Engineering Circuit Analysis
Engineering Circuit Analysis
9th Edition
ISBN: 9780073545516
Author: Hayt, William H. (william Hart), Jr, Kemmerly, Jack E. (jack Ellsworth), Durbin, Steven M.
Publisher: Mcgraw-hill Education,
Question
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Chapter 11, Problem 1E

(a)

To determine

Find the instantaneous power delivered to the 1Ω resistor at t=0,t=1s,andt=2s, when vs=9V.

(a)

Expert Solution
Check Mark

Answer to Problem 1E

The instantaneous power delivered to the 1Ω resistor at t=0,t=1s,andt=2s is 3.24 W.

Explanation of Solution

Calculation:

Refer to the figure given in the question.

Consider the expression of instantaneous power delivered to the resistor.

p1Ω=v1Ω21Ω        (1)

Here,

v1Ω is the voltage across 1Ω resistor.

Find the voltage across 1Ω resistor using voltage divider rule.

v1Ω=vs4+1(1)

v1Ω=vs5        (2)

Substitute 9 V for vs in equation (2).

v1Ω=95V

Substitute 95V for v1Ω in equation (1).

p1Ω=(95V)21Ω=8125W=8125W=3.24W

Since the power value is independent to the time. Hence, the instantaneous power delivered to the resistor is same (3.24 W) for all values of time.

Conclusion:

Thus, the instantaneous power delivered to the 1Ω resistor at t=0,t=1s,andt=2s is 3.24 W.

(b)

To determine

Find the instantaneous power delivered to the 1Ω resistor at t=0,t=1s,andt=2s, when vs=9sin2tV.

(b)

Expert Solution
Check Mark

Answer to Problem 1E

The instantaneous power delivered to the 1Ω resistor at t=0,t=1s,andt=2s are,

t in sp1Ω(t)=81(sin2t)225W
0p1Ω(0s)=0W
1p1Ω(1s)=2.68W
2p1Ω(2s)=1.856W

Explanation of Solution

Calculation:

Substitute 9sin2tV for vs in equation (2).

v1Ω=9sin2t5V

Substitute 9sin2t5V for v1Ω in equation (1).

p1Ω(t)=(9sin2t5V)21Ω=81(sin2t)225W

The power values for various values of time are tabulated below.

Table 1

t in sp1Ω(t)=81(sin2t)225W
0p1Ω(0s)=81(sin2(0))225W=0W
1p1Ω(1s)=81(sin2(1))225W=2.68W
2p1Ω(2s)=81(sin2(2))225W=1.856W

Conclusion:

Thus, the instantaneous power delivered to the 1Ω resistor at t=0,t=1s,andt=2s are calculated.

(c)

To determine

Find the instantaneous power delivered to the 1Ω resistor at t=0,t=1s,andt=2s, when vs=9sin(2t+13°)V.

(c)

Expert Solution
Check Mark

Answer to Problem 1E

The instantaneous power delivered to the 1Ω resistor at t=0,t=1s,andt=2s are,

t in sp1Ω(t)=81(sin(2t+13π180))225W
00.164W
12.03W
22.53W

Explanation of Solution

Calculation:

Substitute 9sin(2t+13°)V for vs in equation (2).

v1Ω=9sin(2t+13°)5V

Substitute 9sin(2t+13°)V for v1Ω in equation (1).

p1Ω(t)=(9sin(2t+13°)5V)21Ω=81(sin(2t+13°))225W

Convert 13° into radians for easy calculation.

p1Ω(t)=81(sin(2t+13×π180))225W=81(sin(2t+13π180))225W

The power values for various values of time are tabulated below.

Table 2

t in sp1Ω(t)=81(sin(2t+13π180))225W
0p1Ω(0s)=0.164W
1p1Ω(1s)=2.03W
2p1Ω(2s)=2.53W

Conclusion:

Thus, the instantaneous power delivered to the 1Ω resistor at t=0,t=1s,andt=2s are calculated.

(d)

To determine

Find the instantaneous power delivered to the 1Ω resistor at t=0,t=1s,andt=2s, when vs=9etV.

(d)

Expert Solution
Check Mark

Answer to Problem 1E

The instantaneous power delivered to the 1Ω resistor at t=0,t=1s,andt=2s are,

t in sp1Ω(t)=81e2t25W
03.24W
10.44W
20.06W

Explanation of Solution

Calculation:

Substitute 9etV for vs in equation (2).

v1Ω=9et5V

Substitute 9et5V for v1Ω in equation (1).

p1Ω(t)=(9et5V)21Ω=81e2t25W

The power values for various values of time are tabulated below.

Table 3

t in sp1Ω(t)=81e2t25W
0p1Ω(0s)=3.24W
1p1Ω(1s)=0.44W
2p1Ω(2s)=0.06W

Conclusion:

Thus, the instantaneous power delivered to the 1Ω resistor at t=0,t=1s,andt=2s are calculated.

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Chapter 11 Solutions

Engineering Circuit Analysis

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