V2 120V V Ix 1 j1 2I x I x I1 0 V V V 120 2 1 1 1 0 1 j1 j1 2V1 V1 j1V1 j1(120) 0 j1(120) V1 1.2 j 3.6V 3 j1 KCL at node 0: 2 I x I 2 20 0 V V V0 2 1 2 20 0 2 j1 1.2 j 3.6 120 V0 2 20 0 2 j1 V0 30.88.97V
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 120V
2V0
I2 +
4
V0
-
The correct answer is a.
I1 I I 2 I I1 I 2 KVL around the left loop: 120 2 I 1 j 2 I 2V0 V0 4I 2 (2 j 2) I 1 (8 j 2) I 2 120 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 2 120 j 2 I 1 (4 j1) I 2 0 It follows that: I 1 4.2445 A
I 2 2.06 30.96 A V0 4( 2.06 30.96) V0 8.24 30.96V
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 60V 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 60 Is 1.32456.34 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
60 V
1
KVL around the outer loop: 60 3(1.32456.34) I 0 (1 j1) I 0 1.4832.92 A
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