(a) What is the value of the peak voltage for 120-V AC power? (b) What is the peak power consumption rate of a 60.0-W AC light bulb?

Strategy
We are told that $V_{ rms }$ is 120 V and $P_{\text {ave }}$ is 60.0 W. We can use $V_{ rms }=\frac{V_{0}}{\sqrt{2}}$ to find the peak voltage, and we can manipulate the definition of power to find the peak power from the given average power.

Question

(a) What is the value of the peak voltage for 120-V AC power? (b) What is the peak power consumption rate of a 60.0-W AC light bulb?

Strategy
We are told that [latex]V_{ rms }[/latex] is 120 V and [latex]P_{ ext {ave }}[/latex] is 60.0 W. We can use [latex]V_{ rms }=\frac{V_{0}}{\sqrt{2}}[/latex] to find the peak voltage, and we can manipulate the definition of power to find the peak power from the given average power.

Accepted Answer

Solution for (a)
Solving the equation [latex]V_{ rms }=\frac{V_{0}}{\sqrt{2}}[/latex] for the peak voltage [latex]V_{0}[/latex] and substituting the known value for [latex]V_{ rms }[/latex] gives

[latex]V_{0}=\sqrt{2} V_{ rms }=1.414(120 V )=170 V[/latex] (20.49)

Discussion for (a)
This means that the AC voltage swings from 170 V to –170 V and back 60 times every second. An equivalent DC voltage is a constant 120 V.

Solution for (b)
Peak power is peak current times peak voltage. Thus,

[latex]P_{0}=I_{0} V_{0}=2\left(\frac{1}{2} I_{0} V_{0} ight)=2 P_{ ext {ave }}[/latex]. (20.50)

We know the average power is 60.0 W, and so

[latex]P_{0}=2(60.0 W )=120 W[/latex]. (20.51)

Discussion
So the power swings from zero to 120 W one hundred twenty times per second (twice each cycle), and the power averages 60 W.

Question 20.9: (a) What is the value of the peak voltage for 120-V AC power...

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