 E+01
i.e.
V 1 = 19.5583.32 V, V 2 = 68.4746.4 V,
I o = 443.4–92.6 mA.
magnitude of the inductances.
Chapter 13, Solution 83.
The schematic is shown below. In the AC Sweep box, we set Total Pts = 1, Start Freq =
0.1592, and End Freq = 0.1592. After simulation, the output file includes
FREQ
1.592 E–01
IM(V_PRINT1)
1.080 E+00
IP(V_PRINT1)
3.391 E+01
FREQ
1.592 E–01
VM(\$N_0001)
1.514 E+01
VP(\$N_0001)
–3.421 E+01
i.e.
i X = 1.0833.91 A, V x = 15.14–34.21 V.
This is most likely incorrect and needs to have the values of
turns changed.
Checking with hand calculations.
Loop 1.
–6 + 1I 1 + V 1 = 0 or I 1 + V 1 = 6
Loop 2.
–V 2 – j10I 2 + 8(I 2 –I 3 ) = 0 or (8–j10)I 2 – 8I 3 – V 2 = 0
Loop 3.
8(I 3 –I 2 ) + 6I 3 +2V x +V 3 = 0 or –8I 2 + 14I 3 + V 3 + 2V x = 0 but
V x = 8(I 2 –I 3 ), therefore we get 8I 2 – 2I 3 + V 3 = 0
(3)
–V 4 + (4+j2)I 4 = 0 or (4+j2)I 4 – V 4 = 0
(4)
Loop 4.
(1)
(2)
We also need the constraint equations, V 2 = 2V 1 , I 1 = 2I 2 , V 3 = 2V 4 , and I 4 = 2I 3 . Finally,
I x = I 2 and V x = 8(I 2 – I 3 ).
We can eliminate the voltages from the equations (we only need I 2 and I 3 to obtain the required
(1)+0.5(2) = I 1 + (4–j5)I 2 – 4I 3 = 6 and
0.5(3) + (4) = 4I 2 – I 3 + (4+j2)I 4 = 0.
Next we use I 1 = 2I 2 and I 4 = 2I 3 to end up with the following equations,
(6–j5)I 2 – 4I 3 = 6 and 4I 2 + (7+j4)I 3 = 0 or I 2 = –[(7+j4)I 3 ]/4 = (–1.75–j)I 3
= (2.01556 –150.255&deg;)I 3
This leads to (6–j5)(–1.75–j)I 3 – 4I 3 = (–10.5–5–4+j(8.75–6))I 3 = (–19.5+j2.75)I 3 = 6 or
I 3 = 6/(19.69296 171.973&deg;) = 0.304677 –171.973&deg; amps
= –0.301692–j0.042545.
I 2 = (–1.75–j)(0.304677 –171.973&deg;)
= (2.01556 –150.255&deg;)(0.304677 –171.973&deg;)
= 614.096 37.772&deg; mA = 0.48541+j0.37615
and I 2 – I 3 = 0.7871+j0.4187 = 0.89154 28.01&deg;.
Therefore,
V x = 8(0.854876 22.97&deg;) = 7.132 28.01&deg; V
I x = I 2 = 614.1 37.77&deg; mA.
Checking with MATLAB we get A and X from equations (1) – (4) and the four constraint
equations.
&gt;&gt; A = [1 0 0 0 1 0 0 0;0 (8-10j) -8 0 0 -1 0 0;0 8 -2 0 0 0 1 0;0 0 0 (4+2j) 0 0 0 -1;0 0 0 0 -2 1 0
0;1 -2 0 0 0 0 0 0;0 0 0 0 0 0 1 -2;0 0 -2 1 0 0 0 0]
A=
1.0000
0
0
0
1.0000
0
0
0
0
8.0000 -10.0000i -8.0000
0
8.0000
0
0
-1.0000
0
0
-2.0000
0
0
0
1.0000
0
0
1.0000
0
0
1.0000
0
0
2.0000
0
0
0
0
0
0
-2.0000
X=
6
0
0
0
0
0
0
0
0
-2.0000
&gt;&gt; X = [6;0;0;0;0;0;0;0]
4.0000 + 2.0000i
0
-2.0000
0
0
1.0000
0
1.0000
0
0
0
0
0
0
0
0
1.0000
0
-
0
0
0
0
0
&gt;&gt; Y = inv(A)*X
Y=
0.9708 + 0.7523i = I 1 = 1.2817 37.773&deg; amps
0.4854 + 0.3761i = I 2 = 614.056 37.769&deg; mA = I x
-0.3017 - 0.0425i = I 3 = 0.30468 –171.982&deg; amps
-0.6034 - 0.0851i = I 4
5.0292 - 0.7523i = V 1
10.0583 - 1.5046i = V 2
-4.4867 - 3.0943i = V 3
-2.2434 - 1.5471i = V 4
I x = 614.1 37.77&deg; mA
Finally, V x = 8(I 2 – I 3 ) = 8(0.7871+j0.4186) = 8(0.891489 28.01&deg;)
= 7.132 28.01&deg; volts
Chapter 13, Solution 84.
The schematic is shown below. we set Total Pts = 1, Start Freq = 0.1592, and End
Freq = 0.1592. After simulation, the output file includes
FREQ
1.592 E–01
IM(V_PRINT1)
4.028 E+00
IP(V_PRINT1)
–5.238 E+01
FREQ
1.592 E–01
IM(V_PRINT2)
2.019 E+00
IP(V_PRINT2)
–5.211 E+01
FREQ
1.592 E–01
IM(V_PRINT3)
1.338 E+00
IP(V_PRINT3)
–5.220 E+01
i.e.
I 1 = 4.028–52.38 A, I 2 = 2.019–52.11 A,
I 3 = 1.338–52.2 A.
Dot convention is wrong.
Chapter 13, Solution 85.
Z1
VS
+

Z L /n2
For maximum power transfer,
Z 1 = Z L /n2 or n2 = Z L /Z 1 = 8/7200 = 1/900
n = 1/30 = N 2 /N 1 . Thus N 2 = N 1 /30 = 3000/30 = 100 turns.
Chapter 13, Solution 86.
n = N 2 /N 1 = 48/2400 = 1/50
Z Th = Z L /n2 = 3/(1/50)2 = 7.5 k
Chapter 13, Solution 87.
Z Th = Z L /n2 or n =
Z L / Z Th  75 / 300 = 0.5
Chapter 13, Solution 88.
n = V 2 /V 1 = I 1 /I 2 or I 2 = I 1 /n = 2.5/0.1 = 25 A
p = IV = 25x12.6 = 315 watts
Chapter 13, Solution 89.
n = V 2 /V 1 = 120/240 = 0.5
S = I 1 V 1 or I 1 = S/V 1 = 10x103/240 = 41.67 A
S = I 2 V 2 or I 2 = S/V 2 = 104/120 = 83.33 A
Chapter 13, Solution 90.
(a)
n = V 2 /V 1 = 240/2400 = 0.1
(b)
n = N 2 /N 1 or N 2 = nN 1 = 0.1(250) = 25 turns
(c)
S = I 1 V 1 or I 1 = S/V 1 = 4x103/2400 = 1.6667 A
S = I 2 V 2 or I 2 = S/V 2 = 4x104/240 = 16.667 A
Chapter 13, Solution 91.
(a)
The kVA rating is S = VI = 25,000x75 = 1.875 MVA
(b)
Since S 1 = S 2 = V 2 I 2 and I 2 = 1875x103/240 = 7.812 kA
Chapter 13, Solution 92.
(a)
V 2 /V 1 = N 2 /N 1 = n, V 2 = (N 2 /N 1 )V 1 = (28/1200)4800 = 112 V
(b)
I 2 = V 2 /R = 112/10 = 11.2 A and I 1 = nI 2 , n = 28/1200
I 1 = (28/1200)11.2 = 261.3 mA
(c)
p = |I 2 |2R = (11.2)2(10) = 1254 watts.
Chapter 13, Solution 93.
(a)
For an input of 110 V, the primary winding must be connected in parallel, with
series aiding on the secondary. The coils must be series opposing to give 14 V. Thus,
the connections are shown below.
110 V
14 V
(b)
To get 220 V on the primary side, the coils are connected in series, with series
aiding on the secondary side. The coils must be connected series aiding to give 50 V.
Thus, the connections are shown below.
220 V
50 V
Chapter 13, Solution 94.
V 2 /V 1 = 110/440 = 1/4 = I 1 /I 2
There are four ways of hooking up the transformer as an auto-transformer. However it is
clear that there are only two outcomes.
V1
V1
V1
V2
V2
(1)
V1
(2)
V2
(3)
V2
(4)
(1) and (2) produce the same results and (3) and (4) also produce the same results.
Therefore, we will only consider Figure (1) and (3).
(a)
For Figure (3), V 1 /V 2 = 550/V 2 = (440 – 110)/440 = 330/440
Thus,
(b)
V 2 = 550x440/330 = 733.4 V (not the desired result)
For Figure (1), V 1 /V 2 = 550/V 2 = (440 + 110)/440 = 550/440
Thus,
V 2 = 550x440/550 = 440 V (the desired result)
Chapter 13, Solution 95.
(a)
n = V s /V p = 120/7200 = 1/60
(b)
I s = 10x120/144 = 1200/144
S = VpIp = VsIs
I p = V s I s /V p = (1/60)x1200/144 = 139 mA
*Chapter 13, Solution 96.
Problem,
Some modern power transmission systems now have major, high voltage DC
transmission segments. There are a lot of good reasons for doing this but we will not go
into them here. To go from the AC to DC, power electronics are used. We start with
three-phase AC and then rectify it (using a full-wave rectifier). It was found that using a
delta to wye and delta combination connected secondary would give us a much smaller
ripple after the full-wave rectifier. How is this accomplished? Remember that these are
real devices and are wound on common cores. Hint, using Figures 13.47 and 13.49, and
the fact that each coil of the wye connected secondary and each coil of the delta
connected secondary are wound around the same core of each coil of the delta connected
primary so the voltage of each of the corresponding coils are in phase. When the output
leads of both secondaries are connected through full-wave rectifiers with the same load,
you will see that the ripple is now greatly reduced. Please consult the instructor for more
help if necessary.
Solution,
This is a most interesting and very practical problem. The solution is actually quite easy,
you are creating a second set of sine waves to send through the full-wave rectifier, 30˚
out of phase with the first set. We will look at this graphically in a minute. We begin by
showing the transformer components.
The key to making this work is to wind the secondary coils with each phase of the
primary. Thus, a-b is wound around the same core as A 1 -N 1 and A 2 -B 2 . The next thing
we need to do is to make sure the voltages come out equal. We need to work the number
of turns of each secondary so that the peak of V A1 – V B1 is equal to V A2 –V B2 . Now, let
us look at some of the equations involved.
If we let v ab (t) = 100sin(t) V, assume that we have an ideal transformer, and the turns
ratios are such that we get v A1-N1 (t) = 57.74sin(t) V and V A2-B2 (t) = 100sin(t) V. Next, let
us look at V bc (t) = 100sin(t+120˚) V. This leads to V B1-N1 (t) = 57.74sin(t+120˚) V. We
now need to determine V A1-B1 (t).
V A1-B1 (t) = 57.74sin(t) – 57.74sin(t+120˚) = 100sin(t–30˚) V.
This then leads to the output per phase voltage being equal to v out (t) = [100sin(t) +
100sin(t–30˚)] V. We can do this for each phase and end up with the output being sent to
the full-wave rectifier. This looks like v out (t) = [|100sin(t)| + |100sin(t–30˚)| +
|sin(t+120˚)| |100sin(t+90˚)| + |100sin(t–120˚)| + |100sin(t–150˚)|] V. The end result will
be more obvious if we look at plots of the rectified output.
A1
N1
B1
a
C1
b
A2
c
C2
B2
In the plot below we see the normalized (1 corresponds to 100 volts) ripple with only one
of the secondary sets of windings and then the plot with both. Clearly the ripple is
greatly reduced!
Chapter 14, Solution 1.
Vo
R
jRC


Vi R  1 jC 1  jRC
j  0
1
H () 
,
where  0 
1  j  0
RC
H () 
H  H () 
 0
  H () 
1  ( 0 ) 2


 tan -1  
2
 0 
This is a highpass filter. The frequency response is the same as that for P.P.14.1 except that
0  1 RC . Thus, the sketches of H and  are shown below.
H
1
0.7071
0
0 =


90
45
0
0 =

Chapter 14, Solution 2.
Using Fig. 14.69, design a problem to help other students to better understand how to
determine transfer functions.
Although there are many ways to solve this problem, this is an example based on the
same kind of problem asked in the third edition.
Problem
Obtain the transfer function V o /V i of the circuit in Fig. 14.66.
10 
+
2
+
_
Vo
1
F
8
Figure 14.66
Vo
_
For Prob. 14.2.
Solution
V
H ( s)  o 
Vi
2
1
s/8
10  20 
1
s/8

2  8/ s
1 s4

12  8 / s 6 s  0.6667
Chapter 14, Solution 3.
0.2 F


0.1F


1
1
5


jC s (0.2) s
1
10

s (0.1) s
The circuit becomes that shown below.
2
V1
5
s
+
Vi
+
_
5
10
s
Vo
_
10
5
10 1  s
(5  )
5(
)
10
5
s  s
s  10( s  1)
Let Z  //(5  )  s
15
5
s
s
s ( s  3)
5
(3  s )
s
s
Z
V1 
Vi
Z 2
5
s
s
Z

Vi
Vo 
V1 
V1 
55/ s
s 1
s 1 Z  2
10( s  1)
s
10s
5s
s ( s  3)
V


 2
H (s)  o 
Vi s  1
10( s  1) 2 s ( s  3)  10( s  1) s  8s  5
2
s( s  3)
H(s) = 5s/(s2+8s+5)
Chapter 14, Solution 4.
(a)
R ||
1
R

jC 1  jRC
R
Vo
R
1  jRC
H () 


R
Vi
R  jL (1  jRC)
jL 
1  jRC
(b)
H () 
R
-  RLC  R  jL
H () 
jC (R  jL)
R  jL

R  jL  1 jC 1  jC (R  jL)
H () 
-  2 LC  jRC
1   2 LC  jRC
2
Chapter 14, Solution 5.
(a) Let Z  R // sL 
Vo 
sRL
R  sL
Z
Vs
Z  Rs
sRL
Vo
Z
sRL
H (s)  
 R  sL 
Vs Z  Rs R  sRL
RRs  s ( R  Rs ) L
s
R  sL
1
Rx
1
sC  R
(b) Let Z  R //

sC R  1 1  sRC
sC
Z
Vo 
Vs
Z  sL
V
Z
H(s)  o 

Vi Z  sL
R
R
1  sRC 
R
s 2 LRC  sL  R
sL 
1  sRC
Chapter 14, Solution 6.
The 2 H inductors become jω2 or 2s.
Let Z = 2s||2 = [(2s)(2)/(2s+2)] = 2s/(s+1)
We convert the current source to a voltage source as shown below.
2
Is  2
2S
+
+
_
Vo
Z
_
V o = [(Z)/(Z+2s+2)](2I s ) =
H(s) = I o /I s = [2s/(s2+3s+1)].
or
Chapter 14, Solution 7.
(a)
0.05  20 log10 H
2.5  10 -3  log10 H
H  10 2.510  1.005773
-3
(b)
- 6.2  20 log10 H
- 0.31  log10 H
H  10 -0.31  0.4898
(c)
104.7  20 log10 H
5.235  log10 H
H  10 5.235  1.718  105
Chapter 14, Solution 8.
Design a problem to help other students to better calculate the magnitude in dB and phase in
degrees of a variety of transfer functions at a single value of ω.
Although there are many ways to solve this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
Determine the magnitude (in dB) and the phase (in degrees) of H() at  = 1 if H()
equals
(a) 0.05
(b) 125
10 j
(c)
2  j
3
6
(d)

1  j 2  j
Solution
(a)
(b)
(c)
H  0.05
H dB  20 log10 0.05  - 26.02 ,
φ = 0
H  125
H dB  20 log10 125  41.94 ,
φ = 0
H dB
(d)
j10
 4.47263.43
2 j
 20 log10 4.472  13.01 ,
H(1) 
φ = 63.43
3
6

 3.9  j2.7  4.743 - 34.7
1 j 2  j
H dB  20 log10 4.743  13.521,
φ = –34.7˚
H(1) 
Chapter 14, Solution 9.
H ( ) 
10
10(1  j )(1  j 10)
H dB  20 log 10 1 - 20 log 10 1  j  20 log 10 1  j / 10
  - tan -1 ()  tan -1 ( / 10)
The magnitude and phase plots are shown below.
20
H dB
0.1
1
10

100
20 log 10
-20
1
1  j / 10
20 log10
-40
1
1  j

0.1
-45
1
10

100
arg
1
1  j / 10
-90
arg
-135
-180
1
1  j
Chapter 14, Solution 10.
Design a problem to help other students to better understand how to determine the Bode
magnitude and phase plots of a given transfer function in terms of jω.
Although there are many ways to solve this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
Sketch the Bode magnitude and phase plots of:
H  j  
50
j  5  j 
Solution
H( j) 
50

j(5  j)
10
j 

1 j1  
5 

H dB
40
20 log1
20
10
0.1
-20
1
100


1
20 log

j
 1
5

 1 

20 log

 j 
-40

0.1
-45







1
10

100
arg
1
1  j / 5
-90
arg
-135
-180
1
j
Chapter 14, Solution 11.
H ( ) 
0.2 x10 (1  j 10)
2[ j (1  j 2)]
H dB  20 log10 1  20 log10 1  j 10  20 log10 j  20 log10 1  j 2
  -90  tan -1  10  tan -1  2
The magnitude and phase plots are shown below.
H dB
40
20
0.1
1
10
100

1
10
100

-20
-40

90
45
0.1
-45
-90
Chapter 14, Solution 12.
T ( ) 
10(1  j )
j (1  j / 10)
To sketch this we need 20log 10 |T(ω)| = 20log 10 |10| + 20log 10 |1+jω| – 20log 10 |jω| –
20log 10 |1+jω/10| and the phase is equal to tan–1(ω) – 90&deg; – tan–1(ω/10).
The plots are shown below.
|T|
(db)
20

0
0.1
1
10
100
-20
-40
arg T
90o

0
0.1
-90o
1
10
100
Chapter 14, Solution 13.
G ( ) 
0.1(1  j )
(1 100)(1  j )

2
( j ) (10  j ) ( j ) 2 (1  j 10)
GdB  40  20 log10 1  j  40 log10 j  20 log10 1  j 10
  -180  tan -1  tan -1  10
The magnitude and phase plots are shown below.
G dB
40
20
0.1
-20
1
10
100

1
10
100

-40

90
0.1
-90
-180
Chapter 14, Solution 14.
250
25
H ( ) 
1  j
2

j10  j  


j 1 
 

25
5

 

H dB  20 log10 10  20 log10 1  j  20 log10 j
 20 log10 1  j2 5  ( j 5) 2
 10 25 

  -90  tan -1   tan -1 
1  2 5 
The magnitude and phase plots are shown below.
H dB
40
20
0.1
-20
1
10
100

1
10
100

-40

90
0.1
-90
-180
Chapter 14, Solution 15.
H ( ) 
2 (1  j )
0.1(1  j )

(2  j )(10  j ) (1  j 2)(1  j 10)
H dB  20 log10 0.1  20 log10 1  j  20 log10 1  j 2  20 log10 1  j 10
  tan -1   tan -1  2  tan -1  10
The magnitude and phase plots are shown below.
H dB
40
20
0.1
-20
1
10
100

1
10
100

-40

90
45
0.1
-45
-90
Chapter 14, Solution 16.
H(ω) =
H db = 20log 10 |0.1| – 20olg 10 |jω| – 20log 10 |1+jω+(jω/4)2|
The magnitude and phase plots are shown below.
H
20
20 log (j)
1
10
4
40
100

0.1
–20
 j 
20 log 1  j  

 4 
–40
–60


0.4
-90
1
4
10
40
90
100
-tan-1

2
1
-180
16
2
Chapter 14, Solution 17.
G () 
(1 4) j
(1  j)(1  j 2) 2
G dB  -20log10 4  20 log10 j  20 log10 1  j  40 log10 1  j 2
  -90 - tan -1  2 tan -1  2
The magnitude and phase plots are shown below.
G dB
20
0.1
1
10
100

-12
-20
-40

90
0.1
-90
-180
1
10
100

Chapter 14, Solution 18.
The MATLAB code is shown below.
&gt;&gt; w=logspace(-1,1,200);
&gt;&gt; s=i*w;
&gt;&gt; h=(7*s.^2+s+4)./(s.^3+8*s.^2+14*s+5);
&gt;&gt; Phase=unwrap(angle(h))*57.23;
&gt;&gt; semilogx(w,Phase)
&gt;&gt; grid on
60
40
H (jw ) P h a s e
20
0
-2 0
-4 0
-6 0
-1
10
10
w
0
Now for the magnitude, we need to add the following to the above,
&gt;&gt; H=abs(h);
&gt;&gt; HdB=20*log10(H);
&gt;&gt; semilogx(w,HdB);
&gt;&gt; grid on
10
1
0
-5
HdB
-1 0
-1 5
-2 0
-2 5
-1
10
10
w
0
10
1
Chapter 14, Solution 19.
H(ω) = 80jω/[(10+jω)(20+jω)(40+jω)]
= [80/(10x20x40)](jω)/[(1+jω/10)(1+jω/20)(1+jω/40)]
H db = 20log 10 |0.01| + 20log 10 |jω| – 20log 10 |1+jω/10| – 20log 10 |1+jω/20| –
20log 10 |1+jω/40|
The magnitude and phase plots are shown below.
20 log j
20 db
0 db
1
0.1
100
10
ω
–20 log |1+jω/40|
–20 db
–20 log 1 
20 log |1/80|
–40 db
j
10
–20 log |1+jω/20|
jω
90˚
0˚
0.1
1
100
10
ω
(1+jω/40)
–90˚
(1+jω/10)
–180˚
(1+jω/20)
Chapter 14, Solution 20.
Design a more complex problem than given in Prob. 14.10, to help other students to
better understand how to determine the Bode magnitude and phase plots of a given
transfer function in terms of jω. Include at least a second order repeated root.
Although there are many ways to solve this problem, this is an example based on the
same kind of problem asked in the third edition.
Problem
Sketch the magnitude phase Bode plot for the transfer function
Solution
20log(1/100) = -40
For the plots, see the next page.
The magnitude and phase plots are shown below.
20 log j
40
1
20 log
j
10
1
20

0.1
1
5
10
50
100
20 log
-20
20 log
1
100
1
1  j
-40
20 log
-60
180˚
1
j 

1 

5 

2
jω
(1+jω/10)
90˚
(1+jω)
0.1

1
5
10
50
100
–90˚
(1+jω/5)2
–180˚
–270˚
Chapter 14, Solution 21.
H(ω) = 10(jω)(20+jω)/[(1+jω)(400+60jω–ω2)]
= [10x20/400](jω)(1+jω/20)/[(1+jω)(1+(3jω/20)+(jω/20)2)]
H dB
j
j 3  j 
 20 log(0.5)  20 log j  20 log 1 
 20 log 1  j  20 log 1 


20
20  20 
2
The magnitude plot is as sketched below. 20log 10 |0.5| = –6 db
db
40
20log|jω|
20 log |1+jω/20|
20
1
20 log 0.5
10
20

0.1
100
–20
–40
–20 log 1  j
–60
–20 log
–80
Chapter 14, Solution 22.
20  20 log10 k

 k  10
A zero of slope  20 dB / dec at   2 
 1  j 2
1
A pole of slope - 20 dB / dec at   20 

1  j 20
1

A pole of slope - 20 dB / dec at   100 
1  j 100
Hence,
H () 
10 (1  j 2)
(1  j 20)(1  j 100)
10 4 ( 2  j)
H () 
( 20  j)(100  j)
Chapter 14, Solution 23.
A zero of slope  20 dB / dec at the origin


j
1
1  j 1
1
A pole of slope - 40 dB / dec at   10 

(1  j 10) 2

A pole of slope - 20 dB / dec at   1 
Hence,
H () 
j
(1  j)(1  j 10) 2
H () 
100 j
(1  j)(10  j) 2
(It should be noted that this function could also have a minus sign out in
front and still be correct. The magnitude plot does not contain this
information. It can only be obtained from the phase plot.)
Chapter 14, Solution 24.
40  20 log10 K

 K  100
There is a pole at =50 giving 1/(1+j/50)
There is a zero at =500 giving (1 + j/500).
There is another pole at =2122 giving 1/(1 + j/2122).
Thus,
H(jω) = 100(1+jω)/[(1+jω/50)(1+jω/2122)]
= [100(50x2122)/500](jω+500)/[(jω+50)(jω+2122)]
or
H(s) = 21220(s+500)/[(s+50)(s+2122)].
Chapter 14, Solution 25.
0 
1
LC

1
(40  10 -3 )(1  10 -6 )
Z(0 )  R  2 k
 0
4 

Z(0 4)  R  j  L 
0 C 
 4

 5  10 3
4

Z(0 4)  2000  j 
 40  10 -3 
3
-6
(5  10 )(1  10 ) 
 4
Z(0 4)  2000  j (50  4000 5)
Z(0 4)  2  j0.75 k
 0
2 

Z(0 2)  R  j  L 
0 C 
 2

 (5  10 3 )
2

Z(0 2)  2000  j 
(40  10 -3 ) 
3
-6
(5  10 )(1  10 ) 
2

Z(ω 0 /2) = 200+j(100-2000/5)
Z(0 2)  2  j0.3 k

1 

Z(20 )  R  j  20 L 
20 C 



1

Z(20 )  2000  j  (2)(5  10 3 )(40  10 -3 ) 
3
-6
(2)(5  10 )(1  10 ) 

Z(20 )  2  j0.3 k

1 

Z(40 )  R  j  40 L 
40 C 



1

Z(40 )  2000  j  (4)(5  10 3 )(40  10 -3 ) 
3
-6
(4)(5  10 )(1  10 ) 

Z(40 )  2  j0.75 k
Chapter 14, Solution 26.
Design a problem to help other students to better understand ω o , Q, and B at resonance in
series RLC circuits.
Although there are many ways to solve this problem, this is an example based on the
same kind of problem asked in the third edition.
Problem
A coil with resistance 3  and inductance 100 mH is connected in series with a capacitor
of 50 pF, a resistor of 6 , and a signal generator that gives 110V-rms at all frequencies.
Calculate  o , Q, and B at resonance of the resultant series RLC circuit.
Solution
Consider the circuit as shown below. This is a series RLC resonant circuit.
6
50 pF
3
+
_
100 mH
R=6+3=9
o 
Q
1
1

3
LC
100 x10 x50 x1012
o L
R
o

447.21x103 x100 x103
 4969
9
447.21x103
B

Q
4969
Chapter 14, Solution 27.
o 
1
 40
LC

 LC 
1
402
R
 10

 R  10 L
L
If we select R =1 , then L = R/10 = 100 mH and
B
C
1
1
 2
 6.25 mF
2
40 L 40 x0.1
Chapter 14, Solution 28.
R  10  .
L
R 10

 0.5 H
B 20
C
1
1

 2 F
2
0 L (1000) 2 (0.5)
Q
0 1000

 50
B
20
Therefore, if R  10  then
L  500 mH , C  2 F ,
Q  50
Chapter 14, Solution 29.
We convert the voltage source to a current source as shown below.
12 k
is
is 
45 k
1 F
20
cos t , R = 12//45= 12x45/57 = 9.4737 k
12
1
1
o 

3
LC
60 x10 x1x106
B
1
1

RC 9.4737 x103 x106

4082
Q o 
 38.674 = 38.67
B 105.55
60 mH
Chapter 14, Solution 30.
(a) f o = 15,000 Hz leads to ω o = 2πf o = 94.25 krad/s = 1/(LC)0.5 or
LC = 1/8.883x109 or C = 1/(8.883x109x10–2) = 11.257x10–9 F = 11.257 pF.
(b) since the capacitive reactance cancels out the inductive reactance at resonance, the
current through the series circuit is given by
I = 120/20 = 6 A.
(c) Q = ω o L/R = 94.25x103(0.01)/20 = 47.12.
Chapter 14, Solution 31.
R  10  .
R
10

 0.05 H  50 mH
0 Q (10)(20)
1
1
C 2 
 0.2 F
0 L (100)(0.05)
1
1
B

RC (10)(0.2)
L
Chapter 14, Solution 32.
Design a problem to help other students to better understand the quality factor, the
resonant frequency, and bandwidth of a parallel RLC circuit.
Although there are many ways to solve this problem, this is an example based on the
same kind of problem asked in the third edition.
Problem
A parallel RLC circuit has the following values:
R = 60 , L = 1 mH, and C = 50 F
Find the quality factor, the resonant frequency, and the bandwidth of the RLC circuit.
Solution
1
1

3
LC
10 x50 x106
1
1
B

RC 60 x50 x106

4472
Q o 
 13.42
B 333.33
o 
Chapter 14, Solution 33.
B = ω o /Q = 6x106/120 = 50 krad/s.
ω 1 = ω o – B = 5.95x106 rad/s and ω 2 = ω o + B = 6.05x106 rad/s.
Chapter 14, Solution 34.
Q   o RC
Q
R
o L




C
L
Q
80

 56.84 pF
2f o R 2x5.6x10 6 x40x10 3
R
40 x10 3

= 14.21 &micro;H
2f o Q 2x 5.6 x10 6 x80
Chapter 14, Solution 35.
1
o 
(b)
B
(c)
Q   o RC  1.443x10 3 x5x10 3 x60x10 6  432.9
LC

1
(a)
8x10  3 x60x10  6
1
1

3
RC 5x10 x60x10  6
Chapter 14, Solution 36.
At resonance,
1
1

 40 
Y 25  10 -3
Q
80
Q  0 RC 
 C 

 10 F
0 R (200  10 3 )(40)
1
1
1
0 

 L  2 
 2.5 H
10
0 C (4  10 )(10  10 -6 )
LC
Y
1
R

 R 
0 200  10 3
B

Q
80
B
1  0   200  1.25  198.75 krad/s
2
B
 2  0   200  1.25  201.25 krad/s
2
Chapter 14, Solution 37.
0 
1
LC
Y(0 ) 
1
R

 Z(0 )  R  2 k

1
4 
  0.5  j18.75 mS
 j  0 C 
R
0 L 
 4
1
Z(0 4) 
 (1.4212  j53.3) 
0.0005  j0.01875
Y(0 4) 

1
2 
  0.5  j7.5 mS
 j  0 C 
R
0 L 
 2
1
Z(0 2) 
 (8.85  j132.74) 
0.0005  j0.0075
Y(0 2) 

1
1 
  0.5  j7.5 mS
 j  20 L 
R
20 C 

Z(20 )  (8.85  j132.74) 
Y ( 2 0 ) 

1
1 
  0.5  j18.75 mS
 j  40 L 

R
4
C
0 

Z(40 )  (1.4212  j53.3) 
Y ( 4 0 ) 
Chapter 14, Solution 38.
1 
L

)
  jLR  R  j(L 
1
C
C 



Z  jL //( R 
)

1
1 2
jC
R
R 2  (L 
)
 jL
j C
C
jL(R 
1
)
jC
L
1 
 L 

C
C 
0
1 2
2
R  ( L 
)
C
LR 2 
Im(Z) 


Thus,

1
LC  R 2 C 2
 2 ( LC  R 2 C 2 )  1
Chapter 14, Solution 39.
Y
1
R  j L
 jC  jC 
R  jL
R 2   2 L2
At resonance, Im(Y)  0 , i.e.
0 L
0 C  2
0
R  02 L2
L
R 2  02 L2 
C
0 
1
R2


LC L2
 50 

 
(40  10 -3 )(1  10 -6 )  40  10 -3 
1
2
Chapter 14, Solution 40.
(a)
B   2  1  2(f 2  f1 )  2(90  86) x10 3  8krad / s
1
(1   2 )  2(88) x10 3  176X10 3
2
1
1
1
B

 C 

 19.89nF
RC
BR 8x10 3 x 2x10 3
o 
1
(b)
o 
(c )
o  176  552.9krad / s
(d)
B  8  25.13krad / s
(e)
Q
LC


o 176

 22
B
8
L
1
2 o C

1
(176X10 3 ) 2 x19.89 x10  9
= 164.45 &micro;H
Chapter 14, Solution 41.
Using Fig. 14.80, design a problem to help other students to better understand the quality factor,
the resonant frequency, and bandwidth of an RLC circuit.
Although there are many ways to solve this problem, this is an example based on the same kind
of problem asked in Example 14.9.
Problem
For the circuits in Fig. 14.80, find the resonant frequency  0 , the quality factor Q, and the
bandwidth B. Let C = 0.1 F, R 1 = 10 Ω, R 2 = 2 Ω, and L = 2 H.
R2
C
R1
L
Figure 14.80
For Prob. 14.41.
Solution
To find ω o , we need to find the input impedance or input admittance and set imaginary
component equal to zero. Finding the input admittance seems to be the easiest approach.
Y = jω0.1 + 0.1 + 1/(2+jω2) = jω0.1 + 0.1 + [2/(4+4ω2)] – [jω2/(4+4ω2)]
At resonance,
0.1ω = 2ω/(4+4ω2) or 4ω2 + 4 = 20 or ω2 = 4 or ω o = 2 rad/s
and,
Y = 0.1 + 2/(4+16) = 0.1 + 0.1 = 0.2 S
The bandwidth is define as the two values of ω such that |Y| = 1.4142(0.2) = 0.28284 S.
I do not know about you, but I sure would not want to solve this analytically. So how about
using MATLAB or excel to solve for the two values of ω?
Using Excel, we get ω 1 = 1.414 rad/s and ω 2 = 3.741 rad/s or B = 2.327 rad/s
We can now use the relationship between ω o and the bandwidth.
Q = ω o /B = 2/2.327 = 0.8595
Chapter 14, Solution 42.
(a)
This is a series RLC circuit.
R  26  8,
L 1H,
1
0 
(b)
LC

1
0.4
Q
 0 L 1.5811

 0.1976
R
8
B
R
L
C  0.4 F
This is a parallel RLC circuit.
(3)(6)
 2 F
3 6
R  2 k ,
L  20 mH
3 F and 6 F 

C  2 F ,
0 
1
LC

1
(2  10 -6 )(20  10 -3 )
Q
R
2  10 3

 20
0 L (5  10 3 )(20  10 -3 )
B
1
1

3
RC (2  10 )(2  10 -6 )
Chapter 14, Solution 43.
(a)
Z in  (1 jC) || (R  jL)
Z in 
R  jL
jC

R  jL
1  2 LC  jRC
1
jC
(R  jL)(1  2 LC  jRC)
Z in 
(1  2 LC) 2  2 R 2 C 2
R  jL 
At resonance, Im(Z in )  0 , i.e.
0   0 L(1   02 LC)   0 R 2 C
02 L2 C  L  R 2 C
0 
(b)
L  R 2C
L2 C

1
R2

LC L2
Z in  R || ( jL  1 jC)
R ( jL  1 jC)
R (1  2 LC)

R  jL  1 jC (1   2 LC)  jRC
R (1  2 LC)[(1  2 LC)  jRC]
Z in 
(1  2 LC) 2  2 R 2 C 2
Z in 
At resonance, Im(Z in )  0 , i.e.
0  R (1  2 LC) RC
1  2 LC  0
1
0 
LC
Chapter 14, Solution 44.
Consider the circuit below.
1/jC
Z in
jL
R1
(a)
R2
Z in  (R 1 || jL) || (R 2  1 jC)
 R 1 jL  
1 
 ||  R 2 

Z in  
jC 
 R 1  jL  
jR 1 L 
1 

 R 2 
R 1  jL 
jC 
Z in 
jR 1L
1
R2 

jC R 1  jL
jR 1 L (1  jR 2 C)
Z in 
(R 1  jL)(1  jR 2 C)  2 LCR 1
- 2 R 1 R 2 LC  jR 1 L
Z in 
R 1  2 LCR 1  2 LCR 2  j (L  R 1 R 2 C)
(-2 R 1 R 2 LC  jR 1 L)[R 1  2 LCR 1  2 LCR 2  j (L  R 1 R 2 C)]
Z in 
(R 1  2 LCR 1  2 LCR 2 ) 2  2 (L  R 1 R 2 C) 2
At resonance, Im(Z in )  0 , i.e.
0  3 R 1 R 2 LC (L  R 1 R 2 C)  R 1 L (R 1  2 LCR 1  2 LCR 2 )
0  3 R 12 R 22 LC 2  R 12 L  3 R 12 L2 C
0  2 R 22 C 2  1  2 LC
2 (LC  R 22 C 2 )  1
0 
0 
1
LC  R 22 C 2
1
(0.02)(9  10 )  (0.1) 2 (9  10 -6 ) 2
0  2.357 krad / s
-6
(b)
At   0  2.357 krad / s ,
jL  j(2.357  10 3 )(20  10 -3 )  j47.14
R 1 || jL 
R2 
j47.14
 0.9996  j0.0212
1  j47.14
1
1
 0.1 
 0.1  j47.14
j (2.357  10 3 )(9  10 -6 )
jC
Z in (0 )  (R 1 || jL) || (R 2  1 jC)
(0.9996  j0.0212)(0.1  j47.14)
(0.9996  j0.0212)  (0.1  j47.14)
Z in (0 )  1 
Z in (0 ) 
Chapter 14, Solution 45.
Convert the voltage source to a current source as shown below.
Is
30 k
50 F
R = 30//50 = 30x50/80 = 18.75 k
This is a parallel resonant circuit.
1
1
o 

LC
10 x103 x50 x106
1
1
B

RC 18.75 x103 x50 x106
 447.21
Q o 
 419.13
B
1.067
10 mH
50 k
Chapter 14, Solution 46.
(a)
j
,
1  j
1 || j 
1 ||
1 j
1
1


j  1  1 j 1  j
Transform the current source gives the circuit below.
j
I
1  j
+

j
1  j
1
1
1  j
+
Vo

1
j
1  j
Vo 
I

1
j 1  j 

1
1  j 1  j
(b)
H () 
Vo
j

I
2 (1  j) 2
H (1) 
1
2 (1  j) 2
H (1) 
1
2 ( 2)2
 0.25
Chapter 14, Solution 47.
H () 
Vo
R
1


Vi R  jL 1  jL R
H(0)  1 and H()  0 showing that this circuit is a lowpass filter.
1
At the corner frequency, H(c ) 
, i.e.
2
c L
1
1
R



1

or


c
2
L
R
2
c L 

1 
 R 
Hence,
c 
R
 2f c
L
fc 
1 R
1 10  10 3
 

 796 kHz
2 L 2 2  10 -3
Chapter 14, Solution 48.
R ||
H () 
1
jC
1
jC
R jC
R  1 jC
H () 
R jC
jL 
R  1 jC
R
H () 
R  jL   2 RLC
jL  R ||
H(0)  1 and H()  0 showing that this circuit is a lowpass filter.
Chapter 14, Solution 49.
Design a problem to help other students to better understand lowpass filters described by transfer
functions.
Although there are many ways to solve this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
Determine the cutoff frequency of the lowpass filter described by
H( ) 
4
2  j10
Find the gain in dB and phase of H() at  = 2 rad/s.
Solution
At dc, H(0) 
Hence,
H() 
2
2

1
2
4
 2.
2
H(0) 
2
2
4
4  100c2
4  100c2  8 
 c  0.2
H(2) 
4
2

2  j20 1  j10
H(2) 
2
101
 0.199
In dB, 20 log10 H(2)  - 14.023
arg H(2)  -tan -110  - 84.3 or ω c = 1.4713 rad/sec.
Chapter 14, Solution 50.
H () 
Vo
jL

Vi R  jL
H(0)  0 and H()  1 showing that this circuit is a highpass filter.
H (c ) 
or
fc 
1
2
c 

1
 R 

1 
 c L 
2

 1 
R
 2f c
L
1 R
1 200
 

 318.3 Hz
2 L 2 0.1
R
c L
Chapter 14, Solution 51.
The lowpass RL filter is shown below.
L
+
+
R
vs
vo
-
-
H
c 
R
 2 f c
L
Vo
R
1


Vs R  jL 1  jL / R


R  2f c L  2x5x10 3 x40x10  3  1.256k
Chapter 14, Solution 52.
Design a problem to help other students to better understand passive highpass filters.
Although there are many ways to solve this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
In a highpass RL filter with a cutoff frequency of 100 kHz, L = 40 mH. Find R.
Solution
c 
R
 2f c
L
R  2f c L  (2)(10 5 )(40  10 -3 )  25.13 k
Chapter 14, Solution 53.
1  2f 1  20  10 3
2  2f 2  22  10 3
B  2  1  2  10 3
2  1
0 
 21  10 3
2
0 21
Q

 10.5
B
2
0 
L
1
LC

 L 
1
02 C
1
 2.872 H
(21  10 ) (80  10 -12 )
3 2
R

 R  BL
L
R  (2  10 3 )(2.872)  18.045 k
B
Chapter 14, Solution 54.
We start with a series RLC circuit and the use the equations related to the circuit and the
values for a bandstop filter.
Q = ω o L/R = 1/(ω o CR) = 20; B = R/L = ω o /Q = 10/20 = 0.5; ω o = 1/(LC)0.5 = 10
(LC)0.5 = 0.1 or LC = 0.01. Pick L = 10 H then C = 1 mF.
Q = 20 = ω o L/R =10x10/R or R = 100/20 = 5 Ω.
Chapter 14, Solution 55.
o 
Therefore,
1
LC

1
(25  10 )(0.4  10  6 )
3
B
R
10

L 25  10 -3
Q
10
 25
0.4
1  o  B 2  10  0.2  9.8 krad / s
or
2  o  B 2  10  0.2  10.2 krad / s
or
9.8
 1.56 kHz
2
10.2
f2 
 1.62 kHz
2
f1 
1.56 kHz  f  1.62 kHz
Chapter 14, Solution 56.
(a)
From Eq 14.54,
R
s
sRC
L
H (s) 


1 1  sRC  s 2 LC
R
1
R  sL 
s2  s 
sC
L LC
R
R
1
,
and 0 
L
LC
sB
H (s)  2
s  sB   02
Since B 
(b)
From Eq. 14.56,
H (s) 
sL 
1
sC
1
R  sL 
sC
s 2   02
H (s)  2
s  sB   02

s2 
s2  s
1
LC
R
1

L LC
Chapter 14, Solution 57.
(a)
Consider the circuit below.
R
I
I1
1/sC
+
Vs
+

1/sC
R
Vo

1 
1
R  
1 
1
sC 
sC 
Z(s)  R 
||  R    R 
2
sC 
sC 
R
sC
1  sRC
Z(s)  R 
sC (2  sRC)
Z(s) 
I
1  3sRC  s 2 R 2 C 2
sC (2  sRC)
Vs
Z
I1 
Vs
1 sC
I
2 sC  R
Z (2  sRC)
Vo  I 1 R 
H (s) 
R Vs
sC (2  sRC)

2  sRC 1  3sRC  s 2 R 2 C 2
Vo
sRC

Vs 1  3sRC  s 2 R 2 C 2

3
s
1
RC
H (s)  
3
1
3 2
s 
s 2 2

RC
R C
1
or
R C2
3
B
RC
Thus, 02 
2





0 
1
RC
(b)
Similarly,
Z(s)  sL  R || (R  sL)  sL 
Z(s) 
I
R (R  sL)
2R  sL
R 2  3sRL  s 2 L2
2R  sL
Vs
,
Z
Vo  I 1  sL 
I1 
R Vs
R
I
2R  sL
Z (2R  sL)
sLR Vs
2R  sL
 2
2R  sL R  3sRL  s 2 L2
1  3R 

s
Vo
sRL
3 L 


H (s) 
R2
3R
Vs R 2  3sRL  s 2 L2
s 2
s2 
L
L
R
L
3R
B
L
Thus, 0 
Chapter 14, Solution 58.
(a)
(b)
0 
1
LC

1
(0.1)(40  10 )
-12
 0.5  10 6 rad / s
R 2  10 3

 2  10 4
L
0.1
0 0.5  10 6

 25
Q
B
2  10 4
B
As a high Q circuit,
B
1  0   10 4 (50  1)  490 krad / s
2
B
2  0   10 4 (50  1)  510 krad / s
2
(c)
As seen in part (b),
Q  25
Chapter 14, Solution 59.
Consider the circuit below.
Ro
+
1/sC
Vi
+

R
Vo
sL

where L = 1 mH, C = 4 &micro;F, R o = 6 Ω, and R = 4 Ω.

1  R (sL  1 sC)
Z(s)  R || sL   

sC  R  sL  1 sC
R (1  s 2 LC)
Z(s) 
1  sRC  s 2 LC
H
Vo
R (1  s 2 LC)
Z


Vi Z  R o R o  sRR o C  s 2 LCR o  R  s 2 LCR
Z in  R o  Z  R o 
Z in 
R (1  s 2 LC)
1  sRC  s 2 LC
R o  sRR o C  s 2 LCR o  R  s 2 LCR
1  sRC  s 2 LC
s  j
R o  jRR o C  2 LCR o  R  2 LCR
Z in 
1  2 LC  jRC
(R o  R  2 LCR o  2 LCR  jRR o C)(1   2 LC  jRC)
Z in 
(1  2 LC) 2  (RC) 2
Im(Z in )  0 implies that
- RC [R o  R  2 LCR o  2 LCR ]  RR o C (1  2 LC)  0
R o  R  2 LCR o  2 LCR  R o  2 LCR o  0
2 LCR  R
1
0 
H
LC
1

(1  10 )(4  10 -6 )
-3
R (1  2 LC)
R o  jRR o C  R  2 LCR o  2 LCR
H max  H(0) 
or
H max
R
Ro  R
 1

R  2  LC 
R


 H()  lim

  R o  R
RR o C
j
 LC (R  R o ) R  R o
2


At 1 and 2 , H 
1
2
H mzx
R (1  2 LC)

R o  R   2 LC (R o  R )  jRR o C
2 (R o  R )
R
1
2
1
2
0


(R o  R )(1  2 LC)
(RR o C) 2  (R o  R  2 LC(R o  R )) 2
10 (1  2  4  10 -9 )
(96  10 -6 ) 2  (10  2  4  10 -8 ) 2
10 (1  2  4  10 -9 )
(96  10 )  (10    4  10 )
-6
2
2
-8 2

1
2
(10  2  4  10 -8 )( 2 )  (96  10 -6 ) 2  (10  2  4  10 -8 ) 2  0
(2)(10  2  4  10 -8 ) 2  (96  10 -6 ) 2  (10  2  4  10 -8 ) 2
(96  10 -6 ) 2  (10  2  4  10 -8 ) 2  0
1.6  10 -15 4  8.092  10 -7 2  100  0
4  5.058  10 8  6.25  1016  0
 2.9109  10 8
2  
 2.1471  10 8
Hence,
1  14.653 krad / s
2  17.061 krad / s
B  2  1  17.061  14.653  2.408 krad / s
Chapter 14, Solution 60.
H () 
jRC
j

1  jRC j  1 RC
(from Eq. 14.52)
This has a unity passband gain, i.e. H()  1 .
1
 c  50
RC
j10
H ^ ()  10 H () 
50  j
j10
H () 
50  j
Chapter 14, Solution 61.
(a)
V 
1 jC
V,
R  1 jC i
V  Vo
Since V  V ,
1
V  Vo
1  jRC i
H () 
(b)
V 
Vo
1

Vi 1  jRC
R
V,
R  1 jC i
Since V  V ,
jRC
V  Vo
1  jRC i
H () 
Vo
jRC

Vi 1  jRC
V  Vo
Chapter 14, Solution 62.
This is a highpass filter.
jRC
1

1  jRC 1  j RC
1
1
H () 
,
c 
 2 (1000)
RC
1  j c 
1
1
H () 

1  j f c f 1  j1000 f
H () 
(a)
H (f  200 Hz) 
Vo 
(b)
1  j5
 23.53 mV
H (f  2 kHz) 
Vo
1

1  j0.5 Vi
120 mV
 107.3 mV
Vo 
(c)
120 mV
Vo
1

1  j5 Vi
1  j0.5
H (f  10 kHz) 
Vo 
120 mV
1  j0.1
Vo
1

1  j0.1 Vi
 119.4 mV
Chapter 14, Solution 63.
For an active highpass filter,
H(s)  
sC i R f
1  sC i R i
(1)
H(s)  
10s
1  s / 10
(2)
But
Comparing (1) and (2) leads to:
C i R f  10


Rf 
10
 10M
Ci
C i R i  0.1


Ri 
0.1
 100k
Ci
Chapter 14, Solution 64.
Z f  R f ||
Rf
1

jC f 1  jR f C f
Zi  R i 
1  jR i C i
1

jC i
jC i
Hence,
H () 
Vo - Z f
- jR f C i


Vi
Zi
(1  jR f C f )(1  jR i C i )
This is a bandpass filter. H () is similar to the product of the transfer function of a
lowpass filter and a highpass filter.
Chapter 14, Solution 65.
V 
R
jRC
Vi 
V
R  1 jC
1  jRC i
V 
Ri
V
Ri  Rf o
Since V  V ,
Ri
jRC
Vo 
V
Ri  Rf
1  jRC i
H () 
Vo 
R f   jRC 


 1 
Vi 
R i   1  jRC 
It is evident that as    , the gain is 1 
Rf
1
and that the corner frequency is
.
Ri
RC
Chapter 14, Solution 66.
(a)
Proof
(b)
When R 1 R 4  R 2 R 3 ,
R4
s
H (s) 

R 3  R 4 s  1 R 2C
(c)
When R 3   ,
H (s) 
- 1 R 1C
s  1 R 2C
Chapter 14, Solution 67.
DC gain 
Rf 1

Ri 4

 R i  4R f
Corner frequency  c 
1
 2 (500) rad / s
R f Cf
If we select R f  20 k , then R i  80 k and
1
 15.915 nF
C
(2)(500)(20  10 3 )
Therefore, if R f  20 k , then R i  80 k and C  15.915 nF
Chapter 14, Solution 68.
Design a problem to help other students to better understand the design of active highpass filters
when specifying a high-frequency gain and a corner frequency.
Although there are many ways to solve this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
Design an active highpass filter with a high-frequency gain of 5 and a corner frequency
of 200 Hz.
Solution
Rf

 R f  5R i
Ri
1
 2 (200) rad / s
Corner frequency  c 
R i Ci
High frequency gain  5 
If we select R i  20 k , then R f  100 k and
1
 39.8 nF
C
(2)(200)(20  10 3 )
Therefore, if R i  20 k , then R f  100 k and C  39.8 nF
Chapter 14, Solution 69.
This is a highpass filter with f c  2 kHz.
1
c  2f c 
RC
1
1
RC 

2f c 4  103
10 8 Hz may be regarded as high frequency. Hence the high-frequency gain is
 R f  10

or
R f  2 .5 R
R
4
If we let R  10 k , then R f  25 k , and C 
1
 7.96 nF .
4000  10 4
Chapter 14, Solution 70.
(a)
Vo (s)
Y1 Y2

Vi (s) Y1 Y2  Y4 (Y1  Y2  Y3 )
1
1
where Y1 
 G 1 , Y2 
 G 2 , Y3  sC1 , Y4  sC 2 .
R1
R2
H (s) 
H (s) 
(b)
G 1G 2
G 1 G 2  sC 2 (G 1  G 2  sC1 )
G 1G 2
 1,
H()  0
G 1G 2
showing that this circuit is a lowpass filter.
H ( 0) 
Chapter 14, Solution 71.
R  50  , L  40 mH , C  1 F
L 
Km
K
L 
 1  m  (40  10 -3 )
Kf
Kf
25K f  K m
C 
C
KmKf
10 6 K f 
(1)

 1 
1
Km
Substituting (1) into (2),
1
10 6 K f 
25K f
K f  2x10–4
K m  25K f  5  10 -3
10 -6
KmKf
(2)
Chapter 14, Solution 72.
Design a problem to help other students to better understand magnitude and frequency scaling.
Although there are many ways to solve this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
What values of K m and K f will scale a 4-mH inductor and a 20-F capacitor to 1 H and 2 F
respectively?
Solution
LC 
K f2 
LC
K f2

 K f2 
LC
L C
(4  10 -3 )(20  10 -6 )
 4  10 -8
(1)(2)
K f  2  10 -4
L L 2
 K
C C m
K 2m 

 K 2m 
L C

C L
(1)(20  10 -6 )
 2.5  10 -3
(2)(4  10 -3 )
K m  5  10 -2
Chapter 14, Solution 73.
R   K m R  (12)(800  10 3 )  9.6 M
L 
Km
800
L
(40  10 -6 )  32 F
Kf
1000
C
300  10 -9
C 

 0.375 pF
K m K f (800)(1000)
Chapter 14, Solution 74.
R'1  K m R 1  3x100  300
R' 2  K m R 2  10x100  1 k
Km
10 2
L' 
L  6 ( 2)  200 H
Kf
10
1
C
C' 
 108  1 nF
K m K f 10
Chapter 14, Solution 75.
R'  K m R  20x10  200 
L' 
Km
10
L  5 (4)  400 H
Kf
10
C' 
C
1

 1 F
K m K f 10x10 5
1
Chapter 14, Solution 76.
R '  K m R  500 x5 x103  25 M
K
500
L '  m L  5 (10mH )  50  H
Kf
10
C'
C
20 x106

 0.4 pF
K m K f 500 x105
Chapter 14, Solution 77.
L and C are needed before scaling.
B
R
L
0 

 L 
1
LC
R 10

2H
B 5

 C 
1
1

 312.5 F
2
0 L (1600)(2)
(a)
L   K m L  (600)(2)  1.200 kH
C
3.125  10 -4
C 

 0.5208 F
Km
600
(b)
L 
L
2
 3  2 mH
K f 10
C
3.125  10 -4
C 

 312.5 nF
Kf
10 3
(c)
L 
Km
(400)(2)
L
 8 mH
Kf
10 5
C
3.125  10 -4
C 

 7.81 pF
KmKf
(400)(10 5 )
Chapter 14, Solution 78.
R   K m R  (1000)(1)  1 k
Km
10 3
L  4 (1)  0.1 H
L 
10
Kf
C
1
C 

 0.1 F
3
K m K f (10 )(10 4 )
The new circuit is shown below.
1 k
+
I
1 k
0.1 H
0.1 F
1 k
Vx

Chapter 14, Solution 79.
(a)
Insert a 1-V source at the input terminals.
Ro
Io
R
V1
V2
+ 
+

1V
1/sC
+
sL
3V o
Vo

There is a supernode.
1  V1
V2

R
sL  1 sC
But
V1  V2  3Vo
Also,
Vo 

 V2  V1  3Vo
sL
V
sL  1 sC 2
Combining (2) and (3)
V2  V1  3Vo 
(1)


Vo
V2

sL sL  1 sC
Substituting (3) and (4) into (1) gives
1  V1 Vo
sC
V


R
sL 1  4s 2 LC 1
sRC
1  4s 2 LC  sRC

1  V1 
V
V1
1  4s 2 LC 1
1  4s 2 LC
1  4s 2 LC
V1 
1  4s 2 LC  sRC
1  V1
sRC

R
R (1  4s 2 LC  sRC)
Z in 
(3)
sL  1 sC
Vo
sL
s 2 LC
Vo 
V
1  4s 2 LC 1
Io 
(2)
1 1  sRC  4s 2 LC

sC
Io
(4)
Z in  4sL  R 
1
sC
(5)
When R  5 , L  2 , C  0.1 ,
10
Z in (s)  8s  5 
s
At resonance,
Im(Z in )  0  4L 
or
(b)
0 
1
2 LC

1
C
1
2 (0.1)(2)
After scaling,
R 
 K m R
4 
 40 
5 
 50 
L 
Km
10
L
( 2 )  0 .2 H
Kf
100
C 
C
0.1

 10 -4
K m K f (10)(100)
From (5),
Z in (s)  0.8s  50 
0 
1
2 LC

10 4
s
1
2 (0.2)(10 -4 )
Chapter 14, Solution 80.
(a)
R   K m R  (200)(2)  400 
K m L (200)(1)

 20 mH
Kf
10 4
C
0.5
C 

 0.25 F
K m K f (200)(10 4 )
L 
The new circuit is shown below.
20 mH
a
Ix
0.25 F
400 
0.5 I x
b
(b)
Insert a 1-A source at the terminals a-b.
a
sL
V1
V2
Ix
1A
1/(sC)
R
0.5 I x
b
At node 1,
1  sCV1 
V1  V2
sL
(1)
At node 2,
V1  V2
V
 0.5 I x  2
sL
R
But, I x  sC V1 .
V1  V2
V2
 0.5sC V1 
sL
R
(2)
Solving (1) and (2),
sL  R
V1  2
s LC  0.5sCR  1
Z Th 
V1
sL  R
 2
1 s LC  0.5sCR  1
At   10 4 ,
Z Th 
( j10 4 )(20  10 -3 )  400
( j10 4 ) 2 (20  10 -3 )(0.25  10 -6 )  0.5( j10 4 )(0.25  10 -6 )(400)  1
Z Th 
400  j200
 600  j200
0.5  j0.5
Z Th  632.5 - 18.435 ohms
Chapter 14, Solution 81.
(a)
1
1
(G  jC)(R  jL)  1
 G  jC 

Z
R  j L
R  j L
Z
jL  R
2
  LC  j(RC  LG)  GR  1
 R

C
LC
Z() 
R
G  GR  1

 2  j   
LC
L C
j
(1)
We compare this with the given impedance:
Z() 
1000( j  1)
(2)
  2  2 j  1  2500
Comparing (1) and (2) shows that
1
 1000
C
R G
 2
L C




C  1 mF,
R/L  1


RL
G  C  1 mS
GR  1 10 3 R  1
2501 

LC
10 3 R

R  0.4  L
Thus,
R = 0.4Ω, L = 0.4 H, C = 1 mF, G = 1 mS
(b) By frequency-scaling, K f =1000.
R’ = 0.4 Ω, G’ = 1 mS
L' 
L
0.4
 3  0.4mH ,
K f 10
C' 
C
10 3
  3  1F
K f 10
Chapter 14, Solution 82.
C 
C
KmKf
Kf 
c 200

 200

1
Km 
C 1
1
1

 -6 
 5000
C K f 10 200
R   K m R  5 k,
thus,
R f  2R i  10 k
Chapter 14, Solution 83.
1
10 6
C
 0.1 pF
K mK f
100x10 5
1F


C' 
5F


C'  0.5 pF
10 k


R'  K m R  100x10 k  1 M
20 k


R '  2 M
Chapter 14, Solution 84.
The schematic is shown below. A voltage marker is inserted to measure v o . In the AC
sweep box, we select Total Points = 50, Start Frequency = 1, and End Frequency = 1000.
After saving and simulation, we obtain the magnitude and phase plots in the probe menu
as shown below.
Chapter 14, Solution 85.
We let I s  10 o A so that Vo / I s  Vo . The schematic is shown below. The circuit is
simulated for 100 &lt; f &lt; 10 kHz.
Chapter 14, Solution 86.
Using Fig. 14.103, design a problem to help other students to better understand how to use
PSpice to obtain the frequency response (magnitude and phase of I) in electrical circuits.
Although there are many ways to solve this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
Use PSpice to provide the frequency response (magnitude and phase of i) of the circuit in
Fig. 14.103. Use linear frequency sweep from 1 to 10,000 Hz.
Figure 14.103
Solution
The schematic is shown below. A current marker is inserted to measure I. We set Total
Points = 101, start Frequency = 1, and End Frequency = 10 kHz in the AC sweep
box. After simulation, the magnitude and phase plots are obtained in the Probe menu as
shown below.
Chapter 14, Solution 87.
The schematic is shown below. I n the AC Sweep box, we set Total Points = 50, Start
Frequency = 1, and End Frequency = 100. After simulation, we obtain the magnitude
response as shown below. It is evident from the response that the circuit represents a
high-pass filter.
Chapter 14, Solution 88.
The schematic is shown below. We insert a voltage marker to measure V o . In the AC
Sweep box, we set Total Points = 101, Start Frequency = 1, and End Frequency =
100. After simulation, we obtain the magnitude and phase plots of V o as shown below.
Chapter 14, Solution 89.
The schematic is shown below. In the AC Sweep box, we type Total Points = 101, Start
Frequency = 100, and End Frequency = 1 k. After simulation, the magnitude plot of
the response V o is obtained as shown below.
Chapter 14, Solution 90.
The schematic is shown below. In the AC Sweep box, we set
Total Points = 1001, Start Frequency = 1, and End Frequency = 100k. After
simulation, we obtain the magnitude plot of the response as shown below. The response
shows that the circuit is a high-pass filter.
Chapter 14, Solution 91.
The schematic is shown below. In the AC Sweep box, we select Total Points = 101, Start
Frequency = 10, and End Frequency = 10 k. After simulation, the magnitude plot of the
frequency response is obtained. From the plot, we obtain the resonant frequency f o is
approximately equal to 800 Hz so that  o = 2f o = 5026 rad/s.
Chapter 14, Solution 92.
The schematic is shown below. We type Total Points = 101, Start Frequency = 1, and
End Frequency = 100 in the AC Sweep box. After simulating the circuit, the magnitude
plot of the frequency response is shown below.
Chapter 14, Solution 93.
Consider the circuit as shown below.
R
Vs
+
+
_
V1

C
V1 
1
sC
Vs 
V
1  sRC
1
sC
R
sRC
V2 
Vs 
Vs
R  sC
1  sRC
R
Vo  V1  V2 
Hence,
H (s) 
1  sRC
Vs
1  sRC
Vo 1  sRC

V s 1  sRC
Vo
_

C
V2
R
Chapter 14, Solution 94.
c 
1
RC
We make R and C as small as possible. To achieve this, we connect 1.8 k  and 3.3 k  in
parallel so that
R
1.8x 3.3
 1.164 k
1.8  3.3
We place the 10-pF and 30-pF capacitors in series so that
C = (10x30)/40 = 7.5 pF
Hence,
c 
1
1

3
RC 1.164x10 x7.5x10 12
Chapter 14, Solution 95.
(a)
f0 
1
2 LC
When C  360 pF ,
f0 
1
2 (240  10 -6 )(360  10 -12 )
 0.541 MHz
When C  40 pF ,
f0 
1
2 (240  10 -6 )(40  10 -12 )
 1.624 MHz
Therefore, the frequency range is
0.541 MHz  f 0  1.624 MHz
(b)
Q
2fL
R
At f 0  0.541 MHz ,
Q
(2 )(0.541  10 6 )(240  10 -6 )
 67.98
12
At f 0  1.624 MHz ,
Q
(2 )(1.624  10 6 )(240  10 -6 )
 204.1
12
Chapter 14, Solution 96.
Ri
L
V1
Vo
+
Vi
+

C1
C2
RL
Vo

Z2
Z1  R L ||
Z1
RL
1

sC 2 1  sR 2 C 2
1
1  sL  R L  s 2 R L C 2 L 

Z2 
|| (sL  Z1 ) 
|| 
sC1
sC1 
1  sR L C 2

1 sL  R L  s 2 R L C 2 L

sC1
1  sR L C 2
Z2 
sL  R L  s 2 R L C 2 L
1

sC1
1  sR L C 2
Z2 
sL  R L  s 2 R L LC 2
1  sR L C 2  s 2 LC1  sR L C1  s 3 R L LC1C 2
V1 
Z2
V
Z2  R i i
Vo 
Z2
Z1
Z1
V
V1 

Z 2  R 2 Z1  sL i
Z1  sL
Vo
Z2
Z1


Vi Z 2  R 2 Z1  sL
where
Z2

Z2  R 2
sL  R L  s 2 R L LC 2
sL  R L  s 2 R L LC 2  R i  sR i R L C 2  s 2 R i LC1  sR i R L C1  s 3 R i R L LC1C 2
Z1
RL

and
Z1  sL R L  sL  s 2 R L LC 2
Therefore,
Vo

Vi
R L (sL  R L  s 2 R L LC 2 )
(sL  R L  s 2 R L LC 2  R i  sR i R L C 2  s 2 R i LC 1  sR i R L C 1
 s 3 R i R L LC 1 C 2 )( R L  sL  s 2 R L LC 2 )
where s  j .
Chapter 14, Solution 97.
Ri
L
V1
Vo
+
Vi
+

C1
C2
RL
Vo

Z2
Z1

1  sL (R L  1 sC 2 )

Z  sL ||  R L 
,
sC 2  R L  sL  1 sC 2

s  j
V1 
Z
V
Z  R i  1 sC1 i
Vo 
RL
RL
Z
V1 

V
R L  1 sC 2
R L  1 sC 2 Z  R i  1 sC1 i
H () 
Vo
RL
sL (R L  1 sC 2 )


Vi R L  1 sC 2 sL (R L  1 sC 2 )  (R i  1 sC1 )(R L  sL  1 sC 2 )
H () 
s 3 LR L C 1C 2
(sR i C 1  1)(s 2 LC 2  sR L C 2  1)  s 2 LC 1 (sR L C 2  1)
where s  j .
Chapter 14, Solution 98.
B  2  1  2 (f 2  f 1 )  2 (454  432)  44
0  2f 0  QB  (20)(44 )
f0 
(20)(44)
 (20)(22)  440 Hz
2
Chapter 14, Solution 99.
Xc 
C
1
1

C 2f C
1
1
10 -9


2f X c (2 )(2  10 6 )(5  10 3 ) 20
X L  L  2f L
L
f0 
B
XL
300
3  10 -4


2f (2 )(2  10 6 )
4
1
2 LC

1
3  10 -4 10 -9

2
4
20
 8.165 MHz
 4 
R
  4.188  10 6 rad / s
 (100) 
 3  10 -4 
L
Chapter 14, Solution 100.
c  2f c 
R
1
RC
1
1

 15.91 
2f c C (2)(20  10 3 )(0.5  10 -6 )
Chapter 14, Solution 101.
c  2f c 
R
1
RC
1
1

 1.061 k
2f c C (2)(15)(10  10 -6 )
Chapter 14, Solution 102.
(a)
When R s  0 and R L   , we have a low-pass filter.
c  2f c 
fc 
(b)
1
RC
1
1

 994.7 Hz
2RC (2)(4  10 3 )(40  10 -9 )
We obtain R Th across the capacitor.
R Th  R L || (R  R s )
R Th  5 || (4  1)  2.5 k
fc 
1
1

2R Th C (2 )(2.5  10 3 )(40  10 -9 )
f c  1.59 kHz
Chapter 14, Solution 103.
H () 
H (s) 
Vo
R2
,

Vi R 2  R 1 || 1 jC
s  j
R2
R 2 (R 1  1 sC)

R (1 sC) R 1R 2  (R 1  R 2 )(1 sC)
R2  1
R 1  1 sC
H(s) 
R 2 (1  sCR 1 )
R 1  R 2  sCR 1 R 2
Chapter 14, Solution 104.
The schematic is shown below. We click Analysis/Setup/AC Sweep and enter Total
Points = 1001, Start Frequency = 100, and End Frequency = 100 k. After simulation,
we obtain the magnitude plot of the response as shown.
Chapter 15, Solution 1.
e at  e - at
2
1 1
1 
s
L  cosh(at )   

 2

2  s  a s  a  s  a2
(a)
cosh(at ) 
(b)
sinh(at ) 
e at  e - at
2
1 1
1 
a
L  sinh(at )   

 2

2 s  a s  a  s  a2
Chapter 15, Solution 2.
(a)
f ( t )  cos(t ) cos()  sin(t ) sin()
F(s)  cos() L  cos(t )   sin() L  sin(t ) 
s cos()   sin()
F(s) 
s 2  2
(b)
f ( t )  sin(t ) cos()  cos(t ) sin()
F(s)  sin() L  cos(t )   cos() L  sin(t ) 
s sin()   cos()
F(s) 
s 2  2
Chapter 15, Solution 3.
(a)
L  e -2t cos(3t ) u ( t )  
s2
(s  2 ) 2  9
(b)
L  e -2t sin(4 t ) u ( t )  
4
(s  2) 2  16
(c)
Since L  cosh(at )  
2
(d)
Since L  sinh(at )  
2
(e)
L  e - t sin( 2t )  
s
s  a2
s3
L  e -3t cosh(2 t ) u ( t )  
(s  3 ) 2  4
a
s  a2
1
L  e -4t sinh( t ) u ( t )  
(s  4) 2  1
2
(s  1) 2  4
f (t) 
 F(s)
-d
t f (t) 

F(s)
ds
-d
-1
Thus, L  t e - t sin(2 t )  
2  (s  1) 2  4
ds
2

 2 (s  1)
((s  1) 2  4) 2
4 (s  1)
L  t e -t sin( 2t )  
((s  1) 2  4) 2
If


Chapter 15, Solution 4.
Design a problem to help other students better understand how to find the Laplace transform of
different time varying functions.
Although there are many ways to solve this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
Find the Laplace transforms of the following:
(a) g(t) = 6cos(4t – 1)
(b) f(t) = 2tu(t) + 5e-3(t – 2)u(t – 2)
Solution
s
6se  s
s
e

s 2  42
s 2  16
(a)
G (s )  6
(b)
2
e 2 s
F(s )  2  5
s3
s
Chapter 15, Solution 5.
(a)
s cos(30)  2 sin(30)
s2  4
d 2  s cos(30)  1 
L  t 2 cos(2t  30)   2 
ds  s 2  4 
L  cos(2t  30)  


-1 
d d  3
s  1 s 2  4 

ds ds  2



 3

-1
-2 
d  3 2
s  1 s 2  4 
 s  4  2s 
ds  2

 2

 3

 3
 3

3

 2s 
 (8s 2 ) 



2
s
1
s
1
- 2s   2

 2 
 2


 



 2

2
2
2
3
2
2
2
2
s 4
s 4
s 4
s 4







 3

(8s 2 ) 
s  1
- 3s  3s  2  3s
 2



2
3
2
2
s 4
s 4


L t

2


(-3 3 s  2)(s  4)
2
s
2
4

3
8  12
cos(2t  30)  

4!


4 3 s  8s
3
s
2
4
2

3
3 s  6s 2  3s 3
 s 2  4 3
72
(s  2 ) 5
(b)
L 3 t 4 e - 2t  3 
(c)

 2
d
2
L  2t u ( t )  4 ( t )   2  4(s  1  0)  2  4s

 s
s
dt
(s  2)
5

(d)
2 e -(t-1) u ( t )  2 e -t u ( t )
2e
L  2 e -(t-1) u ( t )  
s1
(e)
Using the scaling property,
1
1
1
5

 5 2 
L  5 u ( t 2)   5 
2s s
1 2 s (1 2)
(f)
L  6 e -t 3 u ( t )  
18
6

s  1 3 3s  1

(g)
Let f ( t )  ( t ) . Then, F(s)  1 .
 dn

 dn

L  n ( t )   L  n f ( t )   s n F(s)  s n 1 f (0)  s n  2 f (0)  
 dt

 dt

 dn

 dn

L  n ( t )   L  n f ( t )   s n  1  s n 1  0  s n  2  0  
 dt

 dt

n
 d

L  n ( t )   s n
 dt

Chapter 15, Solution 6.
f(t) = 5t[u(t)–u(t–1)] – 5t[u(t–1)–u(t–2)] = 5[tu(t)–tu(t–1) – tu(t–1) + tu(t–2)]
= 5[tu(t) – 2tu(t–1) + tu(t–2)]
= 5[tu(t) – 2(t–1)u(t–1) – 2u(t–1) + (t–2)u(t–2) + 2u(t–2)] which leads to
F(s) =5[(1/s2) – (1/s2)e–s – (2/s)e–s + (1/s2)e–2s + (2/s)e–2s]
Chapter 15, Solution 7.
(a)
F (s) 
(b) G ( s ) 
2 4

s2 s
4
3

s s2
(c ) H(s )  6
3
s
8s  18
8 2
 2
s 9
s 9 s 9
2
(d) From Problem 15.1,
s
L{cosh at}  2
s  a2
s2
s2
X (s) 
 2
2
2
( s  2)  4
s  4s  12
(a )
2 4
4
3
8s  18
s2
, (c ) 2
, (d ) 2
 , (b ) 
2
s
s s2
s
s 9
s  4s  12
Chapter 15, Solution 8.
(a) 2t=2(t-4) + 8
f(t) = 2tu(t-4) = 2(t-4)u(t-4) + 8u(t-4)
2
8
 2 8
F ( s )  2 e 4 s  e 4 s   2   e 4 s
s
s
s
s


(b) F ( s )   f (t )e dt   5cos t (t  2)e st dt 5cos te st
0
(c)
 st
0
t2
 5 5cos(2)e
cos 2e2 s –2s
e  t  e  ( t  ) e 
f (t )  e  e (t  )u (t   )
F ( s )  e  e  s
1
e  ( s 1)

s 1
s 1
(d) sin 2t  sin[2(t   )  2 ]  sin 2(t   ) cos 2  cos 2(t   )sin 2
f (t )  cos 2 sin 2(t   )u (t   )  sin 2 cos 2(t   )u (t   )
2
s
F ( s )  cos 2 e  s 2
 sin 2 e  s 2
s 4
s 4
Chapter 15, Solution 9.
(a)
f ( t )  ( t  4) u ( t  2)  ( t  2) u ( t  2)  2 u ( t  2)
e -2s 2 e -2s
F(s)  2  2
s
s
(b)
g( t )  2 e -4t u ( t  1)  2 e -4 e -4(t -1) u ( t  1)
2 e -s
G (s)  4
e (s  4)
(c)
h ( t )  5 cos(2 t  1) u ( t )
cos(A  B)  cos(A) cos(B)  sin(A) sin(B)
cos(2t  1)  cos(2t ) cos(1)  sin(2t ) sin(1)
h ( t )  5 cos(1) cos(2 t ) u ( t )  5 sin(1) sin(2t ) u ( t )
2
s
 5 sin(1)  2
s 4
s 4
2.702 s 8.415
H(s)  2

s  4 s2  4
H(s)  5 cos(1) 
(d)
2
p( t )  6u ( t  2)  6u ( t  4)
P(s) 
6 - 2s 6 -4s
e  e
s
s
Chapter 15, Solution 10.
(a)
By taking the derivative in the time domain,
g( t )  (-t e -t  e -t ) cos( t )  t e -t sin( t )
g( t )  e -t cos( t )  t e -t cos( t )  t e -t sin( t )
G (s) 

s 1
d  s 1  d 
1
 
 

2
2
2
(s  1)  1 ds  (s  1)  1 ds  (s  1)  1
G (s) 
s 1
s 2  2s
2s  2


2
2
2 
2
s  2s  2 (s  2s  2)
(s  2s  2) 2
s 2 (s  2)
(s 2  2s  2) 2
(b)
By applying the time differentiation property,
G (s)  sF(s)  f (0)
where f ( t )  t e -t cos( t ) , f (0)  0
- d  s 1 
(s)(s 2  2s)

G (s)  (s)  

ds  (s  1) 2  1  (s 2  2s  2) 2
s 2 (s  2)
(s 2  2s  2) 2
Chapter 15, Solution 11.
(a)
Since L  cosh(at )  
2
(b)
Since L  sinh(at )  
2
s
s  a2
6 (s  1)
6 (s  1)
 2
F(s) 
2
(s  1)  4 s  2s  3
a
s  a2
(3)(4)
12
L  3 e -2t sinh(4t )  
 2
2
(s  2)  16 s  4s  12
F(s)  L  t  3 e -2t sinh(4t )  
-d
 12 (s 2  4s  12) -1 
ds
24 (s  2)
F(s)  (12)(2s  4)(s 2  4s  12) -2  2
(s  4s  12) 2
(c)
1
 (e t  e - t )
2
1
f ( t )  8 e -3t   (e t  e - t ) u ( t  2)
2
-2t
 4 e u ( t  2)  4 e-4t u ( t  2)
 4 e-4 e-2(t - 2) u ( t  2)  4 e-8 e-4(t - 2) u ( t  2)
cosh( t ) 
L  4 e -4 e -2(t -2) u ( t  2)  4 e -4 e -2s  L  e -2 u ( t )
4 e -(2s 4)
L  4 e -4 e -2(t -2) u ( t  2) 
s2
Similarly, L  4 e -8 e -4(t -2) u ( t  2) 
4 e -(2s8)
s4
Therefore,
4 e -(2s 4) 4 e -(2s8) e -(2s 6)  (4 e 2  4 e -2 ) s  (16 e 2  8 e -2 )
F(s) 


s2
s4
s 2  6s  8
Chapter 15, Solution 12.
G(s) 
s2
s2
 2
2
2
( s  2)  4
s  4 s  20
Chapter 15, Solution 13.


(a) tf (t )

d
F (s)
ds
If f(t) = cost, then F(s)
s
s2  1
and -
L (t cos t ) 
d
(s 2  1)(1)  s(2s)
F(s) 
ds
(s 2  1) 2
s2  1
(s 2  1) 2
(b) Let f(t) = e-t sin t.
F (s) 
1
1
 2
2
( s  1)  1 s  2s  2
dF ( s 2  2s  2)(0)  (1)(2s  2)

ds
( s 2  2s  2) 2
dF
2(s  1)
L (e  t t sin t )  
 2
ds (s  2s  2) 2
(c )
f (t )
t



 F (s)ds
s
Let f (t )  sin t , then F ( s ) 


s 2
2

1
s
 sin  t 
L
 2
ds   tan 1
2



 t  s s 

s

s


 tan 1  tan 1
2
s

Chapter 15, Solution 14.
Taking the derivative of f(t) twice, we obtain the figures below.
f’(t)
5
0
t
2
4
6
-2.5
f’’(t)
5  (t)
0
2.5(t-6)
2
6
-7.5(t-2)
f” = 5δ(t) – 7.5δ(t–2) + 2.5δ(t–6)
Taking the Laplace transform of each term,
s2F(s) = 5 – 7.5e–2s + 2.5e–6s or F(s ) 
5
e 2 s
e 6 s
 7.5 2  2.5 2
s
s
s
Please note that we can obtain the same answer by representing the function as,
f(t) = 5tu(t) – 7.5u(t–2) + 2.5u(t–6).
Chapter 15, Solution 15.
This is a periodic function with T=3.
F ( s)
F ( s )  1 3 s
1 e
To get F 1 (s), we consider f(t) over one period.
f 1 (t)
f 1 ’(t)
5
f 1 ’’(t)
5
5(t)
0
1
t
0
1
t
–5(t-1)
0
1
t
–5(t-1)
–5’(t-1)
f 1 ” = 5δ(t) –5δ(t–1) – 5δ’(t–1)
Taking the Laplace transform of each term,
s2F 1 (s) = 5 –5e–s – 5se–s or F 1 (s) = 5(1 – e–s – se–s)/s2
Hence,
F(s) = 5
1  e  s  se  s
s 2 (1  e  3s )
Alternatively, we can obtain the same answer by noting that f 1 (t) = 5tu(t) – 5tu(t–1) –
5u(t–1).
Chapter 15, Solution 16.
f ( t )  5 u ( t )  3 u ( t  1)  3 u ( t  3)  5 u ( t  4)
F(s) 
1
 5  3 e -s  3 e - 3 s  5 e - 4 s 
s
Chapter 15, Solution 17.
Using Fig. 15.29, design a problem to help other students to better understand the
Laplace transform of a simple, non-periodic waveshape.
Although there are many ways to solve this problem, this is an example based on the
same kind of problem asked in the third edition.
Problem
Find the Laplace transform of f(t) shown in Fig. 15.29.
f(t)
2
1
0
1
2
Figure 15.29
t(s)
For Prob. 15.17.
Solution
Taking the derivative of f(t) gives f’(t) as shown below.
f’(t)
2(t)
t
-(t-1) – (t-2)
f’(t) = 2δ(t) – δ(t–1) – δ(t–2)
Taking the Laplace transform of each term,
sF(s) = 2 – e–s – e–2s which leads to
F(s) = [2 – e–s – e–2s]/s
We can also obtain the same answer noting that f(t) = 2u(t) – u(t–1) – u(t–2).
Chapter 15, Solution 18.
(a)
g ( t )  u ( t )  u ( t  1)  2  u ( t  1)  u ( t  2)  3  u ( t  2)  u ( t  3)
 u ( t )  u ( t  1)  u ( t  2)  3 u ( t  3)
1
G (s)  (1  e -s  e - 2s  3 e - 3s )
s
(b)
h ( t )  2 t  u ( t )  u ( t  1)  2  u ( t  1)  u ( t  3)
 (8  2 t )  u ( t  3)  u ( t  4)
 2t u ( t )  2 ( t  1) u ( t  1)  2 u ( t  1)  2 u ( t  1)  2 u ( t  3)
 2 ( t  3) u ( t  3)  2 u ( t  3)  2 ( t  4) u ( t  4)
 2t u ( t )  2 ( t  1) u ( t  1)  2 ( t  3) u ( t  3)  2 ( t  4) u ( t  4)
H(s) 
2
2 - 3s 2 - 4 s
2
-s
 2 e  2 (1  e -s  e - 3s  e -4s )
2 (1  e )  2 e
s
s
s
s
Chapter 15, Solution 19.
Since L ( t )  1 and T  2 , F(s) 
1
1  e - 2s
Chapter 15, Solution 20.
Using Fig. 15.32, design a problem to help other students to better understand the Laplace
transform of a simple, periodic waveshape.
Although there are many ways to solve this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
The periodic function shown in Fig. 15.32 is defined over its period as
sin  t ,
g (t )  
0,
0  t 1
1 t  2
Find G(s).
Figure 15.32
Solution
Let
g 1 ( t )  sin(t ), 0  t  1
 sin( t )  u ( t )  u ( t  1) 0 &lt; t &lt; 2
 sin(t ) u ( t )  sin(t ) u ( t  1)
Note that sin(( t  1))  sin(t  )  - sin(t ) .
So,
g1 ( t )  sin( t) u(t)  sin( ( t - 1)) u(t - 1)
G 1 (s) 

(1  e -s )
s  2
2
G 1 (s)
 (1  e -s )
G (s) 

1  e -2s (s 2   2 )(1  e - 2s )
Chapter 15, Solution 21.
T  2
Let
t 

f1 ( t )  1    u ( t )  u ( t  2)
 2 
t
1
f1 ( t )  u ( t ) 
u(t) 
( t  2) u ( t  2)
2
2

1
1
e - 2s 2 s  - 1  e -2s
F1 (s)  


s 2s 2 2s 2
2s 2
F(s) 
F1 (s)
2s  1  e 2 s

1  e -Ts 2s 2 (1  e -2 s )

Chapter 15, Solution 22.
(a)
Let
g1 ( t )  2t, 0  t  1
 2 t  u ( t )  u ( t  1)
 2t u ( t )  2 ( t  1) u ( t  1)  2 u ( t  1)
2 2 e -s 2 -s
G 1 (s)  2  2  e
s
s
s
(b)
G (s) 
G 1 (s)
, T 1
1  e -sT
G (s) 
2 (1  e -s  s e -s )
s 2 (1  e -s )
Let h  h 0  u ( t ) , where h 0 is the periodic triangular wave.
Let h 1 be h 0 within its first period, i.e.
 2t
0  t 1
h 1 (t)  
 4  2t 1  t  2
h 1 ( t )  2 t u ( t )  2 t u ( t  1)  4u ( t  1)  2 t u ( t  1)  2 ( t  2) u ( t  2)
h 1 ( t )  2 t u ( t )  4 ( t  1) u ( t  1)  2 ( t  2) u ( t  2)
2 4 -s 2 e -2s
2
H 1 (s)  2  2 e  2  2 (1  e -s ) 2
s
s
s
s
H 0 (s) 
H(s) 
2 (1  e -s ) 2
s 2 (1  e -2s )
1 2 (1  e -s ) 2

s s 2 (1  e - 2s )
Chapter 15, Solution 23.
(a)
Let
1 0  t 1
f1 ( t )  
- 1 1  t  2
f 1 ( t )   u ( t )  u ( t  1)   u ( t  1)  u ( t  2)
f 1 ( t )  u ( t )  2 u ( t  1)  u ( t  2)
1
1
F1 (s)  (1  2 e -s  e -2s )  (1  e -s ) 2
s
s
F1 (s)
, T2
(1  e -sT )
(1  e -s ) 2
F(s) 
s (1  e - 2s )
F(s) 
(b)
Let
h 1 ( t )  t 2  u ( t )  u ( t  2)  t 2 u ( t )  t 2 u ( t  2)
h 1 ( t )  t 2 u ( t )  ( t  2) 2 u ( t  2)  4 ( t  2) u ( t  2)  4 u ( t  2)
H 1 (s) 
4
4
2
- 2s
)  2 e -2s  e -2s
3 (1  e
s
s
s
H 1 (s)
, T2
(1  e -Ts )
2 (1  e -2s )  4s e -2s (s  s 2 )
H(s) 
s 3 (1  e - 2s )
H(s) 
Chapter 15, Solution 24.
Design a problem to help other students to better understand how to find the initial and
final values of a transfer function.
Although there are many ways to solve this problem, this is an example based on the
same kind of problem asked in the third edition.
Problem
Given that
F ( s) 
s 2  10s  6
s ( s  1)2 ( s  2)
Evaluate f(0) and f(  ) if they exist.
Solution
0
f(0) =
f ()  lim sF ( s )  lim
s 0
s 0
s  10s  6
6

 3= 3
2
( s  1) ( s  2) (1)(2)
2
Chapter 15, Solution 25.
5s ( s  1)
5(1  1/ s )
 lim
5
s 
s  ( s  2)( s  3)
s  (1  2 / s )(1  3 / s )
5s ( s  1)
f ()  lim sF ( s )  lim
0
s 0
s  0 ( s  2)( s  3)
(a)
f (0)  lim sF ( s )  lim
(b) F ( s ) 
5( s  1)
A
B


( s  2)( s  3) s  2 s  3
5(1)
5(2)
 5,
B
 10
1
1
5
10


 f (t )  5e 2t  10e 3t
F (s) 
s2 s3
A
f(0) = -5 + 10 = 5
f(  )= -0 + 0 = 0
Chapter 15, Solution 26.
(a)
5s 3  3s
f (0)  lim sF ( s )  lim 3
5
s 
s  s  4 s 2  6
Two poles are not in the left-half plane.
f () does not exist
(b)
s 3  2s 2  s
s 
s  4( s  2)( s 2  2 s  4)
2 1
1  2
s s
 lim
 0.25
s  
2  2 4 
1   1   2 
 s  s s 
f (0)  lim sF ( s )  lim
One pole is not in the left-half plane.
f () does not exist
Chapter 15, Solution 27.
(a)
f ( t )  u(t )  2 e -t u(t )
(b)
G (s) 
3 (s  4)  11
11
 3
s4
s4
g( t )  3 (t )  11 e -4t u(t )
(c)
4
A
B


(s  1)(s  3) s  1 s  3
A  2,
B  -2
2
2
H(s) 

s 1 s  3
H(s) 
h ( t )  [2 e -t  2 e -3t ] u(t)
(d)
12
A
B
C


2
2 
(s  2) (s  4) s  2 (s  2)
s4
12
12
B
 6, C
3
2
(-2) 2
12  A (s  2) (s  4)  B (s  4)  C (s  2) 2
J (s) 
Equating coefficients :
s2 :
0 AC 
 A  -C  -3
s1 :
s0 :
0  6A  B  4C  2A  B 
 B  -2A  6
12  8A  4B  4C  -24  24  12  12
J (s) 
-3
6
3

2 
s  2 (s  2)
s4
j( t )  [3 e -4t  3 e -2t  6 t e -2t ] u(t)
Chapter 15, Solution 28.
Design a problem to help other students to better understand how to find the inverse
Laplace transform.
Although there are many ways to solve this problem, this is an example based on the
same kind of problem asked in the third edition.
Problem
Find the inverse Laplace transform of the following functions:
20( s  2)
(a) F ( s ) 
s ( s 2  6s  25)
(b) P ( s ) 
6 s 2  36 s  20
( s  1)( s  2)( s  3)
Solution
(a) F ( s ) 
20( s  2)
A
Bs  C
  2
2
s ( s  6s  25) s s  6s  25
20( s  2)  A( s 2  6s  25s )  Bs 2  Cs
Equating components,
s2 :
0 = A + B or B= - A
s:
20 = 6A + C
constant: 40 – 25 A or A = 8/5, B = -8/5, C= 20 – 6A= 52/5
8
52
8
24 52
 s
 ( s  3)  
8
8
5
5
F (s)   5 2 5 2   5
2
2
5s ( s  3)  4
5s
( s  3)  4
8
8
19
f (t )  u (t )  e 3t cos 4t  e 3t sin 4t
5
5
5
6 s 2  36 s  20
A
B
C



( s  1)( s  2)( s  3) s  1 s  2 s  3
6  36  20
A
 5
(1  2)(1  3)
24  72  20
B
 28
(1)(1)
54  108  20
C
 17
(2)(1)
(b) P ( s ) 
P( s) 
5
28
17


s 1 s  2 s  3
p (t )  (5e  t  28e 2t  17e 3t )u (t )
Chapter 15, Solution 29.
V(s) 
2
As  B

; 2s 2  8s  26  As 2  Bs  2s  26  A  2 and B  6
2
2
s (s  2)  3
V(s) 
2
2(s  2)
2
3


2
2
s (s  2)  3
3 (s  2) 2  3 2
2
v(t) = ( 2  2e  2t cos 3t  e 2 t sin 3t )u(t ), t  0
3
Chapter 15, Solution 30.
(a)
A
Bs  C
6 s 2  8s  3
  2
F1 ( s ) 
2
s ( s  2 s  5) s s  2s  5
6 s 2  8s  3  A( s 2  2 s  5)  Bs 2  Cs
We equate coefficients.
6=A+B
s2 :
s:
8= 2A + C
constant: 3=5A or A=3/5
B=6-A = 27/5,
C=8-2A = 34/5
F1 ( s ) 
3 / 5 27 s / 5  34 / 5 3 / 5 27( s  1) / 5  7 / 5
 2


s
s  2s  5
s
( s  1) 2  22
7
 3 27

f1 (t )    e  t cos 2t  e t sin 2t  u (t )
10
5 5

s 2  5s  6
A
B
C



2
2
( s  1) ( s  4) s  1 ( s  1)
s4
2
s  5s  6  A( s  1)( s  4)  B ( s  4)  C ( s  1) 2
Equating coefficients,
(b) F2 ( s ) 
1=A+C
s2 :
s:
5=5A+B+2C
constant: 6=4A+4B+C
Solving these gives
A=7/9, B= 2/3, C=2/9
F2 ( s ) 
7/9
2/3
2/9


2
s  1 ( s  1) s  4
2
2
7

f 2 (t )   e  t  te  t  e 4t  u (t )
3
9
9

10
A
Bs  C

 2
2
( s  1)( s  4s  8) s  1 s  4 s  8
2
10  A( s  4 s  8)  B ( s 2  s )  C ( s  1)
s2 :
0 = A + B or B = -A
s:
0=4A+ B + C
constant:
10=8A+C
Solving these yields
(c ) F 3 ( s ) 
A=2, B= -2, C= -6
2
2 s  6
2
2( s  1)
4
F 3 (s) 
 2



2
2
s  1 s  4s  8 s  1 ( s  1)  2 ( s  1) 2  22
f 3 (t) = (2e–t – 2e–tcos(2t) – 2e–tsin(2t))u(t).
Chapter 15, Solution 31.
(a)
F(s) 
10s
A
B
C



(s  1)(s  2)(s  3) s  1 s  2 s  3
- 10
 -5
2
- 20
B  F(s) (s  2) s -2 
 20
-1
- 30
C  F(s) (s  3) s -3 
 -15
2
A  F(s) (s  1) s -1 
F(s) 
-5
20
15


s 1 s  2 s  3
f ( t )  (-5 e -t  20 e -2t  15 e -3t )u(t )
(b)
F(s) 
2s 2  4s  1
A
B
C
D


3 
2 
(s  1)(s  2)
s  1 s  2 (s  2)
(s  2) 3
A  F(s) (s  1) s -1  -1
D  F(s) (s  2) 3
s  -2
 -1
2s 2  4s  1  A(s  2)(s 2  4s  4)  B(s  1)(s 2  4s  4)
 C(s  1)(s  2)  D(s  1)
Equating coefficients :
s3 :
0 AB 
 B  -A  1
s2 :
s1 :
s0 :
F(s) 
2  6A  5B  C  A  C 
 C  2  A  3
4  12A  8B  3C  D  4A  3C  D
4  6A D 
 D  -2  A  -1
1  8A  4B  2C  D  4A  2C  D  -4  6  1  1
-1
1
3
1


2 
s  1 s  2 (s  2)
(s  2) 3
t 2 -2t
e
2

t2 
f ( t )  (-e -t   1  3 t   e - 2t )u(t )
2

f(t)  -e - t  e -2t  3 t e -2t 
(c)
F(s) 
s 1
A
Bs  C

 2
2
(s  2)(s  2s  5) s  2 s  2s  5
A  F(s) (s  2) s -2 
-1
5
s  1  A (s 2  2s  5)  B (s 2  2s)  C (s  2)
Equating coefficients :
s2 :
s1 :
s0 :
F(s) 
1
5
1  2A  2B  C  0  C 
 C  1
1  5A  2C  -1  2  1
0 AB 
 B  -A 
-1 5
1 5 s 1
-1 5
1 5 (s  1)
45


2
2 
2
2 
s  2 (s  1)  2
s  2 (s  1)  2
(s  1) 2  2 2
f ( t )  (-0.2 e -2t  0.2 e -t cos( 2t )  0.4 e -t sin( 2t ))u(t )
Chapter 15, Solution 32.
(a)
F(s) 
8 (s  1)(s  3) A
B
C
 

s (s  2)(s  4) s s  2 s  4
(8)(3)
3
(2)(4)
(8)(-1)
2
B  F(s) (s  2) s-2 
(-4)
(8)(-1)(-3)
3
C  F(s) (s  4) s-4 
(-4)(-2)
A  F(s) s s 0 
F(s) 
3
2
3


s s2 s4
f ( t )  3 u(t )  2 e -2t  3 e -4t
(b)
F(s) 
s 2  2s  4
A
B
C


2 
(s  1)(s  2)
s  1 s  2 (s  2) 2
s 2  2s  4  A (s 2  4s  4)  B (s 2  3s  2)  C (s  1)
Equating coefficients :
1 A B 
 B  1  A
s2 :
1
- 2  4A  3B  C  3  A  C
s :
0
s :
4  4A  2B  C  -B  2 
 B  -6
A  1 B  7
F(s) 
C  -5 - A  -12
7
6
12


s  1 s  2 (s  2) 2
f ( t )  7 e -t  6 (1  2 t ) e -2t
(c)
s2 1
A
Bs  C
F(s) 

 2
2
(s  3)(s  4s  5) s  3 s  4s  5
s 2  1  A (s 2  4s  5)  B (s 2  3s)  C (s  3)
Equating coefficients :
1 A B 
 B  1  A
s2 :
s1 :
0  4A  3B  C  3  A  C 
 A  C  -3
s0 :
1  5A  3C  -9  2A 
 A  5
B  1  A  -4
F(s) 
C  -A  3  -8
4 (s  2)
5
4s  8
5



2
s  3 (s  2)  1 s  3 (s  2) 2  1
f ( t )  5 e -3t  4 e -2t cos(t )
Chapter 15, Solution 33.
(a)
F(s) 
6 (s  1)
6
As  B
C



s4 1
(s 2  1)(s  1) s 2  1 s  1
6  A (s 2  s)  B (s  1)  C (s 2  1)
Equating coefficients :
s2 :
0 AC 
 A  -C
s1 :
0 AB 
 B  -A  C
0
6  B  C  2B 
 B  3
s :
A  -3 ,
F(s) 
B  3,
C3
3
- 3s  3
3
- 3s
3
 2

 2
 2
s 1 s 1 s 1 s 1 s 1
f ( t )  ( 3 e -t  3 sin( t )  3 cos(t ))u(t )
(b)
F(s) 
s e - s
s2 1
f ( t )  cos(t   ) u(t   )
(c)
8
A
B
C
D


3 
2 
s (s  1)
s s  1 (s  1)
(s  1) 3
A  8,
D  -8
F(s) 
8  A (s 3  3s 2  3s  1)  B (s 3  2s 2  s)  C (s 2  s)  D s
Equating coefficients :
s3 :
0 AB 
 B  -A
s2 :
0  3A  2B  C  A  C 
 C  -A  B
s1 :
s0 :
0  3A  B  C  D  A  D 
 D  -A
A  8, B  8, C  8, D  8
F(s) 
8
8
8
8


2 
s s  1 (s  1)
(s  1) 3
f ( t )  8  1  e -t  t e -t  0.5 t 2 e -t  u(t )


(a) (3 e-t  3 sin( t )  3 cos(t ))u ( t ) , (b) cos( t  ) u ( t  ) , (c) 8 1  e -t  t e-t  0.5 t 2 e-t u ( t )
Chapter 15, Solution 34.
(a)
s2  4  3
3
F(s)  10  2
 11  2
s 4
s 4
f ( t )  11 (t )  1.5 sin( 2t )
(b)
e -s  4 e -2s
G (s) 
(s  2)(s  4)
1
A
B


(s  2)(s  4) s  2 s  4
Let
A 1 2
G (s) 
B 1 2
 1
1 
e -s  1
1 

  2 e -2s 



s  2 s  4
2 s  2 s  4
g( t )  0.5  e -2(t -1)  e -4(t -1)  u(t  1)  2  e -2(t - 2)  e -4(t - 2)  u(t  2)
(c)
Let
s 1
A
B
C
 

s (s  3)(s  4) s s  3 s  4
A  1 12 ,
B  2 3,
C  -3 4
1 1 23
3 4  -2s
e
H(s)    

12 s s  3 s  4 
1 2

3
h ( t )    e - 3(t - 2)  e -4(t - 2)  u(t  2)
 12 3

4
Chapter 15, Solution 35.
(a)
G (s) 
Let
A  2,
G (s) 
s3
A
B


(s  1)(s  2) s  1 s  2
B  -1
2
1

s 1 s  2

 g( t )  2 e - t  e -2t
F(s)  e -6s G (s) 
 f ( t )  g( t  6) u ( t  6)
f ( t )   2 e -(t -6)  e -2(t -6)  u(t  6)
(b)
Let
G (s) 
A  1 3,
1
A
B


(s  1)(s  4) s  1 s  4
B  -1 3
G (s) 
1
1

3 (s  1) 3 (s  4)
g( t ) 
1 -t
 e  e -4t 
3
F(s)  4 G (s)  e -2t G (s)
f ( t )  4 g( t ) u ( t )  g ( t  2) u ( t  2)
4
1
f ( t )   e -t  e -4t  u(t )   e -(t - 2)  e -4(t - 2)  u(t  2)
3
3
(c)
Let
G (s) 
s
A
Bs  C

 2
2
(s  3)(s  4) s  3 s  4
A  - 3 13
s  A (s 2  4)  B (s 2  3s)  C (s  3)
Equating coefficients :
0 AB 
 B  -A
s2 :
1
s :
1  3B  C
s0 :
0  4A  3C
A  - 3 13 ,
13 G (s) 
B  3 13 ,
C  4 13
- 3 3s  4

s  3 s2  4
13 g( t )  -3 e -3t  3 cos(2t )  2 sin(2t )
F(s)  e -s G (s)
f ( t )  g( t  1) u ( t  1)
1
 - 3 e -3(t-1)  3 cos( 2 (t  1))  2 sin( 2 (t  1)) u(t  1)
f (t) 
13
Chapter 15, Solution 36.
(a)
X ( s)  3
B  1 6,
C
D 
1
A B
 3  2 


s ( s  2)( s  3)
s  2 s  3
s s
2
C 1 4,
D  -1 9
1  A (s 3  5s 2  6s)  B (s 2  5s  6)  C (s 3  3s 2 )  D (s 3  2s 2 )
Equating coefficients :
0  ACD
s3 :
2
s :
0  5A  B  3C  2D  3A  B  C
1
0  6 A  5B
s :
0
s :
1  6B 
 B  1 6
A  - 5 6 B  - 5 36
19 
 - 5 36 1 6 1 4
 2 

X ( s )  3

s
s  2 s  3
 s
1
3
1
5

x(t)  
u(t )  t  e- 2t  e- 3t u(t )
4
3
2
 12

(b)
A
B
C 
1

 2 

2
2 
s ( s  1)
 s s  1 ( s  1) 
A  1,
C  -1
Y (s)  2
1  A (s 2  2s  1)  B (s 2  s)  C s
Equating coefficients :
s2 :
0 AB 
 B  -A
s1 :
s0 :
0  2A  B  C  A  C 
 C  -A
1  A, B  -1, C  -1
1
1
1 


Y ( s )  2 
2 
 s s  1 ( s  1) 


y( t )  2  2e-t  2t e-t u(t )
(c)
B
Cs  D 
A
Z ( s )  5 
 2

 s s  1 s  6s  10 
A  1 10 ,
B  -1 5
1  A (s 3  7s 2  16s  10)  B (s 3  6s 2  10s)  C (s 3  s 2 )  D (s 2  s)
Equating coefficients :
0  A BC
s3 :
2
0  7 A  6 B  C  D  6 A  5B  D
s :
1
s :
0  16A  10B  D  10A  5B 
 B  -2A
s0 :
1  10A 
 A  1 10
A  1 10 ,
B  -2A  - 1 5 ,
D  4A 
C  A  1 10 ,
s4
10
1
2
Z ( s)  
 2
s s  1 s  6s  10
5
s3
1
2
1

2 Z ( s)  

2
s s  1 ( s  3)  1 ( s  3) 2  1


z( t )  0.5 1  2 e-t  e-3t cos(t )  e-3t sin(t ) u(t )
4
10
Chapter 15, Solution 37.
(a) H ( s ) 
s4
A
B
 
s ( s  2) s s  2
s+4 =A(s+2) + Bs
Equating coefficients,
s:
1=A+B
constant: 4= 2 A
A =2, B=1-A = -1
2
1

s s2
h(t )  2u (t )  e 2t u (t )  (2  e 2t )u (t )
H (s) 
(b)
G ( s) 
A
Bs  C
 2
s  3 s  2s  2
s 2  4s  5  ( Bs  C )( s  3)  A( s 2  2 s  2)
Equating coefficients,
s2:
1= B + A
(1)
s:
4 = 3B + C + 2A
(2)
Constant: 5 =3C + 2A
(3)
Solving (1) to (3) gives
2
3
7
A ,
B , C
5
5
5
0.4 0.6s  1.4
0.4 0.6( s  1)  0.8
G ( s) 
 2


s  3 s  2s  2 s  3
( s  1) 2  1
g (t )  0.4e 3t  0.6e t cos t  0.8e  t sin t u(t)
(c) f (t )  e 2(t  4)u (t  4)
(d) D( s ) 
10s
As  B Cs  D
 2
 2
2
( s  1)( s  4) s  1
s 4
2
10 s  ( s 2  4)( As  B)  ( s 2  1)(Cs  D)
Equating coefficients,
0=A+C
s3 :
s2 :
0=B+D
s:
10 = 4A + C
constant: 0 = 4B+D
A = -10/3, B = 0, C = -10/3, D = 0
10s / 3 10s / 3

s2  1 s2  4
10
10
d (t )  cos t  cos 2t u(t)
3
3
D( s ) 
Chapter 15, Solution 38.
(a)
s 2  4s
s 2  10s  26  6s  26
F(s)  2

s  10s  26
s 2  10s  26
6s  26
F(s)  1  2
s  10s  26
6 (s  5)
4
F(s)  1 
2
2 
(s  5)  1
(s  5) 2  12
f ( t )  (t )  6 e -t cos(5t )  4 e -t sin( 5t )
(b)
F(s) 
5s 2  7s  29
A
Bs  C
  2
2
s (s  4s  29) s s  4s  29
5s 2  7s  29  A (s 2  4s  29)  B s 2  C s
Equating coefficients :
29  29A 
 A  1
s0 :
s1 :
2
s :
A  1,
7  4A  C 
 C  7  4A  3
5 AB 
 B  5  A  4
B  4,
C3
4 (s  2)
1
4s  3
1
5
F(s)   2
 
2
2 
s s  4s  29 s (s  2)  5
(s  2) 2  5 2
f ( t )  u(t )  4 e -2t cos(5t )  e -2t sin( 5t )
Chapter 15, Solution 39.
(a)
2s 3  4s 2  1
As  B
Cs  D
F(s)  2


(s  2s  17)(s 2  4s  20) s 2  2s  17 s 2  4s  20
s 3  4s 2  1  A(s 3  4s 2  20s)  B(s 2  4s  20)
 C(s 3  2s 2  17s)  D(s 2  2s  17)
Equating coefficients :
s3 :
2 AC
2
4  4 A  B  2C  D
s :
1
s :
0  20A  4B  17C  2D
0
1  20B  17 D
s :
Solving these equations (Matlab works well with 4 unknowns),
A  -1.6 ,
B  -17.8 ,
C  3.6 ,
D  21
- 1.6s  17.8
3.6s  21
 2
2
s  2s  17 s  4s  20
(-1.6)(s  1)
(-4.05)(4)
(3.6)(s  2)
(3.45)(4)
F(s) 
2
2 
2
2 
2
2 
(s  1)  4
(s  1)  4
(s  2)  4
(s  2) 2  4 2
F(s) 
f (t) 
[ - 1.6 e -t cos(4t )  4.05 e -t sin( 4t )  3.6 e -2t cos(4t )  3.45 e -2t sin( 4t ) ]u(t)
(b)
s2  4
As  B
Cs  D
 2
F(s)  2
 2
2
(s  9)(s  6s  3) s  9 s  6s  3
s 2  4  A (s 3  6s 2  3s)  B (s 2  6s  3)  C (s 3  9s)  D (s 2  9)
Equating coefficients :
s3 :
0 AC 
 C  -A
s2 :
1  6A  B  D
1
0  3A  6B  9C  6B  6C 
 B  -C  A
s :
0
4  3B  9D
s :
Solving these equations,
A  1 12 ,
B  1 12 ,
12 F(s) 
s 1
-s 5
 2
2
s  9 s  6s  3
C  - 1 12 ,
D  5 12
s 2  6s  3  0 

Let
G (s) 
-s5
s  5.449
-s5
F
s  0.551
E
G (s) 
- 6  36 - 12
 -0.551, - 5.449
2
-s5
E
F


s  6s  3 s  0.551 s  5.449
2
s  -0.551
 1.133
s  -5.449
 - 2.133
1.133
2.133

s  0.551 s  5.449
12 F(s) 
s
1
3
1.133
2.133
 2

2 
2 
s 3
3 s 3
s  0.551 s  5.449
2
f (t) 
[ 0.08333 cos( 3t )  0.02778 sin( 3t )  0.0944 e -0.551t  0.1778 e -5.449t ]u(t)
Chapter 15, Solution 40.
 4s 2  7s  13 
A
Bs  C

Let H(s)  

2
2
 (s  2)(s  2s  5)  s  2 s  2s  5
4s 2  7s  13  A(s 2  2s  5)  B(s 2  2s)  C(s  2)
Equating coefficients gives:
s2 :
4AB
s:
7  2A  2B  C


C  1
13  5A  2C


5A  15 or A  3, B  1
constant :
H(s) 
3
s 1
3
(s  1)  2



s  2 s 2  2s  5 s  2 (s  1) 2  2 2
Hence,
h ( t )  3e 2 t  e  t cos 2t  e  t sin 2t  3e 2 t  e  t (A cos  cos 2t  A sin  sin 2t )
where A cos   1,
A sin   1


  45 o
A  2,
Thus,
h( t ) 


2e  t cos( 2t  45 o )  3e 2t u(t )
Chapter 15, Solution 41.
We fold x(t) and slide on y(t). For t&lt;0, no overlapping as shown below. x(t) =0.
y(  )
4
2
4
6
8

t 2
4
6
8

0
t
-4
For 0 &lt; t &lt; 2, there is overlapping, as shown below.
y(  )
4
0
-4
t
z (t )   (2)(4)dt  8t
0
For 2 &lt; t &lt; 6, the two functions overlap, as shown below.
y(  )
4
0
t4
6
8

2
4
6
t 8

2
 8
2
t
-4
2
t
0
0
z (t )   (2)(4)d    (2)(4)d   16  8t
For 6&lt;t&lt;8, they overlap as shown below.
y(  )
4
0
-4
2
z (t ) 

t 6
6
t
2
6
(2)(4)d    (2)(4)d    (2)(4)d   8
For 8&lt; t &lt;12, they overlap as shown below.
t 6
6
2
 8
t
6
 16
y(  )
4
0
2
4
6
8
10 t 12 
4
6
8
10
-4
6

z (t ) 
t 6
8
(2)(4)d    (2)(4)d   8
6
6
8
 8  8t  80
t 6
6
For 12 &lt; t &lt; 14, they overlap as shown below.
y(  )
4
0
2
-4
8
z (t ) 
8
 (2)(4)d   8 t  6  112  8t
t 6
Hence,
z(t) = 8t,
16–8t,
–16,
8t–80,
112–8t,
0,
0&lt;t&lt;2
2&lt;t&lt;6
6&lt;t&lt;8
8&lt;t&lt;12
12&lt;t&lt;14
otherwise.
12 t 
1
Chapter 15, Solution 42.
Design a problem to help other students to better understand how to convolve two
functions together.
Although there are many ways to solve this problem, this is an example based on the
same kind of problem asked in the third edition.
Problem
Suppose that f(t) = u(t) – u(t-2). Determine f(t)*f(t).
Solution
For 0&lt;t&lt;2, the signals overlap as shown below.
1
t-2
0
t

2
t
y (t )  f (t ) * f (t )   (1)(1)d   t
0
For 2 &lt; t&lt; 4, they overlap as shown below.
1
0
2
y (t ) 
t 2
Thus,
2
 (1)(1)d   t t  2  4  t
t-2
2
t

2
 t,
0t 2

y (t )  4  t ,
2t 4
 0, otherwise

Chapter 15, Solution 43.
(a)
For 0  t  1 , x ( t  ) and h () overlap as shown in Fig. (a).
t
2 t t 2

y( t )  x ( t )  h ( t )  0 (1)() d 
2 0 2
x(t - )
1
1
h()
t-1
0 t
1

t-1 1
0
(a)
t

(b)
For 1  t  2 , x ( t  ) and h () overlap as shown in Fig. (b).
1
t
2 1
-1 2
t
y( t )  t 1 (1)() d  1 (1)(1) d 
t  2t  1
t 1   1 
2
2
For t  2 , there is a complete overlap so that
y( t )  t 1 (1)(1) d   tt 1  t  ( t  1)  1
t
Therefore,

t 2 2,
0t1
 2
- (t 2)  2t  1, 1  t  2
y( t )  
1,
t2


0,
otherwise
(b)
For t  0 , the two functions overlap as shown in Fig. (c).
y( t )  x ( t )  h ( t )  0 (1) 2 e - d  -2 e -
t
x(t-)
2
t
0
h() = 2e-
1
0
t
(c)
Therefore,
y( t )  2 (1  e -t ), t  0

(c)
For - 1  t  0 , x ( t  ) and h () overlap as shown in Fig. (d).
t 1
2 t 1 1
y( t )  x ( t )  h ( t )  0 (1)() d 
 ( t  1) 2
2 0
2
x(t - )
1
t-1 -1
h()
t 0
t+1 1
2

(d)
For 0  t  1 , x ( t  ) and h () overlap as shown in Fig. (e).
y( t )  0 (1)() d  1 (1)(2  ) d
t 1
1
y( t ) 
2
2
1
0

2 
-1
1
  2   1t 1  t 2  t 
2
2
2

1
-1 t-1
0 t
1 t+1 2

(e)
For 1  t  2 , x ( t  ) and h () overlap as shown in Fig. (f).
y( t )  t 1 (1)() d  1 (1)(2  ) d
1
y( t ) 
2
2
2
1
t 1

2 
-1
1
  2   12  t 2  t 
2
2
2

1
0
t-1 1
t
(f)
2 t+1

For 2  t  3 , x ( t  ) and h () overlap as shown in Fig. (g).

2
2  2
9
1

y( t )  t 1 (1)(2  ) d  2   t 1   3t  t 2
2
2
2

1
0
1 t-1 2
t
(g)
Therefore,
 (t 2 2)  t  1 2, - 1  t  0
 2
- ( t 2 )  t  1 2 , 0  t  2
y( t )   2
(t 2)  3t  9 2, 2  t  3

0,
otherwise
t+1

Chapter 15, Solution 44.
(a)
For 0  t  1 , x ( t  ) and h () overlap as shown in Fig. (a).
y( t )  x ( t )  h ( t )  0 (1)(1) d  t
t
x(t - )
h()
1
0 t
t-1
1
2

-1
(a)
For 1  t  2 , x ( t  ) and h () overlap as shown in Fig. (b).
y( t )  t 1 (1)(1) d  1 (-1)(1) d   1t 1   1t  3  2 t
1
t
For 2  t  3 , x ( t  ) and h () overlap as shown in Fig. (c).
y( t )  t 1 (1)(-1) d  -
2
2
t 1
1
 t3
1
0
t-1 1
t
2

0
1 t-1
-1
-1
(b)
Therefore,
0t 1
 t,
 3  2t , 1  t  2

y( t )  
2t3
 t  3,
 0,
otherwise
(c)
2 t

(b)
For t  2 , there is no overlap. For 2  t  3 , f1 ( t  ) and f 2 () overlap,
as shown in Fig. (d).
y( t )  f 1 ( t )  f 2 ( t )  2 (1)( t  ) d
t

2
  t 
2

 t t2
 2   2 t  2
2

f 1 (t - )
f 2 ( )
1
1 t-1 2
0
3
t
4
5

4
5

(d)
1
0
2 t-1 3
1
t
(e)
For 3  t  5 , f1 ( t  ) and f 2 () overlap as shown in Fig. (e).

t
2 
1
y( t )  t 1 (1)( t  ) d   t   tt 1 
2
2

For 5  t  6 , the functions overlap as shown in Fig. (f).

5
2 
-1
y( t )  t 1 (1)( t  ) d   t   5t 1  t 2  5t  12
2
2

1
0
1
2
3
(f)
4 t-1 5
t

Therefore,
 (t 2 2)  2t  2,
2t3

1 2,
3t5

y( t )   2
- (t 2)  5t  12, 5  t  6

0,
otherwise
Chapter 15, Solution 45.
y (t )  h(t ) * x(t )   4e 2t u (t )  *  (t )  2e2t u (t ) 
t
 4e2t u (t ) *  (t )  4e2t u (t ) * 2e2t u (t )  4e2t u (t )  8e2t  eo d 
0
2 t
2 t
 4e u (t )  8te u (t )
Chapter 15, Solution 46.
(a) x(t ) * y (t )  2 (t ) * 4u (t )  8u (t )
(b) x(t ) * z (t )  2 (t ) * e 2t u (t )  2e 2t u (t )
t
(c ) y (t ) * z (t )  4u (t ) * e 2t u (t )  4  e 2  d  
0
4e 2  t
 2(1  e2t )
2 0
(d) y (t ) *[ y (t )  z (t )]  4u (t ) *[4u (t )  e2t u (t )]  4  [4u ( )  e2  u ( )]d 
t
 4  [4  e2  ]d   4[4t 
0
e 2  t
]  16t  2e2t  2
2 0
Chapter 15, Solution 47.
s
A
B


( s  1)( s  2) s  1 s  2
s=A(s+2) + B(s+1)
We equate the coefficients.
(a) H ( s ) 
s:
1= A+B
constant: 0 =2A +B
Solving these, A = -1, B= 2.
1
2

s 1 s  2
h(t )  (e  t  2e 2t )u (t )
H (s) 
Y ( s)
s
1

 Y (s)  H (s) X (s) 
( s  1)( s  2) s
X ( s)
1
C
D
Y (s) 


( s  1)( s  2) s  1 s  2
(b) H ( s ) 
C=1 and D=-1 so that
1
1
Y (s) 

s 1 s  2
y (t )  (e  t  e 2t )u (t )
Chapter 15, Solution 48.
(a)
Let G (s) 
2
2

s  2s  5 (s  1) 2  2 2
2
g( t )  e -t sin(2 t )
F(s)  G (s) G (s)
f ( t )  L -1  G (s) G (s)  0 g () g ( t  ) d
t
f ( t )  0 e - sin(2) e -( t ) sin( 2( t  )) d
t
sin(A) sin(B) 
1
 cos(A  B)  cos(A  B)
2
1 - t t -
e  e  cos(2t )  cos(2( t  2)) d
2 0
t
e -t
e -t t
f (t) 
cos(2 t ) 0 e -2 d  0 e -2 cos(2 t  4) d
2
2
-t
-t
-2 
e
e t e t -2 
f (t) 
cos(2 t ) 

 e  cos(2t ) cos(4)  sin(2t ) sin(4) d
2
-2 0 2 0
t
1
e -t
f ( t )  e -t cos(2 t ) (-e -2 t  1)  cos(2 t ) 0 e -2  cos(4) d
4
2
-t
t
e
 sin( 2 t ) 0 e -2 sin( 4) d
2
f (t) 
f (t) 
1 -t
e cos(2t ) (1  e -2 t )
4
 e -2

e -t
- 2cos(4)  4 sin(4) 0t
 cos(2t ) 
2
 4  16


 e -2

e -t
- 2sin(4)  4 cos(4) 0t
sin(2t ) 
2
 4  16

e -t
e -3t
e -t
e -3t
cos( 2t ) 
cos( 2t ) 
cos( 2t ) 
cos( 2t ) cos(4t )
f (t) 
2
4
20
20
e -3t
e -t
cos( 2t ) sin( 4t ) 

sin( 2t )
10
10
e -t
e -t

sin( 2t ) sin( 4t ) 
sin( 2t ) cos(4t )
20
10
(b)
Let
X(s) 
2
,
s 1
Y(s) 
s
s4
y( t )  cos(2t ) u ( t )
x ( t )  2 e -t u ( t ) ,
F(s)  X(s) Y(s)
f ( t )  L -1  X(s) Y (s)  0 y() x ( t  ) d

f ( t )  0 cos(2)  2 e -(t ) d
t
f ( t )  2 e -t 
e
cos(2)  2 sin(2) 0t
1 4
2 -t t
e  e  cos(2t )  2 sin(2t )  1 
5
2
4
2
f ( t )  cos( 2t )  sin( 2t )  e -t
5
5
5
f (t) 
Chapter 15, Solution 49.
(a) t*eαtu(t) =
e a
e
(
t
)
d
t




0
a
t
t
a
0
t
e a
t
1 e at
 2 (a  1)  (e at  1)  2  2 (at  1) u(t)
a
a
a
a
0
t
t
0
0
(b) cos t *cos tu (t )   cos  cos(t   )d    {cos t cos  cos   sin t sin  cos }d 
t
t

 1
1
sin 2 t
cos 
]  sin t
 cos t  [1  cos 2 ]d   sin t  cos  d ( cos  )    cos t[ 
0
2
2
2
0
0

  2
=[ 0.5cos(t)(t+0.5sin(2t)) – 0.5sin(t)(cos(t) – 1)]u(t).
t

0 
Chapter 15, Solution 50.
Take the Laplace transform of each term.
s
2
V(s)  s v(0)  v (0)  2  s V(s)  v(0)  10 V(s) 
3s
s 4
2
3s
s 4
3
3s
s  7s
(s 2  2s  10) V (s)  s  2
 2
s 4 s 4
3
s  7s
As  B
Cs  D
V(s)  2
 2
 2
2
(s  4)(s  2s  10) s  4 s  2s  10
s 2 V(s)  s  2  2s V(s)  2  10 V(s) 
2
s 3  7s  A (s 3  2s 2  10s)  B (s 2  2s  10)  C (s 3  4s)  D (s 2  4)
Equating coefficients :
1 AC 
 C  1  A
s3 :
2
s :
0  2A  B  D
1
7  10A  2B  4C  6A  2B  4
s :
0
0  10B  4D 
 D  -2.5 B
s :
Solving these equations yields
9
12
A
,
B
,
26
26
C
17
,
26
D
- 30
26
1  9s  12
17s  30 
 2
2

26  s  4 s  2s  10 

1  9s
2
s 1
47
 6 2
 17 

V(s)   2
26  s  4
s 4
(s  1) 2  3 2 (s  1) 2  3 2 
V(s) 
v( t ) 
9
6
17
47
cos( 2t ) 
sin( 2t )  e -t cos( 3t )  e -t sin( 3t )
26
26
26
78
Chapter 15, Solution 51.
Taking the Laplace transform of the differential equation yields
s V(s)  sv(0)  v' (0) 5sV(s)  v(0)]  6V(s)  s10 1
s  5s  6V(s)  2s  4  10  s10 1  V(s)  (s 2s1)(s 16 2s)(s24 3)
2
or
2
2
Let V(s) 
A
B
C


,
s 1 s  2 s  3
A  5,
B  0,
C  3
Hence,


v(t) = 5e  t  3e 3 t u(t ) .
Chapter 15, Solution 52.
Take the Laplace transform of each term.
s
2
I(s)  s i(0)  i (0)  3  s I(s)  i(0)  2 I(s)  1  0
(s 2  3s  2) I(s)  s  3  3  1  0
I(s) 
s5
A
B


(s  1)(s  2) s  1 s  2
A  4,
I(s) 
B  -3
4
3

s 1 s  2
i( t )  (4 e -t  3 e -2t ) u(t )
Chapter 15, Solution 53.
Transform each term.
We begin by noting that the integral term can be rewritten as,
t
0 x ()e
 (t  )
d which is convolution and can be written as e–t*x(t).
Now, transforming each term produces,
X(s) =
X(s) 
s
2
s 1
s 1
2
s 1


1
s
 s  1 1
X(s)  
X(s)  2
s 1
 s 1 
s 1
s
2
s 1

1
2
s 1
x(t) = cos(t) + sin(t).
If partial fraction expansion is used we obtain,
x(t) = 1.4142cos(t–45˚).
This is the same answer and can be proven by using trigonometric identities.
Chapter 15, Solution 54.
Design a problem to help other students to better understand solving second order
differential equations with a time varying input.
Although there are many ways to solve this problem, this is an example based on the
same kind of problem asked in the third edition.
Problem
Using Laplace transform, solve the following differential equation for t &gt;0
d 2i
di
 4  5i  2e 2t
2
dt
dt
subject to i(0)=0, i’(0)=2.
Solution
Taking the Laplace transform of each term gives
2
 s 2 I ( s )  si (0)  i '(0)   4  sI ( s )  i (0)  5I ( s ) 
s2
2
 s 2 I ( s)  0  2   4  sI ( s )  0  5I ( s ) 
s2
2
2s  6
2
s2
s2
2s  6
A
Bs  C
I ( s) 

 2
2
( s  2)( s  4s  5) s  2 s 4s  5
2s  6  A( s 2  4s  5)  B( s 2  2 s) C ( s  2)
I ( s)( s 2  4 s  5) 
We equate the coefficients.
s2 : 0 = A+ B
s: 2= 4A + 2B + C
constant: 6 = 5A + 2C
Solving these gives
A = 2, B= -2, C = -2
I ( s) 
2
2s  2
2
2( s  2)
2
 2



2
s  2 s 4 s  5 s  2 ( s  2) 1 ( s  2) 2 1
Taking the inverse Laplace transform leads to:
i (t )   2e 2t  2e 2t cos t  2e 2t sin t  u (t )  2e 2t (1  cos t  sin t )u (t )
Chapter 15, Solution 55.
Take the Laplace transform of each term.
s
3
Y(s)  s 2 y(0)  s y(0)  y(0)  6  s 2 Y(s)  s y(0)  y(0)
s 1
 8  s Y(s)  y(0) 
(s  1) 2  2 2
Setting the initial conditions to zero gives
(s 3  6 s 2  8s) Y(s) 
Y(s) 
A
s 1
s 2  2s  5
(s  1)
A
B
C
Ds  E
 

 2
2
s (s  2)(s  4)(s  2s  5) s s  2 s  4 s  2s  5
1
,
40
B
1
,
20
C
-3
,
104
D
-3
,
65
E
-7
65
Y(s) 
3s  7
1 1 1
1
3
1
1
 



 
40 s 20 s  2 104 s  4 65 (s  1) 2  2 2
Y(s) 
3 (s  1)
1 1 1
1
3
1
1
1
4
 



 

2
2 
40 s 20 s  2 104 s  4 65 (s  1)  2
65 (s  1) 2  2 2
1
3 -4t 3 -t
2
 1

y( t )    e - 2t 
e  e cos( 2t )  e -t sin( 2t )  u(t )
104
65
65
 40 20

Chapter 15, Solution 56.
Taking the Laplace transform of each term we get:
4  s V(s)  v(0) 
12
V(s)  0
s

12 
 4 s  s  V(s)  8
V(s) 
8s
2s
 2
4s  12 s  3
2
v( t )  2 cos

3t

Chapter 15, Solution 57.
Although there is no correct way to work this problem, this is an example based on the
same kind of problem asked in the third edition.
Problem
Solve the following integrodifferential equation using the Laplace transform
method:
t
dy (t )
 9 y (t )dt  cos 2t ,
0
dt
y (0)  1
Solution
Take the Laplace transform of each term.
 s Y(s)  y(0)  9 Y(s) 
s
s
s 4
2
s2  9 
s
s2  s  4

 Y(s)  1  2
 2
s 4
s 4
 s 
Y(s) 
s 3  s 2  4s
As  B Cs  D

 2
2
2
(s  4)(s  9) s  4 s 2  9
s 3  s 2  4s  A (s 3  9s)  B (s 2  9)  C (s 3  4s)  D (s 2  4)
Equating coefficients :
s0 :
0  9B  4D
1
4  9 A  4C
s :
2
s :
1 B D
3
s :
1 AC
Solving these equations gives
A  0,
Y(s) 
B  - 4 5,
C  1,
D9 5
-4 5 s9 5 -4 5
95
s
 2
 2
 2
 2
2
s 4 s 9 s 4 s 9 s 9
y( t )  [ - 0.4 sin( 2t )  cos( 3t )  0.6 sin( 3t ) ]u(t)
Chapter 15, Solution 58.
We take the Laplace transform of each term.
5
4
[ sV ( s )  v(0)]  2V ( s)  V ( s ) 
s
s
5
4
4s
[ sV ( s)  1]  2V ( s)  V ( s ) 

 V (s)  2
s
s
s  2s  5
V ( s) 
( s  1)  5
( s  1)
2

5
2
2
2
2
2 ( s  1) 2  22
( s  1)  2
( s  1)  2
v(t )  (e  t cos 2t  2.5e  t sin 2t )u (t )
Chapter 15, Solution 59.
Take the Laplace transform of each term of the integrodifferential equation.
 s Y(s)  y(0)  4 Y(s)  3 Y(s) 
s
6
s2
 6

(s 2  4s  3) Y(s)  s 
 1
s  2 
Y(s) 
s ( 4  s)
( 4  s) s

2
(s  2)(s  4s  3) (s  1)(s  2)(s  3)
Y(s) 
A
B
C


s 1 s  2 s  3
A = –2.5,
Y(s) 
C  -10.5
B = 12,
 2.5 12
10.5


s 1 s  2 s  3
y( t )  [–2.5e–t + 12e–2t – 10.5e–3t]u(t)
Chapter 15, Solution 60.
Take the Laplace transform of each term of the integrodifferential equation.
4
4
3
2  s X(s)  x (0)  5 X(s)  X(s)   2
s s  16
s
4s
2s 3  4s 2  36s  64
(2s  5s  3) X(s)  2s  4  2

s  16
s 2  16
2
X(s) 
2s 3  4s 2  36s  64
s 3  2s 2  18s  32

(2s 2  5s  3)(s 2  16) (s  1)(s  1.5)(s 2  16)
X(s) 
A
B
Cs  D

 2
s  1 s  1.5 s  16
A  (s  1) X(s) s -1  -6.235
B  (s  1.5) X(s) s  -1.5  7.329
When s  0 ,
- 32
B D
 A

(1.5)(16)
1.5 16

 D  0.2579
s3  2s 2  18s  32  A (s3  1.5s 2  16s  24)  B (s3  s 2  16s  16)
 C (s 3  2.5s 2  1.5s)  D (s 2  2.5s  1.5)
Equating coefficients of the s3 terms,
1 A BC 
 C  -0.0935
X(s) 
- 6.235 7.329 - 0.0935s  0.2579


s 1
s  1.5
s 2  16
x ( t )  - 6.235 e -t  7.329 e -1.5t  0.0935 cos(4t )  0.0645 sin( 4t )
Chapter 16, Solution 61.
Solve the following differential equations subject to the specified initial
conditions
(a) d2v/dt2 + 4v = 12, v(0) = 0, dv(0)/dt = 2
(b) d2i/dt2 + 5di/dt + 4i = 8, i(0) = -1, di(0)/dt = 0
(c) d2v/dt2 + 2dv/dt + v = 3, v(0) = 5, dv(0)/dt = 1
(d) d2i/dt2 + 2di/dt + 5i = 10, i(0) = 4, di(0)/dt = -2
8.29
Solution
(a)
Converting into the s-domain we get
s2V(s)–sv(0–)–v’(0–)+4V(s) = 12/s = s2V(s)–s0–2+4V(s) or
(s2+4)V(s) = 2+12/s = 2(s+6)/s or V(s) = 2(s+6)/[s(s+j2)(s–j2)]
= [A/s] + [B/(s+j2)] + [C/(s–j2)] where A = 12/4 = 3, B = 2(–j2+6)/[–j2(–j4)]
= 2(6.325–18.43˚)/(8180˚) = 1.5812161.57˚ and C
= s(j2+6)/[j2(j4)] = 2(6.32518.43˚)/(8180˚) = 1.5812–161.57˚
v(t) = [3+1.5812e161.57˚e–j2t+1. 5812e–161.57˚ej2t]u(t) volts
= [3+3.162cos(2t–161.12˚)]u(t) volts.
(b)
Converting into the s-domain we get
s2I(s)–si(0–)–i’(0–)+5sI(s)–5i(0–)+4I(s) = 8/s
= s2I(s)–s(–1)–0+5sI(s)–5(–1)+4I(s) or
(s2+5s+4)I(s) = (–s–5) + 8/s = –(s2+5s–8)/s
I(s) = –(s2+5s–8)/[s(s+1)(s+4)] = [A/s] + [B/(s+1)] + [C/(s+4)] where
A = 8/[(1)(4)] = 2; B = –[(–1)2+5(–1)–8]/[(–1)(–1+4)] = 12/(–3) = –4;
C = –[(–4)2+5(–4)–8]/[(–4)(–4+1)] = 12/(12) = 1 therefore
i(t) = [2–4e–t+e–4t]u(t) amps.
(c)
s2V(s)–sv(0–)–v’(0–)+2sV(s)–2v(0–)+V(s) = 3/s
= s2V(s)–s5–1+2sV(s)–2x5+V(s) = (s2+2s+1)V(s) – (5s+11) or
(s2+2s+1)V(s) = (5s+11) + 3/s = (5s2+11s+3)/s or
V(s) = (5s2+11s+3)/[s(s+1)2] = [A/s] + [B/(s+1)] + [C/(s+1)2] where
A = 3; C = [5(–1)2+11(–1)+3]/(–1) = (–3)/(–1) = 3; going back to the original
and eliminating the denominators we get 5s2+11s+3 = 3(s2+2s+1)+Bs2+Bs+3s or
B = 2, thus,
v(t) = [3+2e–t+3te–t]u(t) volts.
(d)
s2I(s)–si(0–)–i’(0–)+2sI(s)–2i(0–)+5I(s) = 10/s
= s2I(s)–s(4)–(–2)+2sI(s)–2(4)+5I(s) or
(s2+2s+5)I(s) – (4s–2+8) = 10/s or (s2+2s+5)I(s) = (4s2+6s+10)/s or
I(s) = (4s2+6s+10)/[s(s+1+j2)(s+1–j2)] = [A/s] + [B/(s+1+j2)] + [C/(s+1–j2)]
where A = 10/5 = 2; B = [4(–1–j2)2+6(–1–j2)+10]/[(–1–j2)(–j4)]
= [4(1+j4–4)–6–j12+10]/[–8+j4] = [–12+j16–6–j12+10]/(8.944153.43˚)
= [–8+j4]/(8.944153.43˚) = 1; C = [4(–1+j2)2+6(–1+j2)+10]/[(–1+j2)(j4)]
= [4(1–j4–4)–6+j12+10]/[–8–j4] = [–12–j16–6+j12+10]/(8.944–153.43˚)
= [–8–j4]/(8.944153.43˚) = 1 thus
i(t) = [2+e–te–j2t+ e–tej2t]u(t) amps = [2+2e–tcos(2t)]u(t) amps.
(a) [3+3.162cos(2t–161.12˚)]u(t) volts, (b) [2–4e–t+e–4t]u(t) amps,
(c) [3+2e–t+3te–t]u(t) volts, (d) [2+2e–tcos(2t)]u(t) amps
Chapter16, Solution 1.
The current in an RLC circuit is described by
d 2i
di
 10  25i  0
2
dt
dt
If i(0) = 2 and di(0)/dt = 0, find i(t) for t &gt; 0.
Solution
Step 1.
Transform the equation into the s-domain and solve for I(s).
s2I(s) – (di(0–)/dt) – si(0–) + 10sI(s) – 10i(0–) +25 I(s) = 0
(s2+10s+25)I(s) + [–(di(0–)/dt)–si(0–)–10i(0–)] = 0
(s2+10s+25)I(s) + [–2s–20] = 0 or (s2+10s+25)I(s) = 2(s+10) or
I(s) = 2(s+10)/ (s2+10s+25)
Step 2.
Perform a partial fraction expansion and then solve for i(t) in the time
domain.
 10  10  10
= –5, repeated roots.
2
I(s) = 2(s+10)/(s+5)2 = A/(s+5) + B/(s+5)2 = (As+A5+B)/(s+5)2 or
s2 + 10s + 25 = 0, thus s 1,2 =
A = 2 and 5A+B = 20 or B = 20 – 10 = 10 or
I(s) = 2/(s+5) + 10/(s+5)2 or
i(t) = [(2+10t)e–5t]u(t) A
Chapter 16, Solution 2.
The differential equation that describes the voltage in an RLC network is
d 2v
dv
 5  4v  0
2
dt
dt
Given that v(0) = 0, dv(0)/dt = 5, obtain v(t).
Solution
Step 1.
Transform the equation into the s-domain and solve for V(s).
s2V(s) – (dv(0–)/dt) – sv(0–) + 5sV(s) – 5v(0–) +4V(s) = 0 or
(s2 + 5s +4)V(s) – 5 = 0 or V(s) = 5/(s2+5s+4)
Step 2.
Perform a partial fraction expansion of V(s) and then solve for v(t).
s2 + 5s + 4 = 0, thus s 1,2 =
 5  25  16
= –4, –1.
2
Thus, V(s) = 5/[(s+1)(s+4)] = A/(s+1) + B/(s+4) where A = 5/3 and B = –5/3
Thus,
v(t) = [(5/3)e–t – (5/3)e–4t]u(t) V
Chapter 16, Solution 3.
The natural response of an RLC circuit is described by the differential equation
d 2v
dv
2 v 0
2
dt
dt
for which the initial conditions are v(0) = 20 V and dv(0)/dt = 0. Solve for v(t).
Solution
Step 1.
Transform the equation into the s-domain and solve for v(t).
s2V(s) – (dv(0–)/dt) – sv(0–) + 2sV(s) – 2v(0–) + V(s) = 0 or
(s2+2s+1)V(s) – 20s – 40 = 0 or V(s) = 20(s+2)/(s2+2s+1)
Step 2.
Perform a partial fraction expansion and solve for V(s). Inverse transform
into the time-domain and solve for v(t).
2 44
= -1, repeated roots.
2
V(s) = 20(s+2)/(s2+2s+1) = 20(s+2)/(s+1)2 = A/(s+1) + B/(s+1)2
s2 + 2s + 1 = 0, thus s 1,2 =
As+A+B = 20s+40 or A = 20 and A+B = 40 = 20 + B or B = 40 – 20 = 20
Thus,
v(t) = [(20 + 20t)e-t]u(t) V
Chapter 16, Solution 4.
If R = 20 , L = 0.6 H, what value of C will make an RLC series circuit:
(a) overdamped,
(b) critically damped,
(c) underdamped?
Solution
Step 1.
Since we are working with a series RLC circuit, we can express our values
in terms of I(s) and the s equation that multiplies it in the s-domain. From here
we can easily find the values that produce over damped, critically damped, and
underdamped conditions.
Equating the mesh equation we get, RI(s) + LsI(s) + (1/C)I(s)/s – V(s) = 0 or
(0.6s + 20 + 1/(Cs))I(s) = V(s) or [s2+(20/0.6)+(1/(0.6Cs)]I(s) = V(s)/0.6 or
[ss+(20/0.6)s+1/(0.6C)]I(s) = sV(s)/0.6
The roots for the denominator are s 1,2 =
Step 2.
 (20 / 0.6)  (400 / 0.36)  4 /(0.6C)
.
2
To find the values of our roots that produces overdamped, critically
damped, and underdamped conditions, we note that when s 1 and s 2 values that
produces these values,
overdamped is when s 1 and s 2 are real with no complex values
critically damped is when s 1 = s 2
underdamped is when both s 1 and s 2 have complex roots and s 1 = s 2 *
Now all we need to do is to solve for these conditions.
(a)
Overdamped is when [4/(0.6C)] is less than 400/0.36 or C &gt;
4x0.36/(400x0.6) = 6x10-3, or C &gt; 6 mF
(b)
Critically damped is when [4/(0.6C)] is equal to 400/0.36 or C =
4x0.36/(400x0.6) = 6x10-3 = 6 mF
(c)
Underdamped is when [4/(0.6C)] is greater than 400/0.36 or C &lt;
4x0.36/(400x0.6) = 6x10-3 or C &lt; 6mF
Chapter 16, Solution 5.
The responses of a series RLC circuit are
v C (t) = [30 – 10e–20t + 30e–10t]u(t) V
i L (t) = [40e–20t – 60e–10t]u(t) mA
where v C (t) and i L (t) are the capacitor voltage and inductor current, respectively.
Determine the values of R, L, and C.
Solution
Step 1.
We can start with the generalized mesh equation for a series RLC
network. We can lump the initial conditions (v C (0) = 30–10+30 = 50 volts and
i L (0) = 40–60 = 20 amps) with the source in the loop since all we are currently
after are the values of R, L, and C.
RI(s) + Ls(I(s)+20/s) + (1/C)I(s)/s – 50/s – V(s) = 0 or [s2 + (R/L)s + 1/(LC)]I(s)
= (V(s)/L) – 20 + 50/(Ls)
Step 2.
The values of R, L, and C will come from the roots of the denominator s
equation. We already know that they are equal to –10 and –20. We note
however, that this will give us only two equations. Obviously we need a third,
and that will come from knowing the current through the capacitor and the
voltage across it.
 (R / L)  (R / L) 2  4 /(LC)
= –10, –20
s 1,2 =
2
We can simplify our effort by noting that s 1 + s 2 = –R/L[(1/2)+(1/2)] = –30 or R
= 30L.
2 (R / L) 2  4 /(LC)
= 10 or (R/L)2 – 4/(LC) = 100. Since (R/L)
2
= 30, we then get 900 – 100 = 4/(LC) or LC = 4/800 = 1/200.
Next, s 1 – s 2 =
Now we work with i C (t) = Cdv C (t)/dt or 40e–20t – 60e–10t mA =
C[200e–20t – 300e–10t] V or C = 0.2x10–3 = 200 &micro;F. Since LC = 1/200 then L =
1/(200x200x10–6) = 1/0.04 = 25 H. Finally R = 30L = 30x25 = 750 Ω.
750 Ω, 25 H, 200 &micro;F
Chapter 16, Solution 6.
Design a parallel RLC circuit that has the characteristic equation
.s2+100s+106 = 0.
Solution
Step 1.
Develop a general equation for a parallel RLC circuit with initial
conditions lumped into a parallel current source i(t).
[Cs + (1/R) + (1/(Ls))]V(s) – I(s) = 0 or [s2+(1/(RC))s+1/(LC)]V(s) = sI(s)/C
Step 2.
The next step is to equate the unknowns to the parameters in the
characteristic equation. This does become a design problem in that we have two
equations with three unknowns. We need to pick one of the values so that the other
values are realistic.
1/(RC) = 100 and 1/(LC) = 106 or RC = 0.01 and LC = 10–6. We can start with some
values of R and see what happens to the values of L and C.
R
L
1Ω
10 Ω
100 Ω
1kΩ
10 k Ω
100 k Ω
100 &micro;H
1 mH
10 mH
100 mH
1H
10 H
C
10 mF
1 mF
100 &micro;F
10 &micro;F
1 &micro;F
0.1 &micro;F
We now need to pick reasonable values, R = 10 kΩ, L = 1 H, and C = 1 &micro;F
represents an acceptable set since their values are relatively common and inexpensive.
Chapter 16, Solution 7.
The step response of an RLC circuit is given by
d 2i
di
 2  5i  10
2
dt
dt
Given that i(0) = 6 and di(0)/dt = 12, solve for i(t).
Solution
Step 1.
We start by transforming the equation into the s-domain. We then solve for I(s).
s2I(s) – (di(0–)/dt) – si(0–) + 2sI(s) – 2i(0–) + 5 I(s) = 10/s or
s2I(s) – (12) – 6s + 2sI(s) – 2x6 + 5I(s) = 10/s = (s2+2s+5)I(s) – 6(s+4) or
(s2+2s+5)I(s) = [6(s2+4s)+10]/s or I(s) = [6(s2+4s)+10]/[s(s2+2s+5)]
Step 2.
We need to find the roots of (s2+2s+5) and then perform a partial fraction
expansion and then transform back into the time domain and realize i(t).
 2  4  20
s2 + 2s + 5, has the roots s 1,2 =
= –1j2
2
I(s) = [6(s2+4s)+10]/[s(s+1+j2)(s+1–j2)] = [A/s] + [B/(s+1+j2)] + [C/(s+1–j2)]
A = 10/5 = 2; B = [6(1+j4–4–4–j8)+10]/[(–1–j2)(–j4)] = [6(–7–j4)+10]/[–8+j4] =
(–32–j24)/[4(–2+j)] = 4(–8–j6)/[4(–2+j)] = 2(–4–j3)/(–2+j) =
[2(–4–j3)(–2–j)]/[(–2+j)(–2–j)] = 2(8–3+j6+j4)/5 = 2(1+j2); C =
[6(1–j4–4–4+j8)+10]/[(–1+j2)(j4)] = [6(–7+j4)+10]/[–8–j4] =
(–32+j24)/[4(–2–j)] = 4(–8+j6)/[4(–2–j)] = 2(–4+j3)/(–2–j) =
[2(–4+j3)(–2+j)]/[(–2–j)(–2+j)] = 2(8–3–j6–j4)/5 = 2(1–j2).
I(s) = [2/s] + [(2+j4)/(s+1+j2)] + [(2–j4)/(s+1–j2)] or
i(t) = [2+4e–t(cos(2t)+2sin(2t))]u(t) A
Chapter 16, Solution 8.
A branch voltage in an RLC circuit is described by
d2v
dv
 4  8v  48
2
dt
dt
If the initial conditions are v(0) = 0 = dv(0)/dt, find v(t).
Solution
Step 1.
First we transform the equation into the s-domain. Then we solve for
V(s).
s2V(s) – (dv(0–)/dt) – sv(0–) + 4sV(s) – 4v(0–) + 8V(s) = 48/s or
s2V(s) + 4sV(s) + 8V(s) = 48/s = (s2+5s+8)V(s) or
V(s) = 48/[s(s2+4s+8)]
Step 2.
Now we need to solve for the roots of the denominator and perform a
partial fraction expansion. Then we can inverse transform the answer
back into the time domain.
 4  16  32
s2 + 4s + 8 has the roots s 1,2 =
 2  j2 thus,
2
V(s) = 48/[s(s+2+j2)(s+2–j2)] = [A/s] + [B/(s+2+j2)] + [C/(s+2–j2)]
where A = 48/4 = 6; B = 48/[(–2–j2)(–j4)] = 48/(–8+j8) =
48(–1–j)/[8(–1+j)( –1–j)] = 6(–1–j)/2 = 3(–1–j); and C =
48/[(–2+j2)(j4)] = 48/(–8–j8) = 48(–1+j)/[8(–1–j)(–1+j)] = 6(–1+j)/2 =
3(–1+j).
Therefore, V(s) = [8/s] + [3(–1–j)/(s+2+j2)] + [3(–1+j)/(s+2–j2)]
v(t) = [6 – 6e-2t(cos2t + sin2t)]u(t) volts
Chapter 16, Solution 9.
A series RLC circuit is described by
Find the response when L = 0.5 H, R = 4 , and C = 0.2 F. Let i(0–) = 1 A and
[di(0–)/dt] = 0.
Solution
Step 1.
First transform the equation into the s-domain. Then solve for I(s).
0.5s2I(s) – 0.5(di(0–)/dt) – 0.5si(0–) + 4sI(s) – 4i(0–) + 5I(s) = 2/s or
s2I(s) – s + 8sI(s) – 8 + 10I(s) = 4/s or
(s2+8s+10)I(s) = s+8+4/s = (s2+8s+4)/s or
I(s) = (s2+8s+4)/[s(s2+8s+10)]
Step 2.
Next we need to find the roots of (s2+8s+10) and then perform a partial
fraction expansion and then inverse transform back into the time domain.
s 1,2 = –1.5505 and –6.45
(s2+8s+2)/[s(s2+8s+10)] = [A/s] + [B/(s+1.5505)] + [C/(s+6.45)]
A = 0.4; B = 0.7898; and C = –0.1898 thus,
I(s) = [0.4/s] + [0.7898/(s+1.5505)] + [–0.1898/(s+6.45)] and
i(t) = [400+789.8e–1.5505t–189.8e–6.45t] mA.
Chapter 16, Solution 10.
The step responses of a series RLC circuit are
vc  40  10e2000 t  10e4000t V, t&gt;0
iL(t)  3e2000t  6e4000t mA, t&gt;0
(a) Find C. (b) Determine what type of damping exhibited by the circuit.
Solution
(a)
iL(t)  iC(t)  C
dvo
dt
(1)
dv
 2000 x10e2000t  4000 x10e4000t  2 x104(e2000 t  2e4000 t )
dt
(2)
But iL(t)  3[e2000t  2e4000 t ]x10-3
Substituting (2) and (3) into (1), we get
(3)
2 x104 xC  3 x10 3

 C  1.5 x10 7  150 nF
(b) Since s 1 = - 2000 and s 2 = - 4000 are real and negative, it is an overdamped
case.
Chapter 16, Solution 11.
The step response of a parallel RLC circuit is
v =10 + 20e-300t (cos 400t – 2 sin 400t) V, t  0
when the inductor is 50 mH. Find R and C.
Solution
Step 1.
There are different ways to approach this problem so, we will convert
everything into the s-domain and then solve for the unknowns. We should also
note that the steady-state voltage is 10 volts, then the circuit is a step input voltage
across a parallel combination of a capacitor and an inductor all in series with an
output resistor.
The nodal equation for this circuit is given by,
[(V–10/s)/R] + [(V–0)/(0.05s)] + [(V–0)/(1/sC)] + = 0 or
[(1/R)+(1/(0.05s))+sC]V = 10/(Rs) = [(20R+RCs2+s)/(Rs)]V or
V = [10/(Rs)][Rs/(RCs2+s+20R)] = 10/[(RCs)(s2+(1/(RC))s+(20/C))]
Step 2.
From the value of v(t) we can determine the value of the roots of the
polynomial (s2+(1/(RC))s+(20/C)) = (s+300+j400)(s+300–j400) thus, 20/C =
3002+4002 = 90,000 + 160,000 = 250,000 or
C = 20/250,000 = 80 &micro;F
and 1/(RC) = 600 or
R = 1/(600x80x10–6) = 20.83 Ω.
Chapter 16, Solution 12.
Consider the s-domain form of the circuit which is shown below.
1
1/s
I(s)
+

1/s
s
I(s) 
i( t ) 
1s
1
1
 2

1  s  1 s s  s  1 (s  1 2) 2  ( 3 2) 2
 3 
t  u(t) A
e - t 2 sin 
2
3


2
i( t )  1.155 e -0.5t sin (0.866t )u(t ) A
Chapter 16, Solution 13.
Using Fig. 16.36, design a problem to help other students to better understand circuit analysis
using Laplace transforms.
Although there are many ways to solve this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
Find v x in the circuit shown in Fig. 16.36 given v s = 4u(t) V.
Figure 16.36
For Prob. 16.13.
Solution
s
8/s
+
4
s
+

Vx

2
4
4
s  Vx  0  Vx  0  V (4s  8)  (16s  32)  (2s 2  4s)V  s 2 V  0
x
x
x
8
s
s
2
4
s
Vx 
Vx (3s 2  8s  8) 
16s  32
s


s2
0.25
 0.125
 0.125
Vx  16
 16


2
 s
s(3s  8s  8)
4
8
4
8
s  j
s  j

3
3
3
3

v x  (4  2e (1.3333 j0.9428) t  2e (1.3333 j0.9428) t )u ( t ) V

2 2 
v x = 4  4e  4 t / 3 cos
t 
3




 u( t )V







Chapter 16, Solution 14.
In the s-domain, the circuit becomes that shown below.
1
I
4
2
s
2
0.2s
We transform the current source to a voltage source and obtain the circuit shown below.
2
1
I
8
4
s
+
_
0.2s
8
4
20 s  40 A
B

 
I s
3  0.2 s s ( s  15) s s  15
40 8
15 x 20  40 52
 ,
B

15 3
15
3
8 / 3 52 / 3
I

s
s  15
A
i(t) = [(2.667+17.333e–15t]u(t) A
Chapter 16, Solution 15.
For the circuit in Fig. 16.38, calculate the value of R needed to have a critically
damped response.
Figure 16.38
For Prob. 16.15.
8.13
Solution
Step 1.
Let R||60 = R o . Next, convert the circuit into the s-domain and
solve for T(s) = R o + [1/(0.01s)] + 4s = R o + (100/s) + 4s =
[(4s2+R o s+100)/s]. Now to solve for the roots that represent a
critically damped system.
s 1,2 = {-R o &plusmn;[(R o )2-4(400)]0.5}/2.
The system is critically damped when [(R o )2-4(400)] = 0.
Step 2.
(R o )2 = 1600 or R o = 40. Since R o = [Rx60/(R+60)] = 40 or
60R = 40R+2400 or 20R = 2400 or R = 120 Ω.
Chapter 16, Solution 16.
The circuit in the s-domain is shown below.
I
2
4I
+
5
+
_
Vo
1/s
–
I  4I 
But I 
Vo
1/ s

 5 I  sVo
5  Vo
2
 5  Vo 
5
  sVo
 2 

 Vo 
12.5
s 5/ 2
v o (t) = 12.5e–2.5tu(t) V
1
Chapter 16, Solution 17.
Io
1
s2
2
s
2
s
(-1+-sqrt(1-8))/2 = (-1+-(j2.646))/2 = -0.5+-j(1.3229)



1 
2s 
2s
1 
1


V
 2

s  2  1  1  s  s  2  s  s  2  (s  2)(s  0.5  j1.3229)(s  0.5  j1.3229)


s 2 2
Vs
s2
Io 

2 (s  2)(s  0.5  j1.3229)(s  0.5  j1.3229)
(0.5  j1.3229) 2
(0.5  j1.3229) 2
1
(1.5  j1.3229)( j2.646) (1.5  j1.3229)( j2.646)



s  0.5  j1.3229
s  0.5  j1.3229
s2


i o ( t )  e  2 t  0.3779e 90e  t / 2e  j1.3229 t  0.3779e90e  t / 2e j1.3229 t u ( t ) A
or


 e  2 t  0.7559e  0.5 t sin 1.3229 t u ( t ) A

 7 
2  0.5 t
or i o (t) =  e  2t 
e
sin
t  u(t )A


2
7



Chapter 16, Solution 18.
For t&lt;0, v(0) = v s = 20 V
For t&gt;0, the circuit in the s-domain is as shown below.
I
+
10
s
20
s
100mF  0.1F
+
_
10 
v
_


1 10

sC s
20
s  2
s 1
10  10
s
20
V  10 I 
s 1
I
v(t )  20e  t u (t )
Chapter 16, Solution 19.
The switch in Fig. 16.42 moves from position A to position B at t=0 (please note that the
switch must connect to point B before it breaks the connection at A, a make before break
switch). Find v(t) for t &gt;0.
t=0
30 
A
4H
B
+
+
10 
v(t)
0.25 F
20V

–
Figure 16.42
For Prob. 16.19.
Solution
Step 1.
First find all the initial conditions and then transform into the s-domain.
Since the capacitor is not connected to a circuit, we do not know its initial
condition so we can assume it is zero (v(0) = 0). We can find i L (0) by letting the
inductor be a short and i L (0) = 20/40 = 0.5 amp.
0.5/s
30 
t=0
V2
A
V1
4s
B
20V
+

4/s
10 
[(V 1 –0)/(4/s)] + [(V 1 –V 2 )/(4s)] + (0.5/s) = 0 and
[(V 2 –V 1 )/(4s)] + (–0.5/s) + [(V 2 –0)/10] = 0 where V = V 1 . Next, add these
together, [sV 1 /4] + [V 2 /10] = 0 or V 2 = –2.5sV 1 . Now we can solve for V 1 and
V.
Step 2.
[(s/4)+(1/(4s))+(2.5s/(4s))]V 1 = –0.5/s
= [(s +2.5s+1)/(4s)]V 1 or V 1 = –0.5(4)/(s2+2.5s+1) = –2/[(s+0.5)(s+2)]
= [–1.3333/(s+0.5)]+[1.3333/(s+2)] or
v(t) = [–1.3333e–t/2+1.3333e–2t]u(t) volts.
2
Chapter 16, Solution 20.
Find i(t) for t &gt; 0 in the circuit of Fig. 8.43.
20V
Figure 16.43
For Prob. 16.20.
8.16
Step 1.
Convert the circuit into the s-domain and write one loop equation noting that i(0)
= 0 and v c (0) = 16 volts.
60 
I
103/s
40  16/s
+

2.5s
[(1000/s)]I +[16/s] + [2.5s]I + [40+60]I = 0 or
[(2.5s2+100s+1000)/s]I = –16/s or I = –16/[2.5(s2+40s+400)] = –.64/(s+20)2
Step 2.
I = [A/(s+20)] + [B/(s+20)2] where B = –64 so A = 0. Thus,
i(t) = 6.4te–20tu(t) A.
Chapter 16, Solution 21.
In the circuit of Fig. 16.44, the switch moves (make before break switch) from
position A to B at t = 0. Find v(t) for all t  0.
Figure 16.44
For Prob. 16.21.
Solution
Step 1.
First we need to find our initial conditions, clearly i(0) = 0 and v(0) =
4x15 = 60 volts. Next we convert the circuit into the s-domain. We can then
write a mesh equation and solve for v(t).
s/4
I
25/s
10  60/s
+

[10+(s/4)+(25/s)]I + 60/s = 0 and V = (25/s)I + 60/s
Step 2.
I = –(60/s)/[ 10+(s/4)+(25/s)] = –(60/s){4s/[s2+40s+100]
= –240/[(s+2.679)(s+37.32)] = [A/(s+2.679)]+[B/(s+37.32)] where
A = –240/(–2.679+37.32) = –6.928 and B = –240/(–37.32+2.679) = 6.928.
This now leads to V = (25/s)I+60/s
= {(25)(–6.928)/[s(s+2.679)]}+{(25)(6.928)/[s(s+37.32)]}+60/s
= {–173.2/[s(s+2.679)]}+{173.2/[s(s+37.32)]}+60/s
= [a/s]+[b/(s+2.679)]+[c/(s+37.32)] where
a = [–173.2/2.679]+[173.2/37.32]+60 = –64.65+4.641+60 = –0.009 (In practice
and theoretically, this term must be equal to be zero since there will be no
energy in the circuit at t = ∞!);
b = [–173.2/(–2.679)] = 64.65; and c = 173.2/(–37.32) = –4.641 or –4.65 if we
correct the rounding errors. Thus,
v(t) = [64.65e–2.679t–4.65e–37.32t]u(t) volts.
Chapter 16, Solution 22.
Find the voltage across the capacitor as a function of time for t &gt; 0 for the circuit
in Fig. 16.45. Assume steady-state conditions exist at t = 0-.
Figure 16.45
For Prob. 16.22.
Solution
Step 1.
First we need to calculate the initial conditions, v C (0) = 0 and i L (0) = 20/5
= 4 amps. Next we need to convert the circuit into the s-domain and solve for the
node voltage V = V C . Convert this back into the time domain and obtain v C (t).
V
1
4/s
s/4
1/s
[(V–0)/1]+[4/s]+[(V–0)/(s/4)]+[(V–0)/(1/s)] = 0 then solve for V, next complete a
partial fraction expansion, and then convert back into the time domain.
Step 2.
[1+(4/s)+s]V = [(s2+s+4)/s]V = –4/s or
V = –4/[(s+0.5+j1.9365)(s+0.5–j1.9365)]
= [A/(s+0.5+j1.9365)] + [B/(s+0.5–j1.9365)] where A = –4/(–j3.873)
= 1.0328–90˚ and B = –4/(j3.873) = 1.032890˚. Thus,
v C (t) = 1.0328e–t/2[e–(j1.9365t+90˚)+e(j1.9365t+90˚)]u(t)
= 2.066e–t/2cos(1.9365t+90˚)u(t) volts.
Chapter 16, Solution 23.
Obtain v(t) for t &gt; 0 in the circuit of Fig. 16.46.
90 V
Figure 16.46
For Prob. 16.23.
Solution
Step 1.
First we need to calculate the initial conditions. Clearly since the inductor
looks like a short, v(0) = 0 and i L (0) = 90/10 = 9 amps. Next we convert the
circuit into the s-domain and solve for V and then obtain the partial fraction
expansion and convert back into the time domain.
+
1/s
V 9/s

4s
[(V–0)/(1/s)]+(–9/s)+[(V–0)/4s] = 0
Step 2.
[s+(1/(4s))]V = 9/s = [(s2+0.25)/(4s)]V or V = 36/[(s+j0.5)(s–j0.5)]
= [A/(s+j0.5)] + [B/(s–j0.5)] where A = 36/(–j) = 3690˚ and
B = 36/(j) = 36–90˚. Thus,
v(t) = 36[e–(j0.5t–90˚)+e(j0.5t–90˚)]u(t)
= 18cos(0.5t–90˚)u(t) volts.
Chapter 8, Solution 24.
The switch in the circuit of Fig. 16.47 has been closed for a long time but is
opened at t = 0. Determine i(t) for t &gt; 0.
Figure 16.47
For Prob. 16.24.
8.20
Solution
Step 1.
First we solve for the initial conditions and then convert the circuit into the
s-domain and then solve for I, perform a partial fraction expansion, then convert
back into the time domain. We recognize that the capacitor becomes an open
circuit and the inductor becomes a short circuit at t = 0–. Therefore, v(0) = 12
volts and i(0) = 12/2 = 6 amps.
0.5s
2
I
6/s
4/s
+ 
12/s
We can use mesh analysis, –(12/s) + (4/s)I + (0.5s)(I–6/s) + 2I = 0.
Step 2.
[(4/s)+0.5s+2]I = (12/s)+(3) = (3s+12)/s = [(s2+4s+8)/(2s)]I or
I = (6s+24)/[(s+2+j2)(s+2–j2)] = [A/(s+2+j2)] + [B/(s+2–j2)] where
A = (–12–j12+24)/(–j4) = 16.97–45˚/4–90˚= 4.24345˚and
B = (–12+j12+24)/(j4) = 4.243–45˚. Thus,
i(t) = 4.243e–2t[e–(j2t–45˚)+e(j2t–45˚)]u(t)
= 8.486e–2tcos(2t–45˚) amps.
Chapter 16, Problem 25.
Calculate v(t) for t &gt; 0 in the circuit of Fig. 16.48.
Figure 16.48
For Prob. 16.25.
Step 1.
First solve for the initial conditions. Then simplify the circuit and then
convert it into the s-domain and then solve for v(t). Since the capacitor becomes
an open circuit, i L (0) = 0 and v(0) = (24)24/36 = 16 volts.
12 
t=0
6
I
3s
24V
+

+
24 
V

27/s
+

16/s
We can now use mesh analysis to solve for V(s). (30+3s+27/s)I + 16/s = 0 or
[3(s2+10s+9)/s]I = –16/s or I = –(16/3)/[(s+1)(s+9)] and v = (27/s)I + 16/s.
Step 2.
V = –(16/3)(27)/[s(s+1)(s+9)] + 16/s = –144/[s(s+1)(s+9)] + 16/s. Thus,
V = [A/s]+[B/(s+1)]+[C/(s+9)] where A = –(144/9)+16 = 0 (as expected) and
B = –144/[(–1)(–1+9)] = 144/8 = 18 and C = –144/[(–9)(–9+1)] = –144/72 = –2.
v(t) = [18e–t–2e–9t]u(t) volts.
Chapter 16, Problem 26.
The switch in Fig. 16.49 moves from position A to position B at t=0 (please note
that the switch must connect to point B before it breaks the connection at A, a
make before break switch). Determine i(t) for t &gt;0. Also assume that the initial
voltage on the capacitor is zero.
A
t=0
i(t)
B
12 A
20 
10 mF
10  0.25H
Figure 16.49
For Prob. 16.26.
Solution
Step 1.
Determine the initial conditions and then convert the circuit into the sdomain. Then solve for V and then find I. Convert it into the time domain. It is
clear from the circuit that i L (0) = 12 A.
I
V
100/s
10
0.25s
12/s
Applying nodal analysis we get,
[(V–0)/(100/s)]+[(V–0)/10]+[(V–0)/(0.25s)]+12/s = 0 where
I = [(V–0)/(0.25s)]+12/s.
Step 2.
[(s2+10s+400)/(100s)]V = –12/s or
V = –1200/[(s+5+j19.365)(s+5–j19.365)] or
I = –{4800/[s(s+5+j19.365)(s+5–j19.365)]}+12/s
= [A/s]+[B/(s+5+j19.365)]+[C/(s+5–j19.365)] where A = –12+12 = 0, as to be
expected; B = –4800/[(–5–j19.365)(–j38.73)]
= 4800180˚/[(20–104.48)(38.73–90˚)] = 6.19714.48˚; and
C = –4800/[(–5+j19.365)(j38.73)] = 4800180˚/[(20104.48)(38.7390˚)]
= 6.197–14.48˚.
i(t) = [12.394e–5tcos(19.365t+14.48˚)]u(t) amps.
Chapter 16, Problem 27.
Find v(t) for t &gt; 0 in the circuit in Fig. 16.50.
Figure 16.50
For Problem 16.27.
Solution
Step 1.
First we need to determine the initial conditions. We note that the source
on the right is equal to zero until the switch opens. So, all initial conditions come
from the 3-amp source on the left. Since the capacitor looks like an open and the
inductor looks like a short we get,
v(0) = 3[5x10/(5+10)] = 10 volts and i L (0) = 10/5 = 2 amps.
Next we convert the circuit (t &gt; 0) into the s-domain with initial conditions. Then
we can solve for V, perform a partial fraction expansion and solve for v(t).
2/s
5
+
1/(4s)
V

+

s
I
20/s
+

10/s
–[10/s]+[1/(4s)]I+[s(I–(2/s))]+5I+[20/s] = 0 and V = [1/(4s)](–I) + [10/s]
Step 2.
{[1/(4s)]+s+5}I = {[s2+5s+0.25]/(s)}I = 2–10/s = 2(s–5)/s or
I = 2(s–5)/[(s+0.05051)(s+4.949)] and
V = {–2(s–5)/[(4s)(s+0.05051)(s+4.949)]}+10/s
= {–0.5(s–5)/[s(s+0.05051)(s+4.949)]}+10/s
V = [A/s] + [B/(s+0.05051)] + [C/(s+4.949)] where
A = [2.5/[(0.05051)(4.949)]]+10 = 20;
B = –0.5(–0.05051–5)/[(–0.05051)(–0.05051+4.949)]
= 2.52525/(–0.24742) = –10.206; and
C = –0.5(–4.949–5)/[(–4.949)(–4.949+0.05051)] = 4.9745/24.243 = 0.2052.
v(t) = [20–10.206e–0.05051t+0.2052e–4.949t]u(t) volts.
dv/dt = –10.206(–0.05051)+0.2052(–4.949) = 0.5155–1.0155 = –0.5 or
Cdv/dt = –4x0.5 = –2 amps, the answer checks!
Chapter 16, Problem 28.
For the circuit in Fig. 16.51, find v(t) for t &gt; 0.
4u(–t) A
100u(t) V
Figure 16.51
For Prob. 16.28.
Solution
Step 1.
Determine the initial conditions (at t = 0, the 4 amp current source turns off and
the 100 volt voltage source becomes active). Since the capacitor becomes an open
circuit, i L (0) = 0 and v(0) = –4x6 = –24 volts. Now convert the circuit into the s-domain
and solve for V and then convert it into the time domain to obtain v(t),
25/s
24/s
s
 +
+
4
V

I
2
+ 
100/s
Now for the mesh equation, [4+s+(25/s)+2]I–(24/s)–(100/s) = 0. V = (25/s)I–24/s.
Step 2.
[(s2+6s+25)/s]I = 124/s or I = 124/(s2+6s+25) = 124/[(s+3+j4)(s+3–j4)] thus,
V = {3100/[s(s+3+j4)(s+3–j4)]}–24/s = [A/s]+[B/(s+3+j4)]+[C/(s+3–j4)] where
A = (3100/25)–24 = 124–24 = 100; B = 3100/[(–3–j4)(–j8)]
= 3100/[(5–126.87˚)(8–90˚)] = 77.5–143.13˚; and
C = 3100/[(j8)(–3+j4)] = 3100/[(5126.87˚)(890˚)] = 77.5143.13˚.
v(t) = [100+155e–3tcos(4t+143.13˚)]u(t) volts.
Chapter 16, Problem 29.
Calculate i(t) for t &gt; 0 in the circuit in Fig. 16.52.
20u(–t) V
5
Figure 16.52
For Prob. 16.29.
Solution
Step 1.
Calculate the initial conditions and then convert the above circuit into the
s-domain. Then solve for I, perform a partial fraction expansion, and convert into
the time domain. v(0) = 20 volts and i(0) = 0.
16/s
+ 
I
20/s
0.25s
[16/s]I + [20/s] + 0.25sI = 0.
Step 2.
{[16/s]+0.25s}I = –20/s = {[s2+64]/(4s)}I or I = –80/[(s+j8)(s–j8)] or
I = [A/(s+j8)] + [B/(s–j8)] where A = –80/(–j16) = 5–90˚ and
B = –80/(j16) = 590˚ thus,
i(t) = [5e–j8t–90˚+5ej8t+90˚]u(t) = 10cos(8t+90˚)u(t) amps.
Chapter 16, Solution 30.
The circuit in the s-domain is shown below. Please note, i L (0) = 0 and v o (0) = o because
both sources were equal to zero for all t&lt;0.
1
1
Vo
V1
2/s
+
_
s
2/s
1/s
At node 1
[(V 1 –2/s)/1] + [(V 1 –0)/s] + [(V 1 –V 2 )/1] = 0 or [1+(1/s)+1]V 1 –V 2 = 2/s or
[(2s+1)/s]V 1 –V 2 = 2/s
At node o
[(V o –V 1 )/1] + [(V o –0)/(2/s)] – (1/s) = 0 or
–V 1 +[(s+2)/2]V o = 1/s
In matrix form we get,
or
s2+1.5s+1 = (s+0.75+j0.6614)(s+0.75–j0.6614)
V o = s[(2/s)+(2s+1)/s2]/[(s+0.75+j0.6614)(s+0.75–j0.6614)]
= (4s+1)/[s(s+0.75+j0.6614)(s+0.75–j0.6614)]
= [A/s] + [B/(s+0.75+j0.6614)] + [C/(s+0.75–j0.6614] where A = 1;
B = [4(–0.75–j0.6614)+1]/[(–0.75–j0.6614)(–j1.3228)]
= [–3–j2.6456+1]/[(1–138.59˚)(1.3228–90˚)]
= (3.3165–127.09˚)/[(1–138.59˚)(1.3228–90˚)] = 2.507101.5˚
C = [4(–0.75+j0.6614)+1]/[(–0.75+j0.6614)(j1.3228)]
= [–3+j2.6456+1]/[(1138.59˚)(1.322890˚)]
= (3.3165127.09˚)/[(1138.59˚)(1.322890˚)] = 2.507–101.5˚
Therefore,
v o (t) = [1+2.507e–0.75te–(j0.6614t–101.5˚)+2.507e–0.75te(j0.6614t–101.5˚)]u(t) volts or
= [1+5.014e–0.75tcos(0.6614t–101.5˚)]u(t) volts.
An alternate solution is,
Vo 
(4 s  1)
A
Bs  C
  2
s ( s  1.5s  1) s s  1.5s 1
2
4 s  1  A( s 2  1.5s  1)  Bs 2  Cs
We equate coefficients.
s2 :
0 = A+ B or B = –A
s:
4=1.5A + C
constant:
1 = A, B=–1, C = 4–1.5A = 2.5
3.25
7
x
4
7
 s  2.5
s  3/ 4
1
1
4
where
Vo   2
 

2
2
s s  1.5s  1 s
 7
 7
2
2
( s  3 / 4)  
 ( s  3 / 4)  

 4 
 4 
v o (t) = [1–e–3t/4cos(0.6614t)+4.914e–3t/4sin(0.6614t)]u(t) volts.
= 0.6614.
Chapter 16, Solution 31.
Obtain v(t) and i(t) for t &gt; 0 in the circuit in Fig. 16.54.
Figure 16.54
For Prob. 16.31.
Solution
Step 1.
First, find the initial conditions and then transform the above circuit into
the s-domain after converting the current source in parallel with the 5-ohm
resistor into a 15 volts voltage source in series with a 5-ohm resistor. Then solve
for V and I, perform a partial fraction expansion on each and then convert back
into the time domain. The steady state the values are i(0) = 0 and v(0) = 20 volts.
5 I
1
5s
+
5/s
15/s
+

V
20/s
2
20/s
+


+ 
–[15/s]+5I+(5s)I+1I+[5/s]I+[20/s]–[20/s]+2I = 0 and V = {[5/s]I+[20/s].
Step 2.
{5+[5s]+1+[5/s]+2}I = [15/s]–[20/s]+[20/s] = 15/s or
{5[s2+1.6s+1]/s}I = 15/s or I = 3/[(s+0.8+j0.6)(s+0.8–j0.6)] and
V = {[5/s]I+[20/s] = 15/[s(s+0.8+j0.6)(s+0.8–j0.6)] + 20/s
= [A/s] + [B/(s+0.8+j0.6)] + [C/(s+0.8–j0.6)] where A = [15/(0.64+0.36)]+20 =
35; B = 15/[(–0.8–j0.6)(–j1.2)] = 12.590˚/1–143.13˚ = 12.5–126.87˚; and
C = 15/[(–0.8+j0.6)(j1.2)] = 12.5–90˚/(1143.13˚ = 12.5126.87˚.
v(t) = [35+12.5e–0.8t–j0.6t–126.87˚+12. 5e–0.8t+j0.6t+126.87˚]u(t) volts
= [35+25e–0.8tcos(0.6t+126.87˚)]u(t) volts.
I = 3/[(s+0.8+j0.6)(s+0.8–j0.6)] = [A/(s+0.8+j0.6)] + [B/(s+0.8–j0.6)] where
A = 3/(–j1.2) = 2.590˚ and B = 3/(j1.2) = 2.5–90˚. Thus,
i(t) = 2.5e–0.8t[e–j0.6t+90˚+ej0.6t–90˚]u(t)
= 5e–0.8t[cos(0.6t–90˚)]u(t) amps.
Chapter 16, Solution 32.
For the network in Fig. 16.55, solve for i(t) for t &gt; 0.
Figure 16.55
For Prob. 16.32.
Solution
Step 1.
First we need to find all the initial conditions. Then we need to transform
the circuit into the s-domain and solve for I. We then perform a partial fraction
expansion and convert the results into the time domain. The inductor becomes a
short and the capacitor becomes an open circuit. Thus, i(0) = [20/6] + [20/12] = 5
amps and v C (0) = 10 + 10 = 20 volts.
4
I
8/s
0.5s
5/s
10/s
20/s
+

+

Loop equation, –[10/s]–0.5s(I–5/s)–4I–[8/s]I+[20/s] = 0.
Step 2.
[0.5s+4+8/s]I = [(s2+8s+16)/(2s)]I = –[10/s]+2.5+20/s = (s+4)/(0.4s) or
I = 5(s+4)/[(s+4)2] = [A/(s+4)]+[B/(s+4)2] where A(s+4)+B = 5(s+4) or A = 5 and
B = 0. Therefore,
i(t) = [5e–4t]u(t) amps.
Chapter 16, Solution 33.
Using Fig. 16.56, design a problem to help other students to better understand how to use
Thevenin’s theorem (in the s-domain) to aid in circuit analysis.
Although there are many ways to solve this problem, this is an example based on the
same kind of problem asked in the third edition.
Problem
Use Thevenin’s theorem to determine v o (t), t&gt;0 in the circuit of Fig. 16.56.
1
1H
+
10e-2t u(t) V
0.25 F
+
_
vo
2
–
Figure 16.56
For Prob. 16.33.
Solution
1H

 1s and i L (0) = 0 (the sources is zero for all t&lt;0).
1
1 4


 and v C (0) = 0 (again, there are no source
F
4
sC s
contributions for all t&lt;0).
To find Z Th , consider the circuit below.
1
s
Z Th
2
s2
s3
To find V Th , consider the circuit below.
ZTh  1//( s  2) 
1
s
+
10
s2
+
_
V Th
2
-
s  2 10
10


s3 s2 s3
The Thevenin equivalent circuit is shown below
VTh 
Z Th
+
V Th
+
_
4/s
Vo
–
40
4
3
10
40
3
s
.
Vo 
VTh 



4
4 s  2 s  3 s 2  6s  12 ( s  3)3  ( 3) 2
 ZTh

s
s s3
4
s
vo (t )  23.094e 3t sin 3t V
Chapter 16, Solution 34.
In the s-domain, the circuit is as shown below.
1
10
s
+
_
I1
4
1
H
4
I2
10
s
1
 (1  ) I1  sI 2
s
4
4
1
5
 sI1  I 2 (4  s)  0
4
4
In matrix form,
1s
(1)
(2)
1 
 s
 s 
10  1 
 I1 
1 2 9
4
 s  4
    s  s4
   1
I
5
4
4
 0   s 4  s   2 
 4
4 
10
1
s 10
1
 s
5
40 50
4 s
s
4
1 


2 

1
5
2
s
4
 s 0
0 4 s
4
4
40 25

1
50s  160
s
2
I1 


2

0.25s  2.25s  4 s(s 2  9s  16)
I2 
2
2.5
10

 2
2

0.25s  2.25s  4 s  9s  16
Chapter 16, Solution 35.
Find v o (t) in the circuit in Fig. 16.58.
Figure 16.58
For Prob. 16.35.
Solution
Step 1.
First we note that the initial condition on the capacitor and inductor must
be equal to zero since the circuit is unexcited until t = 0. Next we transform the
circuit into the s-domain.
s
10/(s + 1)
+

Vo
1/(2s)
4
3/s
We then can solve for V o using nodal analysis.
Finally we solve for V o , perform a partial fraction expansion and then convert
into the time-domain.
Step 2.
Next
where A =
B=
C=
=
=3.9306 –117.553&deg;
Thus,
v o (t) =
or
.
Chapter 16, Solution 36.
Refer to the circuit in Fig. 16.59. Calculate i(t) for t &gt; 0.
6(1–u(t)) A
Figure 16.59
For Prob. 16.36.
Solution
Step 1.
First we need to determine the initial conditions and then transform the circuit
into the s-domain.
6A
+
i
10 
v
i1
5

10 
Clearly i = 6 A. The current then travels through the parallel combination of the 10 ohm
resistor and the combined 15 ohm resistance. i 1 = 6[(15)(10)/(15+10)]/(15) = 2.4 A.
Therefore, v(0) = 5x2.4 = 12 V and i(0) = 6 A. We also note that the two 10 ohm
resistors are in series and the combination is in parallel with the 5 ohm resistor resulting
in a 100/25 = 4 ohm resistor.
The circuit in the s-domain is shown below.
–[12/s] + [3/s]I + [0.75s](I–6/s) + 4I = 0.
I
3/s
0.75s
7/s
+

6/s
I
4
Step 2.
[(s2+5.333s+4)/(4s/3)]I = 4.5+12/s = 4.5(s+2.667)/s or
I = 6(s+2.667)/[(s+0.903)(s+4.43)] = [A/(s+0.903)]+[B/(s+4.43)] where
A = 6(–0.903+2.667)/(–0.903+4.43) = 6x1.764/3.527 = 3.001 and
B = 6(–4.43+2.667)/(–4.43+0.903) = 6x(–1.763)/(–3.527) = 2.999 or
i(t) = [3.001e–0.903t+2.999e–4.43t]u(t) amps.
Chapter 16, Solution 37.
Determine v(t) for t &gt; 0 in the circuit in Fig. 16.60.
30 
+ v 
20 
+

60u(t)V
0.25H
0.5F
+

30u(t)V
Figure 16.60
For Prob. 16.37.
Solution
Step 1.
First we need to determine the initial conditions for this circuit. Since
both sources were zero (shorts) until t = 0, the initial conditions for this circuit are
equal to zero (v(0) = 0 and i L (0) = 0). Next we transform the circuit into the sdomain. Then we can write node equations and then solve for V. Then perform a
partial fraction expansion and convert back into the time domain.
30 
60/s
s/4
2/s
+ V 
20 
+

+

30/s
From this circuit there are different ways of solving for v(t). Perhaps the easiest is
to replace the circuit seen by the capacitor and inductor with a Thevenin
equivalent circuit. V Thev = [(60/s)/(30+20)]20 – 30/s = (24/s)–(30/s) = –6/s and
R eq = 20x30/(20+30) = 12 Ω. Thus we now have the following circuit where we
can now find I. Once we have I we can find V and then perform a partial fraction
expansion and then convert into the time domain to solve for v(t).
12 
–6/s
+

2/s
+ V 
s/4
I
–[–6/s] + 12I + [2/s]I + [s/4]I = 0 and V = [2/s]I.
Step 2.
[(s/4)+12+(2/s)]I = –6/s = [(s2+48s+8)/(4s)]I or
I = (–6/s)(4s)/ [(s2+48s+8)] = –24/[(s+0.165)(s+47.84)] and
V = –48/[s(s+0.1672)(s+47.84)] = [A/s]+[B/(s+0.1672)]+[C/(s+47.84)] where
A = –48/[0.1672x47.84] = 6; B = –48/[–0.1672(–0.1672+47.84)] = 6.022; and
C = –48/[–47.84(–47.84+0.1672)] = –0.021
Therefore,
v(t) = [–6+6.022e–0.1672t–0.021e–47.84t]u(t) volts.
Chapter 16, Solution 38.
The switch in the circuit of Fig. 16.61 is moved from position a to b (a make
before break switch) at t = 0. Determine i(t) for t &gt; 0.
Figure 16.61
For Prob. 16.38.
Solution
Step 1.
We first determine the initial conditions. We assume that v C (0) = 0 since
we are not given otherwise. i(0) = [4(2x6)/(2+6)]/2 = 3 amps. Next we need to
convert the circuit for t &gt; 0 into the s-domain converting the current source in
parallel with the 6Ω into a voltage source in series with 6Ω.
3/s
3/s
50/s
50/s
14 
I
I
2s
6
24/s
12/s
+

+ 
I
2s
20 
12/s
+ 
Using the simplified circuit on the right, 2s(I–3/s) + [50/s]I –(12/s) +20I = 0.
Now we solve for I, perform a partial fraction expansion, and then convert into
the time domain.
Step 2.
[2s+(50/s)+20]I = 6+12/s = [(s2+10s+25)/(0.5s)]I = 6(s+2)/s or
I = [3(s+2)/(s+5)2] = [A/(s+5)] + [B/(s+5)2] where As+5A+B = 3s+6 or
A = 3 and B = –5A+6 = –15+6 = –9. Thus,
i(t) = [(3–9t)e–5t]u(t) amps.
Chapter 16, Solution 39.
For the network in Fig. 16.62, find i(t) for t &gt; 0.
50 V
Figure 16.62
For Prob. 16.39.
Solution
Step 1.
First determine the initial conditions at t = 0. Clearly i(0) = 0 and v(0) =
[50/(20+5+5)]5 = 25/3 volts. Next simplify and convert the circuit for t &gt; 0 into
the s-domain.
s
4
I
10/s
+

25/s
I
+

25/(3s)
–[10/s] + [4+s+(25/s)]I + [25/(3s)] = 0 Now we need to solve for I, perform a
partial fraction expansion, and then convert into the time domain.
Step 2.
[(s2+4s+25)/s]I = [10/s] – 25/(3s) = [5/(3s)] or
I = 1.6667/[(s+2+j4.583)(s+2–j4.583)] = [A/(s+2+j4.583)]+[B/(s+2–j4.583)]
where A = 1.6667/(–j9.166) = 0.1818290˚ and B = 0.18182–90˚. Therefore,
i(t) = [0.18182e–2t(e–j4.583t+90˚+ej4.583t–90˚)]u(t) amps or
= [363.6e–2tcos(4.583t–90˚)]u(t) mA.
Chapter 16, Solution 40.
Given the network in Fig. 16.63, find v(t) for t &gt; 0.
Figure 16.63
For Prob. 16.40.
Solution
Step 1.
First we determine initial conditions and then simplify the circuit and then
transform it into the s-domain. Just before the switch closes, the capacitor is an
open circuit (i L (0) = 0) with v(0) = 4 –12 = –8 volts.
6

+
s
+
25/s
12/s
8/s
V

+

We can write a node equation at V and then solve for V. Then we perform a
partial fraction expansion and then solve for v(t).
[(V–(–12/s))/(s+6)]+[(V–(–8/s))/(25/s)] = 0.
Step 2.
[(1/(s+6))+s/25]V = [(s2+6s+25)/(25(s+6))]V
= –[12/(s(s+6))]–[8/25] = –[(12+0.32s2+1.92s)/(s(s+6))]
= –0.32[(s2+6s+37.5)/(s(s+6))] or
V = –8[(s2+6s+37.5)/(s(s+3+j4)(s+3–j4))]
= [A/s]+[B/(s+3+j4)]+[C/(s+3–j4)] where A = –8[37.5/25] = –12;
B = –8[((–3–j4)2+6(–3–j4)+37.5)/((–3–j4)(–j8)]
= –8[(–7+j24–18–j24+37.5)/(–32+j24)]
= –8[(12.5)/(40143.13˚)] = 2.536.87˚; and
C = –8[((–3+j4)2+6(–3+j4)+37.5)/((–3+j4)(j8)]
= –8[(–7–j24–18+j24+37.5)/(–32–j24)]
= –8[(12.5)/(40–143.13˚)] = 2.5–36.87˚
V = [–12/s] + [2.536.87˚/(s+3+j4)] + [2.5–36.87˚/(s+3–j4)] or
v(t) = [–12+2.5e–3t(e–j4t+36.87˚+ej4–36.87˚)]u(t) amps
= [–12+5e–3t(cos(4t–36.87˚)]u(t) volts.
Chapter 16, Solution 41.
Find the output voltage v o (t) in the circuit of Fig. 16.64.
Figure 16.64
For Prob. 16.41.
Solution
Step 1.
First we need to determine the initial conditions. We see that v o (0) = 0
since the inductor becomes a short. We also note that the initial current through
the inductor is the same as the current through the 10 Ω resistor or i L (0) =
[3(5x10)/(5+10)]/10 = 1 amp. Then we simplify the circuit and convert it into the
s-domain and solve for V o . We then perform a partial fraction expansion and
convert into the time domain.
Vo
3/s
5
100/s
s
–[3/s] + [(V o –0)/5] + [(V o –0)/(100/s)] + [(V o –0)/s] + [1/s] = 0.
Step 2.
[0.2+(s/100)+(1/s)]V o = 2/s = [(s2+20s+100)/(100s)]V o or
V o = 200/[(s+10)2] and
v o (t) = [200te–10t]u(t) volts.
1/s
Chapter 16, Solution 42.
Given the circuit in Fig. 16.65, find i(t) and v(t) for t &gt; 0.
12 V
Figure 16.65
For Prob. 16.42.
Solution
Step 1.
First we need to find the initial conditions. Since the inductor becomes a short
and the capacitor becomes an open circuit, all the current flows through the 1 Ω and 2 Ω
resistors or i(0) = –12/3 = –4 amps and v(0) = 4x1 = 4 volts. Next we need to convert the
circuit into the s-domain and solve for V and I. Once we have done that, we can perform
partial fraction expansions and convert back into the time domain.
V
4/s
s
1
4/s
4/s
+

[(V–0)/1] + [(V–4/s)/(4/s)] + [(V–0)/s] – [4/s] = 0 and I = [(V–0)/s] – [4/s].
Step 2.
[(1+(s/4)+(1/s)]V = [(s2+4s+4)/(4s)]V = 1+4/s = (s+4)/s or
V = (4s+16)/[(s+2)2] = [A/(s+2)]+[B/(s+2)2] where As+2A + B = 4s+16 and
A = 4 and B = 16–2A =8. I = [4/(s(s+2))]+[8/(s(s+2)2)]–4/s
The partial fraction expansion is straight forward for the first and third terms, but the
second term takes a little work. 8/(s(s+2)2 = [a/s]+[b/(s+2)]+[c/(s+2)2] or
as2+a4s+a4+bs2+b2s+cs = 8 or a = 2, b = –2, and c = –4.
Thus, I = [2/s]+[–2/(s+2)]+[2/s]+[–2/(s+2)]+[–4/(s+2)2]–4/s = –[4/(s+2)] – [4/(s+2)2] and
we finally get,
v(t) = [4e–2t + 8te–2t]u(t) volts and
i(t) = [–4e–2t – 4te–2t]u(t) amps.
Chapter 16, Solution 43.
Determine i(t) for t &gt; 0 in the circuit of Fig. 16.66.
Figure 16.66
For Prob. 16.43.
Solution
Step 1.
First we need to determine the initial conditions. Then we need to
transform the circuit into the s-domain. Once in the s-domain we can calculate V
and I. We then perform a partial fraction expansion on I and convert back into the
time domain. Since the inductor looks like a short just before the switch opens,
v C (0) = 0 and i(0) = (12/4)+3 = 6 amps.
V
I
5s
6/s
20/s
5
3/s
[(V–0)/(5s)] + [6/s] + [(V–0)/(20/s)] + [(V–0)/5] – [3/s] = 0 and I = [(V–0)/(5s)] +
[6/s].
Step 2.
[(1/(5s))+(s/20)+(1/5)]V = [(s2+4s+4)/(20s)]V = –3/s or
V = –60/(s+2)2 and I = –[12/(s(s+2)2)] + 6/s = [A/s] + [B/(s+2)] + [C/(s+2)2] +
(6/s) where A = –3 and A(s2+4s+4) + B(s2+2s) + Cs = –12
= –3s2 – 12s –12 + Bs2 + B(2s) + Cs or –3 B = 0 or B = 3 and
–12 + 6 + C = 0 or C = 6.
i(t) = [3+3e–2t+6te–2t]u(t) amps.
Chapter 16, Solution 44.
For the circuit in Fig. 16.67, find i(t) for t &gt; 0.
Figure 16.67
For Prob. 16.44.
Solution
Step 1.
First we identify the initial conditions. Then we simplify the circuit (for t
&gt; 0) and then transform it into the s-domain. We then solve for the node voltage,
V, and then find I. Finally we perform a partial fraction expansion and convert
the answer into the time domain. For t &lt; 0, the inductor looks like a short circuit
producing v C (0) = 0 and i(0) = 30/10 = 3 amps.
V
I
9/s
100/s
8
4s
3/s
–[9/s] + [(V–0)/(100/s)] + [(V–0)/8] + [(V–0)/(4s)] + [3/s] = 0 and
I = [(V–0)/(4s)] + [3/s].
Step 2.
[(s/100)+(1/8)+1/(4s)]V = [(s2+12.5s+25)/(100s)]V = 6/s or
V = 600/[(s+2.5)(s+10)] and I = 150/[s(s+2.5)(s+10)]+[3/s]
= [A/s] + [B/(s+2.5)] + [C/(s+10)] where A = 6+3 = 9; B = 150/[–2.5(–2.5+10)]
= –8; and C = 150/[–10(–10+2.5)] = 2.
i(t) = [9–8e–2.5t+2e–10t]u(t) amps.
Chapter 16, Solution 45.
Find v(t) for t &gt; 0 in the circuit in Fig. 16.68.
Figure 16.68
For Prob. 16.45.
Solution
Step 1.
First, determine the initial conditions. Next convert the circuit into the sdomain and solve for V. Perform a partial fraction expansion and convert back
into the time domain. For t &lt; 0, the inductor looks like a short circuit so that v(0)
= 0 and i L (0) = i o .
V
1/(Cs)
Ls
i o /s
[(V–0)/(1/(Cs))] + [(V–0)/(Ls)] + [i o /s] = 0.
Step 2.
[Cs+(1/(Ls))]V = [C{s2+(1/(LC))}/s]V = –i o /s or
V = –(i o /C)/[(s2+1/(LC))]. If we let ω2 = 1/(LC) then we get,
V = –(i o /C)/[(s2+ω2)] = –(i o /C)/[(s+jω)(s–jω)] = [A/(s+jω)]+[B/(s–jω)] where
A = –(i o /C)/(–j2ω) = [i o /(2ωC)]–90˚ and B = –(i o /C)/(j2ω) = [i o /(2ωC)]–90˚.
Thus,
v(t) = [i o /(2ωC)][e–jωt–90˚+ejωt+90˚] = [i o /(ωC)]cos(ωt+90˚)u(t) volts.
Chapter 16, Solution 46.
Consider the following circuit.
1/s
2s
Vo
Io
1/(s + 2)
2
1
Applying KCL at node o,
Vo
Vo
1
s 1
V



s  2 2s  1 2  1 s 2s  1 o
2s  1
Vo 
(s  1)(s  2)
Io 
Vo
1
A
B



2s  1 (s  1)(s  2) s  1 s  2
A  1,
Io 
B  -1
1
1

s 1 s  2
i o ( t )   e -t  e -2t  u(t ) A
Chapter 16, Solution 47.
We first find the initial conditions from the circuit in Fig. (a).
1
4
+
5V
+

v c (0)
io

(a)
i o (0  )  5 A , v c (0  )  0 V
We now incorporate these conditions in the s-domain circuit as shown in Fig.(b).
1
4
Vo
Io
15/s
+

2s
(b)
At node o,
Vo  15 s Vo 5 Vo  0

 
0
1
2s s 4  4 s
15 5 
1
s 
V
  1  
s s  2s 4 (s  1)  o
5s 2  6s  2
10 4s 2  4s  2s  2  s 2
Vo
Vo 

4s (s  1)
s
4s (s  1)
40 (s  1)
Vo  2
5s  6s  2
Vo 5
4 (s  1)
5

 
2
2s s s (s  1.2s  0.4) s
5 A
Bs  C
Io    2
s s s  1.2s  0.4
Io 
5/s
4/s
4 (s  1)  A (s 2  1.2s  0.4)  B s s  C s
Equating coefficients :
s0 :
4  0.4A 
 A  10
1
4  1.2A  C 
 C  -1.2A  4  -8
2
0 AB 
 B  -A  -10
s :
s :
5 10
10s  8
  2
s s s  1.2s  0.4
10 (s  0.6)
10 (0.2)
15
Io  
2
2 
s (s  0.6)  0.2
(s  0.6) 2  0.2 2
Io 
i o ( t )   15  10 e -0.6t  cos(0.2 t )  sin( 0.2 t )  u(t ) A
Chapter 16, Solution 48.
First we need to transform the circuit into the s-domain.
s/4
10
Vo
+
3V x
+

Vx
5/s
+

5
s2
5
Vo 
Vo  3Vx Vo  0
s2 0


s/4
5/s
10
5s
5s
 0  (2s 2  s  40)Vo  120Vx 
40Vo  120Vx  2s 2 Vo  sVo 
s2
s2
But, Vx  Vo 
5
5
 Vo  Vx 
s2
s2
We can now solve for V x .
5 
5s

( 2s 2  s  40) Vx 
0
  120Vx 
s 2
s2

(s 2  20)
2(s 2  0.5s  40)Vx  10
s2
Vx   5
(s 2  20)
(s  2)(s 2  0.5s  40)
Chapter 16, Solution 49.
We first need to find the initial conditions. For t  0 , the circuit is shown in Fig. (a).
2
+

1
1F
V o /2
Vo
+

+

1H
3V
io
(a)
To dc, the capacitor acts like an open circuit and the inductor acts like a short circuit. Hence,
-3
 -1 A ,
v o  -1 V
i L (0)  i o 
3
 - 1
v c (0)  -(2)(-1)     2.5 V
2
We now incorporate the initial conditions for t  0 as shown in Fig. (b).
2
+
Vo

1
1/s
s
5/(s + 2)
+

2.5/s
I1
V o /2
+

I2

+
+

Io
(b)
For mesh 1,
2.5 Vo
1
- 5  1

0
  2   I1  I 2 
s
2
s
s
s2 
-1 V
But,
Vo  I o  I 2
 1 1
 1
5
2.5
 2   I1     I 2 

2 s

s2 s
s
For mesh 2,
V

2.5
1
1
1  s   I 2  I1  1  o 
0

2
s
s
s
1
2.5
1
1
1
- I1    s   I 2 
2
s
s
s
(1)
(2)
Put (1) and (2) in matrix form.
 1
2  s

 -1
 s
 5
1 1 
2.5 
   I1  

2 s   s2 s 



1
1    2.5

 s   I 2  
1 
 s

2
s
3
  2s  2  ,
s
Io  I2 
 2  -2 
4
5

s s (s  2)
2
- 2s 2  13
A
Bs  C


 2
2

(s  2)(2s  2s  3) s  2 2s  2s  3
- 2s 2  13  A (2s 2  2s  3)  B (s 2  2s)  C (s  2)
Equating coefficients :
s2 :
s1 :
s0 :
- 2  2A  B
0  2A  2 B  C
13  3A  2C
A  0.7143 , B  -3.429 , C  5.429
0.7143 3.429 s  5.429 0.7143 1.7145 s  2.714



s2
2s 2  2s  3
s2
s 2  s  1.5
0.7143 1.7145 (s  0.5) (3.194)( 1.25 )
Io 


s2
(s  0.5) 2  1.25 (s  0.5) 2  1.25
Io 


i o ( t )  0.7143 e -2t  1.7145 e -0.5t cos(1.25t )  3.194 e -0.5t sin(1.25t ) u(t ) A
Chapter 16, Solution 50.
For the circuit in Fig. 16.73, find v(t) for t &gt; 0. Assume that v(0+) = 4 V and
i(0+) = 2 A.
Figure 16.73
For Prob. 16.50.
Solution
Step 1.
Determine the initial condition of the second capacitor and then convert
the circuit into the s-domain. Finally, solve for V, perform a partial fraction
expansion and convert the answer back into the time domain. Since v(0) = 4 volts
and i(0) = 2 amps then –4–2(2)+v 2 (0) = or v 2 (0) = 8.
2
+
I
10/s
2/s
V
4/s
+
 
I/4
8/s
+

–
[(V–4/s)/(10/s)]–[I/4]+[(V–8/s)/(2+2/s)] = 0 and I = [((8/s)–V)/(2+2/s)]
= [4/(s+1)]–0.5sV/(s+1)
Step 2.
[(V–4/s)/(10/s)]+[0.125sV/(s+1)]–[1/(s+1)]+[(V–8/s)/(2+2/s)]
[(s/10)+(0.125s/(s+1))+(0.5s/(s+1))]V =
[0.4+4/(s+1)]+[1/(s+1)] = (0.4s+5.4)/(s+1)
= [(s2+s+6.25s)/(10(s+1))]V = [s(s+7.25)/(10(s+1))]V or
V = 4(s+13.5)/[s(s+7.25)] = [A/s]+[B/(s+7.25)] where A = 4(13.5)/7.25 = 7.748
and B = 4(–7.25+13.5)/(–7.25) = –3.448 or
v(t) = [7.748–3.448e–7.25t]u(t) volts.
Chapter 16, Solution 51.
In the circuit of Fig. 16.74, find i(t) for t &gt; 0.
50 V
Figure 16.74
For Prob. 16.51.
Solution
Step 1.
First we note that the initial conditions for the capacitor and inductor have to be
equal to zero. Next we simplify the circuit and then convert the circuit into the s-domain
and solve for V. Then we can solve for I and then perform a partial fraction expansion
and convert I back into the time domain.
0.25s
6
V
50/s
+

4
25/s
I
[(V–50/s)/(0.25(s+24))]+[s(V–0)/25]+[(V–0)/4] = 0 and I = [(0–V)/4] = –V/4.
Step 2.
[(4/(s+24))+(s/25)+0.25]V = [(s2+24s+6.25s+100+150)/(25(s+24))]V
= [(s +30.25s+250)/(25(s+24))]V
= [{(s+15.125+j4.608)(s+15.125–j4.608)}/(25(s+24))]V = [200/(s(s+24))] or
V = 5,000/[s(s+15.125+j4.608)(s+15.125–j4.608)] and
I = –1250/[s(s+15.125+j4.608)(s+15.125–j4.608)] = [A/s]+[B/(s+15.125+j4.608)]
+[C/(s+15.125–j4.608)] where A = –1250/250 = –5;
B = –1250/[(–15.125–j4.608)(–j9.216)] = 1250180˚/[(15.811–163.06˚)(9.216–90˚)]
= 8.57873.06˚; and C = 1250180˚/[(15.811163.06˚)(9.21690˚)] = 8.578–73.06˚.
Thus, i(t) = [–5+8.578e–15.125t(e–j4.608t+73.06˚+ej4.608–73.06˚)]u(t) amps
2
i(t) = [–5+17.156e–15.125tcos(4.608t–73.06˚)]u(t) amps.
Chapter 16, Solution 52.
If the switch in Fig. 16.75 has been closed for a long time before t = 0 but is
opened at t = 0, determine i x and v R for t &gt; 0.
Figure 16.75
For Prob. 16.52.
Solution
Step 1.
Fist we need to determine the initial conditions. Just before the switch
opens, v C (0) = 16 volts and i L (0) = 2 amps. Next we convert the circuit into the
s-domain.
12 
+
VR
–
Ix
8
36/s
s
16/s
2/s
+

We can now write a mesh equation (this time going in the counter-clockwise
direction). [s(I x +2/s)]+[8I x ]+[12I x ]+[(36/s)I x ]+(16/s) = 0 and V R = –8I x .
Step 2.
[s+8+12+(36/s)]I x = [(s2+20s+36)/s]I x = –2–16/s = –[2(s+8)/s] or
I x = –2(s+8)/[(s+2)(s+18)] = [A/(s+2)]+[B/(s+18)] where
A = –2(–2+8)/(–2+18) = –2x6/16 = –0.75 and B = –2(–18+8)/(–18+2) = –1.25
thus,
i x (t) = [–0.75e–2t–1.25e–18t]u(t) amps and
v R (t) = –8i x (t) = [6e–2t+10e–18t]u(t) volts.
Chapter 16, Solution 53.
In the circuit of Fig. 16.76, the switch has been in position 1 for a long time but
moved to position 2 at t = 0. Find:
(a) v(0+), dv(0+)/dt
(b) v(t) for t  0.
Figure 16.76
For Prob. 16.53.
Solution
Step 1.
Clearly i L (0) = 0 and v(0) = 4 volts. When the switch moves to 2, i C (0+) =
Cdv(0)/dt = –4/0.5 = –8 volts/second = 1dv(0)/dt. Next we convert the circuit
into the s-domain and solve for V. Then we perform a partial fraction expansion
and then convert back into the time domain.
+
1/s
0.25s
V
0.5 
4/s
+


[(V–0)/(0.25s)]+[(V–0)/0.5]+[(V–4/s)s/1] = 0.
Step 2.
[(4/s)+2+s]V = [(s2+2s+4)/s]V = 4 or V = 4s/[(s+1+j1.7321)(s+1–
j1.7321)]
= [A/(s+1+j1.7321)] + [B/(s+1–j1.7321)] where
A = 4(–1–j1.7321)/(3.464–90˚) = 4(2–120˚)/(3.464–90˚) = 2.309–30˚ and
B = 4(2120˚)/(3.46490˚) = 2.30930˚ or
v(t) = 2.309e–t[e–j1.7321t–30˚+ej1.7321t+30˚]u(t) volts or
v(t) = [4.618e–tcos(1.7321t+30˚)]u(t) volts.
Chapter 16, Solution 54.
The switch in Fig. 16.77 has been in position 1 for t &lt; 0. At t =0, it is moved from
position 1 to the top of the capacitor at t = 0. Please note that the switch is a make
before break switch, it stays in contact with position 1 until it makes contact with
the top of the capacitor and then breaks the contact at position 1. Determine v(t).
Figure 16.77
For Prob. 16.54.
Solution
Step 1.
First determine the initial conditions and then transform the circuit into the
s-domain and solve for V. Then perform a partial fraction expansion and then
find v(t). We will assume that the value of v(0) = 0. i L (0) = 40/20 = 2 amps.
4s
+
V
16/s

2/s
I
16 
[16/s]I + [4s](I–2/s) + 16I = 0 and V = [16/s](–I).
Step 2.
[(16/s)+4s+16]I = [4(s2+4s+4)/s]I = 8 or
I = 8s/[4(s+2)2] = 2s/[(s+2)2] and V = –32/[(s+2)2]
v(t) = [–32te–2t]u(t) volts.
Chapter 16, Solution 55.
Obtain i 1 and i 2 for t &gt; 0 in the circuit of Fig. 16.78.
Figure 16.78
For Prob. 16.55.
Solution
Step 1.
The first thing we do is to determine the initial conditions. Since there is
no excitation of the circuit before t = 0, all initial conditions must be zero. Next
we convert the circuit into the s-domain. Then use nodal analysis and eventually
solve for I 1 and I 2 , then perform a partial fraction expansion and convert back into
the time domain.
3
V1
I2
I1
4/s
2
s
s
–[4/s]+[(V 1 –0)/2]+[(V 1 –0)/s]+[(V 1 –0)/(s+3)] = 0 and I 1 = [(V 1 –0)/s] and
I 2 = [(V 1 –0)/(s+3)].
Step 2.
{[1/2]+[1/s]+[1/(s+3)]}V 1 = 4/s = {[s2+3s+2s+6+2s]/[2s(s+3)]}V 1 or
V 1 = 8(s+3)/[s2+7s+6] = 8(s+3)/[(s+1)(s+6)] and I 1 = 8(s+3)/[s(s+1)(s+6)]
= [A/s]+[B/(s+1)]+[C/(s+6)] where A = 8x3/6 = 4; B = 8(–1+3)/[(–1)(–1+6)]
= –16/5 = –3.2; C = 8(–6+3)/[(–6)(–6+1)] = –24/30 = –0.8. Thus,
i 1 (t) = [4–3.2e–t–0.8e–6t]u(t) amps.
I 2 = [(V 1 –0)/(s+3)] = 8/[(s+1)(s+6)] = [A/(s+1)] + [B/(s+6)] where A = 8/5 = 1.6
and B = 8/(–6+1) = –1.6. Thus,
i 2 (t) = [1.6e–t–1.6e–6t]u(t) amps.
[4–3.2e–t–0.8e–6t]u(t) amps, [1.6e–t–1.6e–6t]u(t) amps
Chapter 16, Solution 56.
We apply mesh analysis to the s-domain form of the circuit as shown below.
2/(s+1)
+ 
I3
1/s
1
s
I1
I2
1
4/s
For mesh 3,
 1
1
2
  s   I 3  I1  s I 2  0
s
s 1  s
For the supermesh,
 1
1 
1   I1  (1  s) I 2    s  I 3  0
 s
s 
(1)
(2)
Adding (1) and (2) we get, I 1 + I 2 = –2/(s+1)
(3)
But
(4)
–I 1 + I 2 = 4/s
Adding (3) and (4) we get, I 2 = (2/s) – 1/(s+1)
(5)
Substituting (5) into (4) yields, I 1 = –(2/s) – (1/(s+1))
(6)
Substituting (5) and (6) into (1) we get,
2
s2

 s2  1
1
s
I 3   2
2

s(s  1)
s  1  s 
s 1
2 1.5  0.5 j 1.5  0.5 j
I3   

s
s j
s j
Substituting (3) into (1) and (2) leads to
2(s 2  2s  2)
 1
 1
-  s   I 2   s   I3 
 s
 s
s 2 (s  1)
(4)
1
4(s  1)

 1
 2  s   I 2   s   I3  
s
s


s2
(5)
We can now solve for I o .
I o = I 2 – I 3 = (4/s) – (1/(s+1)) + ((–1.5+0.5j)/(s+j)) + ((–1.5–0.5)/(s–j))
or
i o (t) = [4 – e–t + 1.5811e–jt+161.57˚ + 1.5811ejt–161.57˚]u(t)A
This is a challenging problem. I did check it with using a Thevenin equivalent circuit and
Chapter 16, Solution 57.
3 e s 3
 (1  e  s )
v s (t) = 3u(t) – 3u(t–1) or V s = 
s
s
s
1
+
Vs
+

1/s
2
Vo

Vo  Vs
V
 sVo  o  0  (s  1.5)Vo  Vs
1
2
Vo 
3
2 
2
s
(1  e  s )   
(1  e  )
s(s  1.5)
s
s

1
.
5


v o (t )  [( 2  2e 1.5 t )u(t )  ( 2  2e 1.5( t 1) )u(t  1)] V
(a) (3/s)[1–e–s], (b) [(2–2e–1.5t)u(t) – (2–2e–1.5(t–1))u(t–1)] V
Chapter 16, Solution 58.
Using Fig. 16.81, design a problem to help other students to better understand circuit analysis in
the s-domain with circuits that have dependent sources.
Although there are many ways to solve this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
In the circuit of Fig. 16.81, let i(0) = 1 A, v o (0) = 2 V, and v s = 4 e-2t u(t) V. Find v o (t)
for t &gt; 0.
Figure 16.81
For Prob. 16.58.
Solution
We incorporate the initial conditions in the s-domain circuit as shown below.
2I
2
Vo
V1
 +
I
4/(s + 2)
+

1/s
1/s
2
s
At the supernode,
(4 (s  2))  V1
V 1
 2  1   sVo
2
s s
 1 1
1
2
 2     V1   s Vo
2 s
s
s2
(1)
But
Vo  V1  2 I and
Vo  V1 
2 (V1  1)
s
I
V1  1
s

 V1 
Vo  2 s s Vo  2

(s  2) s
s2
(2)
Substituting (2) into (1)
2
1  s  2   s 
2 
2 
 s Vo
 
 Vo 
s  2 
s2
s  2s   s  2 
2
1 1  1  
 2       s  Vo
s s  2  
s2
2s  4  2 2s  6

 (s  1 / 2)Vo
(s  2)
s2
2s  6
A
B


Vo 
(s  2)(s  1 / 2) s  1 / 2 s  2
A  (1  6) /(0.5  2)  3.333 , B  (4  6) /(2  1 / 2)  1.3333
3.333 1.3333
Vo 

s  1/ 2 s  2
Therefore,
v o ( t )  (3.333e-t/2 – 1.3333e-2t)u(t) V
Chapter 16, Solution 59.
We incorporate the initial conditions and transform the current source to a voltage source
as shown.
2/s
1
1/s
Vo
+ 
1/(s + 1)
+

1
At the main non-reference node, KCL gives
1 (s  1)  2 s  Vo Vo Vo 1



11 s
1
s s
s
s 1
 2  s Vo  (s  1)(1  1 s) Vo 
s 1
s
s
s 1

 2  (2s  2  1 s) Vo
s 1
s
- 2s 2  4s  1
Vo 
(s  1)( 2s 2  2s  1)
- s  2s  0.5
A
Bs  C

 2
Vo 
2
(s  1)(s  s  0.5) s  1 s  s  0.5
A  (s  1) Vo
s  -1
1
- s 2  2s  0.5  A (s 2  s  0.5)  B (s 2  s)  C (s  1)
Equating coefficients :
s2 :
-1  A  B 
 B  -2
s1 :
s0 :
Vo 
-2  ABC 
 C  -1
- 0.5  0.5A  C  0.5  1  -0.5
2 (s  0.5)
1
2s  1
1
 2


s  1 s  s  0.5 s  1 (s  0.5) 2  (0.5) 2
v o ( t )   e -t  2 e -t 2 cos(t 2) u(t ) V
s
1/s
Chapter 16, Solution 60.
Find the response v R (t) for t &gt; 0 in the circuit in Fig. 16.83. Let R = 3 , L = 2
H, and C = 1/18 F.
Figure 16.83
For Prob. 16.60.
Solution
Step 1.
First convert the circuit into the s-domain. Then use nodal analysis and
eventually solve for V R , then perform a partial fraction expansion and convert
back into the time domain.
3
V1
+

VR
10/s
+

18/s
2s
[(V 1 -10/s)/3]+[(V 1 -0)/(18/s)]+[(V 1 -0)/(2s)] = 0 and V R = (10/s)-V 1 .
Step2.
[(1/3)+(s/18)+1/(2s)]V 1 = 3.333/s = [(s2+6s+9)/(18s)]V 1 or
V 1 = 60/[(s+3)2] and V R = (10/s)-60/[(s+3)2].
Thus,
v R (t) = [10-60te-3t]u(t) volts.
Chapter 16, Solution 61.
The s-domain version of the circuit is shown below.
1
s
V1
+
10/s
2/s
Vo
2
1/s
-
At node 1,
10
s  V1  Vo  s (V  0)  0
1
1
s
2
V1 



 s2
  s  1V1  (  1)Vo  10

2
(1)
At node 2,
Vo  V1 Vo  0

 s(Vo  0)  0
s
2


V1  (s 2  0.5s  1)Vo
Substituting (2) into (1) gives
10  [0.5(s 2  2s  2)(s 2  0.5s  1)Vo  Vo  0.5(s 4  2.5s 3  4s 2  3s  2  2)Vo
Vo 
20
s(s  2.5s 2  4s  3)
3
Use MATLAB to find the roots.
&gt;&gt; p=[1 2.5 4 3]
p=
1.0000 2.5000 4.0000 3.0000
&gt;&gt; r=roots(p)
r=
-0.6347 + 1.4265i
-0.6347 - 1.4265i
-1.2306
(2)
Thus,
20
s(s  1.2306)(s  0.6347  j1.4265)(s  0.6347  j1.4265)
A
B
C
D
 


s (s  1.2306) (s  0.6347  j1.4265) (s  0.6347  j1.4265)
Vo 
Where A = 20/3 = 6.667; B =
20
(1.2306)(1.2306  0.6347  j1.4265)(1.2306  0.6347  j1.4265)
 16.252

 6.8
(0.3551  2.035)
20
C
(0.6347  j1.4265)(0.6347  j1.4265  1.2306)( j2.853)
20
20


 2.904  88.68
(1.5613  113.99)(1.546  67.33)(2.853  90) 6.88688.68
20
(0.6347  j1.4265)(0.6347  j1.4265  1.2306)( j2.853)
20
20


 2.90488.68
(1.5613113.99)(1.54667.33)(2.85390) 6.886  88.68
D
Vo 
 6.8
6.667
2.904  88.68
2.90488.68



or
s
(s  1.2306) (s  0.6347  j1.4265) (s  0.6347  j1.4265)
v o (t) = [6.667–6.8e–1.2306t+2.904e–0.6347t(e–(1.4265t+88.68˚)+e(1.4265t+88.68˚))]u(t) volts or
= [6.667–6.8e–1.2306t+5.808e–0.6347tcos(1.4265t+88.68˚)]u(t) V.
Answer does check for initial values and final values.
Chapter 16, Solution 62.
Using Fig. 16.85, design a problem to help other students better understand solving for node
voltages by working in the s-domain.
Although there are many ways to solve this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
Find the node voltages v 1 and v 2 in the circuit of Fig. 16.85 using Laplace transform
technique. Assume that i s = 12e-t u(t) A and that all initial conditions are zero.
Figure 16.85
For Prob. 16.62.
Solution
The s-domain version of the circuit is shown below.
4s
V1
12
s 1
1
V2
2
3/s
At node 1,
V
V  V2
12
 1 1
s 1 1
4s
At node 2,

12
1 V

 V1 1    2
s 1
 4s  4s
(1)
V1  V2 V2 s

 V2
4s
2 3


4
V1  V2  s 2  2s  1

3
(2)
Substituting (2) into (1),
 4
12
1  1  4
7
3

 V2  s 2  2s  11       s 2  s  V2
2
s 1
3
 4s  4s   3
 3
V2 
9
7
9
(s  1)(s 2  s  )
4
8
9  A(s 2 

A
Bs  C

(s  1) (s 2  7 s  9 )
4
8
7
9
s  )  B(s 2  s)  C(s  1)
4
8
Equating coefficients:
s2 :
0AB
7
3
0  ABC  AC
4
4
9
3
9  A  C A 

8
8
s:
constant :


3
C A
4
A  24, B  -24, C  -18
3
24s  18
24
24(s  7 / 8)



7
23
7
9
7
23
(s  1)
(s  ) 2 
(s 2  s  )
(s  ) 2 
8
64
4
8
8
64
Taking the inverse of this produces:
V2 
24

(s  1)


v 2 (t )  24e  t  24e 0.875 t cos(0.5995t )  5.004e 0.875 t sin( 0.5995t ) u(t )V
Similarly,
4

9 s 2  2s  1
Es  F
3
  D 
V1 
7
9
7
9
(s  1)
(s  1)(s 2  s  )
(s 2  s  )
4
8
4
8
7
9
4

9 s 2  2s  1  D(s 2  s  )  E(s 2  s)  F(s  1)
4
8
3

Equating coefficients:
s2 :
12  D  E
7
3
3
s:
18  D  E  F or 6  D  F


F  6 D
4
4
4
9
3
constant :
9  D  F or 3  D 
 D  8, E  4, F  0
8
8
8
4s
8
4(s  7 / 8)
7/2
V1 




7
9
7
23
7
23
(s  1)
(s  1)
(s 2  s  )
(s  ) 2 
(s  ) 2 
4
8
8
64
8
64
Thus,
v 1 (t )  8e  t  4e 0.875 t cos(0.5995t )  5.838e 0.875 t sin( 0.5995t ) u(t )V


Chapter 16, Solution 63.
The s-domain form of the circuit with the initial conditions is shown below.
V
I
4/s
sL
R
-2/s
1/sC
5C
At the non-reference node,
V V
4 2
  5C  
 sCV
R sL
s s
6  5 sC CV  2
s
1 
s 



s
s 
RC LC 
5s  6 C
V 2
s  (s RC)  (1 LC)
But
1
1

 8,
RC 10 80
V
1
1

 20
LC 4 80
5s  480
5 (s  4 )
( 230)( 2)

2
2 
s  8s  20 (s  4)  2
(s  4 ) 2  2 2
2
v( t )  [5 e -4t cos( 2t )  230 e -4t sin( 2t )]u(t ) V
I
V
5s  480

sL 4s (s 2  8s  20)
I
1.25s  120
A
Bs  C
  2
2
s (s  8s  20) s s  8s  20
A  6,
I
B  -6 , C  -46.75
6 6s  46.75 6
6 (s  4)
(11.375)(2)
 2
 
2
2 
s s  8s  20 s (s  4)  2
(s  4) 2  2 2
i( t )  [6  6 e -4t cos( 2t )  11.375 e -4t sin( 2t )]u( t ) A
Checking, Ldi/dt = 4{24 e–4tcos(2t) + 12 e–4tsin(2t) + 45.5 e–4tsin(2t)
– 22.75 e–4tcos(2t)}u(t) = [5 e–4tcos(2t) + 230 e–4tsin(2t)]u(t). Answer checks.
Chapter 16, Solution 64.
When the switch is position 1, v(0)=12, and i L (0) = 0. When the switch is in position 2,
we have the circuit as shown below.
s/4
+
100/s
v
–
1 100

sC
s
12 / s
48
,

I
s / 4  100 / s s 2  400
10mF  0.01F
+
_
12/s


V  sLI 
s
12s
I 2
4
s  400
v(t) = [12cos(20t)]u(t) V
Chapter 16, Solution 65.
For t  0 , the circuit in the s-domain is shown below.
6
s
I
+
9/s
(2s)/(s2 + 16)
+

V

+

2/s
Applying KVL,
 2s 
9 2
 6  s   I   0
2
s s
s  16 
 32
I
(s 2  6s  9)(s 2  16)
V
9
2 2
 288
I  
s
s s s (s  3) 2 (s 2  16)
2 A
B
C
Ds  E
  

 2
2
s s s  3 (s  3)
s  16
- 288  A (s 4  6s 3  25s 2  96s  144)  B (s 4  3s 3  16s 2  48s)
 C (s 3  16s)  D (s 4  6s 3  9s 2 )  E (s 3  6s 2  9s)
Equating coefficients :
 288  144A
s0 :
1
0  96A  48B  16C  9E
s :
2
0  25A  16B  9D  6E
s :
3
0  6A  3B  C  6D  E
s :
4
0  A  B D
s :
Solving equations (1), (2), (3), (4) and (5) gives
A  2 , B  2.202 , C  3.84 , D  -0.202 ,
V(s) 
(1)
(2)
(3)
(4)
(5)
E  2.766
0.202 s (0.6915)(4)
2.202
3.84



2
s  3 (s  3)
s 2  16
s 2  16
v( t )  {2.202e-3t + 3.84te-3t – 0.202cos(4t) + 0.6915sin(4t)}u(t) V
Chapter 16, Solution 66.
Consider the op-amp circuit below where R 1 = 20 kΩ, R 2 = 10 kΩ, C = 50 &micro;F, and
v s (t) = [3e-5t]u(t) V.
R2
1/sC
R1
Vs
0
+


+
+
Vo

At node 0,
Vs  0 0  Vo

 (0  Vo ) sC
R1
R2
 1

 sC   - Vo 
Vs  R 1 
R2

Vo
-1

Vs sR 1C  R 1 R 2
But
So,
R 1 20

 2,
R 2 10
Vo
-1

Vs s  2
v s (t )  3 e -5t
R 1C  (20  103 )(50  10-6 )  1

 Vs  3 ( s  5)
Vo 
-3
A
B

where A = –1 and B = 1.
=
(s  2)(s  5) s  2 s  5
Vo 
1
1

s5 s2
v o ( t )   e -5t  e -2t  u(t ) V.
Chapter 16, Solution 67.
Given the op amp circuit in Fig. 16.90. If v 1 (0+) = 2 V and v 2 (0+) = 0 V, find v o
for t &gt; 0. Let R = 100 k and C = 1 F. R
C
C
+
v1

+
R

+
v2


+
+
vo

Figure 16.90
For Prob. 16.67.
Solution
Step 1.
Convert the circuit into the s-domain and insert initial conditions. Next,
solve for V o (s), then obtain the partial fraction expansion and convert back into
the time domain.
105
106/s
Va
2/s
+ 
+
V1

+
Vb
106/s

Vd
105
+
V2


+
Vc
Ve
+
Vo

[(V a –(V c +2/s))/(106/s)]+[(V a –V o )/105]+0 = 0; V a = V b = 0 and
[(V d –V c )/105]+[(V d –V o )/(106/s)]+0 = 0; V d = V e = 0.
Step 2.
sV c + 10V o = –2 and 10V c + sV o = 0 or V c = –0.1sV o thus,
(–0.1s2+10)V o = –2 or V o = 20/(s2–100) = [A/(s–10)]+[B/(s+10)] where
A = 20/(10+10) = 1 and B = 20/(–10–10) = –1. This now leads to
v o (t) = [e10t–e–10t]u(t) volts.
It should be noted that this is an unstable circuit!
Chapter 16, Solution 68.
Obtain V o /V s in the op amp circuit in Fig. 8.91.
10 pF
60 k
v S (t)

+
60 k
+

+
20 pF v o (t)

Figure 8.91
For Prob. 8.68.
Solution
Step 1.
Convert the circuit into the s-domain and then solve for V o (s) in terms of V s (s).
Then solve for V o /V s = T(s).
1011/s
60 k
60 k
d
VS
+

a
b

+
c
+
Vo
5x1010/s

At a, V a = V b = V c = V o . At b, [(V b –V d )/60k]+[(V b –0)/(5x1010/s)]+0 = 0 or
[(V o –V d )/60k]+[(V o –0)/(5x1010/s)] = 0 or [(1/60k)V d = [(1/60k)+(s/(5x1010))]V o or
V d = [(1.2x10–6)s+1]V o .
At d, [(V d –V s )/60k]+[(V d –V c )/(1011/s)]+(V d –V b )/60k] = 0 or
[(2/60k)+(s/1011)]V d –(s/1011)V o –(1/60k)V o = (1/60k)V s or
[(2/60k)+(s/1011)] [(1.2x10–6)s+1]V o –(s/1011)V o –(1/60k)V o = (1/60k)V s or
[2+(6x10–7)s] [(1.2x10–6)s+1]V o –(6x10–7)sV o –V o = V s or
[7.2x10–13s2+(2.4x10–6+0.6x10–6–0.6x10–6)s+(2–1)]V o = V s or
T(s) = V o /V s = 1/[(7.2x10–13)s2+(2.4x10–6)s+1].
Chapter 16, Solution 69.
Find I 1 (s) and I 2 (s) in the following circuit.
2H
(10e–3t)u(t) V
+

H
i1
2H
i2
1Ω
1Ω
Solution
Step 1.
We note that the initial conditions in this case are equal to zero. Next, we
need to convert the circuit into the s-domain and use the model for mutually
coupled circuits. Then we can write the mesh equations and solve for I 1 and I 2 .
sI 2
2s
sI 1
2s
+ 
10/(s+3)
Step 2.
+

I1
+ 
1Ω
I2
1Ω
–[10/(s+3)] + 2sI 1 + sI 2 + 1(I 1 –I 2 ) = 0 and
1(I 2 –I 1 ) + 2sI 2 + sI 1 + 1I 2 = 0. Simplifying we get,
(2s+1)I 1 + (s–1)I 2 = 10/(s+3) and (s–1)I 1 + (2s+1)I 2 = 0.
We can solve this directly using substitution or use matrices. Let us use matrices.
 10 
2s  1 s  1   I1  

 s  1 2s  1 I    (s  3)  The matrix inverse

 2 
 0 
 2s  1  s  1
 2s  1  s  1
1
 s  1 2s  1 
 s  1 2s  1 
2s  1 s  1 






 s  1 2s  1
2
2
3
s
(
s

2
)
4s  4s  1  s  2s  1


Therefore,
I 1 = 6.667(s+0.5)/[s(s+2)(s+3)] and
I 2 = –3.333(s–1)/[s(s+2)(s+3)]
6.667(s+0.5)/[s(s+2)(s+3)], –3.333(s–1)/[s(s+2)(s+3)]
Chapter 16, Solution 70.
Using Fig. 16.93, design a problem to help other students better understand how to do circuit
analysis with circuits that have mutually coupled elements by working in the s-domain.
Although there are many ways to solve this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
For the circuit in Fig. 16.93, find v o (t) for t &gt; 0.
Figure 16.93
For Prob. 16.70.
Solution
Consider the circuit shown below.
s
1
+
6/s
+

I1
2s
s
I2
Vo
2

For mesh 1,
6
 (1  2s) I1  s I 2
s
For mesh 2,
0  s I1  (2  s) I 2
 2
I1  - 1   I 2
 s
(1)
(2)
Substituting (2) into (1) gives
 2
- (s 2  5s  2)
6
 -(1  2s)1   I 2  s I 2 
I2
 s
s
s
-6
or
I2  2
s  5s  2
Vo  2 I 2 
- 12
- 12

s  5s  2 (s  0.438)(s  4.561)
2
Since the roots of s 2  5s  2  0 are -0.438 and -4.561,
A
B
Vo 

s  0.438 s  4.561
A
- 12
 -2.91 ,
4.123
Vo (s) 
B
- 12
 2.91
- 4.123
- 2.91
2.91

s  0.438 s  4.561
v o ( t )  2.91  e -4.561t  e 0.438 t  u(t ) V
Chapter 16, Solution 71.
Consider the following circuit.
1
1:2
+

10/(s + 1)
Let
Io
ZL  8 ||
4/s
4 (8)(4 s)
8


s 8  4 s 2s  1
When this is reflected to the primary side,
Z
Zin  1  L2 , n  2
n
2
2s  3
Zin  1 

2s  1 2s  1
10 1
10 2s  1



s  1 Zin s  1 2s  3
10s  5
A
B
Io 


(s  1)(s  1.5) s  1 s  1.5
Io 
A  -10 ,
I o (s) 
B  20
- 10
20

s  1 s  1 .5


i o ( t )  10 2 e -1.5t  e  t u(t ) A
8
Chapter 16, Solution 72.
Y (s)  H (s) X (s) ,
X(s) 
4
12

s  1 3 3s  1
12 s 2
4 8s  4 3

2 
(3s  1)
3 (3s  1) 2
4 8
s
4
1
Y (s)   


3 9 (s  1 3) 2 27 (s  1 3) 2
Y (s) 
Let G (s) 
-8
s

9 (s  1 3) 2
Using the time differentiation property,

-8 d
- 8  -1
g( t ) 
 ( t e -t 3 )   t e -t 3  e -t 3 

9 dt
93
8 -t 3 8 -t 3
g( t ) 
te  e
27
9
Hence,
 4 8 -t 3 8 -t 3 4 -t 3 
y( t )   
te  e 
t e  u(t)
9
27

 3 27
4 -t 3 
4 8
t e  u( t )
y( t )    e -t 3 
27
3 9

Chapter 16, Solution 73.
x(t)  u(t)


y( t )  10 cos( 2 t )
H(s) 
X (s) 


Y(s) 10s 2

X(s) s 2  4
1
s
Y(s) 
10s
s2  4
Chapter 16, Solution 74.
Design a problem to help other students to better understand how to find outputs when given a
transfer function and an input.
Although there are many ways to solve this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
A circuit is known to have its transfer function as
H (s) 
s3
s  4s  5
2
Find its output when:
(a) the input is a unit step function
(b) the input is 6te-2t u(t).
Solution
(a)
Y(s)  H(s) X(s)
s3
1

s  4s  5 s
s3
A
Bs  C
  2

2
s (s  4s  5) s s  4s  5

2
s  3  A (s 2  4s  5)  Bs 2  Cs
Equating coefficients :
3  5A 
 A  3 5
s0 :
s1 :
2
s :
1  4A  C 
 C  1  4A  - 7 5
0 AB 
 B  -A  - 3 5
35 1
3s  7
  2
s 5 s  4s  5
0.6 1 3 (s  2)  1
 
Y(s) 
s 5 (s  2) 2  1
Y(s) 
y( t )   0.6  0.6 e -2t cos(t )  0.2 e -2t sin( t ) u(t )
(b)
x ( t )  6 t e -2t

 X(s) 
6
(s  2) 2
s3
6

2
s  4s  5 (s  2) 2
6 (s  3)
A
B
Cs  D


Y(s) 
2
2
2  2
(s  2) (s  4s  5) s  2 (s  2) s  4s  5
Y(s)  H(s) X(s) 
Equating coefficients :
s3 :
0 AC 
 C  -A
2
0  6 A  B  4C  D  2 A  B  D
s :
1
s :
6  13A  4B  4C  4D  9A  4B  4D
0
18  10A  5B  4D  2A  B
s :
Solving (1), (2), (3), and (4) gives
A6,
B  6,
C  -6 ,
D  -18
6
6
6s  18

2 
s  2 (s  2)
(s  2) 2  1
6
6
6 (s  2)
6

Y(s) 

2 
2
s  2 (s  2)
(s  2)  1 (s  2) 2  1
Y(s) 
y( t )   6 e -2t  6 t e -2t  6 e -2t cos(t )  6 e -2t sin( t ) u(t )
(1)
(2)
(3)
(4)
Chapter 16, Solution 75.
1
s
H(s) 
Y(s)
,
X(s)
Y(s) 
4
1
2s
(3)(4)



2
s 2 (s  3) (s  2)  16 (s  2) 2  16
H(s)  s Y(s)  4 
X(s) 
2 s(s  2)
12 s
s


2
2
2 (s  3) s  4s  20 s  4s  20
Chapter 16, Solution 76.
Consider the following circuit.
2
s
Vo
+
kV s
+

4
10/s
V o (s)

Using nodal analysis,
kVs  Vo Vo Vo


s2
4 10 s
1 s 
1
 1
 1

   Vo  (1 / k )1  (s  2)  (s 2  2s)  Vo
Vs  (1 / k )(s  2) 
10
 s  2 4 10 
 4

1
2s 2  9s  30 Vo
Vs 
20k


Vo
 10k/(s2+4.5s+15)
Vs
Chapter 16, Solution 77.
Consider the following circuit.
2/s
I
s
V1
+
Vs
+

2I
Vo

At node 1,
2I  I 
3
V1
,
s3
where I 
Vs  V1
V
 1
2s
s3
V1
3s
3s
 Vs  V1
s3 2
2
 1
3s 
3s

  V1  Vs
s  3 2 
2
3s (s  3)
V1  2
V
3s  9s  2 s
Vo 
3
9s
V1  2
V
s3
3s  9s  2 s
H (s) 
9s
Vo
 2
Vs 3s  9s  2
Vs  V1
2s
3
Chapter 16, Solution 78.
Taking the inverse Laplace transform of each term gives
h(t )   5e t  3e 2t  6e 4t  u (t )
Chapter 16, Solution 79.
(a)
Consider the circuit shown below.
3
2s
+
Vs
+

I1
2/s
Vx
I2
+

4V x

For loop 1,
 2
2
Vs  3   I1  I 2

s
s
For loop 2,

2
2
4Vx   2s   I 2  I1  0

s
s
But,
So,
(1)
2
Vx  ( I1  I 2 )  
s

8
2
2
(I1  I 2 )   2s   I 2  I1  0

s
s
s
6

-6
0
I1    2s  I 2
s

s
In matrix form, (1) and (2) become
 Vs  3  2 s
- 2 s  I1 
 0    - 6 s 6 s  2s  I 
  
 2 

  6  2 
2  6
  3    2s     

  s  s 
s  s
18
   6s  4
s
6

 1    2s  Vs ,
s

I1 
2 
1
(6 s  2s)

V
 18 s  4  6s s
6
V
s s
(2)
I1
3 ss
s2  3

 2
Vs 9 s  2  3 3s  2s  9
(b)
I2 
2

2  2 
2

( I1  I 2 )   1
s  
s
2 s Vs (6 s  2s  6 s) - 4Vs

Vx 


Vx 
6 s Vs - 3
I2


Vx
- 4Vs
2s
Chapter 16, Solution 80.
(a)
Consider the following circuit.
Is
1
V1
s
Vo
Io
+
Vs
+

1/s
1/s
1
Vo

At node 1,
Vs  V1
V1  Vo
 s V1 
1
s

1
1
Vs  1  s   V1  Vo

s
s
At node o,
V1  Vo
 s Vo  Vo  (s  1) Vo
s
V1  (s 2  s  1) Vo
Substituting (2) into (1)
Vs  (s  1  1 s)(s 2  s  1) Vo  1 s Vo
Vs  (s 3  2s 2  3s  2) Vo
H 1 (s) 
(b)
Vo
1
 3
2
Vs s  2s  3s  2
I s  Vs  V1  (s 3  2s 2  3s  2) Vo  (s 2  s  1) Vo
I s  (s 3  s 2  2s  1)Vo
H 2 (s) 
(c)
Io 
Vo
1
 3
2
Is
s  s  2s  1
Vo
1
H 3 (s) 
I o Vo
1

 H 2 (s)  3
2
Is
Is
s  s  2s  1
(1)
(2)
(d)
H 4 (s) 
I o Vo
1

 H1 (s)  3
2
Vs Vs
s  2s  3s  2
Chapter 16, Solution 81.
For the op-amp circuit in Fig. 16.99, find the transfer function, T(s) = I o (s)/V s (s).
Assume all initial conditions are zero.
C
R

+
i o (t
+

V S (t)
L
Figure 16.99
For Prob. 16.81.
Solution
Step 1.
Convert the circuit into the s-domain. The write the node equations at the
input to the op amp and solve for T(s).
1/(Cs)
R
VS
+

Va

+
Vc
Io
Vb
Ls
[(V a –V s )/R]+[(V a –V c )/(1/(Cs))]+0 = 0; V a = V b = 0 and I o = (V c –0)/(Ls).
Step 2.
CsV c = –V s /R or V c = –V s /(RCs) and I o = –V s /(RLCs2) or
T(s) = –1/(RLCs2).
Chapter 16, Solution 82.
Consider the circuit below.
Va
Vb
Vs
+


+
+
R
Vo
1/sC
Io

Since no current enters the op amp, I o flows through both R and C.

1 
Vo  -I o  R  

sC 
Va  Vb  Vs 
H(s) 
- Io
sC
Vo R  1 sC

 sRC  1
Vs
1 sC
Chapter 16, Solution 83.
(a)
(b)
H (s) 
Vo
R
R L


Vs R  sL s  R L
h(t) 
R - Rt L
e
u( t )
L
v s (t )  u(t ) 
 Vs (s)  1 s
Vo 
R L
R L
A
B
Vs 
 
sR L
s (s  R L ) s s  R L
A  1,
B  -1
1
1
Vo  
s sR L
v o ( t )  u ( t )  e -Rt L u ( t )  (1  e -Rt L ) u(t )
Chapter 16, Solution 84.
Consider the circuit as shown below.
Vo
I2
4
Is
Vo Vo 2
 
4 s s
2
But I S 
s 1
2
1 1 2
 Vo    
s 1
4 s s
s
Is 
Vo 
s4
2
2 4s  2
)
 
4s
s  1 s s ( s  1)
8(2 s  1)
( s  1)( s  4)
IL 
A

 Vo (
Vo
8(2s  1)
A
B
C

 

s s( s  1)( s  4) s s  1 s  4
8(1)
8(2  1)
 2,
B
 8 / 3,
(1)(4)
(1)(2)
V
2 8 / 3 14 / 3
IL  o  

s s s 1 s  4
8
14


iL (t )   2  e t  e 4t  u (t )
3
3


C
8(8  1)
 14 / 3
(4)(3)
2
s
Chapter 16, Solution 85.
s4
A
B
C



2
( s  1)( s  2)
s  1 s  2 ( s  2) 2
s  4  A( s  2) 2  B( s  1)( s  2)  C ( s  1)  A( s 2  2s  4)  B( s 2  3s  2)  C ( s  1)
We equate coefficients.
0=A+B or B=-A
s2 :
s:
1=4A+3B+C=B+C
constant:
4=4A+2B+C =2A+C
Solving these gives A=3, B=-3, C=-2
H (s) 
H (s) 
3
3
2


s  1 s  2 ( s  2) 2
h(t )  (3e t  3e2t  2te 2t )u (t )
Chapter 16, Solution 86.
1Ω
u(t)V
+

i(t)
1F
1H
First select the inductor current i L and the capacitor voltage v C to be the state variables.
Applying KVL we get:
 u ( t )  i  v C  i'  0; i  v 'C
Thus,
vC  i
i  vC  i  u (t )
Finally we get,
1   v C   0
 vC 
 v C   0
 i     1  1  i   1 u(t ) ; i(t )  0 1 i   0u(t )
 
   
  
Chapter 16, Solution 87.
Develop the state equations for the problem you designed in Prob. 16.13.
Although there is no correct way to work this problem, this is an example based on the same
kind of problem asked in the third edition.
Problem
Develop the state equations for Problem 16.13.
Chapter 16, Problem 13.
Find v x in the circuit shown in Fig. 16.36 given v s = 4u(t) V.
Figure 16.36
Solution
1/8 F
1H
+
4u ( t )
+

vx
2Ω
4Ω

First select the inductor current i L and the capacitor voltage v C to be the state variables.
Applying KCL we get:
vx vC

 0; or vC  8iL  4vx
2 8
iL  4u (t )  vx
v
v
vx  vC  4 C  vC  C  vC  4iL  2vx ; or vx  0.3333vC  1.3333iL
8
2
 iL 
vC  8iL  1.3333vC  5.333iL  1.3333vC  2.666iL
iL  4u (t )  0.3333vC  1.3333iL
Now we can write the state equations.
2.666   v C  0
 vC 
 v C    1.3333
 i     0.3333  1.3333  i   4 u(t ); v x  0.3333 1.3333 i 
 L   
 L
 L 
Chapter 16, Solution 88.
First select the inductor current i L (current flowing left to right) and the capacitor
voltage v C (voltage positive on the left and negative on the right) to be the state variables.
Applying KCL we get:
vC vo
  iL  0 or vC  4iL  2vo
4 2
iL  vo  v2

v o   v C  v1
vC  4iL  2vC  2v1
iL  vC  v1  v2
 i L  0  1  i L  1  1  v1 (t ) 
 iL 
 v1 ( t ) 

; v o (t )  0  1   1 0
 







 v 2 ( t )
 vC 
 v C  4  2  v C   2 0   v 2 (t )
Chapter 16, Solution 89.
First select the inductor current i L (left to right) and the capacitor voltage v C to be the
state variables.
Letting v o = v C and applying KCL we get:
vC
 is  0 or vC  0.25vC  iL  is
4
iL  vC  vs
 iL  vC 
Thus,
 v C   0.25 1  vC  0 1  v s 
 ; v o ( t ) 
 i    
 
0  i L  1 0  i s 
 L  1
1   v C   0 0  v s 
 0  i    0 0  i 
 s 
  L  
Chapter 16, Solution 90.
First select the inductor current i L (left to right) and the capacitor voltage v C (+ on the
left) to be the state variables.
Letting i 1 =
vC
and i 2 = i L and applying KVL we get:
4
Loop 1:
 v

 v1  vC  2 C  iL   0 or vC  4iL  2vC  2v1
 4

Loop 2:
v 

2 iL  C   iL  v2  0 or
4

4i  2vC  2v1
iL  2iL  L
 v2  vC  v1  v2
2
i1 
4i L  2v C  2v1
 i L  0.5v C  0.5v1
4
 i L  0  1  i L  1  1  v1 (t )   i1 (t )  1  0.5  i L  0.5 0  v1 (t ) 
  
  v    0 0   v ( t )
  v    2 0   v (t ) ; i (t )  1

4
2

0
v
 2 
  C 




C

2
2






 C
Chapter 16, Solution 91.
Let x 1 = y(t). Thus, x1  y  x2 and x2  y  3x1  4 x2  z (t )
This gives our state equations.
 x 1 
 x  
 2
1   x1  0
 x1 
 0
  3  4  x   1 z(t ); y(t )  1 0 x   0z(t )
 2   

 2
Chapter 16, Solution 92.
Let x1  y (t ) and x2  x1  z  y  z or y  x2  z
Thus,
x2  y  z  9 x1  7( x2  z )  z  2 z  z  9 x1  7 x2  5 z
This now leads to our state equations,
1   x1   1 
 x1 
 x 1   0
 x     9  7  x     5 z(t ); y(t )  1 0  x   0 z(t )
 2   
 2
 2 
Chapter 16, Solution 93.
Let x 1 = y(t), x 2 = x1 , and x3  x2 .
Thus,
x3  6 x1  11x2  6 x3  z (t )
We can now write our state equations.
1
0   x1   0 
 x 1   0
 x1 
   
  

0
1   x 2  0 z (t ); y ( t )  1 0 0 x 2   0 z( t )
x2   0
 x 3    6  11  6  x 3  1 
 x 3 
Chapter 16, Solution 94.
We transform the state equations into the s-domain and solve using Laplace transforms.
1
sX (s)  x (0)  AX(s)  B 
s
Assume the initial conditions are zero.
1
(sI  A ) X (s)  B 
s
s  4  4
X (s)  
s 
 2
1
4  0 
0  1 
s
1
 2 s   2
 2 s  4  ( 2 / s ) 
   s  4s  8 


s4
1
 
2
s(s  4s  8) s s  4s  8
1
1
s4
 (s  2)
2
 
 

s (s  2) 2  2 2 s (s  2) 2  2 2 (s  2) 2  2 2
Y(s)  X1 (s) 
8
2


y(t) = 1  e 2t cos 2t  sin 2t  u(t )
Chapter 16, Solution 95.
Assume that the initial conditions are zero. Using Laplace transforms we get,
1 
s  2
X(s)  

  2 s  4
X1 

1
3s  8
2
2
s((s  3)  1 )

s  4  1  3 / s 
1 1  1 / s 
1

 4 0  2 / s   2
s  2 4 / s 
   s  6s  10  2

0.8  0.8s  1.8

s
(s  3) 2  12
s3
1
0.8
 0.8
 .6
s
(s  3) 2  12
(s  3) 2  12
x1 ( t )  (0.8  0.8e 3t cos t  0.6e 3t sin t )u ( t )
X2 

4s  14
s((s  3) 2  12

1.4  1.4s  4.4

s (s  3) 2  12
1.4
s3
1
 1.4
 0.2
s
(s  3) 2  12
(s  3) 2  12
x 2 ( t )  (1.4  1.4e 3t cos t  0.2e 3t sin t )u ( t )
y 1 (t )  2x 1 (t )  2x 2 (t )  2u(t )
 ( 2.4  4.4e  3 t cos t  0.8e  3 t sin t )u(t )
y 2 ( t )  x 1 ( t )  2u( t )  ( 1.2  0.8e 3 t cos t  0.6e 3 t sin t )u( t )
[–2.4 + 4.4e–3tcos(t) – 0.8e–3tsin(t)]u(t), [–1.2 – 0.8e–3tcos(t) + 0.6e–3tsin(t)]u(t)
Chapter 16, Solution 96.
If Vo is the voltage across R, applying KCL at the non-reference node gives
Is 
Vo
V 1
1
 sC Vo  o    sC   Vo
R
sL  R
sL 
Is
sRL Is

1
1
sL  R  s 2 RLC
 sC 
R
sL
Vo 
Io 
Vo
sL Is
 2
R s RLC  sL  R
H (s ) 
Io
sL
s RC
 2
 2
Is s RLC  sL  R s  s RC  1 LC
The roots
-1
1
1

2 
2RC
(2RC)
LC
both lie in the left half plane since R, L, and C are positive quantities.
s1, 2 
Thus, the circuit is stable.
Chapter 16, Solution 97.
(a)
H1 (s) 
3
,
s 1
H 2 (s) 
H (s )  H 1 (s ) H 2 (s ) 
1
s4
3
(s  1)(s  4)
 A
B 

h ( t )  L-1  H(s)  L-1 
 s  1 s  4 
A  1,
B  -1
h ( t )  (e -t  e -4t ) u(t )
(b)
Since the poles of H(s) all lie in the left half s-plane, the system is stable.
Chapter 16, Solution 98.
Let
Vo1 be the voltage at the output of the first op amp.
Vo1  1 sC
1
,


Vs
R
sRC
H (s) 
Vo
1
 2 2 2
Vs s R C
h(t) 
t
R C2
Vo
1

Vo1 sRC
2
lim h ( t )   , i.e. the output is unbounded.
t 
Hence, the circuit is unstable.
Chapter 16, Solution 99.
1
sL 
1
sC  sL
sL ||

1 1  s 2 LC
sC
sL 
sC
sL
2
V2
sL
 1  s LC  2
sL
V1
s RLC  sL  R
R
2
1  s LC
1
s

V2
RC

1
1
V1
s2  s 

RC LC
Comparing this with the given transfer function,
1
1
,
2
6
RC
LC
If R  1 k ,
1
 500 F
2R
1
L
 333.3 H
6C
C
Chapter 16, Solution 100.
The circuit is transformed in the s-domain as shown below.
1/sC 2
R2
1/sC 1
Vi
–
R1
Vo
+
1
R1
1
sC1
Let Z1  R 1 //


1  sR1C1
sC1 R  1
1
sC1
R1
1
R2
1
sC2
Z 2  R 2 //


sC2 R  1
1  sR2C2
2
sC2
This is an inverting amplifier.
R2
R2
1 
1 


s
s



Z
R RC
C
1  sR2C2
R1C1
R1C1 
V
 1

H (s)  o   2 
 2 1 1 
Vi
R1
Z1
R1 R2C2  s  1  C2  s  1 


R2C2 
R2C2 
1  sR1C1
Comparing this with
( s  1000)
H (s)  
2( s  4000)
we obtain:
C1
 1/ 2

 C2  2C1  20 F
C2
1
1
1
 1000

 R1 
 3
 100
R1C1
1000C1 10 x10 x106
1
1
1
12.5Ω
 4000

 R2 

 12.8
3
4000C2 4 x10 x 20 x106
R2C2

Chapter 16, Solution 101.
We apply KCL at the noninverting terminal at the op amp.
(Vs  0) Y3  (0  Vo )(Y1  Y2 )
Y3 Vs  - (Y1  Y2 )Vo
Vo
- Y3

Vs Y1  Y2
Let
Y1  sC1 ,
Y2  1 R 1 ,
Y3  sC 2
Vo
- sC 2
- sC 2 C1


Vs sC1  1 R 1 s  1 R 1C1
Comparing this with the given transfer function,
C2
1
 1,
 10
C1
R 1 C1
If R 1  1 k ,
C1  C 2 
1
 100 F
10 4
Chapter 16, Solution 102.
Consider the circuit shown below. We notice that V3  Vo and V2  V3  Vo .
Y4
Y1
V in
+

Y2
V2
V1

+
Vo
Y3
At node 1,
(Vin  V1 ) Y1  (V1  Vo ) Y2  (V1  Vo ) Y4
Vin Y1  V1 (Y1  Y2  Y4 )  Vo (Y2  Y4 )
At node 2,
(V1  Vo ) Y2  (Vo  0) Y3
V1 Y2  (Y2  Y3 ) Vo
Y2  Y3
V1 
Vo
Y2
Substituting (2) into (1),
Y2  Y3
Vin Y1 
 (Y1  Y2  Y4 ) Vo  Vo (Y2  Y4 )
Y2
(1)
(2)
Vin Y1 Y2  Vo ( Y1 Y2  Y22  Y2 Y4  Y1 Y3  Y2 Y3  Y3 Y4  Y22  Y2 Y4 )
Vo
Y1 Y2

Vin Y1 Y2  Y1 Y3  Y2 Y3  Y3 Y4
Y1 and Y2 must be resistive, while Y3 and Y4 must be capacitive.
1
1
Y4  sC 2
,
,
Y3  sC1 ,
Let
Y1 
Y2 
R1
R2
1
Vo
R 1R 2

sC1 sC1
1
Vin


 s 2 C1 C 2
R 1R 2 R 1 R 2
1
Vo
R 1 R 2 C1 C 2

 R1  R 2 
Vin
1

s2  s 
 R 1 R 2 C 2  R 1 R 2 C1 C 2
Choose R 1  1 k , then
1
 10 6
R 1 R 2 C1 C 2
and
R1  R 2
 100
R 1R 2 C 2
We have three equations and four unknowns. Thus, there is a family of solutions. One
such solution is
R 2  1 k ,
C1  50 nF ,
C 2  20 F
Chapter 16, Solution 103.
Using the result of Practice Problem 16.14,
Vo
- Y1 Y2

Vi Y2 Y3  Y4 (Y1  Y2  Y3 )
When Y1  sC1 ,
1
,
Y2 
R1
Y3  Y2 ,
Vo
Vi
Vo
Vi
Vo
Vi
Vo
Vi
C1  0.5 F
R 1  10 k
Y4  sC 2 ,
C 2  1 F
- sC1 R 1
- sC1 R 1

1 R  sC 2 (sC1  2 R 1 ) 1  sC 2 R 1 (2  sC1 R 1 )
- sC1 R 1
 2
s C1C 2 R 12  s  2C 2 R 1  1
- s (0.5  10 -6 )(10  10 3 )
 2
s (0.5  10 -6 )(1  10 -6 )(10  10 3 ) 2  s (2)(1  10 -6 )(10  10 3 )  1


2
1
- 100 s
s  400 s  2  10 4
2
Therefore,
a  - 100 ,
b  400 ,
c  2  10 4
Chapter 16, Solution 104.
(a)
Y(s) 
Let
K (s  1)
s3
K (s  1)
K (1  1 s)
 lim
K
s 
s 
s3
1 3 s
0.25  K .
Y ()  lim
i.e.
Hence, Y (s) 
(b)
s1
4 (s  3)
Consider the circuit shown below.
t=0
Vs = 8 V
I
+

Vs  8 u ( t ) 
 Vs  8 s
Vs
8 s  1 2 (s  1)

 Y(s) Vs (s)  
4s s  3 s (s  3)
Z
A
B
I 
s s3
I
A  2 3,
i( t ) 

B = 2(–3+1)/(–3) = 4/3

1
2  4 e - 3t u(t ) A
3
YS
Chapter 16, Solution 105.
The gyrator is equivalent to two cascaded inverting amplifiers. Let V1 be the voltage at
the output of the first op amp.
-R
V1 
V  -Vi
R i
Vo 
- 1 sC
1
V1 
V
R
sCR i
Io 
Vo
Vo

R sR 2 C
Vo
 sR 2 C
Io
Vo
 sL, when L  R 2 C , so if you let L = R2C then V o /I o = sL.
Io
Chapter 17, Solution 1.
(a)
This is periodic with  =  which leads to T = 2/ = 2.
(b)
y(t) is not periodic although sin t and 4 cos 2t are independently
periodic.
(c)
Since sin A cos B = 0.5[sin(A + B) + sin(A – B)],
g(t) = sin 3t cos 4t = 0.5[sin 7t + sin(–t)] = –0.5 sin t + 0.5 sin7t
which is harmonic or periodic with the fundamental frequency
 = 1 or T = 2/ = 2.
(d)
h(t) = cos 2 t = 0.5(1 + cos 2t). Since the sum of a periodic function and
a constant is also periodic, h(t) is periodic.  = 2 or T = 2/ = .
(e)
The frequency ratio 0.6|0.4 = 1.5 makes z(t) periodic.
 = 0.2 or T = 2/ = 10.
(f)
p(t) = 10 is not periodic.
(g)
g(t) is not periodic.
1
Chapter 17, Solution 2.
The function f(t) has a DC offset and is even. We use the following MATLAB code to
plot f(t). The plot is shown below. If more terms are taken, the curve is clearly indicating
a triangular wave shape which is easily represented with just the DC component and
three, cosinusoidal terms of the expansion.
for n=1:100
tn(n)=n/10;
t=n/10;
y1=cos(t);
y2=(1/9)*cos(3*t);
y3=(1/25)*cos(5*t);
factor=4/(pi*pi);
y(n)=0.5- factor*(y1+y2+y3);
end
plot(tn,y)
Chapter 17, Solution 3.
T = 4,  o = 2/T = /2
g(t) = 5,
10,
0,
0&lt;t&lt;1
1&lt;t&lt;2
2&lt;t&lt;4
T
1
0
0
2
a o = (1/T)  g( t )dt = 0.25[  5dt +  10dt ] = 3.75
a n = (2/T)
T
 g( t ) cos(n t )dt
o
0
1
= (2/4)[
0
1
n
t )dt ]
2
2
an =
b n = (2/T)
2
t )dt +  10 cos(
2
n
2
n
sin
t + 10
sin
t ] = (–1/(n))5 sin(n/2)
n
2 0
n
2 1
1
= 0.5[ 5
n
1
 5 cos( 2
(5/(n))(–1)(n+1)/2,
0,
T
 g( t ) sin(no t )dt = (2/4)[
0
1
 5 sin(
0
n = odd
n = even
2
n
n
t )dt +  10 sin( t )dt ]
1
2
2
 2x5
2x10
n
n
cos
cos
= 0.5[
t –
t ] = (5/(n))[3 – 2 cos n + cos(n/2)]
n
n
2 0
2 1
1
n
1
2
3
4
5
6
7
8
an
–1.59
0
0.53
0
–0.32
0
0.23
0
2
bn
7.95
0
2.65
0.80
1.59
0
1.15
0.40
An
8.11
0
2.70
0.80
1.62
0
1.17
0.40
phase
–101.31
0
–78.69
–90
–101.31
0
–78.69
–90
8
An
–78.69˚
90˚
–101.31˚
φ
Figure D. 35
For Prob. 17.3.
0
π 2π 3π 4π 5π 6π 7π 8π
ω
0
π 2π 3π 4π 5π 6π 7π 8π
ω
Chapter 17, Solution 4.
f(t) = 10 – 5t, 0 &lt; t &lt; 2, T = 2,  o = 2/T = 
a o = (1/T)
a n = (2/T)
=
T
2
2
0
0
0
2
 f ( t )dt = (1/2)  (10  5t )dt = 0.5[10t  (5t / 2)] = 5
T
 f ( t ) cos(n t )dt
o
0
2
2
0
0
= (2/2)
0
 (10) cos(nt )dt –  (5t ) cos(nt )dt
5t
5
cos nt +
sin nt = [–5/(n22)](cos 2n – 1) = 0
2 2
n 
n
0
0
2
=
b n = (2/2)
=
2
 (10  5t ) cos(nt )dt
2
2
 (10  5t ) sin( nt )dt
0
2
2
0
0
 (10) sin(nt )dt –  (5t ) sin( nt )dt
5
5t
= 2 2 sin nt +
cos nt = 0 + [10/(n)](cos 2n) = 10/(n)
n
n 
0
0
2
2
Hence
f(t) = 5 
10  1
 sin(nt ) .
 n 1 n
Chapter 17, Solution 5.
T  2,
  2 / T  1
T
ao 
1
1
z (t )dt  [2 x  4 x ]  1

T0
2
T

2
2
1
1
2
an   z (t ) cos(no )dt   2 cos(nt )dt   4 cos(nt )dt 
sin( nt )
T0
n
 0
 
T
bn 

2
2
1
1
2
z (t ) sin( no )dt   2 sin(nt )dt   4 sin( nt )dt  
cos(nt )

T 0
n
0
 
 12

, n  odd
  n
 0, n  even
Thus,
z(t )   1 


n 1
n  odd
12
sin(nt )
n

0

0

4
sin( nt )
n

4
cos(nt )
n
2

2

0
Chapter 17, Solution 6.
T=2,  o =2/T = 1
T

2

1
1 
1
ao   f(t)dt 
(5  10 )  7.5
  5dt   10dt  
T0
2  0

 2
T

2

2
2 


f
t
n

tdt
ntdt
(
)cos
5cos
10cos ntdt   0

o


T0
2  0


T

2
 1 1
 1
2 
2
2 
bn   f(t)sin notdt 

  5 sin ntdt   10 s inntdt     cos nt  cos nt
T0
2  0
0 n
 

   n
 10
, n  odd
5


cos  n  1   n
n
 0, n  even
an 
Thus,
f(t)  7.5 
10
sin nt
n odd n


Chapter 17, Solution 7.
T  3,
o  2 / T  2 / 3
T
2
3
 1
1
1
ao   f(t)dt    2dt   (1)dt   (4  1)  1
T0
3 0
2
 3
an 

T
2
3
2
2n t
2
2n t
2n t 


f
(
t
)cos
dt
2cos
dt
(1)cos
dt 



T0
3
3 0
3
3
2

2 3
2nt
3
2nt
2
sin
sin
1
3  2n
3 0
2n
3

2
3
3
4 n

sin
3
2  n
T
2
3
2
2n t
2
2n t
2n t 


f
(
t
)sin
dt
2
sin
dt
(1)s in
dt 



T0
3
3 0
3
3
2

2
3
2n t 2
3
2n t 3  3
4n
  2 x

cos
cos
(1 2cos
)

3 
2n
3 0 2n
3 2  n
3
1 
4n  3 
4 n 
=  2  3 cos
 1 

1  cos
n 
3
3 
 n 
bn 
Hence,

 3
4n
2nt 3 
4n 
2nt 
cos

f(t) = 1    sin
 sin
 1  cos
3
3
n 
3 
3 
n  0  n
We can now use MATLAB to check our answer,
&gt;&gt; t=0:.01:3;
&gt;&gt; f=1*ones(size(t));
&gt;&gt; for n=1:1:99,
f=f+(3/(n*pi))*sin(4*n*pi/3)*cos(2*n*pi*t/3)+(3/(n*pi))*(1cos(4*n*pi/3))*sin(2*n*pi*t/3);
end
&gt;&gt; plot(t,f)
2 .5
2
1 .5
1
0 .5
0
-0 . 5
-1
-1 . 5
0
0 .5
1
1 .5
2
Clearly we have nearly the same figure we started with!!
2 .5
3
Chapter 17, Solution 8.
Using Fig. 17.51, design a problem to help other students to better understand how to
determine the exponential Fourier Series from a periodic wave shape.
Although there are many ways to solve this problem, this is an example based on the
same kind of problem asked in the third edition.
Problem
Obtain the exponential Fourier series of the function in Fig. 17.51.
f(t)
5
t
0
1
2
Figure 17.51
3
4
5
For Prob. 17.8.
Solution
o  2 / T  
T  2,
5(1 t), 0  t  1
f(t)  
 0, 1  t  2
1
1 T
1
cn   f(t)e jnot dt   5(1 t)e jn t dt
0
20
T

1
1
5 1  jn t
5
5 e jn t 1 5 e jn t
 jn t
( jn t  1)
e
dt
te
dt



2


2 0
20
2  jn 0 2 ( jn )
0
 jn
 1 5 e jn
5 e
5 (1)


( jn  1) 
2 2
2  jn
2 n 
2 n2 2
But e jn  cos  n  j sin n  cos n  0  (1)n
cn 
2.5[1 (1)n ] 2.5(1)n[1 jn ] 2.5

 2 2
jn
n2 2
n
Chapter 17, Solution 9.
f(t) is an even function, b n =0.
T  8,   2 / T   / 4
ao 
1
T
T

0
2
 10 4
2
f (t )dt    10 cos t / 4dt  0  ( ) sin t / 4
8 0
 4 
T /2
2
2
0

10

 3.183
2
4
40
f (t ) cos n o dt  [  10 cos t / 4 cos nt / 4dt 0]  5 cos t (n  1) / 4  cos t (n  1) / 4dt

T 0
8 0
0
For n = 1,
an 
2

2
a1  5 [cos t / 2  1]dt  5 sin t / 2dt  t   10
0

0
For n&gt;1,
2
20
20
(n  1) t
(n  1) t
an 
sin
sin

4
4
(n  1)
(n  1)
2

0
20
20
(n  1)
(n  1)
sin
sin

2
2
(n  1)
(n  1)
20
20
sin 1.5  sin  / 2  2,122066 sin( 270)  6.3662 sin(90)
3

20
10
 2.122066  6.3662  4.244, a 3 
sin 2  sin   0
4

a2 
Thus,
a 0  3.183,
a 1  10,
a 2  4.244,
a 3  0,
b1  0  b 2  b 3
Chapter 17, Solution 10.
T  2 ,
o  2 / T  1
T

Vo
Vo e jnt 
1
 jnot
 jnt
cn   f(t)e
dt 
e
dt
(1
)

T0
2 0
2  jn 0

Vo
jV
 je jn  j   o (cos n  1)
2n
2n
f(t) 

jVo
 2n (cos n  1)e
n
jnt
Chapter 17, Solution 11.
T  4,
 o  2 / T   / 2
T
cn 
1
1
1 0
y (t )e  jnot dt    10(t  1)e  jnt / 2 dt   (10)e  jnt / 2 dt 


0
T0
4  1
cn 
10  e  jnt / 2
2  jnt / 2
e
 2 2 ( jnt / 2  1) 
4  n  / 4
jn

0
1

2  jnt / 2
e
jn
1
0



10  4
2
4
2 jn / 2
2  jn / 2
2 

 2 2 e jn / 2 ( jn / 2  1) 
e

e


2 2
4 n 
jn n 
jn
jn
jn 
But
e jn / 2  cos(n / 2)  j sin( n / 2)  j sin( n / 2),
e  jn / 2  cos(n / 2)  j sin( n / 2)   j sin( n / 2)
10
cn  2 2 1  j ( jn / 2  1) sin(n / 2)  n sin(n / 2)
n
y(t ) 


n  
10
1  j([ jn / 2]  1) sin(n / 2)  n sin(n / 2) e jnt / 2
2 2
n
Chapter 17, Solution 12.
A voltage source has a periodic waveform defined over its period as
v(t) = 10t(2 – t) V, for all 0 &lt; t &lt; 2
Find the Fourier series for this voltage.
v(t) = 10(2 t – t2), 0 &lt; t &lt; 2, T = 2,  o = 2/T = 1
ao =
T
(1/T)  f (t )dt 
0

1
2
10
(t 2  t 3 / 3)
2

2
0
2
0

10(2t  t 2 )dt
40 3
20 2
(1  2 / 3) 
2
3
2
2 T
10  2
2t

a n =  10(2t  t 2 ) cos(nt )dt   2 cos(nt ) 
sin(nt )
T 0
n
 n
0
bn =




10
2nt cos(nt )  2 sin(nt )  n 2t 2 sin(nt )
n 3

20
10
 40
(1  1)  3 4n cos(2n)  2
2
n
n
n

2
0
20 T
10 T
2
(
2
nt

t
)
sin(
nt
)
dt

(2nt  t 2 ) sin( nt )dt


0
0
T

2
2n 10
10

2 2
(sin(
nt
)

nt
cos(
nt
))

(
2
nt
sin(
nt
)

2
cos(
nt
)

1
n
t
cos(
nt
))
0
0
 n2
n 3
 40 40

0
n
n
Hence,
f(t) =
20 2  40
  2 cos(nt )
3
n 1 n
Chapter 17, Solution 13.
Design a problem to help other students to better understand obtaining the Fourier series from a
periodic function.
Although there are many ways to solve this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
A periodic function is defined over its period as
10sin t ,
h(t )  
 20sin(t   ),
0t 
  t  2
Find the Fourier series of h(t).
Solution
T = 2,  o = 1
T
a o = (1/T)  h ( t )dt 
0

a n = (2/T)
2
1 
[  10 sin t dt +  20 sin( t  ) dt ]

2 0


1
30

2
 10 cos t 0  20 cos( t  )  
2

T
 h( t ) cos(n t )dt
o
0
= [2/(2)]   10 sin t cos( nt )dt 
 0


2

20 sin( t  ) cos( nt )dt 

Since sin A cos B = 0.5[sin(A + B) + sin(A – B)]
sin t cos nt = 0.5[sin((n + 1)t) + sin((1 – n))t]
sin(t – ) = sin t cos  – cost sin  = –sin t
sin(t – )cos(nt) = –sin(t)cos(nt)
an =
2
1  
10 [sin([1  n ]t )  sin([1  n ]t )]dt  20 [sin([1  n ]t )  sin([1  n ]t )]dt 


2  0
5
=

 cos([1  n ]t ) cos([1  n ]t )    2 cos([1  n ]t ) 2 cos([1  n ]t )  2  


 
 
 
1 n
1 n
1 n
1 n
0 
  

5 3
3
3 cos([1  n ]) 3 cos([1  n ]) 





 1  n 1  n
1 n
1 n
an =
But,
[1/(1+n)] + [1/(1-n)] = 2/(1–n2)
cos([n–1]) = cos([n+1]) = cos  cos n – sin  sin n = –cos n
a n = (5/)[(6/(1–n2)) + (6 cos(n)/(1–n2))]
= [30/((1–n2))](1 + cos n) = [–60/((n2–1))], n = even
= 0,
n = odd
T
b n = (2/T)  h( t ) sin no t dt
0

2
0

= [2/(2)][  10 sin t sin nt dt +  20(  sin t ) sin nt dt
This is an interesting function which will have a value for b 1 but not for any of the other b n terms
(they will be zero).
1  cos(2 t )
dt  5
2
2
2
+  20( sin t ) sin t dt  20  (sin t ) 2 dt  10   5



b 1 = [2/(2)][   10 sin t sin t dt  10
0
 0


Now we can calculate the rest of the b n for values of n = 2 and greater than 2. We note that,
sin A sin B = 0.5[cos(A–B) – cos(A+B)]
sin t sin nt = 0.5[cos([1–n]t) – cos([1+n]t)]

b n = (5/){[(sin([1–n]t)/(1–n)) – (sin([1+n]t)/ (1  n )] 0
2
+ [(2sin([1-n]t)/(1-n)) – (2sin([1+n]t)/ (1  n )]  }
=
5

 sin([1  n ]) sin([1  n ]) 

 
 = 0
1 n
1 n
{Note, that if we substitute 1 for n, the first term is undefined!}
Thus,
h(t) =
30
60  cos( 2kt )
 5 sin( t )  

 k 1 (4k 2  1)
This does make a very good approximation!
Chapter 17, Solution 14.
Since cos(A + B) = cos A cos B – sin A sin B.

25
 25

cos(n / 4) cos( 2nt )  3
sin(n / 4) sin( 2nt ) 
f(t) = 5    3
n 1

n 1  n  1
Chapter 17, Solution 15.
Dcos t + Esin t = A cos(t - )
(a)
where
f(t) = 10 
A =
D 2  E 2 ,  = tan-1(E/D)
A =
16
1
 6 ,  = tan-1((n2+1)/(4n3))
2
( n  1)
n


n 1
2
2

16
1
1 n  1 




cos
10
nt
tan

( n 2  1) 2 n 6
4n 3 

Dcos t + Esin t = A sin(t + )
(b)
where
f(t) = 10 
A =


n 1
D 2  E 2 ,  = tan-1(D/E)

16
1
4n 3 
1




sin
10
nt
tan

( n 2  1) 2 n 6
n 2  1 

Chapter 17, Solution 16.
If v 2 (t) is shifted by 1 along the vertical axis, we obtain v 2 *(t) shown below, i.e.
v 2 *(t) = v 2 (t) + 1.
v 2 *(t)
2
1
-2 -1
0
1
2
3
4
5
t
Comparing v 2 *(t) with v 1 (t) shows that
v 2 *(t) = 2v 1 ((t + t o )/2)
where (t + t o )/2 = 0 at t = -1 or t o = 1
Hence
v 2 *(t) = 2v 1 ((t + 1)/2)
But
v 2 *(t) = v 2 (t) + 1
v 2 (t) + 1 = 2v 1 ((t+1)/2)
v 2 (t) = -1 + 2v 1 ((t+1)/2)
= -1 + 1 
v 2 (t) = 
8
2
8
2
1


 t  1 1
 t  1
 t  1
cos  2   9 cos 3 2   25 cos 5 2   








1
  t   1

 3t 3 
 5t 5 
cos 2  2   9 cos 2  2   25 cos 2  2   





 

v 2 (t) = 
8
2
  t  1

1
 3 t 
 5 t 
sin 2   9 sin 2   25 sin 2   




  

Chapter 17, Solution 17.
We replace t by –t in each case and see if the function remains unchanged.
(a)
1 – t,
neither odd nor even.
(b)
t2 – 1,
even
(c)
cos n(-t) sin n(-t) = - cos nt sin nt,
odd
(d)
sin2 n(-t) = (-sin t)2 = sin2 t,
even
(e)
e t,
neither odd nor even.
Chapter 17, Solution 18.
(a)
T = 2 leads to  o = 2/T = 
f 1 (-t) = -f 1 (t), showing that f 1 (t) is odd and half-wave symmetric.
(b)
T = 3 leads to  o = 2/3
f 2 (t) = f 2 (-t), showing that f 2 (t) is even.
(c)
T = 4 leads to  o = /2
f 3 (t) is even and half-wave symmetric.
Chapter 17, Solution 19.
o  2 / T   / 2
T  4,
 10t, 0  t  1
f(t)  
10(2  t), 1  t  2
a0 
T
1
2
1
1
1
1 2 1 10
t2 2
(
)
10
10(2
)
5
(2
)  2.5
f
t
dt

tdt


t
dt

t

t

4 0
4 1
4
2 1
T 0
0 4
T
1
2
2
2
2
an   f(t)cos notdt   10t cos notdt   10(2  t)cos notdt
40
41
T0
1 10
2
2
20
t
5
5t
cos not 
sin not 
sin not  2 2 cos not 
sin not
no
no
no
0 no
1 n o
1
20
1
10
5
(cos n / 2  1) 
sin n / 2 
(sin n  sin n / 2)  2 2
cos n

no
no
no
n  /4


5
n  /4
2
2
cos n / 2 
T
bn 


10
5
sin n 
sin n / 2
no
n / 2
1
2
2
2
2
f(t)s innotdt   10t s innotdt   10(2  t)sin notdt

40
41
T0
1 10
1
2
2
5
5
t
sin not 
cos not  2 2 sin not 
cos not
no
0 no
0 n o
1 no
1
5
n
2
2
o
sin n / 2 

10
5
(cos  n  cos n / 2)  2 2 (sin  n  sin n / 2)
no
n o
2
cos  n / 2
cos n 
no
no
Chapter 17, Solution 20.
This is an even function.
b n = 0, T = 6,  = 2/6 = /3
2
T
ao =

T/2
f ( t )dt 
0
3
2 2
( 4 t  4)dt  4 dt 


2
6  1
2
1 2
( 2 t  4 t )  4(3  2) = 2
1

3 
=
4
T
an =

T/4
f ( t ) cos( nt / 3)dt
0
2
= (4/6)[  ( 4 t  4) cos( nt / 3)dt +
1

3
2
4 cos( nt / 3)dt ]
2
3
16  3
3
16  9
 nt 
 nt 
 nt 
 nt  3t
=
sin
sin
sin
cos

 


2 2


6  n  3  2
6 n 
 3  n   3  n  3   1
= [24/(n22)][cos(2n/3)  cos(n/3)]
f(t) = 2 
Thus
24  1

2 n 1 n2
  2n 
 nt 
 n  
cos 3   cos 3   cos 3 





 
At t = 2,
f(2) = 2 + (24/2)[(cos(2/3)  cos(/3))cos(2/3)
+ (1/4)(cos(4/3)  cos(2/3))cos(4/3)
+ (1/9)(cos(2)  cos())cos(2) + -----]
= 2 + 2.432(0.5 + 0 + 0.2222 + -----)
f(2) = 3.756
Chapter 17, Solution 21.
This is an even function.
b n = 0, T = 4,  o = 2/T = /2.
f(t) = 2  2t,
= 0,
0&lt;t&lt;1
1&lt;t&lt;2
1

t2 
2 1
ao =
2(1  t )dt   t   = 0.5
2 0
4 0

an =
4
T

T/2
0
f ( t ) cos( no t )dt 
4 1
 nt 
2(1  t ) cos
dt

4 0
 2 
= [8/(2n2)][1  cos(n/2)]
f(t) =
1

2

8
n 
n1
2
2

 n 
 nt 
1  cos 2  cos 2 
 



Chapter 17, Solution 22.
Calculate the Fourier coefficients for the function in Fig. 16.54.
f(t)
4
-5 -4 -3 -2 -1
0
1
Figure 16.54
2
3
4
5
t
For Prob. 16.15
This is an even function, therefore b n = 0. In addition, T=4 and  o = /2.
ao =
an =
2
T

4
T
T2
0

f ( t )dt 
T2
0
1
2 1
4 tdt  t 2  1

0
4 0
f ( t ) cos(o nt )dt 
4 1
4 t cos( nt / 2)dt
4 0
1
2t

 4
sin( nt / 2)
 4  2 2 cos( nt / 2) 
n
0
n 
an =
16
8
(cos( n / 2)  1) 
sin( n / 2)
2 2
n
n 
Chapter 17, Solution 23.
Using Fig. 17.61, design a problem to help other students to better understand finding the Fourier
series of a periodic wave shape.
Although there are many ways to solve this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
Find the Fourier series of the function shown in Fig. 17.61.
Figure 17.61
Solution
f(t) is an odd function.
f(t) = t, 1&lt; t &lt; 1
a o = 0 = a n , T = 2,  o = 2/T = 
bn =
=
4
T

T/2
0
f ( t ) sin( no t )dt 
4 1
t sin( nt )dt
2 0
2
sin(nt )  nt cos(nt ) 10
2
n 
2
= [2/(n)]cos(n) = 2(1)n+1/(n)
f(t) =
2

( 1) n  1
sin( nt )

n
n 1

Chapter 17, Solution 24.
(a)
This is an odd function.
a o = 0 = a n , T = 2,  o = 2/T = 1
bn =
4
T

T/2
0
f ( t ) sin(o nt )dt
f(t) = 1 + t/,
bn =
0&lt;t&lt;
4 
(1  t / ) sin( nt )dt
2 0

=
2 1
1
t

 cos( nt )  2 sin( nt ) 
cos( nt )

 n
n 
n
0
= [2/(n)][1  2cos(n)] = [2/(n)][1 + 2(1)n+1]
a 2 = 0, b 2 = [2/(2)][1 + 2(1)] = 1/ = 0.3183
(b)
n = n o = 10 or n = 10
a 10 = 0, b 10 = [2/(10)][1  cos(10)] = 1/(5)
Thus the magnitude is A 10 =
and the phase is
 10 = tan1(b n /a n ) = 90

(c)
f(t) =
2
a 210  b10
= 1/(5) = 0.06366
2
 n [1  2 cos(n)] sin(nt ) 
n 1

f(/2) =
2
 n [1  2 cos(n)] sin(n / 2) 
n 1
For n = 1,
f 1 = (2/)(1 + 2) = 6/
For n = 2,
f2 = 0
For n = 3,
f 3 = [2/(3)][1  2cos(3)]sin(3/2) = 6/(3)
For n = 4,
f4 = 0
For n = 5,
f 5 = 6/(5), ----
f(/2) = 6/  6/(3) + 6/(5)  6/(7) ---------
Thus,
= (6/)[1  1/3 + 1/5  1/7 + --------]
f(/2)  1.3824
which is within 8% of the exact value of 1.5.
(d)
From part (c)
f(/2) = 1.5 = (6/)[1  1/3 + 1/5  1/7 + - - -]
(3/2)(/6) = [1  1/3 + 1/5  1/7 + - - -]
or /4 = 1  1/3 + 1/5  1/7 + - - -
Chapter 17, Solution 25.
This is a half-wave (odd) function since f(tT/2) = f(t).
a o = 0, a n = b n = 0 for n = even, T = 3,  o = 2/3.
For n = odd,
4 1.5
4 1
f ( t ) cos n0 tdt   t cos n0 tdt

3 0
3 0
an =
1
4 9
 2nt 
 2nt  3t
sin 
= 
cos



3  4 2 n 2
 3  2n  3  0
 3
= 
2 2
 n
bn =
  2n   2
 2n 
sin 
 cos
  1 

  3   n  3 
4 1.5
4 1
f ( t ) sin( no t )dt   t sin(2nt / 3)dt

0
3
3 0
1
4 9
3t
 2nt 
 2 nt 
=
sin
cos


2 2

3  4 n
 3  2n
 3  0
 3
 2n  2
 2n 
sin 
= 
cos


2 2
 3  n
 3 
 n
  3   2n   2
 2n    2nt  




cos
1
sin



  cos


 2 2


   n   3   n  3    3  
f(t) =  

n 1  
3
2 n  2
2n    2nt 



n  odd 
   2 n 2 sin 3   n cos 3   sin 3 


 

Chapter 17, Solution 26.
T = 4,  o = 2/T = /2
ao =
1 T
1 1
f ( t )dt    1 dt 

T 0
4  0
an =
2 T
f ( t ) cos(no t )dt
T 0
an =
2 2
1 cos( nt / 2)dt 
4  1

3
1

3
2
2 dt   1 dt  = 1

3
4
2 cos( nt / 2)dt   1 cos( nt / 2)dt 

3
4
2
3
4
2
nt
4
nt
2
nt 


= 2  sin
sin
sin

2 1 n
2 2 n
2 3 
 n
=
4
n
n 
 3n
sin 2  sin 2 
bn =
2 T
f ( t ) sin( no t )dt
T 0
=
2 2
nt
1
sin
dt 

4  1
2

3
2
2 sin
nt
dt 
2

4
3
1 sin
nt 
dt 
2

2
3
4
 2
nt
4
nt
2
nt 


= 2
cos
cos
cos

2 1 n
2 2 n
2 3 
 n
=
4
cos(n)  1
n
Hence
f(t) =
1

 n (sin( 3n / 2)  sin(n / 2)) cos(nt / 2)  (cos( n)  1) sin(nt / 2)
n 1
4
Chapter 17, Solution 27.
(a)
(b)
odd symmetry.
a o = 0 = a n , T = 4,  o = 2/T = /2
f(t)
= t,
0&lt;t&lt;1
= 0,
1&lt;t&lt;2
1
nt 
nt 2 t
nt
4 1
 4
bn =
cos
dt   2 2 sin
t sin


2  0
2
n
2
4 0
n 
=
4
n
2
n
sin
cos
0

2
n 
2
n
2
2
= 4(1)(n1)/2/(n22),
n = odd
2(1)n/2/(n),
n = even
a 3 = 0, b 3 = 4(1)/(92) = –0.04503
(c)
b 1 = 4/2, b 2 = 1/, b 3 = 4/(92), b 4 = 1/(2), b 5 = 4/(252)
F rms =
a 2o 
1
 a 2n  b 2n 
2
F rms 2 = 0.5b n 2 = [1/(22)][(16/2) + 1 + (16/(812)) + (1/4) + (16/(6252))]
= (1/19.729)(2.6211 + 0.27 + 0.00259)
F rms =
0.14667 = 0.383
Compare this with the exact value of F rms =
(0.383/0.4082)x100 = 93.83%, close.
2
T
1
 t dt
0
2
 1 / 6 = 0.4082 or
Chapter 17, Solution 28.
This is half-wave symmetric since f(t  T/2) = f(t).
a o = 0, T = 2,  o = 2/2 = 
an =
4
T

T/2
0
f ( t ) cos( no t )dt 
4 1
( 2  2 t ) cos( nt )dt
2 0
1
1
t
1

= 4  sin( nt )  2 2 cos( nt ) 
sin( nt )
n 
n
 n
0
= [4/(n22)][1  cos(n)] =
8/(n22),
0,
n = odd
n = even
1
b n = 4  (1  t ) sin( nt )dt
0
1
1
t
 1

= 4 
cos( nt )  2 2 sin( nt ) 
cos( nt )
n 
n
 n
0
= 4/(n), n = odd

f(t) =

  n
k 1
8
4

cos( nt ) 
sin(nt )  , n = 2k  1
2
n


2
Chapter 17, Solution 29.
This function is half-wave symmetric.
T = 2,  o = 2/T = 1, f(t) = t, 0 &lt; t &lt; 
For odd n,
an =
2
T


bn =
2



0
0
(  t ) cos( nt )dt = 
(  t ) sin( nt )dt  
2
cos(nt )  nt sin(nt ) 0 = 4/(n2)
2
n 
2
sin(nt )  nt cos(nt ) 0 = 2/n
2
n 
Thus,

1

 2
f(t) = 2  2 cos( nt )  sin(nt ) ,
n

k 1  n 
n = 2k  1
Chapter 17, Solution 30.
1
cn 
T
(a)
T/2

T / 2
f ( t )e  jno t dt 
T/2
1  T/2


f
(
t
)
cos
n
tdt
j
f ( t ) sin no tdt 
o



T
/
2
T
/
2


T

The second term on the right hand side vanishes if f(t) is even. Hence
cn 
(b)
(1)
2
T
T/2
 f (t ) cos n tdt
o
0
The first term on the right hand side of (1) vanishes if f(t) is odd. Hence,
j2
cn  
T
T/ 2
 f (t ) sin n tdt
o
0
Chapter 17, Solution 31.
If h(t )  f (t ),
T'  T / 


T'
o ' 
2
2

  o
T' T / 
T'
2
2
an ' 
h(t ) cos n o ' tdt   f (t ) cos n o ' tdt

T' 0
T' 0
Let t   , ,
d t  d /  ,
2
f ( ) cos n o  d /   a n
T 0
T
an ' 
Similarly,
T'  T
bn ' bn
Chapter 17, Solution 32.
When i s = 1 (DC component)
i = 1/(1 + 2) = 1/3
For n  1,
 n = 3n, I s = 1/n20
I = [1/(1 + 2 + j n 2)]I s = I s /(3 + j6n)
1
0
2
1
n

  tan(2n )
=
3 1  4n 2  tan 1 (6n / 3) 3n 2 1  4n 2
Thus,
1
i(t) =

3


n 1
1
3n
2
1  4n
2
cos( 3n  tan 1 ( 2n ))
Chapter 17, Solution 33.
For the DC case, the inductor acts like a short, V o = 0.
For the AC case, we obtain the following:
Vo  Vs
V
jnVo
 o 
0
10
j2n
4

5 

1  j 2.5n   Vo  Vs
n  


Vo 
Vs
5 

1  j 2.5n  
n 

A n  n 
4
n
An 
1
5 

1  j 2.5n  
n 


4
n  j(2.5n 2  2  5)
 2.5n 2  2  5 

;  n   tan 
2 2
2 2
2

n
n   ( 2.5n   5)


4
1

v o (t )   A n sin(nt   n ) V
n 1
Chapter 17, Solution 34.
Using Fig. 17.70, design a problem to help other students to better understand circuit responses
to a Fourier series.
Although there are many ways to solve this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
Obtain v o (t) in the network of Fig. 17.70 if
10
n

cos  nt 
2
4

n 1 n

v(t )  

V

Figure 17.70
Solution
For any n, V = [10/n2](n/4),  = n.
1 H becomes j n L = jn and 0.5 F becomes 1/(j n C) = j2/n
2
jn
+
V
+

j2/n
Vo

V o = {j(2/n)/[2 + jn  j(2/n)]}V = {j2/[2n + j(n2  2)]}[(10/n2)(n/4)]
20((n / 4)   / 2)

n
2
4n  (n 2  2) 2  tan 1 ((n 2  2) / 2n )
2
20

n
2
n 4
4
[(n / 4)  ( / 2)  tan 1 ((n 2  2) / 2n )]

v o (t) =

n 1

n 
n2  2 

  tan 1
cos nt 
4 2
2n 
n4  4

20
n2
Chapter 17, Solution 35.
If v s in the circuit of Fig. 17.72 is the same as function f 2 (t) in Fig. 17.57(b),
determine the dc component and the first three nonzero harmonics of v o (t).
1
1H
+
vS
+

1
1F
vo

Figure 16.64
For Prob. 16.25
f 2 (t)
2
1
-2 -1
0
1
2
Figure 16.50(b)
3
4
5
t
For Prob. 16.25
The signal is even, hence, b n = 0. In addition, T = 3,  o = 2/3.
v s (t)
ao =
an =
=
= 1 for all 0 &lt; t &lt; 1
= 2 for all 1 &lt; t &lt; 1.5
2 1
1dt 
3  0
4
2dt  
 3
1.5

1
4 1
cos(2nt / 3)dt 
3  0
1.5

1
2 cos(2nt / 3)dt 

4 3
6
2
1
1.5 
sin(2nt / 3) 0 
sin(2nt / 3) 1   
sin(2n / 3)

3  2n
2n
n

4 2  1
v s (t) =
  sin(2n / 3) cos(2nt / 3)
3  n 1 n
Now consider this circuit,
j2n/3
1
+
vS
+

-j3/(2n)
1
vo

Let Z = [-j3/(2n)](1)/(1 – j3/(2n)) = -j3/(2n - j3)
Therefore, v o = Zv s /(Z + 1 + j2n/3). Simplifying, we get
vo =
 j9 v s
12n  j( 4 n 2  2  18)
For the dc case, n = 0 and v s = &frac34; V and v o = v s /2 = 3/8 V.
We can now solve for v o (t)
3 

 2nt
v o (t) =    A n cos
  n   volts

 3
 8 n 1
where A n 
6
sin( 2n / 3)
n
3 
 n

and  n  90 o  tan 1 

2

3
2
n
2 2


 4n 

 6 
16n 2  2  
 3

where we can further simplify A n to this, A n 
9 sin( 2n / 3)
n 4n 4  4  81
Chapter 17, Solution 36.
We first find the Fourier series expansion of v s . T  1,
 o  2 / T  2
T
1
1
1
t2 0
a0   f(t)dt   10(1 t)tdt  10(t  )  5
20
2 1
T0
1
T
2
an   f (t ) cos notdt  2 10(1  t ) cos 2n tdt
T 0
0
1
t
 1
1
sin2n t  2 2 cos 2n t 
sin2n t   0
 20 
2n
4n 
 2 n
o
1
T
bn 
2
2
f(t)sin notdt   10(1 t)t sin notdt

20
T0
1
1
 1
 1 10
 20  
cos 2n t  2 2 sin 2n t 
cos 2n t  
4n 
2n
 2n
 0 n

10
sin 2n t
n 1 n
1H


jn L  jn
1
1
 j100
10mF




jn C jn 0.01
n
vs (t )  5  
Io 
Vs
5  jn 
j100
n
For dc component, ω 0 = 0 which leads to I 0 = 0.
For the nth harmonic,
10
 0
10
n

In 

 A n  n
j100 5n  j( 2n 2  2  50)
5  j2n 
2n
where
An 
10
25n 2 2  (2n 2 2  50) 2
n   tan 1
,

2n 2 2  50
5n
io (t )   An sin(2n t  n )
n 1
Chapter 17, Solution 37.
We first need to express i s in Fourier series. T  2,
ao 
o  2 / T  
T
1
2
 1
1
1
f
t
dt
dt
(
)
3
1dt   (3  1)  2





T0
2 0
1
 2
T
1
2
 3
1 1
2
2
2
an   f(t)cos notdt    3 cos n tdt   cos n tdt  
sin n t 
sin n t  0
T0
2 0
0 n
1
1
 n
bn 
T
1
2
 3
1 1
2 2
2
2


f
t
inn

tdt
inn

tdt
co s n t 
(
)s
3
s
s inn tdt  
cos n t 
(1 cos n

o


T0
2 0
0 n
1 n
1
 n
2
(1 cos n )sin n t
n1 n

is(t)  2  
By current division,
Io 
Is
1
Is 
1 2  jnL
3  j 3n
jn 3Is
jnIs

3  j3n 1 jn
For dc component (n=0), V o = 0.
For the nth harmonic,
Vo  jnLIo 
Vo 
2(1  cos n)
jn 2
(1  cos n)  90 
(90  tan 1 n  90)
1  jn n
1  n 22

2(1 cos  n)
n1
1 n 
vo(t)  
2
2
cos(n t  tan1 n )
Chapter 17, Solution 38.
1 2  1
v s ( t )    sin nt ,
2  k 1n
Vo 
For dc,  n  0,
j n
Vs ,
1  j n
Vs  0.5,
For nth harmonic, Vs 
Vo 
n  2k  1
 n  n
Vo  0
2
  90 o
n
2
2  tan 1 n
90 o 
1  n 2  2  tan 1 n n
1  n 22
n90 o

v o (t )  
k 1

2
1 n 
2
2
cos(nt  tan 1 n ),
n  2k  1
Chapter 17, Solution 39.
Comparing v s (t) with f(t) in Figure 15.1, v s is shifted by 2.5 and the magnitude is
5 times that of f(t).
Hence
10  1
v s (t) = 5 
n = 2k  1
 sin(nt ),
 k 1 n
T = 2,  o = 2//T = ,  n = n o = n
For the DC component,
i o = 5/(20 + 40) = 1/12
For the kth harmonic,
V s = (10/(n))0
100 mH becomes j n L = jnx0.1 = j0.1n
50 mF becomes 1/(j n C) = j20/(n)
I 20 
VS
40 
Io
j20/(n)
+

j0.1n
Z
j20
( 40  j0.1n)
n

Let Z = j20/(n)||(40 + j0.1n) =
j20

 40  j0.1n
n

=
Z in
 j20( 40  j0.1n
2 n  j800

2 2
40n  j(0.1n 2  2  20)
 j20  40n  j0.1n 
802n  j( 2n 2  2  1200)
= 20 + Z =
40n  j(0.1n 2  2  20)
Vs
400n  j( n 2  2  200)

Z in
n[802n  j( 2n 2  2  1200)]
j20
I

 j20I
n

Io =
2 2
j20

 ( 40  j0.1n) 40n  j(0.1n   20)
n
I =
=
=
 j200
n[802n  j( 2n 2  2  1200)]
200  90  tan 1{(2n 2  2  1200) /(802n)}
n (802) 2  ( 2 n 2  2  1200) 2
Thus
i o (t) =
where
1 200 

 I n sin(n t  n ) ,
20
 k 1
2n 2  2  1200
n  90  tan
802n
1
In =
1
n (804n )  ( 2n 2  2  1200)
2
n = 2k  1
Chapter 17, Solution 40.
T = 2,  o = 2/T = 
1
1
ao =
T
an =
2
T

1 1
t2 
v
(
t
)
dt

(
2

2
t
)
dt

t

 1/ 2

0
2 0
2  0

T
T
1
0
0
 v( t ) cos(nt )dt   2(1  t ) cos(nt )dt
1
1
t
1

= 2  sin( nt )  2 2 cos( nt ) 
sin( nt )
n 
n
 n
0
2
= 2 2 (1  cos n) 
n 
bn =
2
T
n  even
0,
4
4
, n  odd  2
2
n 
 ( 2n  1) 2
2
T
1
0
0
 v( t ) sin(nt )dt  2 (1  t ) sin(nt )dt
1
2
t
1

 1
= 2
cos( nt ) 
cos( nt )  2 2 sin( nt ) 
n
n 
 0 n
 n
v s (t) =
1

2
A
where n  tan 1
n
cos( nt   n )
( 2n  1) 2
, An 
2n
4
16
 4
2
n 
 ( 2n  1) 4
2
For the DC component, v s = 1/2. As shown in Figure (a), the capacitor acts like an
open circuit.
1
Vx
2V x
 +
+
0.5V
i
+

Vo
+
Vx
3

Vo

(a)
1
Vx
 +
2V x
Vo
+
VS
+

3
(1/4)F
Vo

(b)
Applying KVL to the circuit in Figure (a) gives
But
–0.5 – 2V x + 4i = 0
(1)
–0.5 + i + V x = 0 or –1 + 2V x + 2i = 0
(2)
–1.5 + 6i = 0 or i = 0.25
V o = 3i = 0.75
For the nth harmonic, we consider the circuit in Figure (b).
 n = n, V s = A n –, 1/(j n C) = –j4/(n)
At the supernode,
(V s – V x )/1 = –[n/(j4)]V x + V o /3
V s = [1 + jn/4]V x + V o /3
But
(3)
–V x – 2V x + V o = 0 or V o = 3V x
Substituting this into (3),
V s = [1 + jn/4]V x + V x = [2 + jn/4]V x
= (1/3)[2 + jn/4]V o = (1/12)[8 + jn]V o
V o = 12V s /(8 + jn) =
Vo =
12
64  n 
2
2
12A n   
64  n 2  2  tan 1 (n / 8)
4
16
 4
[tan 1 (n / 8)  tan 1 ((2n  1) /(2n ))]
2
4
n 
 (2n  1)
2
Thus
v o (t) =
3

4

V
n 1
n
cos( nt   n ) volts
where
Vn =
12
64  n 2  2
4
16
and
 4
2
n 
 ( 2n  1) 4
2
 n = tan–1(n/8) – tan–1((2n – 1)/(2n))
Chapter 17, Solution 41.
For the full wave rectifier,
T = ,  o = 2/T = 2,  n = n o = 2n
Hence
2 4 
1

v in (t) =    2 cos 2nt  volts

   n1 4n  1
For the DC component,
V in = 2/
The inductor acts like a short-circuit, while the capacitor acts like an open circuit.
V o = V in = 2/
For the nth harmonic,
V in = [–4/((4n2 – 1))]0
2 H becomes j n L = j4n
0.1 F becomes 1/(j n C) = –j5/n
Z = 10||(–j5/n) = –j10/(2n – j)
V o = [Z/(Z + j4n)]V in = –j10V in /(4 + j(8n – 10))

j10
40 

 
= 
4  j(8n  10)  (4n 2  1) 
=
40{90  tan 1 (2n  2.5)}
(4n 2  1) 16  (8n  10) 2
Hence
2 

v o (t) =    A n cos( 2nt  n ) volts
  n 1

where
An =
20
( 4n 2  1) 16n 2  40n  29
 n = 90 – tan–1(2n – 2.5)
Chapter 17, Solution 42.
20  1
vs  5 
 sin nt, n  2k - 1
 k 1n
Vs  0
 j n C(0  Vo )
R
For n = 0 (dc component),

Vo 
j
Vs ,  n  no  n
 n RC
V o =0.
For the nth harmonic,
Vo 
190 o 20
20
10 5
  90 o 

nRC n
n 2  2 x10 4 x 40 x10 9 2n 2  2
Hence,
v o (t ) 
10 5
2 2

1
n
k 1
2
cos nt , n  2k - 1
Alternatively, we notice that this is an integrator so that
v o (t )  
10 5
1

v
dt
s
RC 
2 2

1
n
k 1
2
cos nt , n  2k - 1
Chapter 17, Solution 43.
(a)
V rms =
(b)
I rms =
(c)
a 02 
1  2
1
(a n  b 2n )  30 2  (20 2  10 2 ) = 33.91 V

2 n 1
2
1
6 2  (4 2  2 2 ) = 6.782 A
2
1
P = V dc I dc +  Vn I n cos( n   n )
2
= 30x6 + 0.5[20x4cos(45o-10o) – 10x2cos(-45o+60o)]
= 180 + 32.76 – 9.659 = 203.1 W
Chapter 17, Solution 44.
Design a problem to help other students to better understand how to find the rms voltage across
and the rms current through an electrical element given a Fourier series for both the current and
the voltage. In addition, have them calculate the average power delivered to the element and the
power spectrum.
Although there are many ways to solve this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
The voltage and current through an element are respectively
v(t) = [30 cos (t + 35) + 10cos(2t – 55˚) + 4 cos(3t - 10)] V
i(t) = [2 cos (t–10˚) + cos(2t – 10)] A
(a) Find the average power delivered to the element
(b) Plot the power spectrum.
Solution


1
60 cos(25 o )  10 cos 45 o  0  27.19  3.536 = 30.73 W.
2
(a)
p  vi 
(b)
The power spectrum is shown below.
p
27.19
3.536
0
1
2
3

Chapter 17, Solution 45.
 n = 1000n
j n L = j1000nx2x10–3 = j2n
1/(j n C) = –j/(1000nx40x10–6) = –j25/n
Z = R + j n L + 1/(j n C) = 10 + j2n – j25/n
I = V/Z
For n = 1, V 1 = 100, Z = 10 + j2 – j25 = 10 – j23
I 1 = 100/(10 – j23) = 3.98773.89
For n = 2, V 2 = 50, Z = 10 + j4 – j12.5 = 10 – j8.5
I 2 = 50/(10 – j8.5) = 3.8140.36
For n = 3, V 3 = 25, Z = 10 + j6 – j25/3 = 10 – j2.333
I 3 = 25/(10 – j2.333) = 2.43513.13
I rms =
0.5(3.987 2  3.812  2.435 2 ) = 4.263 A
p = R(I rms )2 = 181.7W
Chapter 17, Solution 46.
(a)The MATLAB commands are:
t=0:0.01:5;
y=5*cos(3*t) – 2*cos(3*t-pi/3);
plot(t,y)
5
4
3
2
1
0
-1
-2
-3
-4
-5
0
0.5
1
1.5
2
(b) The MATLAB commands are:
t=0:0.01:5;
&raquo; x=8*sin(pi*t+pi/4)+10*cos(pi*t-pi/8);
&raquo; plot(t,x)
&raquo; plot(t,x)
2.5
3
3.5
4
4.5
5
20
15
10
5
0
-5
-10
-15
-20
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
Chapter 17, Solution 47.
T  2,
o  2 / T  
T
1
2
 1
1
1
ao   f(t)dt    4dt   (2)dt   (4  2)  1
T0
2 0
1
 2
T
1
2

R 2
R 2
2
P  Rirms   f (t)dt    4 dt   (2)2 dt   10R
2 0
T0
1

The average power dissipation caused by the dc component is
P 0  Rao2  R  10% of P
Chapter 17, Solution 48.
(a) For the DC component, i(t) = 20 mA. The capacitor acts like an open circuit so that
v = Ri(t) = 2x103x20x10–3 = 40
For the AC component,
 n = 10n, n = 1,2
1/(j n C) = –j/(10nx100x10–6) = (–j/n) k
Z = 2||(–j/n) = 2(–j/n)/(2 – j/n) = –j2/(2n – j)
V = ZI = [–j2/(2n – j)]I
For n = 1,
V 1 = [–j2/(2 – j)]1645 = 14.311–18.43 mV
For n = 2,
V 2 = [–j2/(4 – j)]12–60 = 5.821–135.96 mV
(b)
v(t) = 40 + 0.014311cos(10t – 18.43) + 0.005821cos(20t – 135.96) V
1 
p = V DC I DC +  Vn I n cos( n   n )
2 n 1
= 20x40 + 0.5x10x0.014311cos(45 + 18.43)
+0.5x12x0.005821cos(–60 + 135.96)
= 800.1 mW
Chapter 17, Solution 49.
(a)
Z
2
rms
T

2
 1
1 2
1 
(20 )  10
  z (t )dt 
  4dt   16dt  
T0
2  0

 2
Z rms = 3.162
(b)
Z 2 rms  a02 
1
1  144
72 
1
1  2

(an  bn2 )  1   2 2  1  2 1  0   0   ...   9.396

 
25
2 n1 n 
9
2 n1

n odd
Z rms = 3.065
(c )
 3.065 
%error  1 
 x100  3.068%
 3.162 
Chapter 17, Solution 50.
cn =
=
1
T

T
0
f ( t )e  jo nt dt,
o 
2n

1
1 1  jnt
te
dt
2 
Using integration by parts,
u = t and du = dt
dv = e–jntdt which leads to v = –[1/(2jn)]e–jnt
t
cn = 
e  jnt
2 jn

1

1
1 1  jnt
e
dt
2 jn 1

1
j  jn
 e jnt 
e  jnt
e
=
2 2
2
2n  (  j)
n
1
1
= [j/(n)]cos(n) + [1/(2n22)](e–jn – ejn)
cn =
j( 1) n
j( 1) n
2j



sin(
n
)
n
n
2n 2  2
Thus

f(t) =
 c n e jnot =
n 

 (1)
n 
n
j jnt
e
n
Chapter 17, Solution 51.
Design a problem to help other students to better understand how to find the exponential Fourier
series of a given periodic function.
Although there are many ways to solve this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
Given the periodic function
f(t) = t2,
0&lt;t&lt;T
obtain the exponential Fourier series for the special case T = 2.
Solution
T  2,
o  2 / T  


1
1
1 e  jnt
2
c n   f ( t )e  jno t dt   t 2 e  jnt dt 
 n 2  2 t 2  2 jnt  2 0
3
T
2
2 ( jn)
0
0
T
cn 
2
1
3 3
j2n 
(4n 2  2  j4n) 
f (t ) 

n 
2
n
n  
2
2 2
2
2
(1  jn)
(1  jn )e jnt
Chapter 17, Solution 52.
cn =
=
1
T

T
0
f ( t )e  jo nt dt,
o 
2n

1
1 1  jnt
te
dt
2 
Using integration by parts,
u = t and du = dt
dv = e–jntdt which leads to v = –[1/(2jn)]e–jnt
t
cn = 
e  jnt
2 jn
1

1

1 1  jnt
e
dt
2 jn 1

1
j  jn
 e jnt 
e  jnt
e
=
2 2
2
2n  (  j)
n
1
1
= [j/(n)]cos(n) + [1/(2n22)](e–jn – ejn)
cn =
j( 1) n
j( 1) n
2j



sin(
n
)
n
n
2n 2  2
Thus

f(t) =
 c n e jnot =
n  

 ( 1)
n  
n
j jnt
e
n
Chapter 17, Solution 53.
 o = 2/T = 2
cn =

T
0
1
e  t e  jno t dt   e (1 jno ) t dt
0
1
=
e  (1  j2 n ) t
1  j2 n
1

0


1
e  (1  j 2 n  )  1
1  j2n
= [1/(j2n)][1 – e–1(cos(2n) – jsin(2n))]
= (1 – e–1)/(1 + j2n) = 0.6321/(1 + j2n
0.6321e j2 nt

n   1  j2n

f(t) =
Chapter 17, Solution 54.
T = 4,  o = 2/T = /2
cn =
1
T

T
0
f ( t )e  jo nt dt
=
1  1  jnt / 2
2e
dt 
4  0
=
j
2e  jn / 2  2  e  jn  e  jn / 2  e  j2 n  e  jn
2n

2
1
1e  jnt / 2 dt 

4
2
1e  jnt / 2 dt 


=

j
3e  jn / 2  3  2e  jn
2n

f(t) =
c e
n  
n
jn o t


Chapter 17, Solution 55.
T = 2,  o = 2/T = 1
cn =
But
i(t) =
cn =
1
T

T
0
i( t )e  jno t dt
sin( t ),
0,
0t
  t  2
1 
1  1 jt
sin( t )e  jnt dt 
(e  e  jt )e  jnt dt

2 0
2 0 2 j
1  e jt (1 n ) e  jt (1 n ) 




4j  j(1  n )
j(1  n ) 



0
1  e j(1 n )  1 e  j( n 1)  1



4  1  n
1 n


1
e j (1  n )  1  ne j (1  n )  n  e  j (1  n )  1  ne  j (1 n )  n
4( n 2  1)

But ej = cos() + jsin() = –1 = e–j
cn =
1
1  e  jn
 jn
 jn
 jn
 jn

e

e

ne

ne

2

4( n 2  1)
2 (1  n 2 )


Thus

i(t) =

n  
1  e  jn jnt
e
2(1  n 2 )
Chapter 17, Solution 56.
c o = a o = 10,  o = 
c o = (a n – jb n )/2 = (1 – jn)/[2(n2 + 1)]
f(t) = 10 

(1  jn ) jnt
e
2
 1)
 2(n
n  
n0
Chapter 17, Solution 57.
a o = (6/–2) = –3 = c o
c n = 0.5(a n –jb n ) = a n /2 = 3/(n3 – 2)
f(t) =  3 

n
n  
n0
3
3
e j50nt
2
Chapter 17, Solution 58.
c n = (a n – jb n )/2, (–1)n = cos(n),  o = 2/T = 1
c n = [(cos(n) – 1)/(2n2)] – j cos(n)/(2n)
Thus
f(t) =

cos(n )  jnt
 cos(n)  1
 
j
e
2
4
2n 
 2 n
Chapter 17, Solution 59.
For f(t), T = 2,  o = 2/T = 1.
a o = DC component = (1x + 0)/2 = 0.5
For
h(t), T = 2,  o = 2/T = .
a o = (2x1 – 2x1)/2 = 0
Thus by replacing  o = 1 with  o =  and multiplying the magnitude by four,
we obtain
h(t) = 
j4e  j( 2n1) t

n   ( 2n  1)

n0
Chapter 17, Solution 60.
From Problem 17.24,
a o = 0 = a n , b n = [2/(n)][1 – 2 cos(n)], c o = 0
c n = (a n – jb n )/2 = [j/(n)][2 cos(n) – 1], n  0.
Chapter 17, Solution 61.
 o = 1.
(a)
f(t) = a o +
A
n
cos(n o t   n )
= 6 + 4cos(t + 50) + 2cos(2t + 35)
+ cos(3t + 25) + 0.5cos(4t + 20)
= 6 + 4cos(t)cos(50) – 4sin(t)sin(50) + 2cos(2t)cos(35)
– 2sin(2t)sin(35) + cos(3t)cos(25) – sin(3t)sin(25)
+ 0.5cos(4t)cos(20) – 0.5sin(4t)sin(20)
= 6 + 2.571cos(t) – 3.73sin(t) + 1.635cos(2t)
– 1.147sin(2t) + 0.906cos(3t) – 0.423sin(3t)
+ 0.47cos(4t) – 0.171sin(4t)
(b)
f rms =
a o2 
1  2
 An
2 n 1
f rms 2 = 62 + 0.5[42 + 22 + 12 + (0.5)2] = 46.625
f rms = 6.828
Chapter 17, Solution 62.
(a)
f(t)  12  10cos(2ot  90o )  8 cos(4ot  90o )  5cos(6ot  90o )  3cos(8ot  90 o )
(b) f(t) is an even function of t.
Chapter 17, Solution 63.
This is an even function.
T = 3,  o = 2/3, b n = 0.
f(t) =
ao =
an =
1, 0  t  1
2, 1  t  1.5
1.5
2 T/2
2 1
f
(
t
)
dt
1
dt


1 2 dt  = (2/3)[1 + 1] = 4/3
T 0
3  0
1.5
4 T/2
4 1
f ( t ) cos(no t )dt    1cos(2nt / 3)dt   2 cos(2nt / 3)dt 



0
0
1
3
T
4 3
6
 2nt 
 2nt 
= 
sin 
sin 

 
3  2n  3  0 2n  3  1
1
1.5



= [–2/(n)]sin(2n/3)
f 2 (t) =
4 2  1  3n   2nt 
  sin 
 cos

3  n 1 n  3   3 
a o = 4/3 = 1.3333,  o = 2/3, a n = –[2/(n)]sin(2nt/3)
An =
a 2n  b 2n 
2
 2n 
sin 

n  3 
A 1 = 0.5513, A 2 = 0.2757, A 3 = 0, A 4 = 0.1375, A 5 = 0.1103
The amplitude spectra are shown below.
1.333
An
0.551
0.275
0.1378 0.1103
0
0
1
2
3
4
5
n
Chapter 17, Solution 64.
Design a problem to help other students to better understand the amplitude and phase spectra of a
given Fourier series.
Although there are many ways to solve this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
Given that
v(t) = (10/π)[1 + (1/2)cos(2πt) + (2/3)cos(4πt) – (2/15)cos(6πt)] V
draw the amplitude and phase spectra for v(t).
Solution
The amplitude and phase spectra are shown below.
An
3.183
2.122
1.591
0.4244
0
2
4
2
4
6

n
0
6
-180o

Chapter 17, Solution 65.
a n = 20/(n22), b n = –3/(n),  n = 2n
A n = a 2n  b 2n 
=
400
9
 2 2
4 4
n 
n 
3
44.44
1  2 2 , n = 1, 3, 5, 7, 9, etc.
n
n 
n
1
3
5
7
9
An
2.24
0.39
0.208
0.143
0.109
 n = tan–1(b n /a n ) = tan–1{[–3/(n)][n22/20]} = tan–1(–nx0.4712)
n
1
3
5
7
9

n
–25.23
–54.73
–67
–73.14
–76.74
–90
2.24
0
2
6
10
14
18
n
–30
–25.23
An
0.39
–60
0.208
0
2
6
10
–54.73
0.0.143 0.109
14
18
n
–90
n
–67
–73.14
–76.74
Chapter 17, Solution 66.
The schematic is shown below. The waveform is inputted using the attributes of VPULSE. In
the Transient dialog box, we enter Print Step = 0.05, Final Time = 12, Center Frequency = 0.5,
Output Vars = V(1) and click enable Fourier. After simulation, the output plot is shown below.
The output file includes the following Fourier components.
FOURIER COMPONENTS OF TRANSIENT RESPONSE V(1)
DC COMPONENT = 5.099510E+00
HARMONIC
NO
1
2
3
4
5
6
7
8
9
FREQUENCY FOURIER
NORMALIZED
(HZ)
COMPONENT COMPONENT
5.000E-01
1.000E+00
1.500E+00
2.000E+00
2.500E+00
3.000E+00
3.500E+00
4.000E+00
4.500E+00
3.184E+00
1.593E+00
1.063E+00
7.978E-01
6.392E-01
5.336E-01
4.583E-01
4.020E-01
3.583E-01
1.000E+00
5.002E-01
3.338E-01
2.506E-01
2.008E-01
1.676E-01
1.440E-01
1.263E-01
1.126E-01
PHASE
(DEG)
1.782E+00
3.564E+00
5.347E+00
7.129E+00
8.911E+00
1.069E+01
1.248E+01
1.426E+01
1.604E+01
NORMALIZED
PHASE (DEG)
0.000E+00
1.782E+00
3.564E+00
5.347E+00
7.129E+00
8.911E+00
1.069E+01
1.248E+01
1.426E+01
TOTAL HARMONIC DISTORTION = 7.363360E+01 PERCENT
From Prob. 17.4, we know the phase angle should be zero. Why do we have a phase angle equal
to n(1.782)? The answer is actually quite straight forward. The angle comes from the
approximation of the leading edge of the pulse. The graph shows an instantaneous rise
whereas PSpice needs a finite rise time, thus artificially creating a phase shift.
Chapter 17, Solution 67.
The Schematic is shown below. In the Transient dialog box, we type “Print step = 0.01s,
Final time = 36s, Center frequency = 0.1667, Output vars = v(1),” and click Enable
Fourier. After simulation, the output file includes the following Fourier components,
FOURIER COMPONENTS OF TRANSIENT RESPONSE V(1)
DC COMPONENT = 2.000396E+00
HARMONIC FREQUENCY FOURIER NORMALIZED
NO
(HZ) COMPONENT COMPONENT (DEG)
1
2
3
4
5
6
7
8
9
1.667E-01
3.334E-01
5.001E-01
6.668E-01
8.335E-01
1.000E+00
1.167E+00
1.334E+00
1.500E+00
2.432E+00
6.576E-04
5.403E-01
3.343E-04
9.716E-02
7.481E-06
4.968E-02
1.613E-04
6.002E-02
1.000E+00 -8.996E+01
2.705E-04 -8.932E+01
2.222E-01 9.011E+01
1.375E-04 9.134E+01
3.996E-02 -8.982E+01
3.076E-06 -9.000E+01
2.043E-02 -8.975E+01
6.634E-05 -8.722E+01
2.468E-02 9.032E+01
PHASE
NORMALIZED
PHASE (DEG)
0.000E+00
6.467E-01
1.801E+02
1.813E+02
1.433E-01
-3.581E-02
2.173E-01
2.748E+00
1.803E+02
TOTAL HARMONIC DISTORTION = 2.280065E+01 PERCENT
Chapter 17, Solution 68.
Since T=3, f =1/3 = 0.333 Hz. We use the schematic below.
We use VPWL to enter in the signal as shown. In the transient dialog box, we enable
Fourier, select 15 for Final Time, 0.01s for Print Step, and 10ms for the Step Ceiling.
When the file is saved and run, we obtain the Fourier coefficients as part of the output file
as shown below.
Why is this problem wrong? Clearly the source is not
periodic. The DC value must be +1!!!!!!!!!!
FOURIER COMPONENTS OF TRANSIENT RESPONSE V(1)
DC COMPONENT = -1.000000E+00
HARMONIC FREQUENCY FOURIER NORMALIZED
NORMALIZED
NO
(HZ) COMPONENT COMPONENT (DEG)
1
2
3
4
5
6
7
8
9
3.330E-01
6.660E-01
9.990E-01
1.332E+00
1.665E+00
1.998E+00
2.331E+00
2.664E+00
2.997E+00
1.615E-16
5.133E-17
6.243E-16
1.869E-16
6.806E-17
1.949E-16
1.465E-16
3.015E-16
1.329E-16
1.000E+00
3.179E-01
3.867E+00
1.158E+00
4.215E-01
1.207E+00
9.070E-01
1.867E+00
8.233E-01
PHASE
PHASE (DEG)
1.762E+02 0.000E+00
2.999E+01 -3.224E+02
6.687E+01 -4.617E+02
7.806E+01 -6.267E+02
1.404E+02 -7.406E+02
-1.222E+02 -1.179E+03
-4.333E+01 -1.277E+03
-1.749E+02 -1.584E+03
-9.565E+01 -1.681E+03
Chapter 17, Solution 69.
The schematic is shown below. In the Transient dialog box, set Print Step = 0.05 s, Final
Time = 120, Center Frequency = 0.5, Output Vars = V(1) and click enable Fourier. After
simulation, we obtain V(1) as shown below. We also obtain an output file which
includes the following Fourier components.
FOURIER COMPONENTS OF TRANSIENT RESPONSE V(1)
DC COMPONENT = 5.048510E-01
HARMONIC FREQUENCY FOURIER NORMALIZED
NO
(HZ) COMPONENT COMPONENT (DEG)
1
2
3
4
5
6
7
8
9
PHASE
NORMALIZED
PHASE (DEG)
5.000E-01 4.056E-01 1.000E+00 -9.090E+01 0.000E+00
1.000E+00 2.977E-04 7.341E-04 -8.707E+01 3.833E+00
1.500E+00 4.531E-02 1.117E-01 -9.266E+01 -1.761E+00
2.000E+00 2.969E-04 7.320E-04 -8.414E+01 6.757E+00
2.500E+00 1.648E-02 4.064E-02 -9.432E+01 -3.417E+00
3.000E+00 2.955E-04 7.285E-04 -8.124E+01 9.659E+00
3.500E+00 8.535E-03 2.104E-02 -9.581E+01 -4.911E+00
4.000E+00 2.935E-04 7.238E-04 -7.836E+01 1.254E+01
4.500E+00 5.258E-03 1.296E-02 -9.710E+01 -6.197E+00
TOTAL HARMONIC DISTORTION = 1.214285E+01 PERCENT
Chapter 17, Solution 70.
Design a problem to help other students to better understand how to use PSpice to solve circuit
problems with periodic inputs.
Although there are many ways to solve this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
Rework Prob. 17.40 using PSpice.
Chapter 17, Problem 40.
The signal in Fig. 17.77(a) is applied to the circuit in Fig. 17.77(b). Find v o (t).
Figure 17.77
Solution
The schematic is shown below. In the Transient dialog box, we set Print Step = 0.02 s, Final
Step = 12 s, Center Frequency = 0.5, Output Vars = V(1) and V(2), and click enable Fourier.
After simulation, we compare the output and output waveforms as shown. The output includes
the following Fourier components.
FOURIER COMPONENTS OF TRANSIENT RESPONSE V(1)
DC COMPONENT = 7.658051E-01
HARMONIC FREQUENCY FOURIER NORMALIZED
NO
(HZ) COMPONENT COMPONENT (DEG)
1
2
3
4
5
6
7
8
9
5.000E-01 1.070E+00
1.000E+00 3.758E-01
1.500E+00 2.111E-01
2.000E+00 1.247E-01
2.500E+00 8.538E-02
3.000E+00 6.139E-02
3.500E+00 4.743E-02
4.000E+00 3.711E-02
4.500E+00 2.997E-02
1.000E+00
3.512E-01
1.973E-01
1.166E-01
7.980E-02
5.738E-02
4.433E-02
3.469E-02
2.802E-02
1.004E+01
-3.924E+01
-3.985E+01
-5.870E+01
-5.680E+01
-6.563E+01
-6.520E+01
-7.222E+01
-7.088E+01
PHASE
NORMALIZED
PHASE (DEG)
0.000E+00
-4.928E+01
-4.990E+01
-6.874E+01
-6.685E+01
-7.567E+01
-7.524E+01
-8.226E+01
-8.092E+01
TOTAL HARMONIC DISTORTION = 4.352895E+01 PERCENT
Chapter 17, Solution 71.
The schematic is shown below. In the Transient dialog box, we set Print Step = 0.02 s, Final
Step = 12 s, Center Frequency = 0.5, Output Vars = V(1) and V(2), and click enable Fourier.
After simulation, we compare the output and output waveforms as shown. The output includes
the following Fourier components.
FOURIER COMPONENTS OF TRANSIENT RESPONSE V(1)
DC COMPONENT = 7.658051E-01
HARMONIC FREQUENCY FOURIER NORMALIZED
NO
(HZ) COMPONENT COMPONENT (DEG)
1
2
3
4
5
6
7
8
9
5.000E-01 1.070E+00
1.000E+00 3.758E-01
1.500E+00 2.111E-01
2.000E+00 1.247E-01
2.500E+00 8.538E-02
3.000E+00 6.139E-02
3.500E+00 4.743E-02
4.000E+00 3.711E-02
4.500E+00 2.997E-02
1.000E+00
3.512E-01
1.973E-01
1.166E-01
7.980E-02
5.738E-02
4.433E-02
3.469E-02
2.802E-02
1.004E+01
-3.924E+01
-3.985E+01
-5.870E+01
-5.680E+01
-6.563E+01
-6.520E+01
-7.222E+01
-7.088E+01
PHASE
NORMALIZED
PHASE (DEG)
0.000E+00
-4.928E+01
-4.990E+01
-6.874E+01
-6.685E+01
-7.567E+01
-7.524E+01
-8.226E+01
-8.092E+01
TOTAL HARMONIC DISTORTION = 4.352895E+01 PERCENT
Chapter 17, Solution 72.
T = 5,  o = 2/T = 2/5
f(t) is an odd function. a o = 0 = a n
bn =
4 T/2
4 10
f ( t ) sin(no t )dt   10 sin(0.4nt )dt

T 0
5 0
1
= 
8x 5
20
cos(0.4nt ) =
[1  cos(0.4n)]
n
2 n
0
f(t) =
20  1
 [1  cos(0.4n)]sin(0.4nt )
 n 1 n
Chapter 17, Solution 73.
2
VDC
1 Vn2
 
p =
R
2
R
= 0 + 0.5[(22 + 12 + 12)/10] = 300 mW
Chapter 17, Solution 74.
(a)
An =
a 2n  b 2n ,
 = tan–1(b n /a n )
A1 =
6 2  8 2 = 10,
 1 = tan–1(6/8) = 36.87
A2 =
3 2  4 2 = 5,
 2 = tan–1(3/4) = 36.87
i(t) = {4 + 10cos(100t – 36.87) – 5cos(200t – 36.87)} A
(b)
p = I 2DC R  0.5 I 2n R
= 2[42 +0.5(102 + 52)] = 157 W
Chapter 17, Solution 75.
The lowpass filter is shown below.
R
+
+
C
vs
vo
-
vs 
-
n
A 2A  1

cos no t
sin

T
T
T n 1n
1
j n C
1
Vs ,
Vs 
Vo 
1
1  j n RC
R
j n C
For n=0, (dc component), Vo  Vs 
 n  no  2n / T
A
T
(1)
For the nth harmonic,
Vo 
When n=1, | Vo |
2A
n
sin
  90 o
T
1   2 n R 2 C 2  tan 1  n RC nT
1
2A
n
sin

T
T

1
(2)
4 2 2 2
1
R C
T
From (1) and (2),

A
2A
 50 x
sin
T
T
10
1
1
4 2 2 2
R C
T


1
4 2 2 2 30.9
 3.09 x10 4
R C 

T
1
4 2 2 2
R C  10 10
T


C
T
10 2 x 3.09x10 4
10 5 
 24.59 mF
2R
4x10 3
Chapter 17, Solution 76.
v s (t) is the same as f(t) in Figure 16.1 except that the magnitude is multiplied by
10. Hence
v o (t) = 5 
20  1
 sin(nt ) , n = 2k – 1
 k 1 n
T = 2,  o = 2/T = 2,  n = n o = 2n
j n L = j2n; Z = R||10 = 10R/(10 + R)
V o = ZV s /(Z + j2n) = [10R/(10R + j2n(10 + R))]V s
Vo =
10R  tan 1{(n / 5R )(10  R )}
100R 2  4n 2  2 (10  R ) 2
Vs
V s = [20/(n)]0
The source current I s is
Is =
Vs
Vs

10R
Z  j2n
 j2n
10  R
20
n

10R  j2n(10  R )
(10  R )
20
  tan 1{( n / 3)(10  R )}
n
100R 2  4n 2  2 (10  R ) 2
(10  R )
=
p s = V DC I DC +
1
 Vsn I sn cos( n   n )
2
For the DC case, L acts like a short-circuit.
Is =
5(10  R )
5
, Vs = 5 = Vo

10R
10R
10  R

 1  

tan  (10  R )  
2 (10  R ) cos


25(10  R ) 1  20 

5

ps 
  
2
2
2

10R
2  
100R  4 (10  R )




 2

(10  R ) 2 cos tan 1  (10  R )  

 5

 10 

 
 
2
2
2


 
100R  16 (10  R )


2
ps =
=
VDC 1  Von
 
R
2 n 1 R

25 1 
100R
100R
 

 
2
2
2
2
2
2
R 2 100R  4 (10  R )
100R  10 (10  R )

We want p o = (70/100)p s = 0.7p s . Due to the complexity of the terms, we
consider only the DC component as an approximation. In fact the DC component
has the latgest share of the power for both input and output signals.
25 7 25(10  R )
 x
R 10
10R
100 = 70 + 7R which leads to R = 30/7 = 4.286 
Chapter 17, Solution 77.
(a) For the first two AC terms, the frequency ratio is 6/4 = 1.5 so that the highest
common factor is 2. Hence  o = 2.
T = 2/ o = 2/2 = 
(b) The average value is the DC component = –2
(c)
V rms =
ao 
1  2
(a n  b 2n )

2 n 1
1
2
Vrms
 (2) 2  (10 2  8 2  6 2  3 2  12 ) = 121.5
2
V rms = 11.02 V
Chapter 17, Solution 78.
2
(a)
V
V2
V2 V2
1
p = DC   n  DC   n ,rms
R
2
R
R
R
= 0 + (402/5) + (202/5) + (102/5) = 420 W
(b)
5% increase = (5/100)420 = 21
p DC = 21 W =
V DC = 10.25 V
2
VDC
2
 21R  105
R
Chapter 17, Solution 79.
From Table 17.3, it is evident that a n = 0,
b n = 4A/[(2n – 1)], A = 10.
A Fortran program to calculate b n is shown below. The result is also shown.
C
10
FOR PROBLEM 17.79
DIMENSION B(20)
A = 10
PIE = 3.142
C = 4.*A/PIE
DO 10 N = 1, 10
B(N) = C/(2.*FLOAT(N) – 1.)
PRINT *, N, B(N)
CONTINUE
STOP
END
n
1
2
3
4
5
6
7
8
9
10
bn
12.731
4.243
2.546
1.8187
1.414
1.1573
0.9793
0.8487
0.7498
0.6700
Chapter 17, Solution 80.
From Problem 17.55,
c n = [1 + e–jn]/[2(1 – n2)]
This is calculated using the Fortran program shown below. The results are also
shown.
C
10
FOR PROBLEM 17.80
COMPLEX X, C(0:20)
PIE = 3.1415927
A = 2.0*PIE
DO 10 N = 0, 10
IF(N.EQ.1) GO TO 10
X = CMPLX(0, PIE*FLOAT(N))
C(N) = (1.0 + CEXP(–X))/(A*(1 – FLOAT(N*N)))
PRINT *, N, C(N)
CONTINUE
STOP
END
n
0
1
2
3
4
5
6
7
8
9
10
cn
0.3188 + j0
0
–0.1061 + j0
0
–0.2121x10–1 + j0
0
–0.9095x10–2 + j0
0
–0.5052x10–2 + j0
0
–0.3215x10–2 + j0
Chapter 17, Solution 81.
(a)
A
0
2T
T
f(t) =
2A 4A 
1
cos(no t )


2

 n 1 4n  1
The total average power is
p avg = F rms 2R = F rms 2 since R = 1 ohm.
P avg = F rms 2 =
(b)
3T
1 T 2
f ( t )dt = 0.5A2

0
T
From the Fourier series above
|c o | = 2A/π, |c n | = |a n |/2 = 2A/[(4n2 – 1)]
n
0
1
2
3
4
o
0
2 o
4 o
6 o
8 o
|c n |
2A/
2A/(3)
2A/(15)
2A/(35)
2A/(63)
(c)
81.1%
(d)
0.72%
|c o |2 or 2|c n |2
4A2/(2)
8A2/(92)
8A2/(2252)
8A2/(12252)
8A2/(39692)
% power
81.1%
18.01%
0.72%
0.13%
0.04%
Chapter 17, Solution 82.
2
VDC
1  Vn2
 
P =
R
2 n 1 R
Assuming V is an amplitude-phase form of Fourier series. But
|A n | = 2|C n |, c o = a o
|A n |2 = 4|C n |2
Hence,

c o2
c 2n
 2
P =
R
n 1 R
Alternatively,
2
Vrms
P =
R
where
2
Vrms
 a o2 


1  2
2
2
A

c

2
c

c 2n



n
o
n
2 n 1
n 1
n  
= 102 + 2(8.52 + 4.22 + 2.12 + 0.52 + 0.22)
= 100 + 2x94.57 = 289.14
P = 289.14/4 = 72.3 W
Chapter 18, Solution 1.
f ' ( t )  ( t  2)  ( t  1)  ( t  1)  ( t  2)
jF()  e j2   e j  e  j  e  j2
 2 cos 2  2 cos 
F() =
2[cos 2  cos ]
j
Chapter 18, Solution 2.
Using Fig. 18.27, design a problem to help other students to better understand the Fouier
transform given a wave shape.
Although there are many ways to solve this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
What is the Fourier transform of the triangular pulse in Fig. 18.27?
Figure 18.27
Solution
t,
f (t)  
0,
0  t 1
otherwise
f ”(t)
f ‘(t)
1
(t)
0
t
1
–’(t-1)
-(t-1)
f&quot;(t) = (t) - (t - 1) - '(t - 1)
Taking the Fourier transform gives
-2F() = 1 - e-j - je-j
F() =
t
(1  j)e  j  1
2
-(t-1)
1
or F()   t e  jt dt
0
eax
But  x e dx  2 (ax  1)  c
a
1
e  j
( jt  1) 10  2 1  je  j  1
F() 
2

 j
ax


Chapter 18, Solution 3.
f (t) 
1
t ,  2  t  2,
2
1
f ' (t)  ,  2  t  2
2
1 j t
e  j t
( jt  1) 2 2
t e dt 
2 2
2( j) 2
1
  2 e  j2 ( j2  1)  e j2 ( j2  1)
2
1

 j2 e  j2  e j2  e j2  e  j2
2
2
1
  2  j4 cos 2  j2 sin 2
2
F()  
2



F() =


j
( 2 cos 2  sin 2)
2

Chapter 18, Solution 4.
We can solve the problem by following the approach demonstrated in Example 18.5.
2δ(t+1)
g’
2
–1
0
1
t
–2
–2δ(t–1)
4δ(t)
2δ’(t+1)
g”
–1
0
–2δ(t+1)
1
t
–2
–2δ(t–1)
–2δ’(t–1)
g   2( t  1)  2( t  1)  4( t )  2( t  1)  2( t  1)
( j) 2 G ()  2e j  2 je j  4  2e  j  2 je  j
 4 cos   4 sin   4
G ( ) 
4
(cos    sin   1)
2
Chapter 18, Solution 5.
h’(t)
1
0
–1
t
1
–2δ(t)
h”(t)
1
δ(t+1)
1
t
0
–1
–2δ’(t)
–δ(t–1)
h ( t )  ( t  1)  ( t  1)  2( t )
( j) 2 H()  e j  e  j  2 j  2 j sin   2 j
H(ω) =
2j 2j

sin 
 2
Chapter 18, Solution 6.
(a) The derivative of f(t) is shown below.
f’(t)
5(t)
0
5(t-1)
1
2
t
-10(t-2)
f '(t )  5 (t )  5 (t  1)  10 (t  2)
Taking the Fourier transform of each term,
j F ( )  5  5e  j  10e  j 2
F ( ) 
5  5e  j  10e  j 2
j
(b) The derivative of g(t) is shown below.
g’(t)
10(t)
0
1
2
-5
-5(t-1)
The second derivative of g(t) is shown below.
g’’(t)
10’(t)
0
5(t-2)
1
2
t
-5’(t-1)
-5(t-1)
g”(t) = 10δ’(t) – 5δ’(t–1) – 5δ(t–1) + 5δ(t–2)
Take the Fourier transform of each term.
(jω)2G(jω) = 10jω – 5jωe–jω – 5e–jω + 5e–j2ω which leads to
G(jω) = (–10jω + 5jωe–jω + 5e–jω – 5e–j2ω)/ω2
Chapter 18, Solution 7.
(a) Take the derivative of f 1 (t) and obtain f 1 ’(t) as shown below.
2(t)
0
1
2
t
-(t-1) -(t-2)
f1' (t )  2 (t )   '(t  1)   (t  2)
Take the Fourier transform of each term,
j F1 ( )  2  e  j  e  j 2
F1 ( ) 
2  e  j  e  j 2
j
(b) f 2 (t) = 5t
F2 ( ) 



f 2 (t )e
 j
2
dt   5te  j dt 
0
F2 ( ) 
2
5
 j t
e

j


(
1)
0
( j ) 2
5e  j 2

2
(1  j 2) 
5
2
Chapter 18, Solution 8.
1
F()   2e
 jt
dt   (4  2 t )e  jt dt
0
(a)

1
2  jt 1
4  jt 2
2  jt
2
e

e

e
( jt  1) 1
0
1
2
 j
 j

F(  ) 
(b)
2
2
4  j2
2
2  j 2


e
 2 (1  j2)e  j2 
e 
2
j j
j


g(t) = 2[ u(t+2) – u(t-2) ] - [ u(t+1) – u(t-1) ]
G ( ) 
4 sin 2 2 sin 



Chapter 18, Solution 9.
(a)
y(t) = u(t+2) – u(t-2) + 2[ u(t+1) – u(t-1) ]
Y( ) 
1
(b) Z()   ( 2t )e  jt dt 
0
2
4
sin 2  sin 


 2e  jt
(  jt  1)
 2
1
0

2 2e  j

(1  j)
2
2
Chapter 18, Solution 10.
x(t) = e2tu(t)
(a)
X() = 1/(–2 + j)
(b)
e  t , t  0
e ( t )   t
e , t  0
1
0
1
1
1
0
Y()   y( t )e jt dt   e t e jt dt   e  t e  jt dt

e (1 j) t
1  j

0
1

e  (1 j) t
 (1  j)
1
0
 cos   jsin  cos   jsin  
2

 e 1 

2
1  j
1  j
1 


Y() =

2
1  e 1 (cos    sin )
2
1 

Chapter 18, Solution 11.
f(t) = sin  t [u(t) - u(t - 2)]
2
F()   sin t e  jt dt 
0



1 2 j t
e  e  j t e  jt dt

0
2j
1  2  j(     ) t
(e
 e  j(  ) t )dt 



0
2j 
e  j(    ) t 2 
1 
1
e  j(   ) t 02 

0
 j(  ) 
2 j   j(  )
1  1  e  j2  1  e  j2  

 

2 
   



1
2  2e  j2 
2
2(   )
2

F() =



e  j 2  1
2
 
2
Chapter 18, Solution 12.
(a) F1 ( ) 
10
(3  j ) 2  100
(b) F2 ( ) 
4  j
(4  j ) 2  100
Chapter 18, Solution 13.
(a) We know that F[cos at ]  [(  a )  (  a )] .
Using the time shifting property,
F[cos a(t   / 3a )]  e  j / 3a [(  a )  (  a )]  e  j / 3 (  a )  e j / 3 (  a )
(b) sin ( t  1)  sin t cos   cos t sin    sin t
g(t) = -u(t+1) sin (t+1)
Let x(t) = u(t)sin t, then X() 
1
2
( j)  1

1
1  2
Using the time shifting property,
G ()  
e j
1
j
e

1  2
2  1
(c ) Let y(t) = 1 + Asin at, then Y()  2()  jA[(  a )  (  a )]
h(t) = y(t) cos bt
Using the modulation property,
1
H()  [Y(  b)  Y(  b)]
2
H()  (  b )  (  b ) 
4
(d) I( )   (1  t )e
0
 jt
jA
(  a  b)  (  a  b)  (  a  b)  (  a  b)
2
e  jt e  jt
dt 

(  jt  1)
 j   2
4
0
1 e  j4  e  j4 
 2
 2 ( j4  1)
j


Chapter 18, Solution 14.
Design a problem to help other students to better understand finding the Fourier transform of a
variety of time varying functions (do at least three).
Although there are many ways to solve this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
Find the Fourier transforms of these functions:
(a)
(b)
(c)
(d)
(e)
f(t) = e-t cos (3t + ) u(t)
g(t) = sin  t [ u(t + 1) - u(t-1)]
h(t) = e-2t cos  t u(t-1)
p(t) = e-2t sin 4t u(-t)
q(t) = 4 sgn (t - 2) + 3(t) - 2 u(t - 2)
Solution
(a)
cos(3t  )  cos 3t cos   sin 3t sin   cos 3t (1)  sin 3t (0)   cos(3t )
f ( t )  e  t cos 3t u ( t )
 1  j
F() =
1  j2  9
(b)
g(t)
1
-1
1
t
-1
g’(t)

-1
1
t
-
g ' ( t )   cos tu ( t  1)  u ( t  1)
g&quot; ( t )   2 g( t )  ( t  1)  ( t  1)
  2 G ()    2 G ()  e j  e  j
 2  2 G()  (e j  e  j )  2 j sin 
2 j sin 
G() =
2   2
Alternatively, we compare this with Prob. 17.7
f(t) = g(t - 1)
F() = G()e-j

e  j  e j
2
 
 j2 sin 

2   2
2 j sin 
G() =
 2  2
G   F()e j 
2


cos ( t  1)  cos t cos   sin t sin   cos t (1)  sin t (0)   cos t
Let x ( t )  e 2( t 1) cos ( t  1)u ( t  1)  e 2 h ( t )
y( t )  e 2 t cos(t )u ( t )
2  j
Y() 
(2  j) 2   2
y( t )  x ( t  1)
Y()  X()e  j
(c)
and
X() 
2  je j
2  j2   2
X()  e 2 H()
H()  e 2 X()
=
 2  je j 2
2  j2   2
Let x ( t )  e 2 t sin( 4t )u ( t )  y( t )
p( t )   x ( t )
where y( t )  e 2 t sin 4t u ( t )
2  j
Y () 
2  j2  4 2
2  j
X()  Y() 
2  j2  16
(d)
p()  X() 
(e)
j  2
j  2 2  16

1 
8  j 2
 3  2 ()  e  j2
e
j 
j

6 j 2
Q() =
e  3  2()e  j 2
j
Q() 
Chapter 18, Solution 15.
(a)
F()  e j3  e  j3  2 j sin 3
(b)
Let g ( t )  2( t  1), G ()  2e  j
t
F()  F   g ( t ) dt 
 

G ()

 F(0)()
j

2e  j
 2(1)()
j
2e  j
=
j
(c)
F (2 t ) 
1
1
2
1
1
1 j
F()   1  j  
3
2
3 2
Chapter 18, Solution 16.
(a) Using duality properly
t 
or
2
2
2
 2 
t2
4
  4 
t2
4
F() = F  2    4 
t 
(b)
e
at
2a
a  2
2
2a
a  t2
2 e
a 
8
a  t2
4 e
2 
2
2
 8 
2 
 4 e
G() = F 
2 
4t 
Chapter 18, Solution 17.
1
F  0   F  0 
2
1
where F() = F u t     , 0  2
j
(a) Since H() = F cos 0 t f ( t )  
H 
1
1
1 
   2  
   2  

2
j   2 
j   2 


  2    2  j    2    2 
2
2    2  2 
H() =
(b)

  2    2   2j
2
 4
G() = F sin 0 t f ( t ) 
j
F  0   F  0 
2
1
where F() = F u t     
j

j
1
1
G     10 
   10 
2
j  10
j  10  

j
  10     10  j  j  j 
2
2    10   10 
=
j
  10    10   2 10
2
  100
Chapter 18, Solution 18.

(a) F [ f (t  to )] 

f (t  to )e jt dt

Let t  to  

 t    to ,


F [ f (t  to )] 
dt  d 
f ( )e  j e jto d  e jto F ( )

(b) Given that
f (t )  F 1[ F ( )] 
1
2



F ( )e jt d

j
f '(t ) 
F ( )e jt dt  j F 1[ F ( )]

2 
or
F [ f '(t )]  j F ( )
(c ) This is a special case of the time scaling property when a = –1. Hence,
F [ f (t )] 
1
F ( )  F ( )
| 1|
(d) F ( )  


f (t )e  jt dt
Differentiating both sides respect to  and multiplying by t yields


dF ( )
j
 j  ( jt ) f (t )e  jt dt   tf (t )e  jt dt
d


Hence,
dF ( )
j
 F [tf (t )]
d
Chapter 18, Solution 19.

F   f ( t )e jt dt 

F 



1 1 j2 t
e  e  j2 t e  jt dt
2 0

1 1  j   2   t
e
 e  j 2  t dt

0
2
1

1
1
1
e  j   2   t 
e  j   2   t 
 
2   j   2 
 j  2 
0
1  e  j  2    1 e  j  2    1 
 


2  j   2
j  2  
But
e j2   cos 2  j sin 2  1  e  j2 
1  e  j  1  1
1 

F   


2
j    2   2 
=


j
e  j  1
2
  4
2
Chapter 18, Solution 20.
(a)
F (c n ) = c n ()


F c n e jno t  c n   no 
(b)
cn 

n  
T  2
o 

 c   n 
n
o
2
1
T
1 T
1  
 jnt
f t  e  jno t dt 
 1 e dt  0 


T 0
2  0
1  1 jnt
 e
2  jn
But e  jn
cn 
 

F   c n e jno t  
 n  


j
 
e  jn  1
2

n

 cos n  j sin n  cos n  (1) n


0


j
 1n  1   0,j ,
2n
 n

n  even
n  odd , n  0
for n = 0
cn 
1 
1
1 dt 

0
2
2
Hence
f (t) 

1
j jnt
e
 
2 n   n
n 0
n  odd
F() =

1
j
  
  n 
2
n   n
n0
n  odd

Chapter 18, Solution 21.
Using Parseval’s theorem,

 
f 2 ( t )dt 
1 
| F() | 2 d



2
If f(t) = u(t+a) – u(t+a), then


a
f 2 ( t )dt   (1) 2 dt  2a 
a
or
2
4a 
 sin a 
  a  d  4a 2  a as required.

2
1 
 sin a 
4a 2 
 d



2
 a 
Chapter 18, Solution 22.
F f ( t ) sin o t   



e
f (t)
j o t

 e  jo t  jt
e dt
2j

1 
f ( t )e  j o t dt   e  j o t dt 



2 j   
=
1
F   o   F  o 
2j
Chapter 18, Solution 23.
1
10
30


3 2  j / 35  j / 3 6  j15  j
30
F f  3t  
6  j15  j
(b) f(2t)
1
10
20


2 2  j / 215  j / 2 4  j10  j
20e  j / 2
4  j10  j
1
1
(c) f(t) cos 2t
F  2  F  2 
2
2
5
5
=

2  j  25  j  2 2  j  2 5  j  2
f(2t-1) = f [2(t-1/2)]
(d) F f ' t   j F  
(e)
j10
2  j5  j
F
 F0 
j
10
x10

 
j2  j5  j
2x5
10
=
 
j2  j5  j
 f t  dt
t

Chapter 18, Solution 24.
(a) X  F + F [3]
j
= 6  e  j  1



(b) yt   f t  2 
Y  e  j2 F 
je  j2  j
e 1


(c) If h(t) = f '(t)
H  jF  j



j  j
e  1  1  e  j

3 3 
3 3 
5 
2 
(d) g t   4f  t   10f  t , G ()  4 x F    10x F  
2 2 
5 5 
3 
3 
 6
=
j
3

2

e
 j3 / 2



1 
6 j  j3 / 5
e
1
3

5



j4  j3 / 2
j10  j3 / 5
e
1 
e
1


Chapter 18, Solution 25.
(a) g(t) = 5e2tu(t)
(b) h(t) = 6e–2t
A
B
(c ) X ( ) 

,
s 1 s  2
s  j
10
10
 10,
B
 10
1 2
2 1
10
10

X ( ) 
j  1 j  2
x(t) = (–10etu(t)+10e2t)u(t)
A
(a) 5e2tu(t), (b) 6e–2t , (c) (–10etu(t)+10e2t)u(t)
Chapter 18, Solution 26.
(a) f (t )  e  ( t  2 ) u(t )
(b) h(t )  te 4 t u(t )
(c) If x(t )  u(t  1)  u(t  1)


X()  2
sin 

By using duality property,
G ()  2u(  1)  2u(  1)


g( t ) 
2 sin t
t
Chapter 18, Solution 27.
(a) Let Fs  
100
A
B
 
, s  j
s s  10 s s  10
100
100
A
 10, B 
 10
 10
10
10
10
F 

j j  10
f(t) = (5 sgnt   10e 10 t )ut 
(b) G s  
10s
A
B


, s  j
2  s 3  s  2  s s  3
20
 30
A
 4, B 
 6
5
5
4
6
G  

  j  2 j   3
g(t) = 4e 2 t u t   6e 3 t ut 
60
60
(c) H 

2
 j  j40  1300  j  202  900
h(t) = 2e 20 t sin( 30t ) ut 
(d) yt  
1  e jt d
1 1 1
   

2  2  j j  1 2 2 4
Chapter 18, Solution 28.
() e jt
1 
1 
j t
F
(
)
e
d



d
2 
2  (5  j)(2  j)
1 1
1


 0.05
2 (5)(2) 20
(a)
f (t) 
(b)
1  10(  2) jt
10
e  j2 t
f (t) 
e d 
2  j( j  1)
2 ( j2)( j2  1)

(c)
f (t) 

(d)
j5 e  j2 t
( 2  j)e  j2 t

2
2 1  j2
20
e jt
1  20(  1)e jt
d


2  (2  j)(3  5)
2 (2  j)(3  j)
(1  j)e jt
20e jt


2(5  5 j)
Let
5()
5

 F1 ()  F2 ()
(5  j) j(5  j)
1  5() jt
5 1
f1 ( t ) 
e d 
  0.5

2    5  j
2 5
F() 
5
A
B
 
, A  1, B  1
s(5  s) s s  5
1
1
F2 () 

j j  5
F2 (s) 
f 2 (t) 
1
1
sgn( t )  e 5 t    u ( t )  e 5 t
2
2
f ( t )  f 1 ( t )  f 2 ( t )  u( t )  e 5 t
Chapter 18, Solution 29.
(a)
(b)
f(t) = F -1 [()]  F -1 4(  3)  4(  3)
1
1 4 cos 3t
1  8 cos 3t 


=
2

2
If h ( t )  u ( t  2)  u ( t  2)
H() 
2 sin 2

G ()  4H()
g(t) =
(c)
g( t ) 
4 sin 2t
t
Since
cos(at)
(  a )  (  a )
Using the reversal property,
2 cos 2  ( t  2)  ( t  2)
or F -1 6 cos 2  3(t  2)  3(t  2)
1 8 sin 2 t

2
t
Chapter 18, Solution 30.
(a)
2
,
j
2a
Y() 2(a  j)
H() 
 2

j
X()
j
y( t )  sgn( t )
(b) X() 
1
,
1  j
H( ) 


Y() 
Y() 
X() 


1
a  j
h(t )  2(t )  a[u(t )  u(  t )]
1
2  j
1  j
1
 1
2  j
2  j
(c ) In this case, by definition, h(t ) 


h(t )  (t )  e  2 t u(t )
y (t )  e  at sin bt u(t )
Chapter 18, Solution 31.
(a)
Y() 
1
(a  j) 2
X() 
,
H() 
Y(  )
1

H() a  j
1
a  j


x(t )  e at u(t )
(b)
By definition, x(t )  y(t )  u(t  1)  u(t  1)
(c )
Y() 
X() 
1
(a  j)
,
H() 
Y(  )
j
1
a

 
H() 2(a  j) 2 2(a  j)
2
j


x(t ) 
1
a
(t )  e at u(t )
2
2
Chapter 18, Solution 32.
(a)
e  j
j  1
and F 
e   t 1 u ( t  1)
Since
f(-t)
j
F1  
e
 j  1
f 1 (t) = e t 1 u t  1
(b)
f 1 t   e   t1 u  t  1
From Section 17.3,
2

2e
t 1

If F2   2e , then
2
f 2 (t) =
2
 t 1
2

(b)

By partial fractions
F3  
1
 j  1  j  12
2
1
1
1
1
4
4

 4 
 4
 j  12  j  1  j  12 j  1
1 t
te  e t  te t  e t ut 
4
1
1
= t  1e  t ut   t  1e t ut 
4
4
Hence f 3 t  
(d)
f 4 t  
1
1  e jt
1 
jt





=
F
e
d
1






2
1  j2
2
2
Chapter 18, Solution 33.
(a)
Let x t   2 sin tu t  1  u t  1
From Problem 17.9(b),
4 j sin 
 2  2
Applying duality property,
X 
1
2 j sin  t 
X t   2 2
2
 t
2 j sin t
f(t) = 2
t  2
f t  
(b)
F 
j
cos 2  j sin 2  j cos   j sin 


 j
j
e
e j2 
 e j2   e  j  


j
j
1
1
f t   sgn t  1  sgn t  2
2
2
But sgn( t )  2u ( t )  1
1
1
f t   u t  1   u t  2  
2
2
= ut  1  ut  2 
Chapter 18, Solution 34.
First, we find G() for g(t) shown below.
g t   10u t  2   u t  2   10u t  1  u t  1
g ' t   10t  2   t  2   10t  1  t  1
The Fourier transform of each term gives
g(t)
20
10
–2
–1
0
1
t
2
g ‘(t)
10(t+2)
–2
10(t+1)
–1
0
1
–10(t-1)




2
–10(t-2)
jG   10 e j2  e  j2  10 e j  e  j
 20 j sin 2  20 j sin 
20 sin 2 20 sin 
G  

 40 sinc(2) + 20 sinc()


Note that G() = G(-).
F  2G  
1
f t  
G t 
2
= (20/)sinc(2t) + (10/)sinc(t)
t
Chapter 18, Solution 35.
(a)
x(t) = f[3(t-1/3)]. Using the scaling and time shifting properties,
1
1
e  j / 3
e  j / 3 
3 2  j / 3
(6  j)
X() 
(b)
Using the modulation property,
Y(  ) 

1
1
1
1

[F(  5)  F(  5)]  
2  2  j(  5) 2  j(  5) 
2
j
2  j
(c )
Z()  jF() 
(d)
H()  F()F() 
(e)
I( )  j
1
( 2  j) 2
d
(0  j)
1
F( )  j

2
d
( 2  j)
( 2  j) 2
Chapter 18, Solution 36.
H ( ) 
Y ( )
X ( )


x(t )  vs (t )  e 4t u (t )
Y ( ) 
Y ( )  H ( ) X ( )


X ( ) 
2
2

,
( j  2)(4  j ) ( s  2)( s  4)
1
4  j
s  j
A
B

s2 s4
2
2
A
 1,
B
 1
2  4
4  2
1
1
Y (s) 

s2 s4
y (t )   e2t  e4t  u (t )
Y (s) 
Please note, the units are not known since the transfer function does not give
them. If the transfer function was a voltage gain then the units on y(t) would be
volts.
Chapter 18, Solution 37.
2 j 
j2
2  j
By current division,
j2
I 
j2
2  j
H  o



j
2
I s 
j2  8  j4
4
2  j
H() =
j
4  j3
Chapter 18, Solution 38.
Using Fig. 18.40, design a problem to help other students to better understand using
Fourier transforms to do circuit analysis.
Although there are many ways to solve this problem, this is an example based on the
same kind of problem asked in the third edition.
Problem
Suppose v s (t) = u(t) for t&gt;0. Determine i(t) in the circuit of Fig. 18.40 using Fourier
transform.
1
i
vs
+
_
1H
Figure 18.40
For Prob. 18.38.
Solution
Vs   ( ) 
1
j
Vs
1 
1 

  ( ) 

1  j 1  j 
j 
 ( )
1
Let I ( )  I1 ( )  I 2 ( ) 

1  j j (1  j )
1
A
B
I 2 ( ) 
 
,
s  j
j (1  j ) s s  1
I ( ) 
1
where A   1,
1
1
1

j j  1
 ( )
I1 ( ) 
1  j
I 2 ( ) 
B
1
 1
1
1

 i2 (t )  sgn(t )  e t
2
i1 (t ) 
1
2
 ( ) jt
1 e jt
1


e
d

 1  j
2 1  j   0 2

Hence,
i (t )  i1 (t )  i2 (t ) 
1 1
 sgn(t )  e t
2 2
Chapter 18, Solution 39.

Vs () 
 (1  t )e
 jt

I( ) 
Vs ()
10 3  jx10  3

dt 
1
1
1  j


e
2
j 
2

10 3  1
1
1

 2  2 e  j 
6

10  j  j 

Chapter 18, Solution 40.
v( t )  ( t )  2( t  1)  ( t  2)
  2 V()  1  2e  j  e j2
Now
1  2e  j  e  j2
V 
 2
1 1  j2
Z  2 

j
j
V 2e j  e j2  1
j


2
Z
1  j2

1

0.5  0.5e  j2 e  j
j0.5  j
1
A
B
But
 
A = 2, B = -2
s(s  0.5) s s  0.5
2
2
I 
0.5  0.5e j2  e  j 
0.5  0.5e  j2  e  j
j
0.5  j
1
1
i(t) = sgn( t )  sgn( t  2)  sgn( t  1)  e  0.5t u(t )  e  0.5( t  2 ) u(t  2)  2e  0.5( t 1) u(t  1)
2
2
I






Chapter 18, Solution 41.
2
1
2  j
+

V
+
V
1/s
2
0.5s

1
2V
2  j
 jV 
20
2
j
j  22  4V  j4  2 jj   42 jj9
2
V(ω) =
2 j(4.5  j2)
( 2  j)(4  2 2  j)
Chapter 18, Solution 42.
By current division, I o 
(a)
2
 I
2  j
For i(t) = 5 sgn (t),
10
I 
j
2
10
20
Io 


2  j j j(2  j)
20
A
B
Let I o 
 
, A  10, B  10
s(s  2) s s  2
10
10
I o  

j  2  j
i o (t) = 5 sgn( t )  10e 2 t u(t )A
i(t)
(b)
i’(t)
4
4(t)
1
1
t
t
–4(t–1)
i' ( t )  4( t )  4( t  1)
j I  4  4e  j
I 

4 1  e  j
j



 1
8 1  e  j
1 
 1  e  j
Io 
 4

j(2  j)
 j  2  j 
4
4
4e  j 4e  j




j 2  j
j
2  j
i o (t) = 2 sgn( t )  2 sgn( t  1)  4e 2 t u(t )  4e 2( t 1) ut  1A


Chapter 18, Solution 43.
20 mF


Vo 
Vo 
1
1
50


,

3
jC j20x10  j
50
250
40
,
Is 

50
j (s  1)(s  1.25)
40 
j
i s  5e  t
s  j
A
B
1 
 1

 1000

s  1 s  1.25
 s  1 s  1.25 
v o (t )  1(e 1t  e 1.25 t )u(t ) kV


Is 
5
1  j
Chapter 18, Solution 44.
1H
j
We transform the voltage source to a current source as shown in Fig. (a) and then
combine the two parallel 2 resistors, as shown in Fig. (b).
Io
+
V s /2
2
2 Vo
Io
+
V s /2
1 Vo
j

(a)
(b)
V
1
 s
1  j 2
j Vs
Vo  j I o 
2(1  j)
v s ( t )  10t   10( t  2)
2 2  1, I o 
j Vs   10  10e  j2 
Vs  

10 1  e  j2
j



5 1  e  j2 
5
5


e  j2 
1  j
1  j 1  j
t
v o ( t )  5e u ( t )  5e  ( t  2) u ( t  2)
Hence Vo 

v o (1)  5e 1  0  1.839 V
j
Chapter 18, Solution 45.
We may convert the voltage source to a current source as shown below.
1H
v s /2
2
2
Combining the two 2- resistors gives 1 . The circuit now becomes that
shown below.
I
1H
v s /2
1
1 Vs
1
5
5
 


,
s  j
1  j 2 1  j 2  j ( s  1)( s  2)
A
B


s 1 s  2
B  5 /  1  5
where A  5 /1  5,
5
5
I

s 1 s  2
i (t )  5(e  t  e 2t )u (t ) A
I
Chapter 18, Solution 46.
1
F
4
1
j
j2
3
2H
3( t )
1
4

 j4

1
1  j
e  t u(t)
The circuit in the frequency domain is shown below:
2
Vo
I o ()
–j4/
1/(1+j)
+

+

3
j2
At node V o , KCL gives
1
 Vo
3  Vo
V
1  j

 o
 j4
2
j2

2
j2Vo
 2Vo  j3  jVo  
1  j

2
 j3
1  j
Vo 
j2
2  j 

2  j3  3 2
V
1  j
I o   o 
j2 
j2

j2 2  j  


I o () =
2  j 2  3 2
4  6 2  j(8  2 3 )
Chapter 18, Solution 47.
1
F
2


Io 
1
2

jC j
1
I
2 s
1
j
2
2
8
2
j
Is 
Io 
Vo 
2
2  j 1  j
j
1
j

16
, s  j
(s  1)(s  2)

A
B

s 1 s  2
where A = 16/1 = 16, B = 16/(–1) = –16
Thus,
v o (t) = 16(e–t – e–2t)u(t) V.
Chapter 18, Solution 48.
0.2F
1
j5

jC

As an integrator,
RC  20 x 10 3 x 20 x 10 6  0.4
1 t
v i dt
RC o

1  Vi
Vo  
 Vi (0)

RC  j

vo  

Io 

1 
2
  

0.4  j 2  j



Vo
2
mA  0.125 
   
20
 j 2  j

0.125 0.125

 0.125 
j
2  j
0.125
i o ( t )  0.125 sgn( t )  0.125e  2 t u t  
 e jt dt

2
0.125
 0.125  0.25u ( t )  0.125e 2 t u ( t ) 
2

i o (t) = [0.625  0.25u(t )  0.125e 2 t u(t )] mA
Chapter 18, Solution 49.
Consider the circuit shown below:
j
j2
j
+
VS
+

i1
i2
2
1  vo

Vs     1     2 
For mesh 1,  Vs  2  j2I1  2I 2  jI 2  0
Vs  2 1  j I1  2  jI 2
For mesh 2, 0  3  jI 2  2I1  jI1
3  I 2
I1 
2  
(2)
Substituting (2) into (1) gives
2 1  j3  jI 2
 2  jI 2
2  j
Vs 2    2 3  j4   2  4  j4   2 I 2
Vs  2

 I 2 2  j4   2 
s  2Vs
I2  2
, s  j
s  4s  2
Vo  I 2 
 

 j  2     1     1
 j2  j4  2
1 
v o e jt d



2
1
 j  2 e jt    1d 1  j  2e jt   1d

2
 2
2

 j  j4  2
 j2  j4  2
v o (t) 
(1)
1
 j  2e jt 1  j  2e jt
 2
 2
 1  j4  2  1  j4  2
1
1
2  j1  j4
2  j1  j4e jt
jt
2
2
v o (t) 
e 
17
17
1
1
6  j7 e jt  6  j7 e jt

34
34
 j  t 13.64 
 0.271e
 0.271e j  t 13.64 
v o (t) = 542 cos(t  13.64 ) mV
Chapter 18, Solution 50.
Consider the circuit shown below:
j0.5
1
+
VS
For loop 1,
For loop 2,
+

i2
i1
j
1
vo
j

 2  1  jI1  j0.5I 2  0
(1)
1  jI 2  j0.5I1  0
(2)
From (2),
I1 
1  jI 2
 j0.5
 2
1  jI 2
j
Substituting this into (1),
 21  jI 2 j
2

I2
j
2
3 

2 j   4  j4   2 I 2
2 

2 j
I2 
4  j4  1.5 2
 2 j
Vo  I 2 
2
4  j4  1.5 j
4
j
3
Vo 
8
8
2
j
  j 
3
3
4

 4  j 
3



2
4
  8 
  j   
3
  3 
2
16
3
2
4
  8 
  j   
3
  3 
2
 8 
 8 
Vo ( t )   4e  4t / 3 cos
t u(t ) V
t  u(t )  5.657e  4t / 3 sin
3
3




Chapter 18, Solution 51.
In the frequency domain, the voltage across the 2- resistor is
2
2
10
20
V ( ) 
Vs 


,
s  j
2  j
2  j 1  j ( s  1)( s  2)
A
B
V ( s) 

s 1 s  2
A  20 /1  20,
B  20 /  1  20
V ( ) 
20
20

j  1 j  2
v(t )   20e  t  20e 2t  u (t )
W


1  2
v ( t )dt  0.5 400 e  2 t  e  4 t  3e  3t dt

0
2
 e  2 t e  4 t 2e  3 t
 200


 2
4
3



 = 16.667 J.

0
Chapter 18, Solution 52.

J = 2  f 2 ( t ) dt 
0
1 
2
F() d

 0

1 
1
1
1 
d 
tan 1 ( / 3) 
=
= (1/6)
2
2

3
3 2
 0 9 
0
Chapter 18, Solution 53.
If f(t) = e-2|t|, find J  


J =

  F()
2
F ( ) d  .
2
d  2


f(t) =
t0
e 2t ,
e
2 t
f 2 ( t ) dt
,
t0

 4t 0
 4 t 
0 4t
e
e 4 t 



= 2[(1/4) + (1/4)] = 
J = 2   e dt   e dt   2
0
 4
4 
 


0 

Chapter 18, Solution 54.
Design a problem to help other students better understand finding the total energy in a given
signal.
Although there are many ways to solve this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
Given the signal f(t) = 4 e-t u(t), what is the total energy in f(t)?
Solution
W 1 =





0
0
f 2 ( t ) dt  16  e  2 t dt   8e  2 t
= 8J
Chapter 18, Solution 55.
f(t) = 5e2e–tu(t)
F() = 5e2/(1 + j), |F()|2 = 25e4/(1 + 2)
W 1
1 
25e 4
2
=



F
(
)
d
 0



0
= 12.5e4 = 682.5 J
or
W 1 =




f 2 ( t ) dt  25e 4  e  2 t dt  12.5e4 = 682.5 J
0

25e 4
1
d


tan 1 ()
2

1 
0
Chapter 18, Solution 56.
 2
e 4t
2
(a) W   V (t )dt   t e dt 
(16
t
8
t
2)



 0.0313 J
0 64
(4)3

0


2
2 4 t
(b) In the frequency domain,
1
V ( ) 
(2  j ) 2
1
(4  j ) 2
2
2
1
2
1
2
Wo 
| V ( ) | d  
d


2 2
2 0 (4   2 ) 2
| V ( ) |2  V ( )V * ( ) 

1 1  


 0.5 tan 1 0.5
 2x 4  2  4

Fraction =
2

0
1
1

 0.0256
32 64
Wo 0.0256

 81.79%
W 0.0313
Chapter 18, Solution 57.
W 1 =



0
0


i 2 ( t ) dt   4e 2 t dt  2e 2 t
I() = 2/(1 – j),
= 2 J or
|I()|2 = 4/(1 + 2)

W 1
1 
4 
1
4
4
2
=
= 2J
d  tan 1 () 
I
(
)
d



2






2
2

2
(1   )
0
In the frequency range, –5 &lt;  &lt; 5,
5
4
4
4
tan 1   tan 1 (5)  (1.373) = 1.7487
W =



0
W/ W 1 = 1.7487/2 = 0.8743 or
87.43%
Chapter 18, Solution 58.
 m = 200 = 2f m
which leads to f m = 100 Hz
(a)
 c = x104 = 2f c which leads to f c = 104/2 = 5 kHz
(b)
lsb = f c – f m = 5,000 – 100 = 4,900 Hz
(c)
usb = f c + f m = 5,000 + 100 = 5,100 Hz
Chapter 18, Solution 59.
10
6

V () 2  j 4  j
5
3
H()  o



Vi ()
2
2  j 4  j
 5
3  4

Vo ()  H()Vi ()  

 2  j 4  j  1  j
20
12


, s  j
(s  1)(s  2) (s  1)(s  4)
Using partial fraction,
Vo () 
A
B
C
D
16
20
4






s  1 s  2 s  1 s  4 1  j 2  j 4  j
Thus,


v o (t )  16e  t  20e 2 t  4e 4 t u(t ) V
Chapter 18, Solution 60.
2
+
1/jω
Is
jω
V

V  jI s
1
j
1
 2  j
j

jI s
1   2  j2
Since the voltage appears across the inductor, there is no DC component.
V1 
2908
1  4 2  j4
V2 

50.2790
 1.2418  71.92
 38.48  j12.566
4905
1  16 2  j8

62.8390
 0.3954  80.9
 156.91  j25.13
v(t )  1.2418 cos( 2t  41.92 )  0.3954 cos(4t  129.1 ) mV
Chapter 18, Solution 61.
y (t )  (2  cos o t ) x(t )
We apply the Fourier Transform
Y(ω) = 2X(ω) + 0.5X(ω+ω o ) + 0.5X(ω–ω o ).
Chapter 18, Solution 62.
For the lower sideband, the frequencies range from
10,000,000 – 3,500 Hz = 9,996,500 Hz to
10,000,000 – 400 Hz = 9,999,600 Hz
For the upper sideband, the frequencies range from
10,000,000 + 400 Hz = 10,000,400 Hz to
10,000,000 + 3,500 Hz = 10,003,500 Hz
Chapter 18, Solution 63.
Since f n = 5 kHz, 2f n = 10 kHz
i.e. the stations must be spaced 10 kHz apart to avoid interference.
f = 1600 – 540 = 1060 kHz
The number of stations = f /10 kHz = 106 stations
Chapter 18, Solution 64.
f = 108 – 88 MHz = 20 MHz
The number of stations = 20 MHz/0.2 MHz = 100 stations
Chapter 18, Solution 65.
 = 3.4 kHz
f s = 2 = 6.8 kHz
Chapter 18, Solution 66.
 = 4.5 MHz
f c = 2 = 9 MHz
T s = 1/f c = 1/(9x106) = 1.11x10–7 = 111 ns
Chapter 18, Solution 67.
We first find the Fourier transform of g(t). We use the results of Example 17.2 in
conjunction with the duality property. Let Arect(t) be a rectangular pulse of height A and
width T as shown below.
Arect(t) transforms to Atsinc(2/2)
f(t)
F()
A

t
–T/2
T/2
G()

– m /2
According to the duality property,
Asinc(t/2)
becomes 2Arect()
g(t) = sinc(200t) becomes 2Arect()
where A = 1 and /2 = 200 or T = 400
i.e. the upper frequency  u = 400 = 2f u or f u = 200 Hz
The Nyquist rate = f s = 200 Hz
The Nyquist interval = 1/f s = 1/200 = 5 ms
m/
Chapter 18, Solution 68.
The total energy is
WT =



v 2 ( t ) dt
Since v(t) is an even function,
WT =


0
2500e
4 t
e 4 t
dt  5000
4

= 1250 J
0
V() = 50x4/(4 + 2)
W =
But
 (a
2
1 5 (200) 2
1 5
2



d
|
V
(
)
|
d
2 1 (4   2 ) 2
2 1
1
1  x
1

dx  2  2
 tan 1 ( x / a )  C
2 2
2
a
x )
2a  x  a

5
2x10 4 1  
1

 tan 1 ( / 2)
W =
2

 8 4  
2
1
= (2500/)[(5/29) + 0.5tan-1(5/2) – (1/5) – 0.5tan–1(1/2) = 267.19
W/W T = 267.19/1250 = 0.2137 or 21.37%
Chapter 18, Solution 69.
The total energy is
WT =
=
W =
1 
1  400
2
d
F() d 

2  
2  4 2   2

400
(1 / 4) tan 1 ( / 4)



0

100 
= 50
 2

1 2
400
2
F() d 
(1 / 4) tan 1 ( / 4)

0
2
2

2
0
= [100/(2)]tan–1(2) = (50/)(1.107) = 17.6187
W/W T = 17.6187/50 = 0.3524 or 35.24%
Chapter 19, Solution 1.
To get z 11 and z 21 , consider the circuit in Fig. (a).
2
8
I2 = 0
+
I1
12 
V1
Io
+
4
V2


(a)
z11 
V1
 2  12 || (8  4)  8 
I1
1
I ,
2 1
V
z 21  2  2 
I1
Io 
V2  4 I o  I1
To get z 22 and z 12 , consider the circuit in Fig. (b).
I1 = 0 2 
8
Io'
+
+
12 
V1
4

V2

(b)
z 22 
V2
 4 || (8  12)  3.333 
I2
4
1
I2  I2 ,
4  20
6
V
z12  1  2 
I2
Io 
'
V1  12I o '  2I 2
Hence,
2 
8
[z ]  

 2 3.333 
I2
Chapter 19, Solution 2.
Consider the circuit in Fig. (a) to get z 11 and z 21 .
1
Io'
1
1
1
+
I1
1
V1
1
I2 = 0
Io
+
1
V2


1
1
1
1
(a)
z 11 
V1
 2  1 || [ 2  1 || (2  1) ]
I1
(1)(11 4)

11
3
 2   2.733
z 11  2  1 ||  2    2 

15
1  11 4
4
1
1
Io'  Io'
1 3
4
1
4
Io' 
I1  I1
1  11 4
15
1 4
1
I o   I1  I1
4 15
15
Io 
V2  I o 
z 21 
1
I
15 1
V2
1

 z 12  0.06667
I 1 15
To get z 22 , consider the circuit in Fig. (b).
I1 = 0 1 
1
1
1
+
+
1
V1
1
1

V2

1
1
1
(b)
1
I2
z 22 
V2
 2  1 || (2  1 || 3)  z 11  2.733
I2
Thus,
 2.733 0.06667 
[z ]  

 0.06667 2.733 
Chapter 19, Solution 3.
We can use Figure 19.5 to determine the z-parameters.
z 12 = j12 = z 21
z 11 – z 12 = 8 or z 11 = (8+j12) Ω
z 22 – z 12 = –j20 or z 22 = (–j8) Ω
Chapter 19, Solution 4.
Using Fig. 19.68, design a problem to help other students to better understand how to determine
z parameters from an electrical circuit.
Although there are many ways to solve this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
Calculate the z parameters for the circuit in Fig.19.68.
Figure 19.68
Solution
Transform the  network to a T network.
Z1
Z3
Z2
(12)( j10)
j120

12  j10  j5 12  j5
- j60
Z2 
12  j5
50
Z3 
12  j5
Z1 
The z parameters are
z 12  z 21  Z 2 
(-j60)(12 - j5)
 -1.775 - j4.26
144  25
z 11  Z1  z 12 
( j120)(12  j5)
 z 12  1.775  j4.26
169
z 22  Z 3  z 21 
(50)(12  j5)
 z 21  1.7758  j5.739
169
Thus,
 1.775  j4.26 - 1.775  j4.26 
[z ]  

 - 1.775  j4.26 1.775  j5.739 
Chapter 19, Solution 5.
Consider the circuit in Fig. (a).
s
1
I2 = 0
Io
+
I1
1/s
1
V1
1/s
+
V2


(a)
 1 
1

1  s  

1 
1
1   s  1 
s
|| 1  s   
z 11  1 || || 1  s   
1 
1
s 
s
s  1 
1

 1 s 
s
 s  1
s
2
s  s 1
z 11  3
s  2s 2  3s  1
1
s
Io 
1 ||
1
s
1
s 1
s
s 1
I 
I 
I1
1
1 1
1
1 1
s
2
1 s 
 s  s 1
1 ||  1  s 
s
s
s 1
s
s 1
s
Io  3
I1
2
s  2s  3s  1
I1
1
V2  I o  3
s
s  2s 2  3s  1
z 21 
V2
1
 3
2
I 1 s  2s  3s  1
Consider the circuit in Fig. (b).
s
1
I1 = 0
+
V1
+
1
1/s
1/s

V2

(b)
I2
z 22 
V2 1 
1 1 
1 

 || 1  s  1 ||   || 1  s 
s s 
s  1
I2 s 
z 22
 1 
1 
1
 1  s 

1

s

 s 
s  1
s 1


1
1
s
1 s 
1 s  s2 
s
s 1
s 1
z 22
s 2  2s  2
 3
s  2s 2  3s  1
z 12  z 21
Hence,


s2  s  1
1
 3

s  2s 2  3s  1 s 3  2s 2  3s  1 
[z ]  


1
s 2  2s  2
 3

2
3
2
 s  2s  3s  1 s  2s  3s  1 
Chapter 19, Solution 6.
To find z 11 and z 21 , consider the circuit below.
I1
5
10
4I 1
I 2 =0
Vo
–
+
+
V1
20 
+
_
V2
–
z11 
Vo 
V1 (20  5) I1

 25 
I1
I1
20
V1  20 I1
25
V o 4 I 2  V2  0


V2  Vo  4 I1  20 I1  4 I1  24 I1
V2
 24 
I1
To find z 12 and z 22 , consider the circuit below.
z21 
I 1 =0 5
10
4I 1
I2
– +
+
V1
–
V2  (10  20) I 2  30 I 2
V2
 30 
I1
V1  20 I 2
z22 
20 
+
_
V2
z12 
V1
 20 
I2
Thus,
 25 20 
[ z]  

 24 30 
Chapter 19, Solution 7.
To get z 11 and z 21 , we consider the circuit below.
I1
20 
I 2 =0
100 
+
+
vx
50 
60 
+
V1
-
V2
-
12v x
-
+
V1  Vx Vx Vx  12Vx


20
50
160
V  Vx
81 V1
I1  1
( )

20
121 20



Vx 
z11 
40
V1
121
V1
 29.88
I1
13Vx
57
57 40
57 40 20x121
)  12Vx   Vx   (
)V1   (
)
I1
160
8
8 121
8 121
81
V
 70.37 I1  z 21  2  70.37
I1
V2  60(
To get z 12 and z 22 , we consider the circuit below.
I 1 =0
20 
I2
100 
+
+
50 
vx
60 
+
V1
-
V2
-
12v x
-
+
Vx 
50
1
V2  V2 ,
100  50
3
z 22 
V2
 1 / 0.09  11.11
I2
V1  Vx 
I2 
V2 V2  12Vx

 0.09V2
150
60
1
11.11
V2 
I 2  3.704I 2
3
3


Thus,
 29.88 3.704
[z ]  

  70.37 11.11
V
z12  1  3.704
I2
Chapter 19, Solution 8.
To get z 11 and z 21 , consider the circuit below.
I1
j4 
-j2 

5
I 2 =0

j6 
+
j8 
+
V2
V1
10 
-
-
V1  (10  j2  j6)I1
V2  10I1  j4I1
V
z11  1  10  j4
I1




z 21 
V2
 (10  j4)
I1
To get z 22 and z 12 , consider the circuit below.
j4 
I 1 =0 -j2 

j6 
5
I2

j8 
+
+
V2
V1
10 
-
-
V2  (5  10  j8)I 2
V1  (10  j4)I 2




z 22 
V2
 15  j8
I2
V
z12  1  (10  j4)
I2
Thus,
 (10  j4)  (10  j4)
[z ]  

  (10  j4) (15  j8) 
Chapter 19, Solution 9.
 y  y11 y22  y12 y21  0.5 x0.4  0.2 x0.2  0.16
z11 
y22 0.4

 2.5 Ω
 y 0.16
z12 
 y12 0.2

 1.25 Ω = z 21
0.16
y
z 22 
y11 0.5

 3.125 Ω
 y 0.16
Thus,
 2.5 1.25 
Z  z   

1.25 3.125
Chapter 19, Solution 10.
(a)
This is a non-reciprocal circuit so that the two-port looks like the one
shown in Figs. (a) and (b).
z 11
I1
z 22
I2
+
+
z 12 I 2
V1
+

+

z 21 I 1
V2


(a)
25 
I1
10 
I2
+
+
20 I 2
V1
+

+

5 I1
V2


(b)
(b)
This is a reciprocal network and the two-port look like the one shown in
Figs. (c) and (d).
I1
z 11 – z 12
z 22 – z 12
+
+
z 12
V1

z 22  z 12  2s
1
z 12 
s
V2

(c)
z 11  z 12  1 
I2
1
2
 1
0.5 s
s
I1
1
0.5 F
2H
+
I2
+
1F
V1

V2

(d)
Chapter 19, Solution 11.
This is a reciprocal network, as shown below.
1+j5
3+j
5-j2
1
j5 
3
5
-j2 
j1 
Chapter 19, Solution 12.
V1  10 I1  6 I 2
V2  4 I 2  12 I 2
V2  10 I 2
(1)
(2)
(3)
If we convert the current source to a voltage source, that portion of the circuit becomes
what is shown below.
4
2
I1
+
12 V
V1
+
_
–
12  6 I1  V1  0

 V1  12  6 I1
(4)
Substituting (3) and (4) into (1) and (2), we get
12  6 I 1  10 I1  6 I 2

 12  16 I1  6 I 2
10 I 2  4 I1  12 I 2

 0  4 I1  22 I 2
(5)

 I1  5.5 I 2
From (5) and (6),
12  88I 2  6 I 2  82 I 2

 I 2  0.1463 A
I1  5.5 I 2  0.8049 A
V2  10 I 2  1.463 V
V1  12  6 I1  7.1706 V
(6)
Chapter 19, Solution 13.
Consider the circuit as shown below.
10  I 1
I2
+
500˚ V
+
_
+
V1
[z]
V2
ZL
_
–
V1  40I1  60I2
(1)
V2  80I1  100I2
(2)
V2  I2 ZL  I2(5  j 4)
(3)
50  V1  10I1

 V1  50  10I1
(4)
Substituting (4) in (1)
50  10I1  40I1  60I2

 5  5I1  6I2 (5)
Substituting (3) into (2),
I2(5  j 4)  80I1  100I2

 0  80I1  (105  j4)I2
Solving (5) and (6) gives
I 2 = –7.423 + j3.299 A
We can check the answer using MATLAB.
First we need to rewrite equations 1-4 as follows,
1
0

0

1
0  40  60   V1 
0


0

1  80  100  V2 
 A*X     U
0
1
0
5  j4  I1 
 
 
0 10
0  I2 
50
&gt;&gt; A=[1,0,-40,-60;0,1,-80,-100;0,1,0,(5+4i);1,0,10,0]
A=
1.0e+002 *
0.0100
0
-0.4000
-0.6000
0
0.0100
-0.8000
-1.0000
0
0.0100
0
0.0500 + 0.0400i
0.0100
0
0.1000
0
&gt;&gt; U=[0;0;0;50]
U=
(6)
0
0
0
50
&gt;&gt; X=inv(A)*U
X=
-49.0722 +39.5876i
50.3093 +13.1959i
9.9072 - 3.9588i
-7.4227 + 3.2990i
P = |I 2 |25 = 329.9 W.
Chapter 19, Solution 14.
To find Z Th , consider the circuit in Fig. (a).
I1
I2
+
ZS
+

V1
Vo = 1

(a)
V1  z 11 I 1  z 12 I 2
V2  z 21 I 1  z 22 I 2
(1)
(2)
But
V2  1 ,
V1  - Z s I 1
0  (z 11  Z s ) I 1  z 12 I 2
Hence,

 I 1 
- z 12
I
z 11  Z s 2
 - z 21 z 12

1
 z 22  I 2
 z 11  Z s

Z Th 
V2
z z
1

 z 22  21 12
z 11  Z s
I2 I2
To find VTh , consider the circuit in Fig. (b).
ZS
VS
+

I1
I2 = 0
+
+
V1
V 2 = V Th


(b)
I2  0 ,
Substituting these into (1) and (2),
V1  Vs  I 1 Z s
Vs  I 1 Z s  z 11 I 1
V2  z 21 I 1 

 I 1 
Vs
z 11  Z s
z 21 Vs
z 11  Z s
VTh  V2 
z 21 Vs
z 11  Z s
Chapter 19, Solution 15.
(a) From Prob. 18.12,
ZTh  z 22 
z12z 21
80x 60
 120 
 24
z11  Zs
40  10
ZL  ZTh  24
(b) VTh 
z 21
80
Vs 
(120)  192
z11  Zs
40  10
 V
Pmax   Th
 2R Th
2

 R Th  4 2 x 24  384W

Chapter 19, Solution 16.
As a reciprocal two-port, the given circuit can be represented as shown in Fig. (a).
5
10 – j6 
4 – j6 
+

150 V
j6 
j4 
b
(a)
At terminals a-b,
Z Th  (4  j6)  j6 || (5  10  j6)
j6 (15  j6)
Z Th  4  j6 
 4  j6  2.4  j6
15
Z Th  6.4 
VTh 
j6
(150)  j6  690 V
j6  5  10  j6
The Thevenin equivalent circuit is shown in Fig. (b).
6.4 
+
690 V
+

Vo

(b)
From this,
Vo 
a
j4
( j6)  3.18148
6.4  j4
v o ( t )  3.18 cos( 2t  148) V
j4 
Chapter 19, Solution 17.
To obtain z 11 and z 21 , consider the circuit in Fig. (a).
8
Io'
+
I1
V1
Io
I2 = 0
+
4
V2
16 


12 
(a)
In this case, the 8- and 16- resistors are in series, since the same current, I o , passes
through them. Similarly, the 4- and 12- resistors are in series, since the same current,
I o ' , passes through them.
z11 
Io 
But
V1
(24)(16)
 (8  16) || (4  12)  24 || 16 
 9.6 Ω
40
I1
16
2
I1  I1
16  24
5
Io' 
3
I
5 1
- V2  8 I o  4 I o  0
- 16
12
-4
'
V2  -8 I o  4 I o 
I1  I 1  I1
5
5
5
V -4
z 21  2 
 –0.8 Ω
I1
5
'
To get z 22 and z 12 , consider the circuit in Fig. (b).
8
I1 = 0
+
V1
+
4
V2
16 
I2


12 
(b)
z 22 
(12)(28)
V2
 (8  4) || (16  12)  12 || 28 
 8.4 Ω
40
I2
z12  z 21  –0.8 Ω
Thus,
 9.6 - 0.8 
[z ]  

 - 0.8 8.4 
We may take advantage of Table 18.1 to get [y] from [z].
 z  (9.6)(8.4)  (0.8) 2  80
z
8.4
-z
0.8
 0.105 S
 0.01 S
y 11  22 
y12  12 
80
 z 80
z
z
-z
0.8
9.6
 0.01 S
 0.12 S
y 21  21 
y 22  11 
80
z
 z 80
Thus,
 0.105 0.01 
[y ]  
S
 0.01 0.12 
Chapter 19, Solution 18.
To get y 11 and y 21 , consider the circuit in Fig.(a).
6
I1
3
I2
+
+

V1
6
3
V2 = 0

(a)
V1  (6  6 || 3) I 1  8 I 1
I1 1

y 11 
V1 8
-6
- 2 V1 - V1
I1 

63
3 8
12
I 2 -1


V1 12
I2 
y 21
To get y 22 and y 12 , consider the circuit in Fig.(b).
I1
6
Io
3
I2
+
V1 = 0
6
3

(b)
y 22 
I2
1
1
1



V2 3 || (3  6 || 6) 3 || 6 2
- Io
3
1
,
Io 
I2  I2
2
3 6
3
- I 2  - 1  1  - V2
I1 
   V2  
 6  2  12
6
I1 
+

V2
y 12 
I1
-1

 y 21
V2 12
Thus,
 1 -1 
 8 12 
[y ]  
S
1
1


 12 2 
Chapter 19, Solution 19.
Using Fig. 19.80, design a problem to help other students to better understand how to find y
parameters in the s-domain.
Although there are many ways to solve this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
Find the y parameters of the two-port in Fig.19.80 in terms of s.
Figure 19.80
Solution
Consider the circuit in Fig.(a) for calculating y 11 and y 21 .
1
I1
I2
+
V1
+

1/s
s
1
(a)
2s
1 
2
V1   || 2  I 1 
I1 
I
s 
2  (1 s)
2s  1 1
I
2s  1
y 11  1 
 s  0.5
V1
2
I2 
- I1
- V1
(- 1 s )
I1 

(1 s)  2
2s  1
2
V2 = 0

y 21 
I2
 -0.5
V1
To get y 22 and y 12 , refer to the circuit in Fig.(b).
1
I1
I2
+
V1 = 0
1/s
s

1
(b)
V2  (s || 2) I 2 
y 22 
2s
I
s2 2
I2 s  2
1

 0.5 
s
V2
2s
- V2
-s
-s s 2
I2 

V2 
s2
s  2 2s
2
I1

 -0.5
V2
I1 
y 12
Thus,
 s  0.5
- 0.5 
[y ]  
S
 - 0.5 0.5  1 s 
+

V2
Chapter 19, Solution 20.
To get y 11 and y 21 , consider the circuit below.
3i x
2
I1
I2
+
ix
4
V1
I1
+
6
V 2 =0
-
-
Since 6-ohm resistor is short-circuited, i x = 0
V1  I1(4 // 2) 
I2  
8
I1
6

I
y11  1  0.75
V1
4
2 6
1
I1   ( V1)   V1
42
3 8
2

I
y 21  2  0.5
V1
To get y 22 and y 12 , consider the circuit below.
3i x
2
I1
+
ix
4
V 1 =0
+
6  V2
-
I2
ix 
V2
,
6
V
V
I 2  i x  3i x  2  2
2
6
V
I1  3i x  2  0
2



I
y12  1  0
V2
Thus,
0 
 0.75
[y ]  
S
  0.5 0.1667
I
1
y 22  2   0.1667
V2 6
Chapter 19, Solution 21.
To get y 11 and y 21 , refer to Fig. (a).
0.2 V 1
I1
I2
V1
+
+

V1
5
10 
V2 = 0

(a)
At node 1,
I1 
V1
 0.2 V1  0.4 V1
5
I 2  -0.2 V1

 y 11 

 y 21 
I1
 0.4
V1
I2
 -0.2
V1
To get y 22 and y 12 , refer to the circuit in Fig. (b).
0.2 V 1
I1
V1
I2
+
5
V1 = 0
10 
+

V2

(b)
Since V1  0 , the dependent current source can be replaced with an open circuit.
V2  10 I 2
y 12 

 y 22 
I1
0
V2
Thus,
 0.4
0
[y ]  
S
 - 0.2 0.1
I2
1

 0 .1
V2 10
Consequently, the y parameter equivalent circuit is shown in Fig. (c).
I1
I2
+
V1
+
0.2 V 1
0.4 S

0.1 S
V2

(c)
Chapter 19, Solution 22.
To obtain y 11 and y 21 , consider the circuit below.
5
I2
+
+
I1
5
0.5V 2
2
V 2 =0
V1
–
–
The 2- resistor is short-circuited.
I
I
2
V1  5 1

 y 11 1   0.4
V1 5
2
1
I1
1
I
I2  I1

 y21  2  2  0.2
2
V1 2.5I1
To obtain y 12 and y 22 , consider the circuit below.
5
I1
+
+
5
0.5V 2
2
V2
V 1 =0
–
–
At the top node, KCL gives
V V
I2  0.5V2  2  2  1.2V2
2
5
I1  
V2
 0.2V2
5

 y22 

 y12 
I1
 0.2
V2
Hence,
 0.4 0.2 
[y ]  
 S
 0.2 1.2 
I2
 1.2
V2
I2
Chapter 19, Solution 23.
(a)
1 /(  y 12 )  1 //
1
1

s s1
y 11  y 12  1


y 22  y 12  s


y 12   (s  1)
y 11  1  y 12  1  (s  1)  s  2


y 22  s  y 12
1
s2  s  1
  (s  1) 
s
s
 (s  1) 
 s2

[y ] 
s2  s  1
  (s  1)

s


(b) Consider the network below.
I1
I2
1
+
+
+
[y]
Vs
-
V1
-
V2
-
2
Vs  I1  V1
(1)
V2  2I 2
(2)
I1  y11V1  y12 V2
(3)
I 2  y 21V1  y 22 V2
(4)
From (1) and (3)
Vs  V1  y11V1  y12 V2


Vs  (1  y11 )V1  y12 V2
(5)
From (2) and (4),
 0.5V2  y 21V1  y 22 V2

V1  
1
(0.5  y 22 )V2
y 21
(6)
Substituting (6) into (5),
Vs  

2
s
(1  y 11 )(0.5  y 22 )
V2  y 12 V2
y 21


2/s


1
(1  y 11 )(0.5  y 22 )
 y 12 
y 21


2/s
V2 
 (s  1) 

V2 
2


1
1  s  2 0.5  s  s  1 
s
s1



2/s
 s  s  s  s  (s  3)(0.5s  s 2  s  1)
s(s  1)
3
2
2
2(s  1)
2(s  1)
0.8(s  1)
 2

2
2
2
 s  2s  s  s  1.5s  s  3s  4.5s  3 2.5s  4.5s  3 s 1.8s  1.2
3
2
3
Chapter 19, Solution 24.
Since this is a reciprocal network, a  network is appropriate, as shown below.
Y2
Y1
Y3
(a)
4
1/4 S
1/4 S
1/8 S
4
8
(b)
Y1  y 11  y 12 
Y2  - y 12 
(c)
1 1 1
  S,
2 4 4
1
S,
4
Y3  y 22  y 21 
Z1  4 
Z2  4 
3 1 1
  S,
8 4 8
Z3  8 
Chapter 19, Solution 25.
This is a reciprocal network and is shown below.
0.5 S
0.5S
1S
Chapter 19, Solution 26.
To get y 11 and y 21 , consider the circuit in Fig. (a).
4
2
1
+
V1
+

Vx
I2
2
1
+
2 Vx
V2 = 0


(a)
At node 1,
V1  Vx
V
V
 2 Vx  x  x
2
1
4
But
I1 
Also,
I2 
y 21

 2 V1  -Vx
V1  Vx V1  2 V1

 1.5 V1
2
2
(1)

 y 11 
I1
 1.5
V1
Vx
 2 Vx 
 I 2  1.75 Vx  -3.5 V1
4
I2

 -3.5
V1
To get y 22 and y 12 , consider the circuit in Fig.(b).
4
2
I1
1
2
+
Vx
1
2 Vx
I2
+

V2

(b)
At node 2,
I 2  2 Vx 
V2  Vx
4
(2)
At node 1,
2 Vx 
V2  Vx Vx Vx 3


 Vx
4
2
1
2

 V2  -Vx
Substituting (3) into (2) gives
1
I 2  2 Vx  Vx  1.5 Vx  -1.5 V2
2
I2
 -1.5
y 22 
V2
I1 
- Vx V2

2
2

 y 12 
I1
 0.5
V2
Thus,
 1.5 0.5 
[y ]  
S
 - 3.5 - 1.5 
(3)
Chapter 19, Solution 27.
Consider the circuit in Fig. (a).
4
I1
I2
+
+

V1
0.1 V 2

+
10 
20 I 1
V2 = 0

(a)
V1  4 I 1

 y 11 
I 2  20 I 1  5 V1
I1
I1

 0.25
V1 4 I 1

 y 21 
I2
5
V1
Consider the circuit in Fig. (b).
I1
4
I2
+
V1 = 0
0.1 V 2

+
10 
20 I 1
+

V2

(b)
4 I 1  0.1 V2
I 2  20 I 1 

 y 12 
I 1 0.1

 0.025
V2
4
V2
 0.5 V2  0.1 V2  0.6 V2
10
Thus,
 0.25 0.025 
[y ]  
S
0.6 
 5
Alternatively, from the given circuit,
V1  4 I 1  0.1 V2

 y 22 
I2
 0.6
V2
I 2  20 I 1  0.1 V2
Comparing these with the equations for the h parameters show that
h11  4 ,
h12  -0.1,
h 21  20 ,
h 22  0.1
Using Table 18.1,
as above.
1
1
  0.25 S ,
h 11 4
h
20
 21 
 5S ,
h 11
4
- h 12 0.1

 0.025 S
h 11
4

0.4  2
 h 
 0 .6 S
h 11
4
y 11 
y 12 
y 21
y 22
Chapter 19, Solution 28.
We obtain y 11 and y 21 by considering the circuit in Fig.(a).
2
8
I2
+
I1
+
12 
V1
4

V2 = 0

(a)
Z in = 2 + (12||8) = 6.8 Ω
I
1
 147.06 mS
y 11  1 
V1 Z in
I 2 = (–6/10)I 1 = (–0.6)(V 1 /6.8) = –0.08824
I
y 21  2  –88.24 mS
V1
To get y 22 and y 12 , consider the circuit in Fig. (b).
I1
2
8
Io
+
+
V1 = 0
12 
4

V2
I2

(b)
(1/y 22 ) = [4||(8+(12||2))] = [4||(8+(1.714286))] = 2.833333 = V 2 /I 2
y 22 = 352.9 mS
I 1 = (–12/14)I o = –0.857143I o and I o = [4/(4+(8+1.714286))]I 2
= 0.29166667I 2 = V 2 /9.714286
Thus, I 1 = [(–0.857143)/9.714286]V 2 = –0.088235V 2 or
y 12 = I 1 /V 2 = –88.24 mS
Thus,
147.06 - 88.24
[y ]  
 mS
- 88.24 352.9 
We note that I = YV, I 1 = 1 A, and –I 2 = V 2 /2. We now have the following equations,
1 = 0.14706V 1 – 0.08824V 2 and I 2 = –0.08824V 1 + 0.3529V 2 or
–V 2 /2 = –0.08824V 1 + 0.3529V 2 or 0.08824V 1 = 0.8529V 2 which leads
to
V 1 = 9.6657V 2 .
Substituting this into the first equation we get,
1 = (1.42144–0.08824)V 2 or V 2 = 0.75 V.
Finally we get,
P 2Ω = (0.75)2/2 = 281.2 mW.
The equivalent circuit is shown in Fig. (c). After transforming the current source to a
voltage source, we have the circuit in Fig. (d).
88.24mS
58.82 mS
1A
264.7 mS
2
(c)
17 
11.333 
+
17 V
+

V

(d)
3.778 
2
V
(2 || 3.778)(17)
(1.3077)(17)

 0.75 V
(2 || 3.778)  17  11.333 1.3077  28.333
P
V 2 (0.75) 2

 281.2 mW
R
2
Chapter 19, Solution 29.
(a)
Transforming the  subnetwork to Y gives the circuit in Fig. (a).
1
1
Vo
+
10 A
+
2
V1

-4 A
V2

(a)
It is easy to get the z parameters
z 12  z 21  2 , z 11  1  2  3 ,
z 22  3
 z  z 11 z 22  z 12 z 21  9  4  5
y 11 
z 22 3
  y 22 ,
z 5
y 12  y 21 
- z 12 - 2

z
5
Thus, the equivalent circuit is as shown in Fig. (b).
2/5 S
I1
I2
+
10 A
V1
+
1/5 S
1/5 S

V2
-4 A

(b)
I 1  10 
3
2
V1  V2
5
5

 50  3 V1  2 V2
(1)
-2
3
V1  V2 
 - 20  -2 V1  3 V2
5
5
10  V1  1.5 V2 
 V1  10  1.5 V2
(2)
I 2  -4 
Substituting (2) into (1),
50  30  4.5 V2  2 V2

 V2  8 V
V1  10  1.5 V2  22 V
(b)
For direct circuit analysis, consider the circuit in Fig. (a).
For the main non-reference node,
Vo
10  4 

 Vo  12
2
10 
V1  Vo
1

 V1  10  Vo  22 V
-4
V2  Vo
1

 V2  Vo  4  8 V
Chapter 19, Solution 30.
(a)
Convert to z parameters; then, convert to h parameters using Table 18.1.
z 11  z 12  z 21  60  ,
z 22  100 
 z  z 11 z 22  z 12 z 21  6000  3600  2400
 z 2400

 24 ,
z 22
100
- z 21

 -0.6 ,
z 22
h 11 
h12 
z 12
60

 0.6
z 22 100
h 21
h 22 
1
 0.01
z 22
Thus,
 24 
0.6 
[h]  

 - 0.6 0.01 S 
(b)
Similarly,
z 11  30 
z 12  z 21  z 22  20 
 z  600  400  200
h11 
200
 10
20
h 21  -1
Thus,
 10 
1 
[h]  

 - 1 0.05 S 
20
1
20
1

 0.05
20
h12 
h 22
Chapter 19, Solution 31.
We get h11 and h 21 by considering the circuit in Fig. (a).
1
2
V3
V4
1
I2
+
I1
2
V1
4 I1

(a)
At node 1,
I1 
V3 V3  V4

2
2

 2 I 1  2 V3  V4
(1)
At node 2,
V3  V4
V
 4 I1  4
2
1
8 I 1  -V3  3 V4 
 16 I 1  -2 V3  6 V4
(2)
18 I 1  5 V4 
 V4  3.6 I 1
V3  3 V4  8 I 1  2.8 I 1
V1  V3  I 1  3.8 I 1
V
h11  1  3.8 
I1
I2 
- V4
 -3.6 I 1
1

 h 21 
I2
 -3.6
I1
To get h 22 and h 12 , refer to the circuit in Fig. (b). The dependent current source can be
replaced by an open circuit since 4 I 1  0 .
I1
1
1
2
I2
+
V1
2
4 I1 = 0

(b)
+

V2
V1 
2
2
V2  V2
2  2 1
5
I2 
V2
V2

2  2 1 5

 h 12 

 h 22 
V1
 0 .4
V2
I2 1
  0 .2 S
V2 5
Thus,
 38  0.4 
[h]  

 - 3.6 0.2 S 
Chapter 19, Solution 32.
Using Fig. 19.90, design a problem to help other students to better understand how to find the h
and g parameters for a circuit in the s-domain.
Although there are many ways to solve this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
Find the h and g parameters of the two-port network in Fig.19.90 as functions of s.
Figure 19.90
Solution
(a)
We obtain h11 and h 21 by referring to the circuit in Fig. (a).
1
s
s
+
I1
I2
+
1/s
V1

V2 = 0

(a)


s 
1
I
V1  1  s  s ||  I 1  1  s  2


s  1 1
s
V
s
h11  1  s  1  2
I1
s 1
By current division,
- I1
I2
-1 s
-1

 h 21 
 2
I2 
I1 
s 1 s
s 1
I1 s  1
To get h 22 and h 12 , refer to Fig. (b).
I1 = 0
1
s
s
I2
+
+

1/s
V1
V2

(b)
V1 
V2
V1
1s
1

 h12 
 2
V2  2
s 1 s
s 1
V2 s  1
 1
V2  s   I 2
 s

 h 22 
I2
1
s

 2
V2 s  1 s s  1
Thus,
s

 s  1  s2  1
[h]  
-1

2
s 1

(b)
1 
s 1 
s 

s2  1 
2
To get g11 and g 21 , refer to Fig. (c).
I1
1
s
s
I2 = 0
+
V1
+

1/s
V2

(c)

1
V1  1  s   I 1

s
V2 

 g 11 
I1
1
s

 2
V1 1  s  1 s s  s  1
V1
V2
1s
1
V1  2

 g 21 
 2
1 s 1 s
s  s 1
V1 s  s  1
To get g 22 and g 12 , refer to Fig. (d).
I1
1
s
s
I2
+
+
1/s
V1 = 0
V2
I2


(d)

 1
(s  1) s 

I 2
V2  s  || (s  1)  I 2  s 
 s

 1 s 1 s 
g 22 
I1 
V2
s 1
s 2
I2
s  s 1
-I
I1
-1 s
-1
I2  2 2

 g 12 
 2
1 s 1 s
s  s 1
I2 s  s 1
Thus,

 2
[g ]   s
 2
s
s
-1

2
s1
s s1 
1
s1 

s 2
s1
s s1 
Chapter 19, Solution 33.
To get h 11 and h 21 , consider the circuit below.
4
j6 
+
I1
V1  5 //( 4  j6)I1 
Also, I 2  
-j3 
5
V1
-
5
I1
9  j6
I2
5(4  j6)I1
9  j6
+
V 2 =0
-
V
h11  1  3.0769  j1.2821
I1
I
h 21  2  0.3846  j0.2564
I1


To get h 22 and h 12 , consider the circuit below.
4
j6 
I2
I1
+
-j3 
5
V1
-
V1 
5
V2
9  j6
+
+
V2
-


V2   j3 //(9  j6)I 2
V
5
h12  1 
 0.3846  j0.2564
V2 9  j6


I
1
9  j3
h 22  2 

V2  j3 //(9  j6)  j3(9  j6)
 0.0769  j0.2821
Thus,
( 3.077  j1.2821) 
0.3846  j0.2564 
[h]  

  0.3846  j0.2564 (0.0769  j0.2821) S 
Chapter 19, Solution 34.
Refer to Fig. (a) to get h11 and h 21 .
300 
10 
50 
2
1
+
I1
V1
Vx

I2
+
+

+
100 
10 V x
V2 = 0


(a)
At node 1,
Vx Vx  0


 300 I 1  4 Vx
100
300
300
Vx 
I  75 I 1
4 1
I1 
V1  10 I 1  Vx  85 I 1
But

 h11 
(1)
V1
 85 
I1
At node 2,
0  10 Vx Vx
Vx Vx
75
75




I1 
I  14.75 I 1
50
300
5 300 5
300 1
I2

 14.75
I1
I2 
h 21
To get h 22 and h 12 , refer to Fig. (b).
300 
I 1 = 0 10 
50 
1
+
V1

+
Vx

+
100 

(b)
10 V x
2
I2
+

V2
At node 2,
V2 V2  10 Vx


 400 I 2  9 V2  80 Vx
400
50
V2
100
Vx 
V2 
400
4
400 I 2  9 V2  20 V2  29 V2
I2
29

 0.0725 S
h 22 
V2 400
I2 
But
Hence,
V1  Vx 
V2
4

 h 12 
V1 1
  0.25
V2 4
 85 
0.25 
[h]  

 14.75 0.0725 S 
To get g 11 and g 21 , refer to Fig. (c).
300 
I1
10 
50 
1
+

Vx
I2 = 0
+
+
V1
2
100 

+
10 V x
V2


(c)
At node 1,
I1 
But
or
Vx Vx  10 Vx

100
350

 350 I 1  14.5 Vx
V1  Vx

 10 I 1  V1  Vx
10
Vx  V1  10 I 1
(2)
I1 
Substituting (3) into (2) gives
350 I 1  14.5 V1  145 I 1

 495 I 1  14.5 V1
(3)
g 11 
I 1 14.5

 0.02929 S
V1 495
At node 2,

 11
V2  (50) 
Vx   10 Vx  -8.4286 Vx
 350 
 14.5 
 -8.4286 V1  84.286 I 1  -8.4286 V1  (84.286) 
 V1
 495 
V2  -5.96 V1

 g 21 
V2
 -5.96
V1
To get g 22 and g 12 , refer to Fig. (d).
300 
I1
+
V1 = 0
Io
10 
+
+
Vx

Io
50 

+
100 
10 V x
I2
V2


(d)
10 || 100  9.091
I2 
But
V2  10 Vx
V2

50
300  9.091
309.091 I 2  7.1818 V2  61.818 Vx
(4)
9.091
V  0.02941 V2
309.091 2
(5)
Vx 
Substituting (5) into (4) gives
309.091 I 2  9 V2
V2
g 22 
 34.34 
I2
Io 
34.34 I 2
V2

309.091 309.091
- 34.34 I 2
- 100
Io 
110
(1.1)(309.091)
I1

 -0.101
I2
I1 
g 12
Thus,
 0.02929 S - 0.101 
[g ]  
34.34  
 - 5.96
Chapter 19, Solution 35.
To get h11 and h 21 consider the circuit in Fig. (a).
1
I1
1:2
4
I2
+
+
V1
V2 = 0


(a)
ZR 
4
4
1
2 
n
4
V1  (1  1) I 1  2 I 1

 h11 
V1
 2
I1
I 2 -1
I1 - N 2

 -2 
 h 21 
  -0.5
I1
2
I2
N1
To get h 22 and h 12 , refer to Fig. (b).
I1 = 0 1 
4
1:2
+

+
V1

(b)
Since I 1  0 , I 2  0 .
h 22  0 .
Hence,
I2
V2
At the terminals of the transformer, we have V1 and V2 which are related as
V1 1
V2 N 2

n2 
 h12 
  0 .5
V2 2
V1 N 1
Thus,
 2  0.5 
[h]  

 - 0.5 0 
Chapter 19, Solution 36.
We replace the two-port by its equivalent circuit as shown below.
4
I1
16 
2 I1
+
10 V
+

V1
I2
+
3 V2
+

-2 I 1

100  V 2

100 || 25  20 
V2  (20)(2 I 1 )  40 I 1
(1)
- 10  20 I 1  3 V2  0
10  20 I 1  (3)(40 I 1 )  140 I 1
I1 
1
,
14
V1  16 I 1  3 V2 
V2 
136
14
-8
 100 
I2  
 (2 I 1 ) 
70
 125 
(a)
V2
40

 0.2941
V1 136
(b)
I2
 - 1.6
I1
(c)
I1
1

 7.353  10 -3 S
V1 136
(d)
V2 40

 40 
1
I1
40
14
25 
Chapter 19, Solution 37.
(a)
We first obtain the h parameters. To get h11 and h 21 refer to Fig. (a).
6
3
I2
+
I1
+
6
V1
3
V2 = 0


(a)
3 || 6  2
V1  (6  2) I 1  8 I 1

 h11 
V1
8
I1
-6
-2
I1 
I
3 6
3 1

 h 21 
I2 - 2

3
I1
I2 
To get h 22 and h 12 , refer to the circuit in Fig. (b).
I1 = 0 6 
3
I2
+
6
V1
3

(b)
3 || 9 
V2 
9
4
9
I
4 2

 h 22 
I2 4

V2 9
+

V2
6
2
V2  V2
63
3
V1 

 h 12 
V1 2

V2 3

2 
8  3 
[h]  
-2 4 
S

3 9 
The equivalent circuit of the given circuit is shown in Fig. (c).
8
I1
I2
+
10 V
+

2/3 V 2
+

9/4  V 2
-2/3 I 1
5

(c)
8 I1 
2
V  10
3 2
(1)
2  9  2  45  30
I 5 ||   I   
I
3 1  4  3 1  29  29 1
29
I1 
V
30 2
V2 
(2)
Substituting (2) into (1),
 29 
2
(8)   V2  V2  10
 30 
3
300
V2 
 1.19 V
252
(b)
By direct analysis, refer to Fig.(d).
6
3
+
10 V
+

6
3
V2

(d)
5
10
-A current source. Since
6
6 || 6  3  , we combine the two 6- resistors in parallel and transform
10
the current source back to
 3  5 V voltage source shown in Fig. (e).
6
Transform the 10-V voltage source to a
3
3
+
+

5V
V2

(e)
3 || 5 
V2 
(3)(5) 15

8
8
15 8
75
 1.1905 V
(5) 
63
6  15 8
3 || 5 
Chapter 19, Solution 38.
From eq. (19.75),
h h R
h h R
0.04 x30 x 400
Z in  hie  re fe L  h11  12 21 L  600 
 333.33 
1  hoe RL
1  h22 RL
1  2 x103 x 400
From eq. (19.79),
Rs  hie
Rs  h11
2, 000  600
Z out 


 650 
( Rs  hie )h0 e  hre h fe ( Rs  h11 )h22  h21h12 2600 x 2 x103  30 x0.04
Chapter 19, Solution 39.
We obtain g 11 and g 21 using the circuit below.
R1
I1
R3
I 2 =0
+
V1
R2
+
_
V2
–
I1 
V1
R1  R2

 g11 
By voltage division,
R2
V2 
V1
R1  R2


I1
1

V1 R1  R2
g 21 
V2
R2

V1 R1  R2
We obtain g 12 and g 22 using the circuit below.
I1
R1
R3
+
+
R2
I2
V 1 =0
V2
–
–
By current division,
R2
I1  
I2
R1  R2

 g12 
I1
R2

I2
R1  R2
Also,

V
RR
RR 
V2  I 2 ( R3  R1 // R2 )  I 2  R3  1 2  g 22  2  R3  1 2
I2
R1  R2
R1  R2 

g 11 
R2
1
, g 12  
R1  R 2
R1  R 2
g 21 
R2
R1R 2
, g 22  R 3 
R1  R 2
R1  R 2
Chapter 19, Solution 40.
Using Fig. 19.97, design a problem to help other students to better understand how to find g
parameters in an ac circuit.
Although there are many ways to solve this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
Find the g parameters for the circuit in Fig.19.97.
Figure 19.97
Solution
To get g 11 and g 21 , consider the circuit in Fig. (a).
-j6 
I1
j10 
I2 = 0
+
+

V1
12 
V2

(a)
V1  (12  j6) I 1
g 21 

 g 11 
I1
1

 0.0667  j0.0333 S
V1 12  j6
V2
12 I 1
2


 0.8  j0.4
V1 (12  j6) I 1 2  j
To get g 12 and g 22 , consider the circuit in Fig. (b).
I1
-j6 
j10 
I2
+
V1 = 0
R
R
12 

(b)
I2
R
I1 
- 12
I
12 - j6 2

 g 12 
I1
- 12

 - g 21  -0.8  j0.4
I 2 12 - j6
V2  ( j10  12 || -j6) I 2
V2
(12)(-j6)
 j10 
 2.4  j5.2 
g 22 
12 - j6
I2
 0.0667  j0.0333 S - 0.8  j0.4 
[g ]  
0.8  j0.4
2.4  j5.2  

Chapter 19, Solution 41.
For the g parameters
I 1  g 11 V1  g 12 I 2
V2  g 21 V1  g 22 I 2
But
or
and
V1  Vs  I 1 Z s
V2  - I 2 Z L  g 21 V1  g 22 I 2
0  g 21 V1  (g 22  Z L ) I 2
- (g 22  Z L )
I2
V1 
g 21
Substituting this into (1),
(g g  Z L g 11  g 21 g 12 )
I 1  22 11
I2
- g 21
I2
- g 21
or

I 1 g 11 Z L   g
Also,
V2  g 21 (Vs  I 1 Z s )  g 22 I 2
 g 21 Vs  g 21 Z s I 1  g 22 I 2
 g 21 Vs  Z s (g 11 Z L   g ) I 2  g 22 I 2
But
I2 
- V2
ZL
V 
V2  g 21 Vs  [ g 11 Z s Z L   g Z s  g 22 ] 2 
 ZL 
V2 [ Z L  g 11 Z s Z L   g Z s  g 22 ]
 g 21 Vs
ZL
V2
g 21 Z L

Vs Z L  g 11 Z s Z L   g Z s  g 22
V2
g 21 Z L

Vs Z L  g 11 Z s Z L  g 11 g 22 Z s  g 21 g 12 Z s  g 22
V2
g 21 Z L

Vs (1  g 11 Z s )(g 22  Z L )  g 12 g 21 Z s
(1)
(2)
Chapter 19, Solution 42.
With the help of Fig. 19.20, we obtain the circuit model below.
I1
600 
I2
+
V1
–
+
10-3 V 2
+
–
120I 1
500 k
V2
–
Chapter 19, Solution 43.
(a)
To find A and C , consider the network in Fig. (a).
Z
I1
I2
+
V1
+

V2

(a)
V1  V2

 A 
 C 
I1  0 
V1
1
V2
I1
0
V2
To get B and D , consider the circuit in Fig. (b).
Z
I1
I2
+
V1
+

V2 = 0

(b)
V1  Z I 1 ,
B
- V1 - Z I 1

Z
- I1
I2
D
- I1
1
I2
I 2  - I1
Hence,
1 Z
[T]  

0 1 
(b)
To find A and C , consider the circuit in Fig. (c).
I1
I2
+
V1
+

Z
V2

(c)
V1  V2

 A 
V1  Z I 1  V2
V1
1
V2

 C 
I1
1
 Y
V2 Z
To get B and D , refer to the circuit in Fig.(d).
I2
+
I1
+
Y
V1
V2 = 0


(d)
V1  V2  0
B
- V1
 0,
I2
I 2  - I1
D
- I1
1
I2
Thus,
 1 0
[T]  

Y 1
Chapter 19, Solution 44.
Using Fig. 19.99, design a problem to help other students to better understand how to find the
transmission parameters of an ac circuit.
Although there are many ways to solve this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
Determine the transmission parameters of the circuit in Fig.19.99.
Figure 19.99
Solution
To determine A and C , consider the circuit in Fig.(a).
j15 
Io
I1
-j10 
-j20 
Io'
V1
+

I2 = 0
Io
20 
+
V2

(a)
V1  [ 20  (- j10) || ( j15  j20) ] I 1


10 
(-j10)(-j5) 
V1   20 
I 1   20  j  I 1


3
- j15 

I o  I1
'
 - j10 
2
 I 1    I 1
I o  
3
 - j10  j5 
V2  (-j20) I o  20 I o ' =  j
V1 (20  j10 3) I 1
 0.7692 + j0.3461

40 
V2

 20  j I1
3 

A
C
40 
40

I1  20I1 =  20  j I1
3 
3

I1

V2
1
40
20  j
3
 0.03461 + j0.023
To find B and D , consider the circuit in Fig. (b).
j15 
I1
-j10 
-j20 
I2
+
V1
+

20 
V2 = 0

(b)
We may transform the  subnetwork to a T as shown in Fig. (c).
( j15)(-j10)
 j10
j15  j10  j20
40
(-j10)(-j20)
Z2 
 -j
3
- j15
Z1 
Z3 
( j15)(-j20)
 j20
- j15
I1
j10 
j20 
I2
+
V1
+

20 – j40/3 
V2 = 0

(c)
- I2 
D
20  j40 3
3  j2
I1 
I
20  j40 3  j20
3 j 1
- I1
3 j

 0.5385  j0.6923
I2
3  j2

( j20)(20  j40 3) 
V1   j10 
I
20  j40 3  j20  1

V1  [ j10  2 (9  j7) ] I 1  j I 1 (24  j18)
- V1 - j I 1 (24  j18) 6

 (-15  j55)
- (3 - j2)
I2
13
I1
3 j
B  -6.923  j25.385 
B
(-6.923  j25.38)  
 0.7692  j0.3461
[T]  

 (0.03461  j0.023) S 0.5385  j0.6923 
Chapter 19, Solution 45.
To determine A and C consider the circuit below.
I1
-j2 
I 2 =0
+
V1
4
+
_
V2
–
V1  (4  j 2)I1, V2  4I1
V 4  j2
A 1 
 1 j 0.5
V2
4
C
I1
I
 1  0.25
V2 4I1
To determine B and D, consider the circuit below.
I1
-j2 
I2
+
V1
4
V 2 =0
+
_
–
The 4- resistor is short-circuited. Hence,
I
I2  I1, D   1  1
I2
V1   j 2I1  j 2I2
B
V1
j 2I
  2  2 j
I2
I2
Hence,
1  j0.5  j2  
[T] = 
1 
 0.25 S
Chapter 19, Solution 46.
To get A and C , refer to the circuit in Fig.(a).
1
I1
V1
I2 = 0
2
Ix
+
+

1
1
+
2
Vo
4 Ix
V2


(a)
At node 1,
I1 
Vo Vo  V2

2
1

 2 I 1  3 Vo  2 V2
(1)
At node 2,
Vo  V2
4 Vo
 4Ix 
 2 Vo
1
2

 Vo  -V2
(2)
From (1) and (2),
2 I 1  -5 V2
But
I1 

 C 
V1  Vo
 V1  V2
1
- 2.5 V2  V1  V2
A
I1 - 5

 -2.5 S
V2
2

 V1  -3.5 V2
V1
 -3.5
V2
To get B and D , consider the circuit in Fig. (b).
I1
1
+
V1
+

1
1
Vo
I2
2
Ix
2
+
4 Ix
V2 = 0


(b)
At node 1,
I1 
Vo Vo

2
1

 2 I 1  3 Vo
(3)
At node 2,
Vo
 4Ix  0
1
– I 2  Vo  2 Vo  0 
 I 2  -3 Vo
I2 
2 I1  I 2  0 
 I 1  -0.5 I 2
But
D
- I1
 0.5
I2
I1 
V1  Vo
1

 V1  I 1  Vo
Substituting (5) and (4) into (6),
-1
-1
-5
V1  I 2  I 2 
I
2
3
6 2
B
- V1 5
  0.8333 
I2
6
Thus,
 - 3.5 0.8333  
[T]  
- 0.5 
 - 2.5 S
(4)
(5)
(6)
Chapter 19, Solution 47.
To get A and C, consider the circuit below.
6
I1
1
+
4
+
Vx
V1
-
V1  Vx Vx Vx  5Vx


1
2
10
2
-


V2  4(0.4Vx )  5Vx  3.4Vx
V  Vx
I1  1
 1.1Vx  Vx  0.1Vx
1
I 2 =0
+
5V x
+
V2
-
-
V1  1.1Vx



A
V1
 1.1 / 3.4  0.3235
V2
I
C  1  0.1 / 3.4  0.02941
V2
To get B and D, consider the circuit below.
6
1
I1
4
I2
0V
+
+
Vx
V1
-
-
V1  Vx Vx Vx


1
6
2
I2  
2


+
5V x
-
V1 
10
Vx
6
5Vx Vx
17

  Vx
4
6
12
-
(1)
(2)
V1  I1  Vx
(3)
From (1) and (3)
I1  V1  Vx 
+
V 2 =0
4
Vx
6

I
4 12
D   1  ( )  0.4706
I 2 6 17
V 10 12
B   1  ( )  1.176
I2
6 17
S
Ω
 0.3235 1.176  
[T]  

0.02941 S 0.4706 
Chapter 19, Solution 48.
(a)
Refer to the circuit below.
I1
I2
+
V1
+

[T]
V2
ZL

V1  4 V2  30 I 2
I 1  0.1 V2  I 2
(1)
(2)
When the output terminals are shorted, V2  0 .
So, (1) and (2) become
V1  -30I 2
and
I1  - I 2
Hence,
Z in 
(b)
V1
 30 
I1
When the output terminals are open-circuited, I 2  0 .
So, (1) and (2) become
V1  4 V2
I 1  0.1 V2
or
V2  10 I 1
V1  40 I 1
Z in 
(c)
V1
 40 
I1
When the output port is terminated by a 10- load, V2  -10 I 2 .
So, (1) and (2) become
V1  -40 I 2  30 I 2  -70 I 2
I 1  - I 2  I 2  -2 I 2
V1  35 I 1
Z in 
V1
 35 
I1
Alternatively, we may use Z in 
A ZL  B
CZL  D
Chapter 19, Solution 49.
To get A and C , refer to the circuit in Fig.(a).
1/s
I1
I2 = 0
+
V1
+

1
1/s
1/s
1
V2

(a)
1 ||
1s
1
1


s 11 s s 1
V2 
1 || 1 s
V
1 s  1 || 1 s 1
1
V2
s
s 1
A


1
V1 1
2s  1

s s 1
 1   2s  1 
 1  1
1 

 || 
  I1
 ||  
V1  I 1 
 s  1   s (s  1) 
 s  1  s s  1
 1   2s  1 



V1  s  1   s (s  1) 
2s  1


1
2s  1
I1
(s  1)(3s  1)

s  1 s (s  1)
But
Hence,
2s  1
s
V2 2s  1
2s  1


I1
s
(s  1)(3s  1)
V1  V2 
C
V2 (s  1)(3s  1)

s
I1
To get B and D , consider the circuit in Fig. (b).
1/s
I1
I2
+
V1
+

1
1/s
1
1/s
V2 = 0

(b)
I
 1 1
 1
V1  I 1 1 || ||   I 1 1 ||   1
 s s
 2s  2s  1
-1
I
-s
s 1 1
I2 

I
1
1 2s  1 1

s 1 s
D
- I 1 2s  1
1

 2
s
I2
s
I
 1  2s  1 
I2  2

V1  
 2s  1  - s 
-s

 B 
- V1 1

s
I2

2

2s  1
[T]  
(s  1)( 3s  1)

s


1
2 
s
Thus,
1
s
Chapter 19, Solution 50.
To get a and c, consider the circuit below.
2
I 1 =0
s
I2
+
+
V1
4/s
V2
-
V1 
-
4/s
4
V2 
V2
s  4/s
s2  4
a  V2


V1
 1  0.25s 2
V2  (s  4 / s)I 2 or
I2 
V2
(1  0.25s 2 )V1

s  4/s
s  4/s


I
s  0.25s3
c 2 
V1
s2  4
To get b and d, consider the circuit below.
I1
2
s
I2
+
+
V 1 =0
4/s
V2
-
I1 
 4/s
2I
I2   2
2  4/s
s2
-


I
d   2  1  0.5s
I1
4
(s 2  2s  4)
V2  (s  2 // )I2 
I2
s
s2

(s 2  2s  4)( s  2)
I1
s2
2


V
b   2  0.5s 2  s  2
I1
 0.25s 2  1 0.5s 2  s  2

[t ]   0.25s 2  s
 2
0.5s  1 

 s  4
Chapter 19, Solution 51.
To get a and c , consider the circuit in Fig. (a).
j
I1 = 0
1
-j3 
I2
+
j2 
V1
+

j
V2

(a)
V2  I 2 ( j  j3)  -j2 I 2
V1  -jI 2
a
V2 - j2 I 2

2
V1
- jI 2
c
I2
1

j
V1 - j
To get b and d , consider the circuit in Fig. (b).
j
I1
1
-j3 
I2
+
V1 = 0
j2 
j

(b)
For mesh 1,
or
0  (1  j2) I1  j I 2
I 2 1  j2

 2 j
I1
j
d
- I2
 -2  j
I1
+

V2
For mesh 2,
V2  I 2 ( j  j3)  j I 1
V2  I 1 (2  j)(- j2)  j I 1  (-2  j5) I 1
b
- V2
 2  j5
I1
Thus,
 2 2  j5 
[t ]  

 j -2 j
Chapter 19, Solution 52.
It is easy to find the z parameters and then transform these to h parameters and T
parameters.
 R1  R 2
[z ]  
 R2
R2 
R 2  R 3 
 z  (R 1  R 2 )(R 2  R 3 )  R 22
 R 1R 2  R 2 R 3  R 3 R 1
(a)
 z
z
[h]   22
-z
 21
 z 22
z 12   R 1 R 2  R 2 R 3  R 3 R 1
z 22  
R2  R3

- R2
1
 
z 22  
R2  R3
R2 
R2  R3 

1

R2  R3 
Thus,
h 11  R 1 
R 2R 3
,
R2  R3
h 12 
R2
 - h 21 ,
R2  R3
h 22 
1
R2  R3
as required.
(b)
 z 11
z
[T]   21
1

 z 21
 z   R1  R 2
z 21   R 2

z 22
1
 
z 21   R 2
R 1R 2  R 2 R 3  R 3 R 1 

R2

R2  R3

R2

Hence,
A  1
as required.
R3
R1
R1
1
, B  R3 
(R 2  R 3 ) , C 
, D  1
R2
R2
R2
R2
Chapter 19, Solution 53.
For the z parameters,
V1  z11 I1  z12 I 2
V2  z12 I1  z 22 I 2
(1)
(2)
For ABCD parameters,
V1  A V2  B I 2
I1  C V2  D I 2
(3)
(4)
From (4),
V2 
I1 D
 I
C C 2
Comparing (2) and (5),
1
z 21  ,
C
(5)
z 22 
D
C
Substituting (5) into (3),

A
 B I 2
V1  I1  
 C

C
A
 I1 
I2
C
C
Comparing (6) and (1),
A
z11 
C
(6)
z 12 

C
C
Thus,
A

[Z] =  C
1

C
T 
C
D

C
Chapter 19, Solution 54.
For the y parameters
I 1  y 11 V1  y 12 V2
I 2  y 21 V1  y 22 V2
(1)
(2)
From (2),
I 2 y 22

V
y 21 y 21 2
-y
1
V1  22 V2 
I
y 12
y 21 2
V1 
or
(3)
Substituting (3) into (1) gives
- y 11 y 22
y 11
I
V2  y 12 V2 
I1 
y 21 2
y 21
- y
y 11
or
I1 
V2 
I
y 21
y 21 2
(4)
Comparing (3) and (4) with the following equations
V1  A V2  B I 2
I 1  C V2  D I 2
clearly shows that
A
as required.
- y 22
,
y 21
B
-1
,
y 21
C
- y
y 21
,
D
- y 11
y 21
Chapter 19, Solution 55.
For the z parameters
V1  z11 I1  z12 I 2
V2  z 21 I1  z 22 I 2
(1)
(2)
From (1),
I1 
z
1
V1  12 I 2
z11
z11
(3)
Substituting (3) into (2) gives

z
z z 
V2  21 V1   z 22  21 12  I 2
z11
z11 

z

V2  21 V1  z I 2
or
z11
z11
(4)
Comparing (3) and (4) with the following equations
I1  g11 V1  g12 I 2
V2  g 21 V1  g 22 I 2
indicates that
g 11 
as required.
1
,
z 11
g 12 
- z 12
,
z 11
g 21 
z 21
,
z 11
g 22 
z
z 11
Chapter 19, Solution 56.
Using Fig. 19.20, we obtain the equivalent circuit as shown below.
I1
Rs
h 11
I2
+
Vs
V
+1
_
V1
+
h 12 V o +
–
h 21 I 1
h 22
–
Vo
RL
–
We can solve this using MATLAB. First, we generate 4 equations from the given circuit.
It may help to let V s = 10 V.
–10 + R s I 1 + V 1 = 0 or V 1 + 1000I 1 = 10
–10 + R s I 1 + h 11 I 1 + h 12 V o = 0 or 0.0001V s + 1500 = 10
I 2 = –V o /R L or V o + 2000I 2 = 0
h 21 I 1 + h 22 V o – I 2 = 0 or 2x10–6V o + 100I 1 – I 2 = 0
&gt;&gt; A=[1,0,1000,0;0,0.0001,1500,0;0,1,0,2000;0,(2*10^-6),100,-1]
A=
1.0e+003 *
0.0010
0 1.0000
0
0 0.0000 1.5000
0
0 0.0010
0 2.0000
0 0.0000 0.1000 -0.0010
&gt;&gt; U=[10;10;0;0]
U=
10
10
0
0
&gt;&gt; X=inv(A)*U
X=
1.0e+003 *
0.0032
-1.3459
0.0000
0.0007
Gain = V o /V s = –1,345.9/10 = –134.59.
There is a second approach we can take to check this problem. First, the resistive value
of h 22 is quite large, 500 kΩ versus R L so can be ignored. Working on the right side of
the circuit we obtain the following,
I 2 = 100I 1 which leads to V o = –I 2 x2k = –2x105I 1 .
Now the left hand loop equation becomes,
–V s + (1000 + 500 + 10–4(–2x105))I 1 = 1480I 1 .
Solving for V o /V s we get,
V o /V s = –200,000/1480 = –134.14.
Chapter 19, Solution 57.
 T  (3)(7)  (20)(1)  1
A

[z ]   C
1

C
T 
C   3 1
D   1 7 

C
D

[y ]   -B1

B
- T  
B  
A  
 
B  
B

[h]   -D1

D
1 
 T   20



D  7
7 

C
-1
1 
S
 
D  7
7 
C

[g ]   A
1

A
- T 
A 
B 

A 
D

[t ]   CT

 T
7
20
-1
20
1
3S
 1

3
-1
20
3
20


S

-1 
3 
20 


3
B 
 T   7 20  
A    1 S
3 

T 
Chapter 19, Solution 58.
Design a problem to help other students to better understand how to develop the y parameters
and transmission parameters, given equations in terms of the hybrid parameters.
Although there are many ways to solve this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
A two-port is described by
V 1 = I 1 + 2V 2 , I 2 = -2I 1 + 0.4V 2
Find: (a) the y parameters, (b) the transmission parameters.
Solution
The given set of equations is for the h parameters.
1 
2 
 h  (1)(0.4)  (2)(-2)  4.4
[h]  

 - 2 0.4 S
(a)
 1
h
[y ]   11
h
 21
 h11
(b)


[T ]  


- h
h 21
- h 22
h 21
- h 12
h 11
h
h 11

  1 -2 

S
  - 2 4.4 

- h11
h 21
-1
h 21

  2.2 0.5  
 

  0.2 S 0.5 

Chapter 19, Solution 59.
 g  (0.06)(2)  (-0.4)(0.2)  0.12  0.08  0.2
(a)


[z]  


1
g 11
g 21
g 11
g
(b)


[y ]  


(c)
 g 22
 
g
[h]   - g
 21
  g
- g 12 
2 
 g   10 

g 11   - 1 0.3 S 
 g 
(d)


[T]  


g 22 
10  
g 21   5
 
g
1 
  0.3 S
g 21 
g 22
- g 21
g 22
1
g 21
g 11
g 21
- g 12 
g 11   16.667 6.667 

 g   3.333 3.333  

g 11 
g 12
g 22
1
g 22

  0.1 - 0.2 

S
  - 0.1 0.5 

Chapter 19, Solution 60.
Comparing this with Fig. 19.5,
z11  z12  4  j 3  2  2  j3k
z 22 – z 12 = 5 – j – 2 = 3 – j kΩ
XL  3 x103   L

 L
3 x103
 3mH
106
X C = 1x103 = 1/(ωC) or C = 1/(103x106) = 1 nF
Hence, the resulting T network is shown below.
2 k
3 mH
3 k
2 k
1 nF
Chapter 19, Solution 61.
(a)
To obtain z 11 and z 21 , consider the circuit in Fig. (a).
1
Io
1
1
I2 = 0
+
I1
+
1
V1
V2


(a)
 2 5
V1  I 1 [1  1 || (1  1) ]  I 1 1    I 1
 3 3
V
5
z 11  1 
I1 3
Io 
1
1
I1  I1
1 2
3
- V2  I o  I 1  0
1
4
V2  I 1  I 1  I 1
3
3
z 21 
V2 4

3
I1
To obtain z 22 and z 12 , consider the circuit in Fig. (b).
1
I1
1
1
+
V1
+
1

V2

(b)
I2
Due to symmetry, this is similar to the circuit in Fig. (a).
5
4
z 22  z 11  ,
z 21  z 12 
3
3


[z ]  


(b)


[h]  



z
z 22
- z 21
z 22
(c)


[T]  



z 11
z 21
1
z 21
z 12
z 22
1
z 22
z
z 21
z 22
z 21
 
 

 
 

 
 

 
 

3
4 

5
5 
-4 3 
S
5
5 
5
4
3
S
4
3 

4 
5 

4 
5
3
4
3
4
3

5
3
Chapter 19, Solution 62.
Consider the circuit shown below.
I1
10 k
a
40 k
+

+
I2
+
50 k
b
30 k
V1
Ib
20 k

Since no current enters the input terminals of the op amp,
V1  (10  30)  10 3 I 1
But
Va  Vb 
V2

(1)
30
3
V1  V1
40
4
Vb
3
V
3 
20  10
80  10 3 1
which is the same current that flows through the 50-k resistor.
Ib 
Thus,
V2  40  10 3 I 2  (50  20)  10 3 I b
3
V2  40  10 3 I 2  70  10 3 
V
80  10 3 1
21
V2  V1  40  10 3 I 2
8
V2  105  10 3 I 1  40  10 3 I 2
From (1) and (2),
 40 0 
[z ]  
 k
 105 40 
 z  z 11 z 22  z 12 z 21  16  10 8
(2)

A B 
[T]  

C D 

z 11
z 21
1
z 21
z
z 21
z 22
z 21

  0.381 15.24 k 

0.381 
  9.52 S

Chapter 19, Solution 63.
To get z 11 and z 21 , consider the circuit below.
I1
1:3
I 2 =0

+
4
V1
 +
+
V’ 1
V’ 2
-
-
+
9
-
V2
-
ZR 
9
n2
 1,
V1  (4 // ZR )I1 
n  3
4
I1
5


V2  V2 '  nV1'  nV1  3(4 / 5)I1
V
z11  1  0.8
I1


z 21 
V2
 2.4
I1
To get z 21 and z 22 , consider the circuit below.
I 1 =0
1:3

+
V1
4
I2
 +
+
V’ 1
V’ 2
-
-
+
9
-
-
Z R '  n 2 ( 4 )  36 ,
V2  (9 // ZR ' )I 2 
V2
9x36
I2
45
n 3

z 22 
V2
 7.2
I2
V1 
V2 V2

 2.4I 2
n
3

V
z 21  1  2.4
I2
Thus,
 0.8 2.4
[z ]  

 2.4 7.2
Chapter 19, Solution 64.
1
-j

 - j k
3
jC (10 )(10 -6 )
1 F 

Consider the op amp circuit below.
40 k
I1
20 k
Vx
10 k
1
+

+
2
I2
+
-j k
V1
V2


At node 1,
V1  Vx Vx Vx  0


20
-j
10
V1  (3  j20) Vx
(1)
At node 2,
Vx  0 0  V2

10
40
But
I1 

 Vx 
-1
V
4 2
V1  Vx
20  10 3
Substituting (2) into (3) gives
V1  0.25 V2
I1 
 50  10 -6 V1  12.5  10 -6 V2
3
20  10
Substituting (2) into (1) yields
-1
V1  (3  j20) V2
4
0  V1  (0.75  j5) V2
or
Comparing (4) and (5) with the following equations
I 1  y 11 V1  y 12 V2
I 2  y 21 V1  y 22 V2
(2)
(3)
(4)
(5)
indicates that I 2  0 and that
 50  10 -6
[y ]  
1

12.5  10 -6 
S
0.75  j5 
 y  (77.5  j25.  12.5)  10 -6  (65  j250)  10 -6


[h]  


1
y 11
y 21
y 11
- y 12 
- 0.25 
y 11   2  10 4 


4
y
1.3  j5 S 
  2  10
y 11 
Chapter 19, Solution 65.
The network consists of two two-ports in series. It is better to work with z parameters
and then convert to y parameters. It is obvious that the upper 1 Ω resistor is shorted out
by the top circuit so we are essentially left with 2 Ω connected to 3 Ω. This then
produces the Z parameters
 5 3 
[z ]  

 3 3 
 z  15  9  6
 z 22
 
[y ]   z
-z
 21
 z
- z 12 
z 

z 11

z 
 0.5  0.5 
5 S

 - 0.5
6 
Chapter 19, Solution 66.
Since we have two two-ports in series, it is better to convert the given y parameters to z
parameters.
 y  y 11 y 22  y 12 y 21  (2  10 -3 )(10  10 -3 )  0  20  10 -6


[z a ]  


y 22
y
- y 21
y
- y 12
y
y 11
y

  500 
0 

100  
  0

 500 0   100 100   600 100 
[z ]  



 0 100   100 100   100 200 
i.e.
V1  z 11 I 1  z 12 I 2
V2  z 21 I 1  z 22 I 2
or
V1  600 I 1  100 I 2
V2  100 I 1  200 I 2
(1)
(2)
But, at the input port,
Vs  V1  60 I 1
(3)
and at the output port,
V2  Vo  -300I 2
(4)
From (2) and (4),
100 I 1  200 I 2  -300 I 2
I 1  -5 I 2
(5)
Substituting (1) and (5) into (3),
Vs  600 I 1  100 I 2  60 I 1
 (660)(-5) I 2  100 I 2
 -3200 I 2
(6)
From (4) and (6),
Vo
- 300 I 2

 0.09375
V2 - 3200 I 2
Chapter 19, Solution 67.
We first the y parameters, To find y 11 and y 21 consider the circuit below.
30 
I1
40  I 2
+
10 
1A
V 2 =0
V1
-
V1  I1(30  10 // 40)  38I1


y11 
I1
1

V1 38
By current division,
10
I
0.2I1
1
I2 
I1  0.2I

 y21  2 

50
V1
38I1
190
1
To find y 22 and y 12 consider the circuit below.
I1
30 
40 
I2
+
+
10 
V 1 =0
-
V2
1A
-
V2  (40  10 // 30)I2  47.5I2


y22 
I2
2

y 22 = 2/95
V2 93
By current division,
I1  
10
I
I2   2
30  10
4


1
 I2
I
1
y12  1  4  
V2 47.5I2
190
1/190 
 1/ 38
[y ]  

 1/190 2 / 95 
For three copies cascaded in parallel, we can use MATLAB.
&gt;&gt; Y=[1/38,-1/190;-1/190,2/95]
Y=
0.0263 -0.0053
-0.0053 0.0211
&gt;&gt; Y3=3*Y
Y3 =
0.0789 -0.0158
-0.0158 0.0632
&gt;&gt; DY=0.0789*0.0632-0.0158*0.158
DY =
0.0025
&gt;&gt; T=[0.0632/0.0158,1/0.0158;DY/0.0158,0.0789/0.0158]
T=
4.0000 63.2911
0.1576 4.9937
63.29  
 4
T= 

0.1576 S 4.994 
Chapter 19, Solution 68.
 4 -2
For the upper network N a , [y a ]  

-2 4 
 2 -1 
and for the lower network N b , [y b ]  

1 2 
For the overall network,
 6 -3
[y ]  [y a ]  [y b ]  

 -3 6 
 y  36  9  27


[h]  


1
y 11
y 21
y 11
- y 12   1
y 11   6 

y   1
 
y 11   2
1 
2 
9 
S
2 
Chapter 19, Solution 69.
We first determine the y parameters for the upper network N a .
To get y 11 and y 21 , consider the circuit in Fig. (a).
n
1
,
2
ZR 
1s 4

n2 s
 2s  4 

4
I
V1  (2  Z R ) I 1   2   I 1  
 s  1

s
I
s
y 11  1 
V1 2 (s  2)
- I1
- s V1
 -2 I 1 
n
s2
I2
-s


V1 s  2
I2 
y 21
To get y 22 and y 12 , consider the circuit in Fig. (b).
I1
2
2:1
1/s
I2
+
+
V 1 =0
V2


(b)
1
1
Z R '  (n 2 )(2)    (2) 
4
2
1

 1 1
s  2
I
V2    Z R '  I 2     I 2  
s

 s 2
 2s  2
y 22 
I2
2s

V2 s  2
 - 1  2s 
 -s 
 V2  
V
I 1  - n I 2   
 2  s  2 
s  2 2
I2
y 12 
I1
-s

V2 s  2

s
 2 (s  2)
[y a ]  
-s

 s2
-s
s2
2s
s2





For the lower network N b , we obtain y 11 and y 21 by referring to the network in Fig. (c).
2
I1
I2
+
+

V1
s
V2 = 0

(c)
V1  2 I 1

 y 11 
I 2  - I1 
- V1
2
I1 1

V1 2
I 2 -1

V1 2

 y 21 
To get y 22 and y 12 , refer to the circuit in Fig. (d).
I1
2
I2
+
+
s
V1 = 0

V2
I2

(d)
V2  (s || 2) I 2 
I1  - I 2 
2s
I
s2 2

 y 22 
I2 s  2

V2
2s
- V2
 - s  s  2 
-s

 V2 

s  2  s  2  2s 
2
y 12 
I1 - 1

V2
2
12
-1 2 
[y b ]  

 - 1 2 (s  2) 2s 
 s1
 s2
[y ]  [y a ]  [y b ]  
 - (3s  2)
 2 (s  2)
- (3s  2)
2 (s  2)
5s 2  4s  4
2s (s  2)





Chapter 19, Solution 70.
We may obtain the g parameters from the given z parameters.
 25 20 
 z a  250  100  150
[z a ]  
,
 5 10 
 50 25 
[z b ]  
,
 25 30 


[g ]  


1
z 11
z 21
z 11
- z 12
z 11
z
z 11
 z b  1500  625  875





 0.04 - 0.8 
[g a ]  
,
6 
 0.2
 0.02 - 0.5 
[g b ]  

 0.5 17.5 
 0.06 S - 1.3 
[g ]  [g a ]  [ g b ]  
23.5  
 0.7
Chapter 19, Solution 71.
This is a parallel-series connection of two two-ports. We need to add their g parameters
together and obtain z parameters from there.
For the transformer,
V1 
1
V2 , I1  2I 2
2
Comparing this with
V1  AV2  BI2 ,
I1  CV2  DI 2
shows that
0.5 0
[Tb1]  

 0 2
To get A and C for T b2 , consider the circuit below.
I1
4
+
5
V1
-
V1  9I1,
A
I 2 =0
2
V2  5I1
V1
 9 / 5  1.8,
V2
I
C  1  1 / 5  0.2
V2
+
V2
-
We obtain B and D by looking at the circuit below.
4
I1
I 2 =0
+
V1
-
5
I 2   I1
7
+
5
2


V 2 =0
-
I
D   1  7 / 5  1.4
I2
7
38
V1  4I1  2I 2  4( I 2 )  2I 2   I 2
5
5


1.8 7.6
[Tb 2 ]  

0.2 1.4 
0.9 3.8
[T ]  [Tb1][Tb 2 ]  
,
0.4 2.8
T  1
C / A   T / A  0.4444  1.1111

[g b ]  
B / A  1.1111 4.2222 
 1/ A
From Prob. 19.52,
1.8 18.8
[Ta ]  

0.1 1.6 
C / A   T / A  0.05555  0.5555

[g a ]  
B / A   0.5555 10.4444 
 1/ A
0.4999  1.6667
[g ]  [ g a ]  [ g b ]  

1.6667 14.667 
Thus,
V
B   1  7.6
I2
 1 / g 11
[z ]  
g 21 / g 11
 g 21 / g 11   2
 3.334



 g / g 11   3.334 20.22 
Chapter 19, Solution 72.
Consider the network shown below.
I1
I a1
+
V a1
+
V1
I a2
Na
I b1
+
V b1

I2
+
V a2
+
V2
I b2
Nb
Va1  25 I a1  4 Va 2
I a 2  - 4 I a1  Va 2
Vb1  16 I b1  Vb 2
I b 2  - I b1  0.5 Vb 2
+
V b2

(1)
(2)
(3)
(4)
V1  Va1  Vb1
V2  Va 2  Vb 2
I 2  I a 2  I b2
I 1  I a1
Now, rewrite (1) to (4) in terms of I 1 and V2
Va1  25 I 1  4 V2
I a 2  - 4 I 1  V2
Vb1  16 I b1  V2
I b 2  - I b1  0.5 V2
(5)
(6)
(7)
(8)
V1  25 I 1  16 I b1  5 V2
(9)
I 2  - 4 I 1  I b1  1.5 V2
(10)
I b1  I a1  I 1
(11)
Because the two networks N a and N b are independent,
I 2  - 5 I 1  1.5 V2
or
V2  3.333 I 1  0.6667 I 2
(12)
Substituting (11) and (12) into (9),
25
5
V1  41I 1 
I1 
I
1.5
1.5 2
V1  57.67 I 1  3.333 I 2
(13)
Comparing (12) and (13) with the following equations
V1  z 11 I 1  z 12 I 2
V2  z 21 I 1  z 22 I 2
indicates that
 57.67 3.333 
[z ]  

 3.333 0.6667 
Alternatively,
 25 4 
[h a ]  
,
-4 1
 16 1 
[h b ]  

 - 1 0.5 
 41 5 
[h]  [h a ]  [h b ]  

 - 5 1.5 


[z ]  


as obtained previously.
h
h 22
- h 21
h 22
 h  61.5  25  86.5
h12
h 22
1
h 22

  57.67 3.333 


3
.
333
0
.
6667




Chapter 19, Solution 73.
From Problem 19.6,
 25 20 
[ z]  
z  25 x30  20 x24  270
,
24 30 
A
C
z11 25

,
z21 24
1
z 21

1
,D
24
B
z 270

z21 24
z 22 30

z 21 24
The overall ABCD parameters can be found using MATLAB.
&gt;&gt; T=[25/24,270/24;1/24,30/24]
T=
1.0417 11.2500
0.0417 1.2500
&gt;&gt; T3=T*T*T
T3 =
2.6928 49.7070
0.1841 3.6133
&gt;&gt; Z=[2.693/0.1841,(2.693*3.613-0.1841*49.71)/0.1841;1/0.1841,3.613/0.1841]
Z=
14.6279 3.1407
5.4318 19.6252
14.628 3.141 
[Z] = 

 5.432 19.625
Chapter 19, Solution 74.
From Prob. 18.35, the transmission parameters for the circuit in Figs. (a) and (b) are
1 Z
[Ta ]  
,
0 1 
 1 0
[Tb ]  

1 Z 1 
Z
Z
(a)
(b)
We partition the given circuit into six subcircuits similar to those in Figs. (a) and (b) as
shown in Fig. (c) and obtain [T] for each.
s
s
1
1/s
T1
T2
T3
1
T4
T5
1/s
T6
1 0 
[T1 ]  
,
1 1 
1 s
[T2 ]  
,
 0 1
1 0 
[T3 ]  

s 1
[T4 ]  [T2 ] ,
[T5 ]  [T1 ] ,
[T6 ]  [T3 ]
 1 0  1 0 
[T]  [T1 ][T2 ][T3 ][T4 ][T5 ][T6 ]  [T1 ][T2 ][T3 ][T4 ]


 1 1  s 1 
0
0
 1
1 s 1
T
T
T
 [T1 ][T2 ][T3 ][T4 ] 

[
]
[
]
[
]
1
2
3

 0 1   s 1 1 
 s 1 1 



 1 0   s2  s 1 s 
 [T1 ][T2 ] 


1
 s 1   s 1
s 
 1 s   s2  s 1
 [T1 ] 
 3


2
2
 0 1   s  s  2s  1 s  1 
 1 0   s 4  s 3  3s 2  2s  1 s 3  2s 



3
2
s2 1 
 1 1   s  s  2s  1
 s 4  s 3  3s 2  2s  1

s 3  2s
[T]   4

 s  2s 3  4s 2  4s  2 s 3  s 2  2s  1 
Note that AB  CD  1 as expected.
Chapter 19, Solution 75.
(a) We convert [z a ] and [z b ] to T-parameters. For N a ,  z  40  24  16 .
 z / z 21   2
4 
z / z

[Ta ]   11 21


 1 / z 21 z 22 / z 21  0.25 1.25
For N b ,  y  80  8  88 .
 y 22 / y 21  1 / y 21    5  0.5
[Tb ]  


   y / y 21  y11 / y 21   44  4 
 17 
  186
[T]  [Ta ][Tb ]  

 56.25  5.125
We convert this to y-parameters.  T  AD  BC  3.
 D / B   T / B  0.3015  0.1765
[y ]  

S
A / B  0.0588
10.94 
 1 / B
(b)
The equivalent z-parameters are
A / C  T / C  3.3067 0.0533
[z]  


 1 / C D / C   0.0178 0.0911
Consider the equivalent circuit below.
I1
z 11
z 22
+
I2
+
+
+
Vi
z 12 I 2
ZL
z 21 I 1
-
Vo
-
-
-
Vi  z11I1  z12 I 2
(1)
Vo  z 21I1  z 22 I 2
But Vo  I 2 ZL

(2)
I 2  Vo / ZL
(3)
From (2) and (3) ,
V
Vo  z 21I1  z 22 o
ZL

 1

z
 22 
I1  Vo 
 z 21 ZL z 21 
(4)
Substituting (3) and (4) into (1) gives
Vi  z 11 z 11 z 22


Vo  z 21 z 21 Z L
 z 12
 
 194.3
Z
L



Vo .
 0.0051
Vi
Chapter 19, Solution 76.
To get z 11 and z 21 , we open circuit the output port and let I 1 = 1A so that
V
V
z11  1  V1, z 21  2  V2
I1
I1
The schematic is shown below. After it is saved and run, we obtain
z11  V1  3.849,
z 21  V2  1.122
Similarly, to get z 22 and z 12 , we open circuit the input port and let I 2 = 1A so that
V
z12  1  V1,
I2
z 22 
V2
 V2
I2
The schematic is shown below. After it is saved and run, we obtain
z12  V1  1.122,
z 22  V2  3.849
Thus,
 3.949 1.122 
[z ]  

1.122 3.849
Chapter 19, Solution 77.
We follow Example 19.15 except that this is an AC circuit.
(a)
We set V 2 = 0 and I 1 = 1 A. The schematic is shown below. In the AC Sweep
Box, set Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. After simulation,
the output file includes
FREQ
IM(V_PRINT2)
IP(V_PRINT2)
1.592 E–01
3.163 E–.01
–1.616 E+02
FREQ
VM(\$N_0001)
VP(\$N_0001)
1.592 E–01
9.488 E–01
–1.616 E+02
From this we obtain
h 11 = V 1 /1 = 0.9488–161.6
h 21 = I 2 /1 = 0.3163–161.6.
(b)
In this case, we set I 1 = 0 and V 2 = 1V. The schematic is shown below. In the
AC Sweep box, we set Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592.
After simulation, we obtain an output file which includes
FREQ
VM(\$N_0001)
VP(\$N_0001)
1.592 E–01
3.163 E–.01
1.842 E+01
FREQ
IM(V_PRINT2)
IP(V_PRINT2)
1.592 E–01
9.488 E–01
–1.616 E+02
From this,
h 12 = V 1 /1 = 0.316318.42
h 21 = I 2 /1 = 0.9488–161.6.
Thus,
0.316318.42 
0.9488  161.6 
[h] = 

 0.3163  161.6 0.9488  161.6 S 
Chapter 19, Solution 78
For h 11 and h 21 , short-circuit the output port and let I 1 = 1A. f   / 2  0.6366 . The
schematic is shown below. When it is saved and run, the output file contains the
following:
FREQ
IM(V_PRINT1)IP(V_PRINT1)
6.366E-01
FREQ
1.202E+00
1.463E+02
VM(\$N_0003) VP(\$N_0003)
6.366E-01
3.771E+00
-1.350E+02
From the output file, we obtain
I 2  1.202146.3o ,
V1  3.771  135o
so that
V
h11  1  3.771  135o ,
1
I
h 21  2  1.202146.3o
1
For h 12 and h 22 , open-circuit the input port and let V 2 = 1V. The schematic is shown
below. When it is saved and run, the output file includes:
FREQ
6.366E-01
VM(\$N_0003) VP(\$N_0003)
1.202E+00
-3.369E+01
FREQ
IM(V_PRINT1)IP(V_PRINT1)
6.366E-01
3.727E-01
-1.534E+02
From the output file, we obtain
I 2  0.3727  153.4o ,
V1  1.202  33.69o
so that
V
h12  1  1.202  33.69o ,
1
I
h 22  2  0.3727  153.4o
1
Thus,
 3.771  135 o 
1.202  33.69 o 
[h]  

0.3727  153.4 o S 
 1.202146.3
Chapter 19, Solution 79
(a)
We set I 1 = 1 A and open-circuit the output-port so that I 2 = 0. The schematic
is shown below with two VPRINT1s to measure V 1 and V 2 . In the AC Sweep box, we
enter Total Pts = 1, Start Freq = 0.3183, and End Freq = 0.3183. After simulation, the
output file includes
FREQ
VM(1)
VP(1)
3.183 E–01
4.669 E+00
–1.367 E+02
FREQ
VM(4)
VP(4)
3.183 E–01
2.530 E+00
–1.084 E+02
From this,
z 11 = V 1 /I 1 = 4.669–136.7/1 = 4.669–136.7
z 21 = V 2 /I 1 = 2.53–108.4/1 = 2.53–108.4.
(b)
In this case, we let I 2 = 1 A and open-circuit the input port. The schematic is
shown below. In the AC Sweep box, we type Total Pts = 1, Start Freq = 0.3183, and
End Freq = 0.3183. After simulation, the output file includes
FREQ
VM(1)
VP(1)
3.183 E–01
2.530 E+00
–1.084 E+02
FREQ
VM(2)
VP(2)
3.183 E–01
1.789 E+00
–1.534 E+02
From this,
z 12 = V 1 /I 2 = 2.53–108.4/1 = 2.53–108..4
z 22 = V 2 /I 2 = 1.789–153.4/1 = 1.789–153.4.
Thus,
4.669  136.7 2.53  108.4 
[z] = 
Ω
 2.53  108.4 1.789  153.4
Chapter 19, Solution 80
To get z 11 and z 21 , we open circuit the output port and let I 1 = 1A so that
V
z11  1  V1,
I1
z 21 
V2
 V2
I1
The schematic is shown below. After it is saved and run, we obtain
z11  V1  29.88,
z 21  V2  70.37
Similarly, to get z 22 and z 12 , we open circuit the input port and let I 2 = 1A so that
V
z12  1  V1,
I2
z 22 
V2
 V2
I2
The schematic is shown below. After it is saved and run, we obtain
z12  V1  3.704,
z 22  V2  11.11
Thus,
 29.88 3.704
[z ]  

  70.37 11.11
Chapter 19, Solution 81
(a)
We set V 1 = 1 and short circuit the output port. The schematic is shown below.
After simulation we obtain
y 11 = I 1 = 1.5, y 21 = I 2 = 3.5
(b)
We set V 2 = 1 and short-circuit the input port. The schematic is shown below.
Upon simulating the circuit, we obtain
y 12 = I 1 = –0.5, y 22 = I 2 = 1.5
1.5  0.5
[Y] = 
S
 3.5 1.5 
Chapter 19, Solution 82
(a)
Set V 2 = 0 and I 1 = 1A. The schematic is shown below. After simulation, we
obtain
h 11 = V 1 /1 = 3.8, h 21 = I 2 /1 = 3.6
(b)
Set V 1 = 1 V and I 1 = 0. The schematic is shown below. After simulation, we
obtain
h 12 = V 1 /1 = 0.4, h 22 = I 2 /1 = 0.25
Hence,
0 .4 
 3 .8 
[h] = 

 3.6 0.25 S 
Chapter 19, Solution 83
To get A and C, we open-circuit the output and let I 1 = 1A. The schematic is shown
below. When the circuit is saved and simulated, we obtain V 1 = 11 and V 2 = 34.
A
V1
 0.3235,
V2
I
1
C 1 
 0.02941
V2 34
Similarly, to get B and D, we open-circuit the output and let I 1 = 1A. The schematic
is shown below. When the circuit is saved and simulated, we obtain V 1 = 2.5 and I 2
= -2.125.
V
2.5
B 1 
 1.1765,
I2 2.125
I
1
D 1 
 0.4706
I 2 2.125
Thus,
 0.3235 1.1765  
[T]  

0.02941 S 0.4706 
Chapter 19, Solution 84
(a)
Since A =
V1
V2
and C =
I 2 0
I1
V2
, we open-circuit the output port and let V 1
I 2 0
= 1 V. The schematic is as shown below. After simulation, we obtain
A = 1/V 2 = 1/0.7143 = 1.4
C = I 2 /V 2 = 1.0/0.7143 = 1.4
(b)
To get B and D, we short-circuit the output port and let V 1 = 1. The schematic is
shown below. After simulating the circuit, we obtain
B = –V 1 /I 2 = –1/1.25 = –0.8
D = –I 1 /I 2 = –2.25/1.25 = –1.8
Thus
A B 
 1.4  0.8  
 C D = 1.4 S  1.8 




Chapter 19, Solution 85
(a)
Since A =
V1
V2
and C =
I 2 0
I1
V2
, we let V 1 = 1 V and
I 2 0
open-circuit the output port. The schematic is shown below. In the AC Sweep box, we
set Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. After simulation, we
obtain an output file which includes
FREQ
1.592 E–01
IM(V_PRINT1)
6.325 E–01
IP(V_PRINT1)
1.843 E+01
FREQ
1.592 E–01
VM(\$N_0002)
6.325 E–01
VP(\$N_0002)
–7.159 E+01
From this, we obtain
A =
1
1

 1.58171.59
V2 0.6325  71.59
C =
I1
0.632518.43

 190 = j
V2 0.6325  71.59
(b)
Similarly, since B =
V1
I2
and D = 
V2  0
I1
I2
, we let V 1 = 1 V and shortV2  0
circuit the output port. The schematic is shown below. Again, we set Total Pts = 1, Start
Freq = 0.1592, and End Freq = 0.1592 in the AC Sweep box. After simulation, we get
an output file which includes the following results:
FREQ
1.592 E–01
IM(V_PRINT1)
5.661 E–04
IP(V_PRINT1)
8.997 E+01
FREQ
1.592 E–01
IM(V_PRINT3)
9.997 E–01
IP(V_PRINT3)
–9.003 E+01
From this,
B = 
1
1

 190   j
I2
0.9997   90
D = 
I1
5.661x10 4 89.97

= 5.561x10–4
I2
0.9997  90
 j 
A B 
1.58171.59
 C D = 
jS
5.661x10 4 



Chapter 19, Solution 86
(a)
By definition, g 11 =
I1
V1
, g 21 =
I 2 0
V1
V2
.
I 2 0
We let V 1 = 1 V and open-circuit the output port. The schematic is shown below. After
simulation, we obtain
g 11 = I 1 = 2.7
g 21 = V 2 = 0.0
(b)
Similarly,
g 12 =
I1
I2
, g 22 =
V1  0
V2
I2
V1  0
We let I 2 = 1 A and short-circuit the input port. The schematic is shown below. After
simulation,
g 12 = I 1 = 0
g 22 = V 2 = 0
Thus
 2.727S 0
[g] = 
0
 0
Chapter 19, Solution 87
(a)
Since
a =
V2
V1
and c =
I1  0
I2
V1
,
I1  0
we open-circuit the input port and let V 2 = 1 V. The schematic is shown below. In the
AC Sweep box, set Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. After
simulation, we obtain an output file which includes
FREQ
1.592 E–01
IM(V_PRINT2)
5.000 E–01
IP(V_PRINT2)
1.800 E+02
FREQ
1.592 E–01
VM(\$N_0001)
5.664 E–04
VP(\$N_0001)
8.997 E+01
From this,
a =
c =
(b)
1
5.664 x10  4 89.97
 1765  89.97
0.5180
 882.28  89.97
5.664x10  4 89.97
Similarly,
b = 
V2
I1
and d = 
V1  0
I2
I1
V1  0
We short-circuit the input port and let V 2 = 1 V. The schematic is shown below. After
simulation, we obtain an output file which includes
FREQ
IM(V_PRINT2)
IP(V_PRINT2)
1.592 E–01
5.000 E–01
1.800 E+02
FREQ
1.592 E–01
IM(V_PRINT3)
5.664 E–04
IP(V_PRINT3)
–9.010 E+01
From this, we get
b = 
d = 
Thus
1
4
5.664x10   90.1
= –j1765
0.5180
= j888.28
5.664x10  4   90.1
  j1765  j1765  
[t] = 
j888.2 
 j888.2 S
Chapter 19, Solution 88
To get Z in , consider the network in Fig. (a).
Rs
I1
I2
+
Vs
+

+
Two-Port
V1
RL
V2


(a)
Z in
I 1  y 11 V1  y 12 V2
I 2  y 21 V1  y 22 V2
But
(1)
(2)
- V2
 y 21 V1  y 22 V2
RL
- y 21 V1
V2 
y 22  1 R L
I2 
(3)
Substituting (3) into (1) yields
 - y 21 V1 
,
I 1  y 11 V1  y 12  
 y 22  1 R L 
  y  y 11 YL 
 V1 ,
I1  
 y 22  YL 
1
RL
 y  y 11 y 22  y 12 y 21
Z in 
or
YL 
V1
y 22  YL

I 1  y  y 11 YL
 y   - y 21 V1 
I2
y V  y 22 V2

 21 1
 y 21 Z in   22  
I1
I1
 I 1   y 22  YL 
 y  YL 
y y Z
 y 21  y 22 y 21
 y 21 Z in  22 21 in   22


y 22  YL
y 22  YL
  y  y 11 YL 
Ai 
Ai 
From (3),
y 21 YL
 y  y 11 YL



Av 
V2
- y 21

V1 y 22  YL
To get Z out , consider the circuit in Fig. (b).
I1
I2
+
Rs
V1
+

Two-Port
V2

(b)
Z out 
But
Z out
V2
V2

I 2 y 21 V1  y 22 V2
(4)
V1  - R s I 1
Substituting this into (1) yields
I 1  - y 11 R s I 1  y 12 V2
(1  y 11 R s ) I 1  y 12 V2
y 12 V2
- V1
I1 

1  y 11 R s
Rs
- y 12 R s
V1

or
V2 1  y 11 R s
Substituting this into (4) gives
1
Z out 
y 12 y 21 R s
y 22 
1  y 11 R s
1  y 11 R s

y 22  y 11 y 22 R s  y 21 y 22 R s
Z out 
y 11  Ys
 y  y 22 Ys
Chapter 19, Solution 89
Av 
- h fe R L
h ie  (h ie h oe  h re h fe ) R L
- 72  10 5
Av 
2640  (2640  16  10 -6  2.6  10 -4  72)  10 5
- 72  10 5
Av 
 - 1613
2640  1824
dc gain  20 log A v  20 log (1613)  64.15 dB
Chapter 19, Solution 90
(a)
Z in  h ie 
h re h fe R L
1  h oe R L
1500  2000 
10 -4  120 R L
1  20  10 -6 R L
12  10 -3
500 
1  2  10 -5 R L
500  10 -2 R L  12  10 -3 R L
500  10 2  0.2 R L
R L  250 k
(b)
- h fe R L
h ie  (h ie h oe  h re h fe ) R L
Av 
- 120  250  10 3
2000  (2000  20  10 -6  120  10 -4 )  250  10 3
- 30  10 6
Av 
 - 3333
2  10 3  7  10 3
Av 
Ai 
h fe
120

 20
1  h oe R L 1  20  10 -6  250  10 3
R s  h ie
600  2000

(R s  h ie ) h oe  h re h fe (600  2000)  20  10 -6  10 -4  120
2600

k  65 k
40
Z out 
Z out
(c)
Av 
Vc Vc

Vb Vs

 Vc  A v Vs  -3333  4  10 -3  - 13.33 V
Chapter 19, Solution 91
R s  1.2 k ,
(a)
At 
R L  4 k
- h fe R L
h ie  (h ie h oe  h re h fe ) R L
- 80  4  10 3
1200  (1200  20  10 -6  1.5  10 -4  80)  4  10 3
- 32000
At 
 - 25.64 This is just the gain for the transistor. If we
1248
calculate the gain for the circuit we get A t = V o /V be and V be =
V s [1.2k/(1.2k+2k)] = 0.375, thus, V A = (0.375)(–25.64) = –9.615.
At 
h fe
80

 74.07
1  h oe R L 1  20  10 -6  4  10 3
(b)
Ai 
(c)
Z in  h ie  h re A i
Z in  1200  1.5  10 -4  74.074  1.2 k
(d)
R s  h ie
(R s  h ie ) h oe  h re h fe
1200  1200
2400


 51.28 k
-6
-4
2400  20  10  1.5  10  80 0.0468
Z out 
Z out
(a) –25.64 for the transistor and –9.615 for the circuit, (b) 74.07, (c) 1.2 kΩ, (d) 51.28
kΩ
Chapter 19, Solution 92
Due to the resistor R E  240  , we cannot use the formulas in section 18.9.1. We will
need to derive our own. Consider the circuit in Fig. (a).
Rs
Ib
h ie
Ic
+
+
h re V c
Vs
+

+

h fe I b
Vb
Vc
IE
RE


(a)
Z in
IE  Ib  Ic
(1)
Vb  h ie I b  h re Vc  (I b  I c ) R E
(2)
Vc
RE  1
(3)
I c  h fe I b 
But
h oe
h oe
Vc  - I c R L
(4)
Substituting (4) into (3),
I c  h fe I b 
or
Ai 
RL
RE  1
Ic
h oe
I c h fe (1  R E h oe )

Ib
1  h oe (R L
100(1  240x 30 x10 6 )
1  30  10 -6 (4,000  240)
A i  79.18
Ai 
From (3) and (5),
(5)
RL
Ic 
h fe (1  R E )h oe
Vc
I b  h fe I b 
1  h oe (R L  R E )
RE  1
(6)
h oe
Substituting (4) and (6) into (2),
Vb  (h ie  R E ) I b  h re Vc  I c R E
Vb 
Vc (h ie  R E )
V
 h re Vc  c R E
RL


1   h fe (1  R E h oe )
 
 R E 
 h fe 
h oe  1  h oe (R L  R E )


V
1
 b 
A v Vc 
1
 R E 
h oe

(h ie  R E )

  h fe (1  R E h oe )
 
 h fe 

 1  h oe (R L  R E )
 h re 
RE
RL
(7)
1
(4000  240)
240
-4
10



6
Av 
4000

1
 100(1  240 x30x10 )
 100
 240 

6 
-6
30 x10   1  30  10  4240


1
 6.06x10 3  10 -4  0.06  -0.066
Av
A v  –15.15
From (5),
Ic 
h fe
I
1  h oe R L b
We substitute this with (4) into (2) to get
Vb  (h ie  R E ) I b  (R E  h re R L ) I c
 h (1  R E h oe )

Vb  (h ie  R E ) I b  (R E  h re R L )  fe
I b 
 1  h oe (R L  R E ) 
Z in 
h (R  h re R L )(1  R E h oe )
Vb
 h ie  R E  fe E
1  h oe (R L  R E )
Ib
(8)
Z in  4000  240 
Z in  12.818 k
(100)(240  10 -4  4  10 3 )(1  240x 30x10 6 )
1  30  10 -6  4240
To obtain Z out , which is the same as the Thevenin impedance at the output, we introduce
a 1-V source as shown in Fig. (b).
Rs
h ie
Ib
Ic
+
+
+

h re V c
h fe I b
Vb
h oe
+

Vc
IE
RE

1V

(b)
Z out
From the input loop,
I b (R s  h ie )  h re Vc  R E (I b  I c )  0
But
Vc  1
So,
I b (R s  h ie  R E )  h re  R E I c  0
(9)
From the output loop,
Ic 
or
Vc
RE 
1
h oe
 h fe I b 
h oe
h fe
Ic
Ib 

h fe 1  R E h oe
h oe
 h fe I b
R E h oe  1
(10)
Substituting (10) into (9) gives
h

(R s  R E  h ie ) oe

h
 Ic 
fe 

  h re  R E I c 
(R s  R E  h ie ) 
0
h
1

R
h
fe
E
oe


R s  R E  h ie
R  R E  h ie  h oe 
  h re

Ic  R E Ic  s
h fe
1  R E h oe  h fe 
 R  R E  h ie 
(h oe h fe )  s
  h re
1  R E h oe 

Ic 
R E  (R s  R E  h ie ) h fe
Z out 
240  100  (1200  240  4000)
1200  240  4000 
-6
-4
 1  240 x 30 x10 6   30  10  10  100


24000  5440

 193.7 k
0.152
Z out 
Z out
R E h fe  R s  R E  h ie
1

I c  R s  R E  h ie 
 h oe  h re h fe

 1  R E h oe 
Chapter 19, Solution 93
We apply the same formulas derived in the previous problem.
(h ie  R E )
R
1

 h re  E
Av 
RL

1   h fe (1  R E h oe )
 R E 
 
 h fe 
h oe  1  h oe (R L  R E )


1

Av
(2000  200)
200
 2.5  10 -4 
3800

150(1  0.002)
(200  10 5 ) 
 150

 1  0.04
1
 0.004  2.5  10- 4  0.05263  -0.05638
Av
A v  –17.74
Ai 
h fe (1  R E h oe )
150(1  200x10 5 )

 144.5
1  h oe (R L  R E ) 1  10 -5  (200  3800)
Z in  h ie  R E 
h fe (R E  h re R L )(1  R E h oe )
1  h oe (R L  R E )
(150)(200  2.5  10 -4  3.8  10 3 )(1.002)
Z in  2000  200 
1.04
Z in  2200  28966
Z in  31.17 k
Z out 
Z out 
R E h fe  R s  R E  h ie
 R s  R E  h ie 

 h oe  h re h fe
 1  R E h oe 
200  150  1000  200  2000
33200

- 0.0055
 3200  10 -5 
-4

  2.5  10  150
 1.002 
Z out  –6.148 M
Chapter 19, Solution 94
We first obtain the ABCD parameters.
 200 0 
Given
[h]  
,
 100 10 -6 


[T]  


h
h 21
- h 22
h 21
- h11
h 21
-1
h 21
 h  h11 h 22  h12 h 21  2  10 -4

  - 2  10 -6

-8
  - 10

-2 

- 10 -2 
The overall ABCD parameters for the amplifier are
 - 2  10 -6
- 2  - 2  10 -6
-2
[T]  
-8
-2 
-8
- 10  - 10
- 10 -2
 - 10
  2  10 -8

  10 -10
 T  2  10 -12  2  10 -12  0
B

[h]   D
-1

D
Thus,
h ie  200 ,
Av 
T
D
C
D

0 
  200

4
-6 
  - 10 10 

h re  0 ,
h fe  -10 4 ,
h oe  10 -6
(10 4 )(4  10 3 )
5
-4
3  2  10
200  (2  10  0)  4  10
Z in  h ie 
h re h fe R L
 200  0  200 
1  h oe R L
2  10 -2 

10 -4 
Chapter 19, Solution 95
1
s 4  10s 2  8

Let Z A 
y 22
s 3  5s
Using long division,
ZA  s 
i.e.
5s 2  8
 s L1  Z B
s 3  5s
L1  1 H
and
5s 2  8
ZB  3
s  5s
as shown in Fig (a).
L1
ZB
y 22 =
(a)
1
s 3  5s
YB 

Z B 5s 2  8
Using long division,
YB  0.2s 
where
C 2  0.2 F
3.4s
 sC 2  YC
5s 2  8
and
YC 
3.4s
5s 2  8
as shown in Fig. (b).
L1
C2
Y c = 1/Z C
(b)
1
5s 2  8 5s
8
1
ZC 



 s L3 
YC
3.4s
3.4 3.4s
s C4
i.e. an inductor in series with a capacitor
5
L3 
 1.471 H and
3.4
C4 
3.4
 0.425 F
8
Thus, the LC network is shown in Fig. (c).
425 mF
1.471 H
1H
200 mF
(c)
Chapter 19, Solution 96
This is a fourth order network which can be realized with the network shown in Fig. (a).
L1
L3
C2
1
C4
(a)
 (s)  (s 4  3.414s 2  1)  (2.613s 3  2.613s)
1
2.613s  2.613s
H(s) 
s 4  3.414s 2  1
1
2.613s 3  2.613s
3
which indicates that
-1
2.613s  2.613s
s 4  3.414s  1

2.613s 3  2.613s
y 21 
y 22
3
We seek to realize y 22 . By long division,
2.414s 2  1
y 22  0.383s 
 s C 4  YA
2.613s 3  2.613s
i.e.
C 4  0.383 F
and
2.414s 2  1
YA 
2.613s 3  2.613s
as shown in Fig. (b).
L1
YA
L3
C2
C4
(b)
y 22
ZA 
1
2.613s 3  2.613s

YA
2.414s 2  1
By long division,
Z A  1.082s 
i.e.
1.531s
 s L3  Z B
2.414s 2  1
L 3  1.082 H
and
ZB 
1.531s
2.414s 2  1
as shown in Fig.(c).
L1
ZB
L3
C2
C4
(c)
YB 
i.e.
1
1
1
 1.577s 
 s C2 
s L1
1.531s
ZB
C 2  1.577 F
and
L1  1.531 H
Thus, the network is shown in Fig. (d).
1.531 H
1.577 F
1.082 H
0.383 F
(d)
1
Chapter 19, Solution 97
s3
s
s 3  12s
H(s)  3

6s 2  24
(s  12s)  (6s 2  24)
1 3
s  12s
3
Hence,
y 22
6s 2  24
1
 3

 ZA
s  12s s C 3
(1)
where Z A is shown in the figure below.
C1
C3
L2
ZA
y 22
We now obtain C 3 and Z A using partial fraction expansion.
Let
6s 2  24
A Bs  C
  2
2
s (s  12) s s  12
6s 2  24  A (s 2  12)  Bs 2  Cs
Equating coefficients :
24  12A 
 A  2
s0 :
1
0C
s :
2
s :
6 AB 
 B  4
Thus,
6s 2  24
2
4s
  2
2
s (s  12) s s  12
(2)
Comparing (1) and (2),
1 1
C3   F
A 2
But
3
1
s 2  12 1

 s
ZA
s
4s
4
(3)
1
1
 sC1 
s L2
ZA
(4)
Comparing (3) and (4),
1
C1  F
4
and
L2 
1
H
3
Therefore,
C1  250 mF ,
L 2  333.3 mH ,
C 3  500 mF
Chapter 19, Solution 98
 h  1  0.8  0.2
   h / h 21  h11 / h 21    0.001
[Ta ]  [Tb ]  

6
 h 22 / h 21  1 / h 21   2.5x10
 10 
 0.005
2.6x105
0.06 
[T]  [Ta ][Tb ]  

8
5x105 
1.5x10
We now convert this to z-parameters
A / C  T / C 1.733x103
[z]  

7
 1 / C D / C  6.667 x10
1000
I1
0.0267 

3.33x103 
z 11
+
z 22
+
I2
+
+
Vs
z 12 I 2
ZL
z 21 I 1
-
-
Vo
-
-
Vs  (1000  z11)I1  z12 I 2
(1)
Vo  z 22 I 2  z 21I1
(2)
But Vo  I 2 ZL

I 2  Vo / ZL
(3)
Substituting (3) into (2) gives
 1

z
 22 
I1  Vo 
 z 21 z 21ZL 
We substitute (3) and (4) into (1)
(4)
 1

z
z
 22  Vo  12 Vo
Vs  (1000  z 11 )
ZL
 z 11 z 21 Z L 
 7.653x10  4  2.136x10  5  744V
Chapter 19, Solution 99
Z ab  Z1  Z 3  Z c || (Z b  Z a )
Z c (Z a  Z b )
Z1  Z 3 
Za  Zb  Zc
Z cd  Z 2  Z 3  Z a || (Z b  Z c )
Z a (Z b  Z c )
Z2  Z3 
Za  Zb  Zc
Z ac  Z1  Z 2  Z b || (Z a  Z c )
Z b (Z a  Z c )
Z1  Z 2 
Za  Zb  Zc
(1)
(2)
(3)
Subtracting (2) from (1),
Z1  Z 2 
Z b (Z c  Z a )
Za  Zb  Zc
(4)
Z1 
ZbZc
Za  Zb  Zc
(5)
Subtracting (5) from (3),
Z2 
ZaZb
Za  Zb  Zc
(6)
Subtracting (5) from (1),
Z3 
ZcZa
Za  Zb  Zc
(7)
Using (5) to (7)
Z a Z b Z c (Z a  Z b  Z c )
(Z a  Z b  Z c ) 2
Za ZbZc
Z1Z 2  Z 2 Z 3  Z 3 Z1 
Za  Zb  Zc
Z1Z 2  Z 2 Z 3  Z 3 Z1 
Dividing (8) by each of (5), (6), and (7),
(8)
Za 
Z1Z 2  Z 2 Z 3  Z 3 Z1
Z1
Zb 
Z1Z 2  Z 2 Z 3  Z 3 Z1
Z3
Zc 
Z1Z 2  Z 2 Z 3  Z 3 Z1
Z2
as required. Note that the formulas above are not exactly the same as those in Chapter 9
because the locations of Z b and Z c are interchanged in Fig. 18.122.
Below are answers for the Network Analysis Tutorials. Some of the tutorial pages have random
parameters. For these pages, there are no fixed right answers, and formulas are provided instead.
Introductory Tutorial (Tut22)
1. Orange
2. 12
3. 3.14159
The Physics of Electricity (Tut1A)
1. 36000
2. 32.04
3. 60
4. 4.32
5. 18000
Basic Elements and Circuit Laws (Tut1)
1. 3
2. -2
3. 50
4. 50
5. -60
Resistors in Series and Parallel (Tut2)
1. 4
2. 1
3. 10
4. 17
5. 11
6. Formula: R1&times;R2/(R1 + R2) + (R3 + R4)&times;R5/(R3 + R4 + R5) + R7
Voltage Dividers and Current Dividers (Tut2A)
1. 40
2. 160
3. 12
4. 20
Circuit Solving with Kirchhoff's Laws (Tut3)
1. 3
2. 2
3. 3
4. 5
5. 4
6. 6
7. I6
8. V9
The Node Voltage Method (Tut4)
1. a
2. g
3. -6
4. 5
5. 4
6. g
7. 9
8. Ve
9. 0
11. -2
12. 4
14. I4
15. I4
The Mesh Current Method (Tut5)
1. c
2. -2
3. -4
4. -2
5. Ib
6. 3A
7. Ia
9. 5
Thevenin Laboratory (Tut6)
1. Formula: 1000&times;Voc&times;Vr/[R&times;(Voc - Vr)]
Maximum Power Transfer (DC) (Tut6A)
1. Formula: R2&times;R3/(R2 + R3)
2. Formula: E&times;R3/(R2 + R3)
3. Formula: VT2/(4&times;RT) where VT = E&times;R3/(R2 + R3)
Superposition (Tut6B)
1. 2
2. 5
3. -3
4. 4
Inductors and Capacitors (Tut7)
1. 0.8
2. 0.2
3. 0
4. 12t
5. -20e-4t
6. 16cos(8t)
7. -80
8. 9.79992
9. 3t
10. 2t3
11. -1.25e-4t
12. -0.25cos[8t]
13. 1.5
14. 29.532
First Order Systems (Tut8)
1. 0
2. 0
3. 120
4. 600
5. 120
6. 34.016
7. 0
8. 0
9. 90
10. 450
11. 0.45
12. -200
13. -200
14. 40
15. 450
16. 240
Second Order Systems (Tut9)
1. 0
2. 0
3. -80
4. capacitor
5. 80
6. 20000
7. 0
8. series
9. 6000
10. 5000
11. over
12. -2683
13. A1
14. -A2
15. s
16. 3.0148
17. 188
18. 1.297
The Properties of Sinusoids (Tut10)
1. 86
2. -170
3. 220
4. 23.4
5. -220
6. 440
7. 0
8. 155.6
9. -67
10. 50
11. 7.958
12. 125.7
13. 24.2
Root-mean-square (Tut10A)
1. 212
2. 50
3. 40
4. 0
5. 3
6. 10
7. 2500
8. 0
9. 5
10. -150
11. 25000
12. 450000
13. 900000
14. 466667
15. 32.27
Complex Numbers (Tut10B)
1. -1
2. -1
3. -j
4. 85.9
5. 383
6. 321
7. 655
8. -459
9. 85
10. -47.86
11. 138
12. 141.5
13. 92
14. 35
15. -15
16. 73
17. 17.0
18. 14.13
19. Formula: The equation is in the form M/θ = (-5 + jA)(B/-152&deg;) - (C - j100)/(2/D&deg;). The
√(X2 + Y2)where
X = √(52 + A2)&times;B&times;cos[arctan(-A/5) + 28&deg;] - ( √(C2 + 1002)/2)&times;cos[arctan(-100/C) - D]
Y = √(52 + A2)&times;B&times;sin[arctan(-A/5) + 28&deg;] - ( √(C2 + 1002)/2)&times;sin[arctan(-100/C) - D]
AC Circuits (Tut11)
1. 98
2. 120
3. 140
4. -125
5. 25
6. 30
7. 14.97
8. 200
AC Power (Tut12)
1. 378
2. 466
3. 441
4. B
5. 0
6. 466
7. 578
8. .779
9. 3373
10. D
11. B
12. 166
13. 2732
Maximum Power Transfer (AC) (Tut12A)
1. Formula: |Z1|2&times;R/(R2 + |Z1|2)
2. Formula: -(R2&times;|Z1| - R2&times;|Z2| - |Z1|2&times;|Z2|)/(R2 + |Z1|2)
3. Formula: |E|&times;R/√ R2 + |Z1|2
4. Formula: Divide the result from page 3 by twice the result from page 1.
5. Formula: Square the result from page 4 and multiply by the result from page 1.
Balanced Three-Phase Circuits (Tut13)
1. 133.3
2. 866
3. 866
4. 173.2
5. 200
6. 346.4
7. 400