# In the AC circuit shown in Figure P32.3, R = 70.0 Ω and the output voltage of the AC source is Δ V max sin ωt . (a) If Δ V R = 0.250 Δ V max for the first time at t = 0.0100 s, what is the angular frequency of the source? (b) What is the next value of t for which Δ V R = 0.250 Δ V max ? Figure P32.6 Problem 3 and 5.

### Physics for Scientists and Enginee...

10th Edition
Raymond A. Serway + 1 other
Publisher: Cengage Learning
ISBN: 9781337553292

Chapter
Section

### Physics for Scientists and Enginee...

10th Edition
Raymond A. Serway + 1 other
Publisher: Cengage Learning
ISBN: 9781337553292
Chapter 32, Problem 5P
Textbook Problem
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## In the AC circuit shown in Figure P32.3, R = 70.0 Ω and the output voltage of the AC source is ΔVmax sin ωt. (a) If ΔVR = 0.250 ΔVmax for the first time at t = 0.0100 s, what is the angular frequency of the source? (b) What is the next value of t for which ΔVR = 0.250 ΔVmax?Figure P32.6 Problem 3 and 5.

(a)

To determine

The angular frequency of the source.

### Explanation of Solution

Given Information: The resistance across the circuit is 70.0Ω , the value of time period is 0.0100s .

It is given that the voltage across the resistor is,

ΔVR=ΔVmaxsin(ωt) (1)

Here,

ΔVR is the ΔVR voltage across the resistor.

ΔVmax is the peak voltage across the resistor.

ω is the angular frequency of the source.

t is the value of time period.

It is given that the voltage across the resistor is 0.250 times of the peak voltage across the resistor.

ΔVR=0.250ΔVmax

Substitute 0.250ΔVmax for ΔVR in equation (1) to find ωt ,

0.250ΔVmax=ΔVmaxsin(ωt)ωt=sin1(0

(b)

To determine

The next value of t for which ΔVR=0.250ΔVmax .

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