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2. *An AC voltage source, with a peak output of 262 V, is connected to a 33-2 resistor. What is the rate of energy dissipated due to heat in the resistor? A. 2080 w B. 1040 W C. 520 W D. 1471 W E. 662 W
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Answer #1

Peak output of an AC voltage is more than its RMS (Root Mean Square) Value and you should always calculate power dissipation by RMS Value for an AC current.

Remember, RMS value multiplied by (square root of 2) = Peak Value.

So we have, RMS x sqrt2 = 262

So RMS = 262/sqrt2 = 185.261

Now, P = V^2/R

P = (185.261)^2/33 = 1040W

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