Engineering Circuit Analysis, 7th Edition
Chapter 10
10.1
Practice Problem Solutions
(p371)
v1  120 cos(120 t  40)
10.2
(a)
2.5cos(120 t  20) lags v1 by  40  20   60
(b)
1.4sin (120 t  70)  1.4 cos (120 t  160) lags v1 by  40  (160)  120
(c)
0.8cos (120 t  110)  0.8cos(120 t  70) lags v1 by  40  70   110
(p371)
40 cos (100t  40)  20sin (100t  170)
 A cos 100t  B sin100t
 C cos (100t   )
40 cos (100t  40)  40 cos100t  cos( 40)  sin 100t  sin( 40) 
 30.64 cos100t  25.71sin100t
20sin (100t  170)  20 cos170 sin100t  sin170  cos100t 
 19.70sin100t  3.473cos100t
Thus, A = 30.64  3.473 = 27.17
B = 25.71 + 19.70 = 45.41
C  A2  B 2  52.92
 B 
  tan 1 
   59.11
 A 
10.3
(p376)
 Removing the inductor temporarily,
 3 
voc  40 cos8000t 
  30 cos 8000t V
 1 3 
and RTH  1 k // 3 k  750 
800 

cos  8000t  tan 1

750 

750  800
= 27.36 cos (8000t  46.85) mA
Thus, iL (t ) 
30
2
2
Engineering Circuit Analysis, 7th Edition
Chapter 10
10.4
(a)
iL (0)  27.36 cos(46.85)  18.71 mA
(b)
vL (t )  L
(c)
iR (0) 
(d)
iS (0)  iL (0)  iR (0)  24.03 mA
Practice Problem Solutions
diL
 (100 103 )(0.02736)(8000)sin(8000t  46.85) so vL(0) = 15.97 V
dt
vL (0)
 5.323 mA
3000
(p380)
(a)
 230  5  110 1  j 2    230  5  110  2.23663.43 
 2  5  2.236(30  110  63.43)
 22.36  16.57  21.43  j6.377
(b)
10.5
5  200  420
 4.698  j1.710  3.759  j1.368
=  0.939 + j3.078
(c)
2  j 7 7.280  74.05

 2.302 55.62
3 j
3.162  18.43
(d)
8  j4 
580
 8  j 4  2.560
220
 8  j 4  1.25  j 2.165
 9.25  j1.835  9.4311.22
(p380)
(a)
vcombination
1
 3
10

t
 4e
j 800 t 
dt   2  4 e j 800t

4
e j 800t  8 e j 800t
j800 103
  j 5 e j 800t  8 e j 800t  (8  j 5) e j 800t
 9.434 e  j 32.0 e j 800t
= 9.434 e j(800t  32) V
Engineering Circuit Analysis, 7th Edition
Chapter 10
Practice Problem Solutions
t
(b)
isource
1
100 j 2000t

100 e j 2000t  dt  
e
3 
10 10 
50
10000 j 2000t

e
 2 e j 2000t
j 2000
 (2  j 5) e j 2000t  5.385 e  j 68.2e j 2000t
= 5.385 e j(2000t  68.2) A
10.6
(p382)
(a)
5sin(580t  110)  5cos(580t  110  180  90)
 5cos(580t  20)  520
(b)
3cos 600t  5sin(600t  110)
 3cos 600t  5cos(600t  110  90)
 3cos 600t  5cos 600t cos 20  5sin 600t sin 20
 3cos 600t  4.698cos 600t  1.71sin 600t
 1.698cos 600t  1.71sin 600t
 2.41cos(600t  45.2)
 2.41cos(600t  134.8)  2.41134.8
(c)
8cos(4t  30)  4sin(4t  100)
 8cos(4t  30)  4 cos(4t  190)
 8cos 4t cos(30)  8sin 4t sin(30)  4 cos 4t cos( 190)  4sin 4t sin(190)
 (6.928  3.939) cos 4t  (4  0.6946) sin 4t
 2.989 cos 4t  3.305sin 4t
 4.456 cos(4t  47.87)  4.456 47.87
10.7
(p383)
 = 2000 rad/s and t = 1 ms
(a)
j10 A = 10+ 90 A so i(t) = 10 cos (2000t + 90) A and i(10-3) =  9.093 A
(b)
20 + j10 A = 22.36 26.57 so i(t) = 22.36 cos (2000t + 26.57) A
and i(103) =  17.42 A
(c)
20 + j(1020) A = 20 + (1 90) (10 20) A
= 20 + 10 110 A
= 19.06 29.54 A
so i(t) = 19.06 cos (2000t + 29.54) A and i(103) =  15.45 A
10.8
Engineering Circuit Analysis, 7th Edition
Chapter 10
(p386)
Practice Problem Solutions
The inductor is represented by a j(10  103) (1200) = j12  impedance and the capacitor by a
j
  j 33.33  impedance.
(1200) (25  106 )
(a)
The voltage across the 20 resistor is then
(1.228) (  j 33.33)  40  62 V and the current through it is
Thus, I s  2  62  1.228
= 2.33231.04 A
(b)
(by KCL)
Vs  10I R  ( j12)I L
 10(I L  I S )  (1290)I L
 10(353  2.332  31.04)  (1290) (353)
 34.8674.55 V
(c)
I R  I L  I S  353  2.332  31.04  3.98617.42
 iR (t )  3.986 cos (1200t + 17.42) A
10.9
(p389)
5 mH  j5 ; 20 mH  j 20 
200  F   j5 ; 100  F   j10 
(a)
Zin (a, g )  ( j 5 //10) //[  j 5  (  j10 // j 20)]
 (2  j 4) //[  j 25]
 2.809 + j4.494 
(b)
Zin (b, g )  ( j 20 //  j10) //[  j 5  (10 // j5)]
  j 20 //(2  j )  1.798  j1.124 
(c)
Zin (a, b)   j 5 //[( j 5 //10)  ( j10 // j 20)]
=  j 5 //(2  j 4  (  j 20)]
=  j 5 //(2  j16)  0.1124  j3.820 
40  62
 2  62 A
20
Engineering Circuit Analysis, 7th Edition
Chapter 10
10.10 (p391)
Practice Problem Solutions
100
100
100


 j 5  5 // j5 2.5  j 2.5 3.536  45
= 28.2845 A
(a)
I1 
(b)
I 2  I1
(c)
I 3  I1  I 2  28.2845  2090
j5
 590 
 28.2845 

5  j5
 7.07145 
= 2090 A
 20  0.009 A  200 A
10.11 (p393)
(a)
(b)
Z  1000  j 400   107721.80 
1
 Y   928.5  21.8  S
Z
= 862.1  j 344.8 S
at  = 106 rad/s, 1 mH  j 106 , 2 nF   j500 
Y
(c)
1
1
1


 1.25 + j2 mS
800 j106 j 500
Z  800  j106  j 500  999.5  103 89.95 
1
Y   1.0005  89.95 μS
Z
= 800.8  j106 pS
10.12 (p394)
50  90  20  j50V1  j 25(V1  V2 )
50  90  40 V2  j 25(V2  V1 )
[1]
[2]
 rewrite, grouping terms:
20  j50  j 25V1  j 25V2
 j 50  j 25V1  (40  j 25)V2
 Solving,
V1  0.9756  j 0.4195  1.06223.27 V
and
V2  1.024  j1.2195  1.593 49.97 V
[1]
[2]
Engineering Circuit Analysis, 7th Edition
Chapter 10
10.13 (p395)
10  j5I1  3I1  3I 2  1590  0
1590  3I 2  3I1  j 4I 2  20  0
Practice Problem Solutions
[1]
[2]
 rewrite, grouping terms:
(3  j5)I1  3I 2  10  j15
3I1  (3  j 4)I 2  20  j15

[1]
[2]
Solving,
I1  4.694  j1.296  4.87164.6 A
I 2  5.868  j 4.12  7.17144.9 A
10.14 (p399)
1
(a)
 1
1 
Yeq  j 50   
  33.9470.67 mS
 40 j 25 
1
 Z eq 
 29.47  70.67 
Yeq
and V1  20 103 Z eq  0.5893  70.67 V
= 0.1951  j0.5561 V
1
(b)
 1
1 
Yeq   j 25   
  25  12.68 mS
 40 j50 
1
 Z eq 
 4012.68 
Yeq
and V1  V2   j 50 103 Z eq  2  77.32
 0.4390  j1.951 V
 ( j 50) 1 
V1  (V1  V2 )  1
 0.7804 + j0.9755 V
1 
 40  ( j50) 
10.15 (p400)
(a)
Vab  (330)   j5 //(10  j5)   16.7733.43 V
(b)
Z ab   j 5 //(10  j 5)  5.59  63.43 
V
I SC  ab  330 A  2.598 + j1.5 A
Z ab
(c)
Z ab  5.59  63.43   2.5  j5 
Engineering Circuit Analysis, 7th Edition
Chapter 10
10.16 (p402)

Practice Problem Solutions
Since the two sources do not operate at the same frequency, we must use superposition in
the time domain.
30o
(4 // j 2)
j 6  (4 // j 2)
0.8  j1.6
3
0.8  j 7.6
 0.7022  20.56 V
V4  3
j2 
j6 
I
1
V4  175.6  20.56 mA
4
so i(t )  175.6 cos(2t  20.56) mA
I 
j15 
I
V4  4
j5 
40o V
I 
(4 // j15)
(3.734  j 0.9959)
4
V
j 5  (4 // j15)
3.734  j 5.996
 2.188  43.15 V
1
V4  547.1  43.15 A
4
so i(t )  547.1cos(5t  43.15) mA
and since i (t )  i(t )  i(t ), i (t )  175.6 cos (2t  20.56) + 547.1 cos (5t  43.15) mA
10.17 (p407)
VR = 2IC
V2 = -jIC + VR = IC-90o + VR
V

V1 = 2  2  I C  = V2 -90o + 2I C
 j2

Vs = V1 + V2
Using a ruler on the actual size
graph (10 small squares = 25.5 mm):
Vs = 10.8 mm, V1 = 5.7 mm,
V2 = 5.7 mm, VR = 5.1 mm.
(a) Vs / V1 = 1.90
(b) V1/ V2 = 1.00
(c) Vs / VR = 2.12