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Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.1
SOLUTION:
8.1 Given i(t) = 5 cos (400t − 120°) A, determine the period of the current and the frequency in Hertz.
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.2
SOLUTION:
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
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.3
SOLUTION:
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).
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.4
SOLUTION:
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.
1 2 3 4 5
24 V
– 24 V
12 V
6 7 8 9 10 11 12 ms–1–2–3–4
υ(t)
Figure P8.4
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.5
SOLUTION:
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.
2 Ω
+
−
i(t)
υ(t)
Figure P8.5
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.6
SOLUTION:
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.
υ(t)
i(t)
C = 1 μF
+
−
Figure P8.6
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.7
SOLUTION:
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
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.8
SOLUTION:
8.8 Find the frequency-domain impedance. Z, as shown in Fig. P8.8.
Z 3 Ω j4 Ω
Figure P8.8
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.9
SOLUTION:
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)
L = 1 mH
+
−
υ(t)
Figure P8.9
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.10
SOLUTION:
8.10 Find the frequency-domain impedance, Z, in the network in Fig. P8.10.
j1 Ω −j2 Ω
1 Ω
Z
Figure P8.10
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.11
SOLUTION:
8.11 Find Z in the network in Fig. P8.11.
1 Ω2 Ω
2 Ω 2 Ω
−j1 Ω
j2 Ωj2 ΩZ
Figure P8.11
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.12
SOLUTION:
8.12 Find the impedance, Z, shown in Fig. P8.12 at a frequency of 400 Hz.
1 Ω
2 Ω10 mH
10 μFZ
Figure P8.12
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.13
SOLUTION:
8.13 Find the frequency-domain impedance, Z, as shown in Fig. P8.13.
−j2 Ωj1 Ω2 ΩZ
Figure P8.13
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.14
SOLUTION:
8.14 Find the impedance, Z, shown in Fig. P8.14 at a frequency of 60 Hz.
Z 1 Ω 10 μF
2 Ω10 mH
Figure P8.14
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.15
SOLUTION:
8.15 Find Y in the network in Fig. P8.15.
j2 S
j1 S
–j1 S
–j2 S
1 S
2 S
2 S
Y
Figure P8.15
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.16
SOLUTION:
8.16 Find the equivalent impedance for the circuit in Fig. P8.16.
Zeq
8 Ω
j10 Ω
j5 Ω
4 Ω
10 Ω
−j2 Ω
−j8 Ω
Figure P8.16
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.17
SOLUTION:
8.17 Find the frequency-domain impedance, Z, shown in Fig. P8.17.
Z
−j1 Ω
−j2 Ω
1 Ω
j2 Ω
2 Ω
Figure P8.17
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.18
SOLUTION:
8.18 Find the impedance, Z, shown in Fig. P8.18 at a frequency of 60 Hz.
Z 10 mH 500 μF
4 Ω2 Ω
Figure P8.18
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.19
SOLUTION:
8.19 Find the frequency-domain impedance, Z, shown in Fig. P8.19.
Z
1 Ω
4 Ω
2 Ω
j2 Ω
j4 Ω
j2 Ω
j1 Ω
–j1 Ω
6 Ω
Figure P8.19
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.20
SOLUTION:
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.
+–12 cos (1000t + 75°) V
100 μF
4 Ω
L
Figure P8.20
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.21
SOLUTION:
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.
Z C
1 Ω 5 mH
Figure P8.21
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.22
SOLUTION:
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?
6 Ω
10 mH
CZ
Figure P8.22
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.23
SOLUTION:
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
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.24
SOLUTION:
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
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.25
SOLUTION:
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).
30 Ω
250 nF
i1(t) i2(t) +−υo(t)
+
−
−
+
υS(t)
υC(t)
Figure P8.25
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.25 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.26
SOLUTION:
8.26 The impedance of the circuit in Fig. P8.26 is real at f = 60 Hz. What is the value of L?
2 Ω
L
10 mFZ
Figure P8.26
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.27
SOLUTION:
8.27 Find the frequency at which the circuit shown in Fig. P8.27 is purely resistive.
5 mH 1 mF1 ΩZ
Figure P8.27
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.28
SOLUTION:
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
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.29
SOLUTION:
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).
20 Ω
i(t)
+− υR(t)
+
−
+ −
υS(t)
υC(t)
66.67 μF
Figure P8.29
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.29 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.30
SOLUTION:
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).
10 Ω υo(t)+
−
iS(t)
iC(t) iR(t)
20 μF
Figure P8.30
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.30 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.31
SOLUTION:
8.31 Find iC(t) and i(t) in the network in Fig. P8.31.
+−
υ(t) = 120cos (5000t) V
16 mH
60 Ω
2.5 μF
iC(t)i(t)
Figure P8.31
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.32
SOLUTION:
8.32 If υs(t) = 20 cos 5t volts, find υo(t) in the network in Fig. P8.32.
υo(t)
+
−
3 Ω
10 Ω+−
0.5 H
0.02 F1 Hυs(t)
Figure P8.32
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.33
SOLUTION:
8.33 Find υo(t) in the circuit in Fig. P8.33.
υo(t)
+
−
20 Ω
15 Ω10 Ω+
−20 mH
30 mH200 μF 100 μF
170 cos 377t V
Figure P8.33
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.33 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.34
SOLUTION:
8.34 Find υo(t) in the network in Fig. P8.34.
5 Ω5 Ω
0.4 H 0.02 F
0.1 F
5 cos 10t Aυo(t)+
−
Figure P8.34
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.35
SOLUTION:
8.35 Find io(t) in the circuit in Fig. P8.35 if υ (t) = 50 cos 100t V.
3 Ω
6 Ω
5 Ω
4 Ω 20 mH
50 mH
2000 μF
1000 μFio(t)υ(t) +–
Figure P8.35
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.35 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.36
8.36 Find υo(t) and io(t) in the network in Fig. P8.36.
5 Ω5 Ω
0.4 H 0.02 F
0.01 F
25 cos 20t Vυo(t)
+
−
+−
io(t)
Figure P8.36
SOLUTION:
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.37
SOLUTION:
8.37 Calculate υo(t) in Fig. P8.37.
3 Ω5 Ω4 Ω
6 Ω
2 H 2 Ω
1 H
0.05 F
50 cos 5t V 0.2 F+−
υo(t)+
−
Figure P8.37
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.37 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.38
8.38 Find i2(t) in the circuit in Fig. P8.38.
8 Ω
6 Ω
2 Ω5 Ω
4 Ω
25 mH
10 mH625 μF 2 cos 400t A
i2(t)
Figure P8.38
SOLUTION:
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.39
SOLUTION:
8.39 Find the voltage Vo shown in Fig. P8.39.
10 120° V
2 Ω
−j12 Ω
j10 Ω
+− Vo
+
−
Figure P8.39
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.40
SOLUTION:
8.40 Find the frequency-domain current I shown in Fig. P8.40.
10 0° A −j60 Ω80 Ω
I
Figure P8.40
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.41
SOLUTION:
8.41 Find the frequency-domain voltage Vo shown in Fig. P8.41.
Vo
+
−
0.5 25° A −j5 Ωj2 Ω 10 Ω
Figure P8.41
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.42
SOLUTION:
8.42 Find the voltage V shown in Fig. P8.42.
V
+
−
1 30° A −j2 Ωj1 Ω 1 ΩZ
Figure P8.42
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.43
SOLUTION:
8.43 Find the frequency-domain current I shown in Fig. P8.43.
37 −145° V
j6 Ω
−j4 Ω
5 Ω
10 Ω+−
I
Figure P8.43
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.44
SOLUTION:
8.44 Find υx(t) in the circuit in Fig. P8.44.
10 Ω
5 Ω
+−
+−
0.1 H
0.2 H
0.005 F
100 cos 40t V 40 cos (40t − 30°) Vυx(t)
+
−
Figure P8.44
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.44 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.45
SOLUTION:
8.45 Find υo(t) in the network in Fig. P8.45.
+–
5 Ω
10 Ω 0.1 H
0.2 H
0.008 F
50 cos 25t V
υx(t)+ −
2υx(t)υo(t)
+–
Figure P8.45
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.45 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.46
SOLUTION:
8.46 Find υ1(t) and υ2(t) in the circuit in Fig. P8.46.
2 Ω 0.2 H
0.02 F υ1(t)
+
−
υ2(t)
+−
4 cos 10t A 2 cos (10t + 15°) A
12 cos (10t − 25°) V
Figure P8.46
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.46 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.47
SOLUTION:
8.47 Find the voltage V shown in Fig. P8.47.
V100 0° V
1 Ω
−j1 Ω+−
+
−
Figure P8.47
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.48
SOLUTION:
8.48 Find the voltage Vo shown in Fig. P8.48.
10 30° V
2 Ω
−j12 Ω
j10 Ω
+− Vo
+
−
Figure P8.48
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.49
SOLUTION:
8.49 Find the frequency-domain voltage Vo shown in Fig. P8.49.
Vo
1 Ω
5 30° A15 Ω −j12 Ω
+
−
Figure P8.49
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.50
SOLUTION:
8.50 Find Vo in the network in Fig. P8.50.
2 Ω
2 Ω
j1 Ω
−j1 Ω
4 Ω−+ Vo
+
−
12 0° V
Figure P8.50
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.51
SOLUTION:
8.51 Determine Io in the network shown in Fig. P8.51 if VS = 12 0° V.
2 Ω 2 Ω 2 Ω
j2 Ω j2 Ω V2VS Io
+
−
V1
+
−
−+
Figure P8.51
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.52
SOLUTION:
8.52 Given the network in Fig. P8.52, determine the value of Vo if VS = 24 0° V.
VoVS
2 Ω j2 Ω
2 Ω−j1 Ω
+
−
V1
+
−
+−
Figure P8.52
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.53
SOLUTION:
8.53 Find VS in the network in Fig. P8.53 if V1 = 4 0° V.
V1
VSj1 Ω
2 Ω1 Ω 2 Ω −j1 Ω
−j1 Ω+−
+
−
Figure P8.53
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.54
8.54 If V1 = 4 0° V, find Io
in Fig. P8.54.
−j2 Ω
2 Ω
j1 Ω 1 ΩIS
V1
Io
+−
Figure P8.54
SOLUTION:
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.55
SOLUTION:
8.55 In the network in Fig. P8.55, Io = 4 0° A. Find Ix.
1 Ω
1 Ω
1 Ω
j1 Ω −j1 Ω
1 Ω
IoIx
+−
2 0° A
12 0° V
Figure P8.55
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.56
SOLUTION:
8.56 If Io = 4 0° A in the circuit in Fig. P8.56, find Ix.
1 Ω
1 Ω
1 Ω
j1 Ω −j1 Ω 1 Ω
IoIx
1 Ω+– 2 0° A12 0° V
+− 4 0° V
Figure P8.56
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.56 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.57
SOLUTION:
8.57 If Io = 4 0° A in the network in Fig. P8.57, find Ix.
1 Ω 1 Ω
1 Ω
1 Ω
1 Ω
−j1 Ω
j1 Ω
Io
Ix
R2
1 Ω
+−
2 0° A
12 0° V
Figure P8.57
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.57 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.58
8.58 In the network in Fig. P8.58, Vo is known to be 4 45° V. Find Z.
2 Ω −j1 Ω
1 Ω+−12 0° V Z Vo
+
−
Figure P8.58
SOLUTION:
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.59
SOLUTION:
8.59 In the network in Fig. P8.59, V1 = 5 −120° V. Find Z.
Z
2 Ω
0.25 Ω
j1 Ω
−j0.25 Ω
V1+ −
6 0° A
Figure P8.59
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.59 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.60
SOLUTION:
8.60 Find Vo in the circuit in Fig. P8.60.
Vo
+
−
24 0° V2 90° A
2 Ω+−
−j2 Ω2 Ω
j2 Ω
Figure P8.60
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.60 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.61
SOLUTION:
8.61 Find Vo in the network in Fig. P8.61.
2 60° A 10 30° V30 Ω
10 Ω
j10 ΩVo
+
−
−+
Figure P8.61
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.61 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.62
SOLUTION:
8.62 Using nodal analysis, find Io in the circuit in Fig. P8.62.
Io
V1 V2
+− 4 0° A
2 0° A12 0° V1 Ω
1 Ω 1 Ω
−j1 Ω
Figure P8.62
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.62 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.63
SOLUTION:
8.63 Use nodal analysis to find Io in the circuit in Fig. P8.63.
Io
V
+−
+− 6 0° V
4 0° A
12 0° V
2 Ω
2 Ω
−j1 Ω
j2 Ω
Figure P8.63
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.64
8.64 Find Vo in the network in Fig. P8.64 using nodal analysis.
2 0° A 4 0° V12 0° V
2 Ω
2 Ω 1 Ω
V
−j1 Ω
j2 Ω
Vo
+
−
+−
+−
Figure P8.64
SOLUTION:
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.65
SOLUTION:
8.65 Use the supernode technique to find Io in the circuit in Fig. P8.65.
12 0° V2 Ω
2 Ω1 Ω
Io
+−
j2 Ω −j2 Ω
−j1 Ω
Figure P8.65
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.65 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.66
SOLUTION:
8.66 Find I1 and Vo in the network in Fig. P8.66.
Vo
+
−
6 0° V12 45° V 1 Ω2 Ω −j1 Ω −j1 Ωj4 Ω
−+
+−
I1
Figure P8.66
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.67
8.67 Use nodal analysis to find Vo in the network in Fig. P8.67.
Vo
+
−
2 0° A
12 0° V
2 Ω
−j2 Ω
j4 Ω
+−
Figure P8.67
SOLUTION:
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.68
SOLUTION:
8.68 Use nodal analysis to find Io in the circuit in Fig. P8.68.
12 0° V2 0° A
j1 Ω
−j1 Ω
1 Ω +−
Io
1 Ω
Figure P8.68
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.68 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.69
SOLUTION:
8.69 Use nodal analysis to find Vo in the network in Fig. P8.69.
Vo
+
−
+−
6 0° V 2 0° A
−j1 Ω
j1 Ω
1 Ω
1 Ω
Figure P8.69
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.69 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.70
SOLUTION:
8.70 Use nodal analysis to find Vo in the circuit in Fig. P8.70.
1 Ω 1 Ω
1 Ω Vo
+
−2 0° A
12 0° V+−
−j1 Ω
Figure P8.70
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.70 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.71
SOLUTION:
8.71 Use nodal analysis to find Vo in the network in Fig. P8.71.
+
−
Vo4 0° V
6 0° V
12 0° V
−j1 Ω
1 Ω
1 Ω
+−
+−
+−
Figure P8.71
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.72
SOLUTION:
8.72 Use nodal analysis to find Io in the network in Fig. P8.72.
1 Ω
1 Ω 1 Ω
Io
−j1 Ω
12 0° V 2 0° A
+−
Figure P8.72
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.72 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.73
SOLUTION:
8.73 Use nodal analysis to find Vo in the circuit in Fig. P8.73.
6 0° V
2 0° A
12 0° V1 Ω
1 Ω1 Ω −j1 Ω j2 Ω
−+ −+
Vo
+
−
Figure P8.73
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.73 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.74
SOLUTION:
8.74 Find Io in the circuit in Fig. P8.74 using nodal analysis.
2 0° A
6 0° V12 0° V
−j2 Ω
2 Ω 2 Ω 1 Ω
j1 Ω
Io
+−
+−
Figure P8.74
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.74 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.75
SOLUTION:
8.75 Use nodal analysis to find Io in the circuit in Fig. P8.75.
6 0° V 4 0° A
2 0° A
12 0° V
j1 Ω
−j1 Ω 1 Ω1 Ω
1 Ω
1 Ω
Io
+−
−+
Figure P8.75
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.75 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.76
SOLUTION:
8.76 Use nodal analysis to find Vo in the network in Fig. P8.76.
Vo1 Ω–j1 Ω
1 Ω +
−
+−
12 0° V
+ −2Vx
Vx+ −
Figure P8.76
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.76 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.77
SOLUTION:
8.77 Use nodal analysis to find Vo in the network in Fig. P8.77.
2 0° A
6 0° V1 Ω Vo
+
−
+–
−j1 Ω
1 Ω
+− 2Vx
Vx+ −
Figure P8.77
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.78
SOLUTION:
8.78 Use nodal analysis to find Io in the network in Fig. P8.78.
1 Ω
1 Ω 1 Ω
+–
−j1 Ω
12 0° V+−
2Vx
Vx− +
Io
Figure P8.78
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.79
SOLUTION:
8.79 Use nodal analysis to find Vo in the network in Fig. P8.79.
+
−
1 Ω
j1 Ω
1 Ω1 Ω
+–
+–
4 0° V
Vo
4Vo
Figure P8.79
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.79 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.80
SOLUTION:
8.80 Use nodal analysis to find Io in the network in Fig. P8.80.
12 0° V
−j1 Ω
+−
1 Ω
1 Ω
1 ΩIo
1 Ω
+ −
−+
2Vx
Vx
Figure P8.80
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.80 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.81
SOLUTION:
8.81 Use nodal analysis to find Vo in the circuit in Fig. P8.81.
4Vo 1 Ω
−j Ω Vo
+
−
+–
+–
1 Ω
1 Ω
4 0° V
Figure P8.81
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.82
SOLUTION:
8.82 Use nodal analysis to find Vx in the circuit in Fig. P8.82.
12 0° V
1 Ω1 Ω −j1 Ω
j1 Ω
2Vx
−+
Vx
+
−
Figure P8.82
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.82 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.83
SOLUTION:
8.83 Find the voltage across the inductor in the circuit shown in Fig. P8.83 using nodal analysis.
10 30° V 4 Ω
−j2 Ω
j1 Ω 2Ix
V1 V2
Ix
+−
Figure P8.83
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.84
SOLUTION:
8.84 Use nodal analysis to find Vo in the circuit in Fig. P8.84.
Vo
+
−
Vx
4Vx+
−
+−
1 Ω
1 Ω
1 Ω
−j1 Ω
4 0° V
−+
Figure P8.84
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.84 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.85
SOLUTION:
8.85 Use nodal analysis to find Vo in the network in Fig. P8.85.
6 0° V2 0° A
1 Ω
1 Ω 1 Ω
1 Ω−j1 Ω
+− +
−
Vx
+
−
Vo
+
−
4Vx
Figure P8.85
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.85 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.86
SOLUTION:
8.86 Use nodal analysis to find Vo in the circuit in Fig. P8.86.
6 0° V 2 0° A
4 0° A
1 Ω
−j1 Ω
+−Vx
1 Ω
1 Ω
+
−
Vo
+
−
+− 4Vx
Figure P8.86
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.86 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.87
SOLUTION:
8.87 Use nodal analysis to find Vo in the circuit in Fig. P8.87.
6 0° V1 Ω
1 Ω
1 Ω
1 Ω−j1 Ω
+−
Vx
+
−
Vo
Ix
+ −
4Ix
+− 2Vx
Figure P8.87
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.87 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.88
SOLUTION:
8.88 Use nodal analysis to find Io in the circuit in Fig. P8.88.
4 0° V
1 Ω 1 Ω−j1 Ω
+–
Vx
1 Ω1 Ω
+
−
Ix
+– 2Vx4 Ix
Io
Figure P8.88
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.88 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.89
SOLUTION:
8.89 Use mesh analysis to find Vo in the circuit shown in Fig. P8.89.
6 0° V
12 45° V
2 Ω
−j1 Ω
j2 Ω
I1
−+
Vo
+
−
+−
I2
Figure P8.89
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.89 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.90
SOLUTION:
8.90 Solve problem 8.67 using loop analysis.
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.91
SOLUTION:
8.91 Solve problem 8.68 using loop analysis.
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.92
8.92 Solve problem 8.69 using loop analysis.
SOLUTION:
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.93
SOLUTION:
8.93 Find Vo in the network in Fig. P8.93 using loop analysis.
j2 Ω−j1 Ω
12 0° V
2 0° A
1 Ω
1 Ω
Vo
+
−
+–
Figure P8.93
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.94
SOLUTION:
8.94 Find Vo in the network in Fig. P8.94 using loop analysis.
+ −
+−
2 0° A12 0° V
−j1 Ω
1 Ω
1 Ω
1 Ω
Vo
Figure P8.94
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.94 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.95
SOLUTION:
8.95 Find Io in the network in Fig. P8.95 using loop analysis.
2 0° A4 0° A
12 0° V
−j1 Ω j1 Ω
1 Ω
Io
1 Ω
1 Ω
+−
Figure P8.95
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.95 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.96
SOLUTION:
8.96 Find Vo in the circuit in Fig. P8.96 using mesh analysis.
2 0° A
6 0° V
4 0° A
−j1 Ω
−+
2 Ω
1 Ω
2 Ω
j2 Ω
Vo
+
−
Figure P8.96
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.97
SOLUTION:
8.97 Find Vo in the network in Fig. P8.97.
−+
+−
2 0° A16 0° V
12 0° V
2 Ω
2 Ω
−j1 Ω
j1 Ω
Vo
+
−
Figure P8.97
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.97 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.98
SOLUTION:
8.98 Determine Vo in the circuit in Fig. P8.98.
6 0° A 12 0° V
2 Ω 2 Ω
1 Ω
−j1 Ω
j2 Ω Vo
+
−
+−
Figure P8.98
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.99
SOLUTION:
8.99 Use mesh analysis to find Vo in the circuit shown in Fig. P8.99.
12 0° V 4 90° A
j2 Ω4 Ω
2 Ω
−j4 Ω
I1+−
I2
Vo
+
−
Figure P8.99
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.100
SOLUTION:
8.100 Using loop analysis, find Io in the network in Fig. P8.100.
2 0° A
4 0° A
12 0°V
−j2 Ω 2 Ω
Io
2 Ω
j1 Ω
+−
Figure P8.100
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.100 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.101
SOLUTION:
8.101 Use mesh analysis to find Vo in the circuit in Fig. P8.101.
2 0° A
6 0° A 4 0°A
−j2 Ω
2 Ω 2 Ω
1 Ω
j1 Ω
Vo
+
−
Figure P8.101
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.102
SOLUTION:
8.102 Use loop analysis to find Io in the network in Fig. P8.102.
2 0° A
4 30° A
1 Ω
1 Ω
1 Ω
1 Ω2Vx Vx
Io
+
−
−j1 Ω
+−
Figure P8.102
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.102 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.103
SOLUTION:
8.103 Find Vo in the network in Fig. P8.103.
4 0° A
j1 Ω
−j1 Ω 1 Ω
1 Ω2Ix Vo
+
−
Ix
Figure P8.103
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.103 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.104
SOLUTION:
8.104 Use loop analysis to find Vo in the circuit in Fig. P8.104.
4Ix
1 Ω
−j1 Ω
+–
1 ΩVo
Ix
+
−
1 Ω
4 0° V
Figure P8.104
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.104 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.105
SOLUTION:
8.105 Use loop analysis to find Vo in the network in Fig. P8.105.
6 0° V
4 0° A
2 0° A1 Ω
1 Ω
−j1 Ω
+−
1 Ω
Vx
Vx
+
−
Vo
+
−
+−
Figure P8.105
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.106
SOLUTION:
8.106 Use superposition to find Vo in the network in Fig. P8.106.
12 0° V
4 0° A1 Ω
1 Ω
−j1 Ω
−+
Vo
+
−
Figure P8.106
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.107
SOLUTION:
8.107 Solve problem 8.67 using superposition.
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.108
SOLUTION:
8.108 Solve problem 8.68 using superposition.
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.109
SOLUTION:
8.109 Solve problem 8.69 using superposition.
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.109 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.110
SOLUTION:
8.110 Use superposition to determine Vo in the circuit in Fig. P8.110.
6 0° A6 0° V
1 Ω
1 Ω
−j1 Ω j1 Ω
Vo
+
−
−+
Figure P8.110
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.110 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.111
SOLUTION:
8.111 Using superposition, find Vo in the circuit in Fig. P8.111.
2 0° A6 0° V
1 Ω −j1 Ω
j2 Ω
2 Ω
+−
Vo
+
−
Figure P8.111
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.112
SOLUTION:
8.112 Find Vo in the network in Fig. P8.112 using superposition.
4 0° A
16 0° V 2 Ω
2 Ω
1 Ω
−j2 Ω
j3 Ω
Vo
+
−
+−
Figure P8.112
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.112 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.113
SOLUTION:
8.113 Find Vo in the network in Fig. P8.113 using superposition.
2 0° A 12 0° V
2 Ω
−j2 Ω j4 Ω 2 Ω
−+
Vo
+
−
Figure P8.113
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.113 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.114
SOLUTION:
8.114 Use superposition to find Vo in the circuit in Fig. P8.114.
6 0° V
2 0° A 4 0° A
1 Ω−j1 Ω
j1 Ω
+−
1 Ω1 Ω
Vo
+
−
Figure P8.114
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.114 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.115
SOLUTION:
8.115 Use superposition to find Vo in the network in Fig. P8.115.
−j1 Ω 4 0° A
1 Ω1 Ω
1 Ω
Vo
+
−
6 0° V+−
Figure P8.115
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.115 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.116
SOLUTION:
8.116 Solve problem 8.68 using source exchange.
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.117
SOLUTION:
8.117 Use source exchange to find the current Io in the network in Fig. P8.117.
2 0° A
12 0° V
2 Ω
2 Ω
Io
−j1 Ω
1 Ω
4 0° A
+−
Figure P8.117
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.118
SOLUTION:
8.118 Use source exchange to determine Vo in the network in Fig. P8.118.
−+
Vo
+
−
6 0° V2 0° A12 0° V 2 Ω1 Ω
−j1 Ω
+−
Figure P8.118
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.119
SOLUTION:
8.119 Use source transformation to find Vo in the circuit in Fig. P8.119.
12 0° V
1 Ω 1 Ω
1 Ω
12 Ω
+−
2 0° A
Vo
+
−
−j2 Ω
Figure P8.119
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.119 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.120
SOLUTION:
8.120 Use source transformation to find Vo in the circuit in Fig. P8.120.
1 Ω
1 Ω
2 Ω
2 Ω
5 Ω
+−
+−4 0° V
2 0° A
6 0° V
−j2 Ω
Vo
+
−
Figure P8.120
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.120 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.21
SOLUTION:
8.21 Use source transformation to find Io in the circuit in Fig. P8.121.
12 Ω
1 Ω
6 Ω
3 Ω
+−
+−
6 0° V
2 0° A
4 0° A
8 0° V
−j3 Ω 4 Ω
Io
Figure P8.121
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.21 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.122
SOLUTION:
8.122 Solve problem 8.67 using Thévenin’s theorem.
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.123
SOLUTION:
8.123 Solve problem 8.68 using Thévenin’s theorem.
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.124
SOLUTION:
8.124 Solve problem 8.69 using Thévenin’s theorem.
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.125
8.125 Use Thevenin’s theorem to find Vo in the circuit in Fig. P8.125.
12 0° V 4 90° A
2 Ω
4 Ω j2 Ω
−j4 Ω
I1
Vo
+
−
+−
I2
Figure P8.125
SOLUTION:
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.126
SOLUTION:
8.126 Apply Thévenin’s theorem twice to find Vo in the circuit in Fig. P8.126.
6 0° V
1 Ω
1 Ωj1 Ω−j1 Ω
1 Ω2 Ω
+− Vo
+
−
Figure P8.126
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.126 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.127
SOLUTION:
8.127 Use Thévenin’s theorem to find Vo in the network in Fig. P8.127.
1 Ω
–j2 Ω
+−
2 0° A
6 0° V
1 Ω
+
−
Vo
Figure P8.127
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.127 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.128
SOLUTION:
8.128 Use Thévenin’s theorem to find Io in the network in Fig. P8.128.
+−
12 0° V
2 0° A
−j1 Ω
j1 Ω
1 Ω
1 Ω
Io
Figure P8.128
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.128 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.129
SOLUTION:
8.129 Use Thévenin’s theorem to find the voltage across the 2-Ω resistor in the network in Fig. P8.129.
2 Ω
2 0° A1 Ω
12 0° V
+−−j1 Ω
j1 Ω
Figure P8.129
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.129 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.130
SOLUTION:
8.130 Use Thévenin’s theorem to find Io in the network in Fig. P8.130.
+−
2 0° A12 0° V
−j1 Ω
1 Ω
1 Ω
1 ΩIo
Figure P8.130
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.130 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.131
SOLUTION:
8.131 Use Thévenin’s theorem to find Vo in the network in Fig. P8.131.
1 Ω
+
−
Vo
2 0° A
12 0° V
1 Ω 1 Ω
−j1 Ω
−+
Figure P8.131
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.131 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.132
SOLUTION:
8.132 Use Thévenin’s theorem to find Io in the network in Fig. P8.132.
12 0° V
4 0° A
3 0° A
1 Ω –j1 Ω
j1 Ω
Io
1 Ω
1 Ω+−
Figure P8.132
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.132 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.133
SOLUTION:
8.133 Use Thévenin’s theorem to find Vo in the network in Fig. P8.133.
6 0° A
4 0° A
4 0° A
1 Ω
j2 Ω
−j1 Ω
1 Ω
2 Ω Vo
+
−
Figure P8.133
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.133 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.134
SOLUTION:
8.134 Use Thévenin’s theorem to determine Io in the circuit in Fig. P8.134.
+−
4 0° A
2 0° A
6 0° V1 Ω 1 Ω1 Ω
Io
−j2 Ω
Figure P8.134
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.134 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.135
SOLUTION:
8.135 Use Thévenin’s theorem to find Vo in the network in Fig. P8.135.
2 0° A
6 0° V1 Ω
1 Ω
Vo
+
−
+–
−j1 Ω
+− 2Vx
Vx+ −
Figure P8.135
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.135 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.136
SOLUTION:
8.136 Use Thévenin’s theorem to find Io in the network in Fig. P8.136.
1 Ω
1 Ω 1 Ω
+–
−j1 Ω
12 0° V+−
2Vx
Vx− +
Io
Figure P8.136
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.136 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.137
SOLUTION:
8.137 Find Vo in the network in Fig. P8.137 using Thévenin’s theorem.
12 0° V
1 Ω1 Ω −j1 Ω
j1 Ω
2Vx
−+
Vo
+
−
Vx
+
−
Figure P8.137
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.137 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.138
SOLUTION:
8.138 Find the Thévenin’s equivalent for the network in Fig. P8.138 at terminals A–B.
2Vx
4 0° A
1 Ω
−j1 Ω
j1 Ω
A
B
Vx
+
−
Figure P8.138
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.138 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 3
Chapter 8: AC Steady-State Analysis Problem 8.138
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.139
SOLUTION:
8.139 Given the network in Fig. P8.139, find the Thévenin’s equivalent of the network at terminals A–B.
V2
V1 V3
1 Ω
1 Ω A
B
−j1 Ω
j1 Ω+−
−++−
6 0° V
4 0° V
12 0° V
Figure P8.139
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.139 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.140
SOLUTION:
8.140 Use Thévenin’s theorem to determine Io in the network in Fig. P8.140.
4 0° A
12 0° V 1 Ω
1 Ω 1 Ω
Io
1 Ω 1 Ω−j1 Ω
j1 Ω
2 0° A
+− 6 0° V+
−
Figure P8.140
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.140 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.141
SOLUTION:
8.141 Use Thévenin’s theorem to find Io in the network in Fig. P8.141.
12 0° V
−j1 Ω
+−
1 Ω
1 Ω
1 Ω
Io
1 Ω
+ −
−+
2Vx
Vx
Figure P8.141
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.141 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 3
Chapter 8: AC Steady-State Analysis Problem 8.141
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.142
SOLUTION:
8.142 Solve problem 8.68 using Norton’s theorem.
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.143
SOLUTION:
8.143 Solve problem 8.69 using Norton’s theorem.
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.144
SOLUTION:
8.144 Find Vx in the circuit in Fig. P8.144 using Norton’s theorem.
−+
11.3 45° V
2 0° A j4 Ω −j3 Ω10 Ω Vx
+
−
Figure P8.144
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.144 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.145
SOLUTION:
8.145 Find Io in the network in Fig. P8.145 using Norton’s theorem.
6 45° A 2 0° Aj1 Ω
Io
2 Ω
−j2 Ω
Figure P8.145
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.145 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.146
SOLUTION:
8.146 Use Norton’s theorem to find Vo in the network in Fig. P8.146.
4 0° A
2Vx 1 Ω
1 Ω
1 Ω
Vo
+
−
Vx
+
−
−j1 Ω
j1 Ω
Figure P8.146
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.146 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.147
SOLUTION:
8.147 Find Vo using Norton’s theorem for the circuit in Fig. P8.147.
4 0° V 8 0° V
Vx
2Vx
1 Ω
1 Ω
1 Ω
+−
+−
+ −
+− Vo
+
−
−j1 Ω
j1 Ω
Figure P8.147
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.147 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.148
SOLUTION:
8.148 Use Norton’s theorem to find Vo in the circuit in Fig. P8.148.
1 Ω
1 Ω
−j1 Ω
2 Ω1 Ω
Ix
4Ix 2 0° A
4 0° A Vo
+
−
Figure P8.148
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.148 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.149
SOLUTION:
8.149 Apply Norton’s theorem to find Vo in the network in Fig. P8.149.
4 0° A
6 0° V
j1 Ω
1 Ω
1 Ω
1 Ω
1 Ω
−j1 Ω
+−
Vo
+
−
Figure P8.149
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.149 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.150
SOLUTION:
8.150 Find Vo in the circuit in Fig. P8.150.
6 0° V
12 0° V
1 Ω
1 Ω1 Ω
1 Ω1 Ω
2 0° A+−
−+
Vo
+
−
−j1 Ω
j1 Ω
Figure P8.150
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.150 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.151
SOLUTION:
8.151 Find the node voltages in the network in Fig. P8.151.
2 0° A
12 0° V
j1 Ω
2 Ω
2 Ω
1 Ω
1 Ω1 Ω
−j1 Ω
−j2 Ω+−
Figure P8.151
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.151 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.152
SOLUTION:
8.152 Determine Vo in the network in Fig. P8.152.
6 0° V2 0° A
1 Ω−j1 Ω j1 Ω 1 Ω
1 Ω
1 Ω 1 Ω 12 30° V+−
+−
Vo
+
−
Figure P8.152
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.152 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.153
SOLUTION:
8.153 Find Io in the network in Fig. P8.153.
4 0° A
12 0° V
Io 1 Ω
1 Ω1 Ω
1 Ω
1 Ω
−j1 Ω
j1 Ω
2 0° A
+− 6 0° V+
−
Figure P8.153
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.153 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.154
SOLUTION:
8.154 Use both nodal analysis and loop analysis to find Io in the network in Fig. P8.154.
2 0° A
12 0° V
1 Ω
1 Ω1 Ω
1 Ω1 Ω
2Ix
2Vx
Io
Ix
+−
−+
Vx
+
−
−j1 Ω
j1 Ω
Figure P8.154
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.154 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.155
SOLUTION:
8.155 Find Io in the network in Fig. P8.155.
2 0° A
6 0° V 12 0° V
4 0° A
1 Ω 1 Ω
1 Ω
1 Ω
1 Ω−j1 Ω
j1 Ω
−+Io+−
Figure P8.155
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.155 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.156
SOLUTION:
8.156 Find Io in the network in Fig. P8.156.
4 0° V6 0° V
1 Ω1 Ω 1 Ω
1 Ω
1 Ω
1 Ω
j1 Ω
−j1 Ω
2IxIx Io
+−
−+
Figure P8.156
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.156 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.157
SOLUTION:
8.157 Determine Io in the network in Fig. P8.157.
2 0° A
12 0° V
1 Ω
1 Ω1 Ω
1 Ω1 Ω
1 Ω
2Vx
+−−+
Vx
+
−
−j1 Ω
j1 Ω
Io
Figure P8.157
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.157 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 8: AC Steady-State Analysis Problem 8.158
SOLUTION:
8.158 Find Io in the circuit in Fig. P8.158.
2 0° A12 0° V
6 0° V1 Ω
1 Ω
j1 Ω
1 Ω
1 Ω 1 Ω
2Ix
IxIo
+−
+−
−j1 Ω
Figure P8.158
2 Irwin, Basic Engineering Circuit Analysis, 11/E
Problem 8.158 Chapter 8: AC Steady-State Analysis