Ch08 Solution
December 13, 2018 | Author: Soji Adimula | Category: N/A
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CH 8 Soln...
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Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.1 Given i(t) = 5 cos (400t − 120°) A, determine the period of the current and the frequency in Hertz.
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.1
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.2 Determine the relative phase relationship of the two waves
υ1(t) = 10 cos (377t − 30°) V υ2(t) = 10 cos (377t + 90°) V
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.2
Irwin, Basic Engineering Circuit Analysis, 11/E
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8.3 Given the following voltage and current:
i(t) = 5 sin (377t − 20°) V
υ(t) = 10 cos (377t + 30°) V Determine the phase relationship between i(t) and υ(t).
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.3
Irwin, Basic Engineering Circuit Analysis, 11/E
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8.4 Write the expression for the waveform shown in Fig. P8.4 as a cosine function with numerical values for the
amplitude, frequency, and phase. υ(t) 24 V 12 V –4 –3 –2 –1
1 2 3 4 5 6 7 8 9 10 11 12 – 24 V
ms
Figure P8.4
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.4
Irwin, Basic Engineering Circuit Analysis, 11/E
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8.5 Calculate the current in the resistor in Fig. P8.5 if the voltage input is
(a) υ1(t) = 10 cos (377t + 180°) V. (b) υ2(t) = 12 sin (377t + 45°) V. Give the answers in both the time and frequency domains. i(t) + υ(t)
2Ω
− Figure P8.5
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.5
Irwin, Basic Engineering Circuit Analysis, 11/E
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8.6 Calculate the current in the capacitor shown in Fig. P8.6 if the voltage input is
(a) υ1(t) = 10 cos (377t − 30°) V. (b) υ2(t) = 12 sin (377t + 60°) V. Give the answers in both the time and frequency domains. i(t) + υ(t)
C = 1 μF
− Figure P8.6
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.6
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.7 Determine the phase angles by which υ1(t) leads i1(t) and υ1(t) leads i2(t), where
υ1(t) = 4 sin (377t + 25°) V i1(t) = 0.05 cos (377t − 20°) A i2(t) = −0.1 sin (377t + 45°) A
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.7
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.8 Find the frequency-domain impedance. Z, as shown in Fig. P8.8.
Z
3Ω
j4 Ω
Figure P8.8
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.8
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.9 Calculate the current in the inductor shown in Fig. P8.9 if the voltage input is
(a) υ1(t) = 10 cos (377t + 45°) V (b) υ2(t) = 5 sin (377t − 90°) V Give the answers in both the time and frequency domains. i(t) + υ(t)
L = 1 mH
− Figure P8.9
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.9
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.10 Find the frequency-domain impedance, Z, in the network in Fig. P8.10. 1Ω
Z
j1 Ω
−j2 Ω
Figure P8.10
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.10
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.11 Find Z in the network in Fig. P8.11. 2Ω
Z
1Ω
j2 Ω
2Ω
−j1 Ω
2Ω
j2 Ω
Figure P8.11
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.11
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.12 Find the impedance, Z, shown in Fig. P8.12 at a frequency of 400 Hz. 10 mH
Z
2Ω
1Ω
10 μF
Figure P8.12
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.12
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.13 Find the frequency-domain impedance, Z, as shown in Fig. P8.13.
Z
2Ω
j1 Ω
−j2 Ω
Figure P8.13
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.13
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.14 Find the impedance, Z, shown in Fig. P8.14 at a frequency of 60 Hz. 10 mH
Z
2Ω
1Ω
10 μF
Figure P8.14
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.14
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.15 Find Y in the network in Fig. P8.15.
1S
j1 S –j2 S
Y 2S
j2 S
2S
–j1 S
Figure P8.15
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.15
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.16 Find the equivalent impedance for the circuit in Fig. P8.16. 8Ω
−j8 Ω
10 Ω j5 Ω −j2 Ω Zeq
j10 Ω 4Ω
Figure P8.16
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.16
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.17 Find the frequency-domain impedance, Z, shown in Fig. P8.17. 2Ω
−j1 Ω
1Ω
Z
j2 Ω
−j2 Ω Figure P8.17
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.17
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.18 Find the impedance, Z, shown in Fig. P8.18 at a frequency of 60 Hz. 2Ω
Z
4Ω
10 mH
500 μF
Figure P8.18
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.18
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.19 Find the frequency-domain impedance, Z, shown in Fig. P8.19. 6Ω 1Ω
2Ω
Z
j1 Ω
–j1 Ω
4Ω
j2 Ω
j2 Ω
j4 Ω Figure P8.19
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.19
Irwin, Basic Engineering Circuit Analysis, 11/E
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8.20 In the circuit shown in Fig. P8.20, determine the value of the inductance such that the current is in phase with
the source voltage. 4Ω
12 cos (1000t + 75°) V
+ –
L
100 μF Figure P8.20
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.20
Irwin, Basic Engineering Circuit Analysis, 11/E
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8.21 Find the value of C in the circuit shown in Fig. P8.21 so that Z is purely resistive at a frequency of 60 Hz. 1Ω
5 mH
Z
C
Figure P8.21
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.21
Irwin, Basic Engineering Circuit Analysis, 11/E
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8.22 The impedance of the network in Fig. P8.22 is found to be purely real at f = 400 Hz. What is the value of C?
6Ω C
Z 10 mH
Figure P8.22
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.22
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.23 The admittance of the box in Fig. P8.23 is 0.1 + j0.2 S at 500 rad/s. What is the impedance at 300 rad/s?
Y
Figure P8.23
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.23
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.24 The impedance of the box in Fig. P8.24 is 5 + j4 Ω at 1000 rad/s. What is the impedance at 1300 rad/s?
Z
Figure P8.24
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.24
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.25 Draw the frequency-domain network and calculate υo(t) in the circuit shown in Fig. P8.25 if i1(t) is 200 cos (105t + 60°)
mA, i2(t) is 100 sin 105 (t + 90°) mA, and υS(t) = 10 sin (105t) V. Also, use a phasor diagram to determine υC (t). − +
i1(t)
i2(t)
30 Ω
υo(t)
υC(t)
+
250 nF
+ −
υS(t)
− Figure P8.25
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.25
2
Problem 8.25
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.26 The impedance of the circuit in Fig. P8.26 is real at f = 60 Hz. What is the value of L? L
Z
2Ω
10 mF
Figure P8.26
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.26
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.27 Find the frequency at which the circuit shown in Fig. P8.27 is purely resistive.
Z
1Ω
5 mH
1 mF
Figure P8.27
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.27
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.28 Find υS (t) in the circuit in Fig. P8.28. 40 Ω υS(t)
+ −
25 μF
+ υC(t) = 80 cos (1000t − 60°) V −
Figure P8.28
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.28
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.29 Draw the frequency-domain network and calculate i(t) in the circuit shown in Fig. P8.29 if υS(t) is 15 sin
(10,000t) V. Also, using a phasor diagram, show that υC(t) + υR(t) = υS(t). i(t)
υC(t) + − 66.67 μF
υS(t)
+ −
20 Ω
+ υR(t) −
Figure P8.29
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.29
2
Problem 8.29
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.30 Draw the frequency-domain network and calculate υo(t) in the circuit shown in Fig. P8.30 if iS(t) is 1 cos (2500t −
45°) A. Also, using a phasor diagram, show that iC(t) + iR(t) = iS(t). iC(t)
iS(t)
20 μF
iR(t) 10 Ω
+ υo(t) −
Figure P8.30
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.30
2
Problem 8.30
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.31 Find iC(t) and i(t) in the network in Fig. P8.31. i(t)
iC(t) 60 Ω
υ(t) = 120 cos (5000t) V
+ −
2.5 μF 16 mH
Figure P8.31
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.31
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.32 If υs(t) = 20 cos 5t volts, find υo(t) in the network in Fig. P8.32. 3Ω
0.5 H +
υs(t)
+ −
1H
10 Ω
0.02 F
υo(t) −
Figure P8.32
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.32
Irwin, Basic Engineering Circuit Analysis, 11/E
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8.33 Find υo(t) in the circuit in Fig. P8.33. 20 Ω 200 μF
30 mH 100 μF +
170 cos 377t V
+ −
10 Ω 20 mH
15 Ω
υo(t) −
Figure P8.33
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.33
2
Problem 8.33
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
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8.34 Find υo(t) in the network in Fig. P8.34.
0.1 F 5 cos 10t A
5Ω
0.4 H
0.02 F
5Ω
+ υo(t) −
Figure P8.34
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.34
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.35 Find io(t) in the circuit in Fig. P8.35 if υ (t) = 50 cos 100t V. 4Ω
20 mH
6Ω
5Ω
2000 μF υ(t)
+ –
io(t) 3Ω
1000 μF 50 mH
Figure P8.35
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.35
2
Problem 8.35
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.36 Find υo(t) and io(t) in the network in Fig. P8.36. io(t) 0.01 F 25 cos 20t V
+ −
5Ω
0.4 H
0.02 F
5Ω
+ υo(t) −
Figure P8.36
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.36
Irwin, Basic Engineering Circuit Analysis, 11/E
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8.37 Calculate υo(t) in Fig. P8.37. 2H
6Ω
2Ω
+ −
50 cos 5t V
1H
0.2 F
5Ω
3Ω
+ 0.05 F
υo(t)
4Ω
− Figure P8.37
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.37
2
Problem 8.37
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.38 Find i2(t) in the circuit in Fig. P8.38. i2(t) 5Ω 8Ω
625 μF
25 mH
2Ω 2 cos 400t A
10 mH
4Ω
6Ω
Figure P8.38
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.38
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.39 Find the voltage Vo shown in Fig. P8.39. 2Ω
j10 Ω +
10 120° V
+ −
−j12 Ω
Vo −
Figure P8.39
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.39
Irwin, Basic Engineering Circuit Analysis, 11/E
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8.40 Find the frequency-domain current I shown in Fig. P8.40.
10 0° A
80 Ω
−j60 Ω I
Figure P8.40
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.40
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.41 Find the frequency-domain voltage Vo shown in Fig. P8.41. + 0.5 25° A
j2 Ω
−j5 Ω
10 Ω
Vo −
Figure P8.41
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.41
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.42 Find the voltage V shown in Fig. P8.42. + 1 30° A
Z
j1 Ω
−j2 Ω
1Ω
V −
Figure P8.42
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.42
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.43 Find the frequency-domain current I shown in Fig. P8.43. 5Ω
37 −145° V
+ −
j6 Ω
−j 4 Ω
10 Ω I
Figure P8.43
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.43
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.44 Find υx(t) in the circuit in Fig. P8.44. 10 Ω
0.005 F
0.1 H +
100 cos 40t V
+ −
+ 40 cos (40t − 30°) V −
υx(t) − 5Ω
0.2 H
Figure P8.44
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.44
2
Problem 8.44
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.45 Find υo(t) in the network in Fig. P8.45.
50 cos 25t V
0.008 F
0.1 H
10 Ω
υo(t)
+ – +
υx(t)
+ –
2υx(t)
−
5Ω
0.2 H
Figure P8.45
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.45
2
Problem 8.45
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.46 Find υ1(t) and υ2(t) in the circuit in Fig. P8.46. 2Ω
4 cos 10t A
12 cos (10t − 25°) V
+−
0.2 H +
υ2(t)
0.02 F
υ1(t)
2 cos (10t + 15°) A
−
Figure P8.46
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.46
2
Problem 8.46
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.47 Find the voltage V shown in Fig. P8.47. 1Ω + 100 0° V
+ −
−j1 Ω
V −
Figure P8.47
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.47
Irwin, Basic Engineering Circuit Analysis, 11/E
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8.48 Find the voltage Vo shown in Fig. P8.48. 2Ω
j10 Ω +
10 30° V
+ −
−j12 Ω
Vo −
Figure P8.48
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.48
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.49 Find the frequency-domain voltage Vo shown in Fig. P8.49. 1Ω + 15 Ω
5 30° A
−j12 Ω
Vo −
Figure P8.49
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.49
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.50 Find Vo in the network in Fig. P8.50. 2Ω
−j1 Ω +
j1 Ω
12 0° V
− +
4Ω
2Ω
Vo −
Figure P8.50
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.50
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.51 Determine Io in the network shown in Fig. P8.51 if VS = 12 0° V.
2Ω
2Ω
2Ω
+
− +
VS
j2 Ω
V1
+ j2 Ω
−
V2
Io
−
Figure P8.51
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.51
Irwin, Basic Engineering Circuit Analysis, 11/E
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8.52 Given the network in Fig. P8.52, determine the value of Vo if VS = 24 0° V. j2 Ω
2Ω + VS
+ −
−j1 Ω
V1
+ 2Ω
−
Vo −
Figure P8.52
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.52
Irwin, Basic Engineering Circuit Analysis, 11/E
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8.53 Find VS in the network in Fig. P8.53 if V1 = 4 0° V. −j1 Ω
−+
+ 1Ω
V1
VS
2Ω
j1 Ω
−j1 Ω
2Ω
− Figure P8.53
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.53
Irwin, Basic Engineering Circuit Analysis, 11/E
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8.54 If V1 = 4 0° V, find Io in Fig. P8.54. −j2 Ω
−
V1
+
2Ω
j1 Ω
IS
1Ω Io
Figure P8.54
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.54
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.55 In the network in Fig. P8.55, Io = 4 0° A. Find Ix.
j1 Ω
1Ω
−j1 Ω 1Ω
2 0° A 1Ω
+ −
12 0° V
Ix
1Ω Io
Figure P8.55
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.55
Irwin, Basic Engineering Circuit Analysis, 11/E
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8.56 If Io = 4 0° A in the circuit in Fig. P8.56, find Ix.
+ –
1Ω 1Ω
j1 Ω
Ix
1Ω
−j1 Ω
1Ω
2 0° A
12 0° V Io
1Ω
+ −
4 0° V
Figure P8.56
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.56
2
Problem 8.56
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
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8.57 If Io = 4 0° A in the network in Fig. P8.57, find Ix. Ix j1 Ω
1Ω
1Ω 1Ω
R2
1Ω
+ −
12 0° V
2 0° A 1Ω
−j1 Ω
1Ω Io
Figure P8.57
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.57
2
Problem 8.57
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
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8.58 In the network in Fig. P8.58, Vo is known to be 4 45° V. Find Z. 2Ω 12 0° V
+ −
+
−j1 Ω Z
1Ω
Vo −
Figure P8.58
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.58
Irwin, Basic Engineering Circuit Analysis, 11/E
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8.59 In the network in Fig. P8.59, V1 = 5 −120° V. Find Z. 2Ω
+
V1
−
0.25 Ω
j1 Ω −j0.25 Ω
6 0° A
Z
Figure P8.59
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.59
2
Problem 8.59
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.60 Find Vo in the circuit in Fig. P8.60. −j2 Ω
2Ω
+
j2 Ω 24 0° V
+ −
2 90° A
2Ω
Vo −
Figure P8.60
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.60
2
Problem 8.60
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
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8.61 Find Vo in the network in Fig. P8.61. 10 Ω + 30 Ω
Vo
j10 Ω
− +
2 60° A
10 30° V
− Figure P8.61
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.61
2
Problem 8.61
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.62 Using nodal analysis, find Io in the circuit in Fig. P8.62. V1
V2 1Ω
1Ω
+ −
−j1 Ω
1Ω 12 0° V
2 0° A
4 0° A
Io
Figure P8.62
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.62
2
Problem 8.62
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.63 Use nodal analysis to find Io in the circuit in Fig. P8.63. V
−j1 Ω
2Ω
j2 Ω
4 0° A 12 0° V
+ −
+ −
2Ω
6 0° V Io
Figure P8.63
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.63
Irwin, Basic Engineering Circuit Analysis, 11/E
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8.64 Find Vo in the network in Fig. P8.64 using nodal analysis. V 2Ω
1Ω
−j1 Ω j2 Ω
12 0° V
+ −
2 0° A
+ 2Ω
+ −
4 0° V
Vo −
Figure P8.64
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.64
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.65 Use the supernode technique to find Io in the circuit in Fig. P8.65. 12 0° V
−j1 Ω
2Ω
−+ 1Ω
j2 Ω
2Ω
−j2 Ω
Io Figure P8.65
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.65
2
Problem 8.65
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.66 Find I1 and Vo in the network in Fig. P8.66.
+− 12 45° V
+ −
2Ω
−j1 Ω
6 0° V 1Ω
j4 Ω I1
+ −j1 Ω
Vo −
Figure P8.66
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.66
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.67 Use nodal analysis to find Vo in the network in Fig. P8.67. 12 0° V −+
+
−j 2 Ω 2 0° A
j4 Ω
2Ω
Vo −
Figure P8.67
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.67
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.68 Use nodal analysis to find Io in the circuit in Fig. P8.68. 1Ω −j1 Ω 1Ω 2 0° A
Io
j1 Ω
+ −
12 0° V
Figure P8.68
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.68
2
Problem 8.68
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.69 Use nodal analysis to find Vo in the network in Fig. P8.69. −j1 Ω 1Ω
j1 Ω +
+ −
1Ω 6 0° V
2 0° A
Vo −
Figure P8.69
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.69
2
Problem 8.69
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.70 Use nodal analysis to find Vo in the circuit in Fig. P8.70. −+
12 0° V
1Ω
1Ω +
2 0° A
−j1 Ω
1Ω
Vo −
Figure P8.70
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.70
2
Problem 8.70
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.71 Use nodal analysis to find Vo in the network in Fig. P8.71. −j1 Ω 1Ω
+ Vo
4 0° V + −
1Ω
− + −
+ −
12 0° V
6 0° V
Figure P8.71
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.71
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.72 Use nodal analysis to find Io in the network in Fig. P8.72. 1Ω
12 0° V
2 0° A
−+ 1Ω
−j1 Ω
1Ω Io
Figure P8.72
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.72
2
Problem 8.72
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.73 Use nodal analysis to find Vo in the circuit in Fig. P8.73. 6 0° V
12 0° V
+−
−+
1Ω
−j1 Ω
+ j2 Ω
2 0° A
1Ω
1Ω
Vo −
Figure P8.73
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.73
2
Problem 8.73
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.74 Find Io in the circuit in Fig. P8.74 using nodal analysis. Io 2Ω
2Ω −j2 Ω
12 0° V
+ −
1Ω
2 0° A
j1 Ω
+ −
6 0° V
Figure P8.74
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.74
2
Problem 8.74
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.75 Use nodal analysis to find Io in the circuit in Fig. P8.75.
+ −
1Ω
6 0° V
j1 Ω
4 0° A 12 0° V
+− 2 0° A −j1 Ω
1Ω
1Ω
1Ω
Io Figure P8.75
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.75
2
Problem 8.75
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.76 Use nodal analysis to find Vo in the network in Fig. P8.76. + Vx −
+−
1Ω
12 0° V
+ −
–j1 Ω
2Vx + 1Ω
Vo −
Figure P8.76
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.76
2
Problem 8.76
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.77 Use nodal analysis to find Vo in the network in Fig. P8.77. 2 0° A
+
−j1 Ω
Vx −
+
1Ω + − 6 0° V
+ –
2Vx
1Ω
Vo −
Figure P8.77
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.77
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.78 Use nodal analysis to find Io in the network in Fig. P8.78. −j 1 Ω − 1Ω 2Vx
+ –
Vx + 1Ω
1Ω
+ −
12 0° V
Io Figure P8.78
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.78
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.79 Use nodal analysis to find Vo in the network in Fig. P8.79. + 4 0° V
+ –
j1 Ω 1Ω
1Ω
1Ω
+ –
Vo
4Vo −
Figure P8.79
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.79
2
Problem 8.79
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.80 Use nodal analysis to find Io in the network in Fig. P8.80.
1Ω
2Vx
− +
+ V − x
1Ω
−j 1 Ω 1Ω
1Ω Io
+ −
12 0° V
Figure P8.80
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.80
2
Problem 8.80
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.81 Use nodal analysis to find Vo in the circuit in Fig. P8.81.
4Vo
+ –
1Ω
1Ω 1Ω
−j Ω
+
+ –
4 0° V
Vo −
Figure P8.81
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.81
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.82 Use nodal analysis to find Vx in the circuit in Fig. P8.82. 12 0° V
+−
j1 Ω +
2Vx
1Ω
−j1 Ω
Vx
1Ω
− Figure P8.82
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.82
2
Problem 8.82
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.83 Find the voltage across the inductor in the circuit shown in Fig. P8.83 using nodal analysis. V1
10 30° V
+ −
−j2 Ω
4Ω
V2
j1 Ω
2Ix
Ix
Figure P8.83
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.83
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.84 Use nodal analysis to find Vo in the circuit in Fig. P8.84.
−j1 Ω
1Ω
4 0° V
+−
+ Vx
1Ω
+ 1Ω
−
+ −
4Vx
Vo −
Figure P8.84
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.84
2
Problem 8.84
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.85 Use nodal analysis to find Vo in the network in Fig. P8.85. + Vx
+ −j1 Ω
1Ω
−
2 0° A
1Ω
6 0° V
Vo
1Ω
−
1Ω
+ −
+ −
4Vx
Figure P8.85
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.85
2
Problem 8.85
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.86 Use nodal analysis to find Vo in the circuit in Fig. P8.86. + Vx
+ −
1Ω
6 0° V
2 0° A
−
4 0° A
+
1Ω
−j1 Ω 1Ω
+ −
4Vx
Vo −
Figure P8.86
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.86
2
Problem 8.86
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.87 Use nodal analysis to find Vo in the circuit in Fig. P8.87. Ix 4Ix
−j1 Ω + +
Vx
Vo
−
1Ω
1Ω
1Ω 1Ω
+ −
6 0° V
+ −
2Vx
− Figure P8.87
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.87
2
Problem 8.87
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.88 Use nodal analysis to find Io in the circuit in Fig. P8.88. Ix
+ 1Ω
−j1 Ω
Vx −
1Ω Io
1Ω
1Ω 4 Ix
+ –
4 0° V
+ –
2Vx
Figure P8.88
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.88
2
Problem 8.88
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.89 Use mesh analysis to find Vo in the circuit shown in Fig. P8.89. −j1 Ω
12 45° V
−+ 6 0° V
− +
j2 Ω I1
I2
+ 2Ω V o −
Figure P8.89
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.89
2
Problem 8.89
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.90 Solve problem 8.67 using loop analysis.
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.90
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.91 Solve problem 8.68 using loop analysis.
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.91
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.92 Solve problem 8.69 using loop analysis.
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.92
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.93 Find Vo in the network in Fig. P8.93 using loop analysis. − j1 Ω
j2 Ω 2 0° A
1Ω
+ 1Ω
+ –
12 0° V
Vo −
Figure P8.93
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.93
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.94 Find Vo in the network in Fig. P8.94 using loop analysis. + Vo − 1Ω 1Ω
+ −
12 0° V
−j1 Ω
1Ω 2 0° A
Figure P8.94
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.94
2
Problem 8.94
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.95 Find Io in the network in Fig. P8.95 using loop analysis.
1Ω
2 0° A
−j1 Ω
12 0° V
+ −
4 0° A
1Ω
1Ω
j1 Ω
Io Figure P8.95
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.95
2
Problem 8.95
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.96 Find Vo in the circuit in Fig. P8.96 using mesh analysis. j2 Ω
2Ω
+ 2 0° A
2Ω
−j1 Ω
+−
1Ω
Vo −
6 0° V 4 0° A
Figure P8.96
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.96
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.97 Find Vo in the network in Fig. P8.97. j1 Ω + 12 0° V
+ −
2Ω 2Ω
16 0° V
− +
−j1 Ω
Vo
2 0° A −
Figure P8.97
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.97
2
Problem 8.97
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.98 Determine Vo in the circuit in Fig. P8.98. 2Ω
−j1 Ω
2Ω +
6 0° A
+ −
j2 Ω 12 0° V
1Ω
Vo −
Figure P8.98
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.98
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.99 Use mesh analysis to find Vo in the circuit shown in Fig. P8.99. j2 Ω
4Ω
−j4 Ω
12 0° V
+ −
I1
I2
+ 2Ω
4 90° A
Vo −
Figure P8.99
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.99
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.100 Using loop analysis, find Io in the network in Fig. P8.100. 12 0°V
2Ω
−+
2 0° A
−j2 Ω
j1 Ω
2Ω
4 0° A
Io Figure P8.100
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.100
2
Problem 8.100
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.101 Use mesh analysis to find Vo in the circuit in Fig. P8.101.
j1 Ω
6 0° A −j2 Ω
4 0°A 1Ω
+ 2Ω
Vo
2Ω
2 0° A
− Figure P8.101
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.101
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.102 Use loop analysis to find Io in the network in Fig. P8.102. Io 4 30° A
−j1 Ω
1Ω 1Ω
1Ω 2Vx
+ −
+ 1Ω
2 0° A
Vx −
Figure P8.102
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.102
2
Problem 8.102
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.103 Find Vo in the network in Fig. P8.103. 4 0° A
−j1 Ω
1Ω
Ix +
2Ix
j1 Ω
1Ω
Vo −
Figure P8.103
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.103
2
Problem 8.103
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.104 Use loop analysis to find Vo in the circuit in Fig. P8.104. + 4Ix
−j1 Ω
1Ω
1Ω 1Ω
+ –
Vo
4 0° V
Ix
−
Figure P8.104
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.104
2
Problem 8.104
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.105 Use loop analysis to find Vo in the network in Fig. P8.105. + Vx
+ −
1Ω
2 0° A
6 0° V
−
4 0° A
+
1Ω
−j1 Ω 1Ω
+ −
Vx
Vo −
Figure P8.105
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.105
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.106 Use superposition to find Vo in the network in Fig. P8.106. 12 0° V
−+ 1Ω
4 0° A
1Ω −j1 Ω
+ Vo −
Figure P8.106
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.106
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.107 Solve problem 8.67 using superposition.
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.107
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.108 Solve problem 8.68 using superposition.
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.108
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.109 Solve problem 8.69 using superposition.
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.109
2
Problem 8.109
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.110 Use superposition to determine Vo in the circuit in Fig. P8.110. 1Ω
−j1 Ω
j1 Ω +
6 0° V
− +
6 0° A
1Ω
Vo −
Figure P8.110
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.110
2
Problem 8.110
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.111 Using superposition, find Vo in the circuit in Fig. P8.111. 1Ω
−j1 Ω j2 Ω
6 0° V
+ −
2 0° A
+ Vo
2Ω
− Figure P8.111
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.111
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.112 Find Vo in the network in Fig. P8.112 using superposition. 1Ω 16 0° V
+ −
2Ω −j2 Ω
4 0° A
+
2 Ω Vo j3 Ω −
Figure P8.112
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.112
2
Problem 8.112
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.113 Find Vo in the network in Fig. P8.113 using superposition. 2Ω
2 0° A
12 0° V
−+ −j2 Ω
j4 Ω
+ 2 Ω Vo −
Figure P8.113
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.113
2
Problem 8.113
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.114 Use superposition to find Vo in the circuit in Fig. P8.114.
j1 Ω
2 0° A
1Ω −j1 Ω
4 0° A
+
1Ω
+ −
6 0° V
1Ω
Vo −
Figure P8.114
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.114
2
Problem 8.114
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.115 Use superposition to find Vo in the network in Fig. P8.115. +
+ −
1Ω
6 0° V
1Ω
1Ω −j1 Ω
Vo
4 0° A −
Figure P8.115
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.115
2
Problem 8.115
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.116 Solve problem 8.68 using source exchange.
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.116
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.117 Use source exchange to find the current Io in the network in Fig. P8.117. 1Ω
2Ω
12 0° V
+ −
−j1 Ω
2Ω 2 0° A
Io
4 0° A
Figure P8.117
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.117
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.118 Use source exchange to determine Vo in the network in Fig. P8.118.
−+ −j1 Ω 12 0° V
+ −
6 0° V 1Ω
2 0° A
2Ω
+ Vo −
Figure P8.118
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.118
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.119 Use source transformation to find Vo in the circuit in Fig. P8.119. + 12 0° V
+ −
−j2 Ω
1Ω
Vo −
1Ω 12 Ω
1Ω 2 0° A
Figure P8.119
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.119
2
Problem 8.119
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.120 Use source transformation to find Vo in the circuit in Fig. P8.120. −j2 Ω + 2 0° A
5Ω
1Ω
Vo −
1Ω 4 0° V
+ −
2Ω
+ −
6 0° V
2Ω Figure P8.120
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.120
2
Problem 8.120
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.21 Use source transformation to find Io in the circuit in Fig. P8.121. 3Ω 6 0° V
Io
1Ω
+ −
12 Ω 4 0° A
2 0° A
+ −
6Ω
−j3 Ω
8 0° V
4Ω
Figure P8.121
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.21
2
Problem 8.21
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.122 Solve problem 8.67 using Thévenin’s theorem.
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.122
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.123 Solve problem 8.68 using Thévenin’s theorem.
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.123
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.124 Solve problem 8.69 using Thévenin’s theorem.
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.124
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.125 Use Thevenin’s theorem to find Vo in the circuit in Fig. P8.125. j2 Ω
4Ω
−j4 Ω
12 0° V
+ −
I1
I2
+ 2Ω
4 90° A
Vo −
Figure P8.125
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.125
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.126 Apply Thévenin’s theorem twice to find Vo in the circuit in Fig. P8.126. 1Ω
2Ω
1Ω +
6 0° V
+ −
−j1 Ω
j1 Ω
1Ω
Vo −
Figure P8.126
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.126
2
Problem 8.126
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.127 Use Thévenin’s theorem to find Vo in the network in Fig. P8.127. 1Ω
–j2 Ω + −
+ Vo
6 0° V
2 0° A 1Ω
− Figure P8.127
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.127
2
Problem 8.127
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.128 Use Thévenin’s theorem to find Io in the network in Fig. P8.128.
2 0° A 1Ω
j1 Ω −j1 Ω
+ −
12 0° V
1Ω Io
Figure P8.128
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.128
2
Problem 8.128
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.129 Use Thévenin’s theorem to find the voltage across the 2-Ω resistor in the network in Fig. P8.129. 2Ω
−+ −j1 Ω
12 0° V 1Ω
j1 Ω
2 0° A
Figure P8.129
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.129
2
Problem 8.129
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.130 Use Thévenin’s theorem to find Io in the network in Fig. P8.130. Io 1Ω
+ −
12 0° V
1Ω −j1 Ω
1Ω 2 0° A
Figure P8.130
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.130
2
Problem 8.130
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.131 Use Thévenin’s theorem to find Vo in the network in Fig. P8.131.
−+ 1Ω
12 0° V
1Ω +
2 0° A
−j 1 Ω
1Ω
Vo −
Figure P8.131
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.131
2
Problem 8.131
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.132 Use Thévenin’s theorem to find Io in the network in Fig. P8.132.
–j1 Ω
1Ω
4 0° A 1Ω
3 0° A
Io
12 0° V
−+ j1 Ω
1Ω
Figure P8.132
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.132
2
Problem 8.132
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.133 Use Thévenin’s theorem to find Vo in the network in Fig. P8.133. 1Ω 4 0° A 1Ω 6 0° A
−j1 Ω
+ 4 0° A
j2 Ω
2Ω
Vo −
Figure P8.133
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.133
2
Problem 8.133
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.134 Use Thévenin’s theorem to determine Io in the circuit in Fig. P8.134. 2 0° A
−+ −j2 Ω
6 0° V 1Ω
1Ω
1Ω
4 0° A
Io Figure P8.134
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.134
2
Problem 8.134
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.135 Use Thévenin’s theorem to find Vo in the network in Fig. P8.135. 2 0° A
+
−j1 Ω
Vx −
+
1Ω + − 6 0° V
2Vx
+ –
1Ω
Vo −
Figure P8.135
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.135
2
Problem 8.135
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.136 Use Thévenin’s theorem to find Io in the network in Fig. P8.136. −j 1 Ω − 1Ω 2Vx
+ –
Vx + 1Ω
1Ω
+ −
12 0° V
Io Figure P8.136
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.136
2
Problem 8.136
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.137 Find Vo in the network in Fig. P8.137 using Thévenin’s theorem. 12 0° V
+− + 2Vx
1Ω
−j1 Ω
Vx −
j1 Ω
+ 1Ω
Vo −
Figure P8.137
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.137
2
Problem 8.137
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.138 Find the Thévenin’s equivalent for the network in Fig. P8.138 at terminals A–B. A
+ −j1 Ω
4 0° A
1Ω
j1 Ω
Vx −
2Vx B
Figure P8.138
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.138
2
Problem 8.138
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
3
Problem 8.138
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.139 Given the network in Fig. P8.139, find the Thévenin’s equivalent of the network at terminals A–B.
+−
12 0° V
+−
V2
6 0° V
1Ω
A
−j1 Ω
1Ω V1
V3 j1 Ω
+ −
4 0° V B
Figure P8.139
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.139
2
Problem 8.139
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.140 Use Thévenin’s theorem to determine Io in the network in Fig. P8.140.
2 0° A
1Ω
1Ω 1Ω
−j1 Ω
4 0° A 1Ω
Io 12 0° V
+ −
1Ω
j1 Ω
+ −
6 0° V
Figure P8.140
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.140
2
Problem 8.140
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.141 Use Thévenin’s theorem to find Io in the network in Fig. P8.141.
1Ω
2Vx
− +
+ V − x
1Ω
−j1 Ω 1Ω
1Ω Io
+ −
12 0° V
Figure P8.141
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.141
2
Problem 8.141
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
3
Problem 8.141
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.142 Solve problem 8.68 using Norton’s theorem.
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.142
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.143 Solve problem 8.69 using Norton’s theorem.
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.143
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.144 Find Vx in the circuit in Fig. P8.144 using Norton’s theorem. 11.3 45° V
−+ 2 0° A
j4 Ω
+ 10 Ω −j3 Ω
Vx −
Figure P8.144
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.144
2
Problem 8.144
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.145 Find Io in the network in Fig. P8.145 using Norton’s theorem. −j2 Ω
j1 Ω
6 45° A
2Ω
2 0° A
Io
Figure P8.145
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.145
2
Problem 8.145
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.146 Use Norton’s theorem to find Vo in the network in Fig. P8.146. + −j1 Ω
4 0° A
1Ω
Vx
1Ω
− +
j1 Ω
2Vx
1Ω
Vo −
Figure P8.146
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.146
2
Problem 8.146
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.147 Find Vo using Norton’s theorem for the circuit in Fig. P8.147.
4 0° V
+ −
+ −
−j1 Ω 1Ω
Vx
+
8 0° V
− +
1Ω j1 Ω
+ −
2Vx
1Ω
Vo −
Figure P8.147
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.147
2
Problem 8.147
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.148 Use Norton’s theorem to find Vo in the circuit in Fig. P8.148.
−j1 Ω
4Ix
2 0° A
1Ω 1Ω 4 0° A
1Ω Ix
+ 2Ω
Vo −
Figure P8.148
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.148
2
Problem 8.148
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.149 Apply Norton’s theorem to find Vo in the network in Fig. P8.149. −j1 Ω +
1Ω 6 0° V
+ −
1Ω
1Ω
j1 Ω 4 0° A
Vo
1Ω −
Figure P8.149
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.149
2
Problem 8.149
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.150 Find Vo in the circuit in Fig. P8.150.
6 0° V
+ −
2 0° A
1Ω
−j1 Ω
+−
+ 1Ω
1Ω
1Ω
12 0° V
Vo
1Ω
j1 Ω
− Figure P8.150
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.150
2
Problem 8.150
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.151 Find the node voltages in the network in Fig. P8.151. 1Ω
−j1 Ω
2Ω
+ −
1Ω
j1 Ω
1Ω 12 0° V
2 0° A
2Ω
−j2 Ω
Figure P8.151
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.151
2
Problem 8.151
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.152 Determine Vo in the network in Fig. P8.152.
1Ω
2 0° A
+ −
1Ω
6 0° V
−+ −j1 Ω
1Ω 1Ω
j1 Ω
12 30° V
+ 1Ω
Vo −
Figure P8.152
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.152
2
Problem 8.152
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.153 Find Io in the network in Fig. P8.153.
2 0° A
1Ω
4 0° A
1Ω 1Ω Io
−j1 Ω 12 0° V
+ −
1Ω
1Ω j1 Ω
+ −
6 0° V
Figure P8.153
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.153
2
Problem 8.153
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.154 Use both nodal analysis and loop analysis to find Io in the network in Fig. P8.154.
2 0° A
1Ω 1Ω
Io
2Vx
1Ω
2Ix
1Ω
−+ +
+ −
−j1 Ω
12 0° V
Vx −
1Ω
j1 Ω
Ix
Figure P8.154
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.154
2
Problem 8.154
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.155 Find Io in the network in Fig. P8.155.
−j1 Ω
1Ω
1Ω
6 0° V
12 0° V
+−
−+
2 0° A
1Ω
1Ω Io 1Ω
4 0° A
j1 Ω
Figure P8.155
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.155
2
Problem 8.155
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.156 Find Io in the network in Fig. P8.156.
1Ω
1Ω j1 Ω
Ix
1Ω
1Ω
2Ix
Io 1Ω
1Ω
+ −
6 0° V
−j1 Ω
− +
4 0° V
Figure P8.156
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.156
2
Problem 8.156
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.157 Determine Io in the network in Fig. P8.157.
1Ω
j1 Ω
−j1 Ω
1Ω
1Ω
2 0° A
1Ω
12 0° V
−+ 2Vx
Io 1Ω
−+
+ Vx
1Ω
−
Figure P8.157
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.157
2
Problem 8.157
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E
1
8.158 Find Io in the circuit in Fig. P8.158.
12 0° V
−j1 Ω
+ − 1Ω
1Ω
1Ω
2 0° A
Io
Ix 1Ω
6 0° V
1Ω 2Ix
+ −
j1 Ω
Figure P8.158
SOLUTION:
Chapter 8: AC Steady-State Analysis
Problem 8.158
2
Problem 8.158
Irwin, Basic Engineering Circuit Analysis, 11/E
Chapter 8: AC Steady-State Analysis
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