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AC ANALYSIS

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FREQ

IM(V_PRINT1)

4.000E+02

6.058E+00

IP(V_PRINT1) -1.361E+01

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Irwin, Basic Engineering Circuit Analysis, 9/E

1

SOLUTION:

2

j1

120V

+ 40 V

 j1

1

V0 -

The correct answer is d.

j1

+ 80 A

2

 j1

1

V0 -

Chapter 8: AC Steady-State Analysis

Problem 8.FE-1

2

Irwin, Basic Engineering Circuit Analysis, 9/E

Z 

2( j1)  0.4  j 0.8 2  j1 I0 j1

80  A

Z

1

  0.4  j 0.8 80  1.6  j 4.8 A I 0    0.4  j 0.8  1  j1  V0  1.6  j 4.81  5.06  71.6V

Problem 8.FE-1

Chapter 8: AC Steady-State Analysis

Irwin, Basic Engineering Circuit Analysis, 9/E

1

SOLUTION:

text

2I x

1

V1 I1

 j1

2

V2

V0

I2

120V

20 A

Ix The correct answer is c.

Chapter 8: AC Steady-State Analysis

Problem 8.FE-2

2

Irwin, Basic Engineering Circuit Analysis, 9/E

V2  120V V Ix  1  j1 2I x  I x  I1  0  V  V V  120 2 1   1  1 0 1   j1   j1 2V1  V1   j1V1  j1(120)  0  j1(120) V1   1.2  j 3.6V 3  j1 KCL at node 0: 2 I x  I 2  20  0  V  V  V0 2 1   2  20  0 2   j1   1.2  j 3.6  120  V0   2  20  0 2   j1  V0  30.88.97V

Problem 8.FE-2

Chapter 8: AC Steady-State Analysis

Irwin, Basic Engineering Circuit Analysis, 9/E

1

SOLUTION:

I

2

j 2

 j1

I1 120V

2V0

I2 +

 4

V0 

-

The correct answer is a.

I1  I  I 2 I  I1  I 2 KVL around the left loop: 120  2 I 1  j 2 I  2V0 V0  4I 2 (2  j 2) I 1  (8  j 2) I 2  120 KVL around the right loop: 2V0   j 2 I  j1I 2  4 I 2 j 2 I 1  (4  j1) I 2  0

Chapter 8: AC Steady-State Analysis

Problem 8.FE-3

2

Irwin, Basic Engineering Circuit Analysis, 9/E

Two equations and two unknowns: (2  j 2) I 1  (8  j 2) I 2  120 j 2 I 1  (4  j1) I 2  0 It follows that: I 1  4.2445 A

I 2  2.06  30.96 A V0  4( 2.06  30.96) V0  8.24  30.96V

Problem 8.FE-3

Chapter 8: AC Steady-State Analysis

Irwin, Basic Engineering Circuit Analysis, 9/E

1

SOLUTION: The correct answer is b. Z c is small at midband. Vx 5k 5   Vs 5k  1k 6

V0  6k (12k )   40 x10 3    160 Vx  6k  12k  V0  V x  V0    Vs  Vs  V x

 5    160  133.33  6

Chapter 8: AC Steady-State Analysis

Problem 8.FE-4

Irwin, Basic Engineering Circuit Analysis, 9/E

1

SOLUTION:

Is 60V Zeq

The correct answer is c. (1  j1)(1  j3)  3  1.58  18.43  3  4.53  6.34  1  j1  1  j 3 60 Is   1.32456.34 A 4.53  6.34

Z eq 

Chapter 8: AC Steady-State Analysis

Problem 8.FE-5

2

Irwin, Basic Engineering Circuit Analysis, 9/E

3

 j 1

Io j 3

1

60 V

1

KVL around the outer loop: 60  3(1.32456.34)  I 0 (1  j1) I 0  1.4832.92 A

Problem 8.FE-5

Chapter 8: AC Steady-State Analysis