COLLEGE PHYSICS,VOLUME 1
COLLEGE PHYSICS,VOLUME 1
2nd Edition
ISBN: 9781319115104
Author: Freedman
Publisher: MAC HIGHER
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Question
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Chapter 21, Problem 64QAP
To determine

(a)

The peak value of the current and average power delivered when a sinusoidal voltage of 40.0 V rms and a frequency of 100 Hz is applied to a 100 O resistor.

Expert Solution
Check Mark

Answer to Problem 64QAP

When a sinusoidal voltage of 40.0 V rms and a frequency of 100 Hz is applied to a 100 O resistor, the peak current is found to be 0.57 A and the average power delivered is 16 W.

Explanation of Solution

Given:

  R=100 ΩVrms=40.0 Vf=100 Hz

Formula used:

The peak value of current imax in an ac circuit is given by,

  imax=2irms......(1)

Here, the rms value of current irms is given by,

  irms=VrmsR......(2)

The average power dissipated in the circuit is given by,

  Pav=Vrmsirms......(3)

Calculation:

Calculate the rms value of current irms by substituting the values of the variables in equation (2).

  irms=VrmsR=40.0 V100 Ω=0.40 A

Calculate the value of peak current using equation (1).

  imax=2irms=2(0.40 A)=0.57 A

Calculate the average power delivered using equation (3).

  Pav=Vrmsirms=(40.0 V)(1.2 A)=16 W

Conclusion;Thus, when a sinusoidal voltage of 40.0 V rms and a frequency of 100 Hz is applied to a 100 O resistor, the peak current is found to be 0.57 A and the average power delivered is 16 W.

To determine

(b)

The peak value of the current and average power delivered when a sinusoidal voltage of 40.0 V rms and a frequency of 100 Hz is applied to a 0.200 H inductor.

Expert Solution
Check Mark

Answer to Problem 64QAP

When a sinusoidal voltage of 40.0 V rms and a frequency of 100 Hz is applied to a 0.200 H inductor, the peak current is found to be 0.45A and the average power delivered is 0 W.

Explanation of Solution

Given:

  L=0.200 HVrms=40.0 Vf=100 Hz

Formula used:

The peak value of current imax in an ac circuit is given by,

  imax=2irms......(1)

Here, the rms value of current irms is given by,

  irms=Vrms2πfL......(4)

The average power dissipated in the circuit is given by,

  Pav=Vrmsirmscosϕ......(5)

Here, ϕ is the phase difference between current and voltage.

Calculation:

Calculate the rms value of current irms by substituting the values of the variables in equation (4).

  irms=Vrms2πfL=40.0 V2(3.14)(100 Hz)(0.200 H)=0.32 A

Calculate the value of peak current using equation (1).

  imax=2irms=2(0.32 A)=0.45 A

Calculate the average power delivered using equation (5). The phase difference between voltage and current in a ac circuit containing an inductor is 90o

  ϕ=90°

Therefore, since cos90°=0

  Pav=Vrmsirmscosϕ=0

Conclusion;Thus, when a sinusoidal voltage of 40.0 V rms and a frequency of 100 Hz is applied to a 0.200 H inductor, the peak current is found to be 0.45A and the average power delivered is 0 W.

To determine

(c)

The peak value of the current and average power delivered when a sinusoidal voltage of 40.0 V rms and a frequency of 100 Hz is applied to a 50.0 μF capacitor.

Expert Solution
Check Mark

Answer to Problem 64QAP

When a sinusoidal voltage of 40.0 V rms and a frequency of 100 Hz is applied to a 50.0 μF capacitor, the peak current is found to be 1.8 A and the average power delivered is 0 W.

Explanation of Solution

Given:

  L=50.0 μF=50.0×106FVrms=40.0 Vf=100 Hz

Formula used:

The peak value of current imax in an ac circuit is given by,

  imax=2irms......(1)

Here, the rms value of current irms is given by,

  irms=2πfCVrms......(6)

The average power dissipated in the circuit is given by,

  Pav=Vrmsirmscosϕ......(3)

Here, ϕ is the phase difference between current and voltage.

Calculation:

Calculate the rms value of current irms by substituting the values of the variables in equation (6).

  irms=2πfCVrms=2(3.14)(100 Hz)(50.0×106F)(40.0 V)=1.256 A

Calculate the value of peak current using equation (1).

  imax=2irms=2(1.256 A)=1.776 A=1.8 A

Calculate the average power delivered using equation (5). The phase difference between voltage and current in a ac circuit containing a capacitor is 90o

  ϕ=90°

Therefore, since cos90°=0

  Pav=Vrmsirmscosϕ=0

Conclusion;Thus, when a sinusoidal voltage of 40.0 V rms and a frequency of 100 Hz is applied to a 50.0 μF capacitor, the peak current is found to be 1.8 A and the average power delivered is 0 W.

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Chapter 21 Solutions

COLLEGE PHYSICS,VOLUME 1

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