Ch08 Solution

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.



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



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).



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



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



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



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



SOLUTION:

8.8 Find the frequency-domain impedance. Z, as shown in Fig. P8.8.

Z 3 Ω j4 Ω

Figure P8.8



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



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



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



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



SOLUTION:

8.13 Find the frequency-domain impedance, Z, as shown in Fig. P8.13.

−j2 Ωj1 Ω2 ΩZ

Figure P8.13



SOLUTION:


Z 1 Ω 10 μF

2 Ω10 mH

Figure P8.14



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



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



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



SOLUTION:


Z 10 mH 500 μF

4 Ω2 Ω

Figure P8.18



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



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



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



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



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



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



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



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



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



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



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





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





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



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



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





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



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





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:



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





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:



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



SOLUTION:

8.40 Find the frequency-domain current I shown in Fig. P8.40.

10 0° A −j60 Ω80 Ω

I

Figure P8.40



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



SOLUTION:

8.42 Find the voltage V shown in Fig. P8.42.

V

+

−

1 30° A −j2 Ωj1 Ω 1 ΩZ

Figure P8.42



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



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





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





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





SOLUTION:

8.47 Find the voltage V shown in Fig. P8.47.

V100 0° V

1 Ω

−j1 Ω+−

+

−

Figure P8.47



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



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



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



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



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



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



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:



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



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





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





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:



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





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





SOLUTION:


2 60° A 10 30° V30 Ω

10 Ω

j10 ΩVo

+

−

−+

Figure P8.61





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





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



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:



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





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



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:



SOLUTION:


12 0° V2 0° A

j1 Ω

−j1 Ω

1 Ω +−

Io

1 Ω

Figure P8.68





SOLUTION:


Vo

+

−

+−

6 0° V 2 0° A

−j1 Ω

j1 Ω

1 Ω

1 Ω

Figure P8.69





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





SOLUTION:


+

−

Vo4 0° V

6 0° V

12 0° V

−j1 Ω

1 Ω

1 Ω

+−

+−

+−

Figure P8.71



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





SOLUTION:


6 0° V

2 0° A

12 0° V1 Ω

1 Ω1 Ω −j1 Ω j2 Ω

−+ −+

Vo

+

−

Figure P8.73





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





SOLUTION:


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





SOLUTION:


Vo1 Ω–j1 Ω

1 Ω +

−

+−

12 0° V

+ −2Vx

Vx+ −

Figure P8.76





SOLUTION:


2 0° A

6 0° V1 Ω Vo

+

−

+–

−j1 Ω

1 Ω

+− 2Vx

Vx+ −

Figure P8.77



SOLUTION:


1 Ω

1 Ω 1 Ω

+–

−j1 Ω

12 0° V+−

2Vx

Vx− +

Io

Figure P8.78



SOLUTION:


+

−

1 Ω

j1 Ω

1 Ω1 Ω

+–

+–

4 0° V

Vo

4Vo

Figure P8.79





SOLUTION:


12 0° V

−j1 Ω

+−

1 Ω

1 Ω

1 ΩIo

1 Ω

+ −

−+

2Vx

Vx

Figure P8.80





SOLUTION:


4Vo 1 Ω

−j Ω Vo

+

−

+–

+–

1 Ω

1 Ω

4 0° V

Figure P8.81



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





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



SOLUTION:


Vo

+

−

Vx

4Vx+

−

+−

1 Ω

1 Ω

1 Ω

−j1 Ω

4 0° V

−+

Figure P8.84





SOLUTION:


6 0° V2 0° A

1 Ω

1 Ω 1 Ω

1 Ω−j1 Ω

+− +

−

Vx

+

−

Vo

+

−

4Vx

Figure P8.85





SOLUTION:


6 0° V 2 0° A

4 0° A

1 Ω

−j1 Ω

+−Vx

1 Ω

1 Ω

+

−

Vo

+

−

+− 4Vx

Figure P8.86





SOLUTION:


6 0° V1 Ω

1 Ω

1 Ω

1 Ω−j1 Ω

+−

Vx

+

−

Vo

Ix

+ −

4Ix

+− 2Vx

Figure P8.87





SOLUTION:


4 0° V

1 Ω 1 Ω−j1 Ω

+–

Vx

1 Ω1 Ω

+

−

Ix

+– 2Vx4 Ix

Io

Figure P8.88





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





SOLUTION:

8.90 Solve problem 8.67 using loop analysis.



SOLUTION:





SOLUTION:



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



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





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





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



SOLUTION:


−+

+−

2 0° A16 0° V

12 0° V

2 Ω

2 Ω

−j1 Ω

j1 Ω

Vo

+

−

Figure P8.97





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



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



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





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



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





SOLUTION:


4 0° A

j1 Ω

−j1 Ω 1 Ω

1 Ω2Ix Vo

+

−

Ix

Figure P8.103





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





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



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



SOLUTION:

8.107 Solve problem 8.67 using superposition.



SOLUTION:




SOLUTION:






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





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



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





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





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





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





SOLUTION:

8.116 Solve problem 8.68 using source exchange.



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



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



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





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





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





SOLUTION:

8.122 Solve problem 8.67 using Thévenin’s theorem.



SOLUTION:




SOLUTION:




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:



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





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





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





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





SOLUTION:


+−

2 0° A12 0° V

−j1 Ω

1 Ω

1 Ω

1 ΩIo

Figure P8.130





SOLUTION:


1 Ω

+

−

Vo

2 0° A

12 0° V

1 Ω 1 Ω

−j1 Ω

−+

Figure P8.131





SOLUTION:


12 0° V

4 0° A

3 0° A

1 Ω –j1 Ω

j1 Ω

Io

1 Ω

1 Ω+−

Figure P8.132





SOLUTION:


6 0° A

4 0° A

4 0° A

1 Ω

j2 Ω

−j1 Ω

1 Ω

2 Ω Vo

+

−

Figure P8.133





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





SOLUTION:


2 0° A

6 0° V1 Ω

1 Ω

Vo

+

−

+–

−j1 Ω

+− 2Vx

Vx+ −

Figure P8.135





SOLUTION:


1 Ω

1 Ω 1 Ω

+–

−j1 Ω

12 0° V+−

2Vx

Vx− +

Io

Figure P8.136





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





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







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





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





SOLUTION:


12 0° V

−j1 Ω

+−

1 Ω

1 Ω

1 Ω

Io

1 Ω

+ −

−+

2Vx

Vx

Figure P8.141







SOLUTION:

8.142 Solve problem 8.68 using Norton’s theorem.



SOLUTION:

8.143 Solve problem 8.69 using Norton’s theorem.



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





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





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





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





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





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





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





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





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





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





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





SOLUTION:


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





SOLUTION:


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





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





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



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Ch08 Solution