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Irwin, Basic Engineering Circuit Analysis - Solution ch, Assignments of Electrical Engineering

University of North Carolina (UNC) - CharlotteElectrical Engineering

ChapterNine STEADY STATE POWER ANALYSIS

Typology: Assignments

2019/2020

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Irwin, Basic Engineering Circuit Analysis, 11/E 1
Chapter 09: Steady - State Power Analysis Problem 9.1
9.1 The voltage and current at the input of a circuit are given by the expressions
υ
(t) = 170 cos (ωt + 30°) V
i(t) = 5 cos (ωt + 45°) A
Determine the average power absorbed by the circuit.
SOLUTION:
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9.1 The voltage and current at the input of a circuit are given by the expressions

υ ( t ) = 170 cos (ω t + 30 °) V

i ( t ) = 5 cos (ω t + 45 °) A

Determine the average power absorbed by the circuit.

SOLUTION:

9.2 The voltage and current at the input of a network are given by the expressions

υ ( t ) = 6 cos ω t V

i ( t ) = 4 sin ω t A

Determine the average power absorbed by the network.

SOLUTION:

9.4 Determine the equations for the current and the instantaneous power in the network in Fig. P9.4.

12 75° V

j 3 Ω

I

Figure P9.

SOLUTION:

9.5 Determine the instantaneous power supplied by the source in the circuit in Fig. P9.5.

12 60° V

− j 2 Ω

j 2 Ω

Figure P9.

SOLUTION:

9.7 Given υ

s

( t ) = 100 cos 100 t volts, find the average power supplied by the source and the current i

2

( t ) in the network

in Fig. P9.7.

1 mF

50 mH

υ

S

( t )

i

2

( t )

Figure P9.

SOLUTION:

Problem 9.7 Chapter 09: Steady - State Power Analysis

9.9 Given the network in Fig. P9.9, find the power supplied and the average power absorbed by each element.

j 2 Ω

− j 1 Ω

V

T

6 0°A

Figure P9.

SOLUTION:

9.10 Calculate the power absorbed by each element in the circuit in Fig. P9.10.

25 cos 20 t V

0.4 H 0.02 F

0.01 F

Figure P9.

SOLUTION:

Problem 9.11 Chapter 09: Steady - State Power Analysis

9.12 Determine the average power supplied by each source in the network shown in Fig. P9.12.

j 1 Ω

− j 1 Ω

10 0° V 1 Ω 2 30° A

Figure P9.

SOLUTION:

9.14 Determine the average power absorbed by the 4-Ω resistor in the network shown in Fig. P9.14.

− j 4 Ω 4 0° A j 2 Ω 12 90° V

Figure P9.

SOLUTION:

9.15 Find the average power absorbed by the network in Fig. P9.15.

− j 1 Ω

12 0° V

j 2 Ω − j 2 Ω

Figure P9.

SOLUTION:

9.17 If i

g

( t ) = 0.5 cos 2000 t A, find the average power absorbed by each element in the circuit in Fig. P9.17.

12.5 μF

60 mH

i

g

( t )

Figure P9.

SOLUTION:

9.18 Find the average power absorbed by the network shown in Fig. P9.18.

4 60° A

j 2 Ω

− j 4 Ω

Figure P9.

SOLUTION: