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 = 8‫ס‬160° mA
(c ) v(t )  120sin(103 t  50o )  120 cos(103 t  50o  90o )
V = 120‫–ס‬140° V
(d)
i(t )  60 cos(30t  10o )  60 cos(30t  10o  180o )
I = 60‫–ס‬170° mA
Chapter 9, Solution 12.
Let X = 440 and Y = 20–30. Evaluate the following quantities and express
your results in polar form.
(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°)
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 )

  
 100 rad/s
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°) 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°) 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
Vs
–
+
V2
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° 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(41.231‫ס‬14.036°)/(60.828‫ס‬9.462°)
60  j10
= –j50 + (13.5566‫ס‬4.574° = –j50 + 13.51342 + j1.08109
= 13.51342 – j48.9189 = 50.751‫–ס‬74.56°
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
Y
o
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)
Z bn 
 4.5  j3.5
15  j5
(5)(- j10)
Z cn 
 -1  j3
15  j5
Z an 
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
Z eq  7  j9  5.511  j0.2
Z eq  7  j9 
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
6R-XR =30-5X which leads to
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)
The phase shift is leading.
(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 

jL  j (2)(60)(200  10 -3 )  j75.4 
Vo 
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.
1
,
jC 2
Let
Z1  R 1 ,
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
Z2  R 2 
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 µF.
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.
10075 o
 471.413.5 o 
45048 o x10  6
(a) Z o 
Z

Y
(b)  
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