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Asked by Jason

An AC voltage with 169 peak volts is applied across a 506 ohm resistor. To the nearest tenth of a watt, what is the average power?
12 years ago

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Answered by Henry
P=(Vrms)^2/R = (0.707*169)^2/506=28.2 Watts.
12 years ago
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An AC voltage with 169 peak volts is applied across a 506 ohm resistor. To the nearest tenth of a watt, what is the average power?

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