The output voltage of an AC generator is given by Δ v = (170 V) sin (60πt). The generator is connected across a 20.0-Ω resistor. By inspection, what are the (a) maximum voltage and (b) frequency? Find the (c) rms voltage across the resistor, (d) rms current in the resistor, (e) maximum current in the resistor, (f) power delivered to the resistor, and (g) current when t = 0.005 0 s. (h) Should the argument of the sine function be in degrees or radians?

College Physics
11th Edition
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Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
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The output voltage of an AC generator is given by Δ v = (170 V) sin (60πt). The generator is connected across a 20.0-Ω resistor. By inspection, what are the (a) maximum voltage and (b) frequency? Find the (c) rms voltage across the resistor, (d) rms current in the resistor, (e) maximum current in the resistor, (f) power delivered to the resistor, and (g) current when t = 0.005 0 s. (h) Should the argument of the sine function be in degrees or radians?

 

GIVEN:

Output voltage, Δv = (170 V) sin (60πt) 

Resistance, R = 20.0-Ω

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Step 2 Part (a)

Part (a) 

Standard equation can be represented as, 

Δv = Vmsin(ωt)

Where, V= maximum voltage. 

ω = angular frequency. 

Compare the given equation with standard equation. 

We get, V= 170 V      Answer

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Step 3 Part (b)

Part (b) 

Angular frequency, ω = 60π    ... (1) 

Also, ω = 2πf     .... (2) 

Equate both, 

2πf = 60π

f = 60π/2π

f = 30 Hz     Answer

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Step 4 Part (c)

Part (c) 

RMS voltage is equal to 1/√2 times of maximum voltage. 

Vrms = Vm/√2

Vrms = (170 V)/√2

Vrms = 120.208 V      Answer

 

 

Part (e)

the maximum current through the resistor

Imax=I2Imax=(6.010)2Imax=8.499A

Part(f)

 power delivered to the resistor

P=I2rmsRP=(8.499)2(20) P=722.4 W

 

 

 Missing g and h

 

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