Peak Voltage and Power for AC

Problema e solução em inglês, como no livro original.

(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?

Resolva no papel primeiro. Depois abra a solução um passo de cada vez e pare assim que conseguir terminar sozinho.

Estratégia
We are told that $V_{rms}$ is 120 V and $P_{ave}$ 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.
Solução para (a)
Solving the equation $V_{rms} = \frac{V_{0}}{\sqrt{2}}$ for the peak voltage $V_{0}$ and substituting the known value for $V_{rms}$ gives

$V_{0} = \sqrt{2} V_{rms} = 1 . 414 (120 V) = 170 V .$
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Discussão para (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.
Solução para (b)
Peak power is peak current times peak voltage. Thus,

$P_{0} = I_{0} V_{0} = 2 (\frac{1}{2} I_{0} V_{0}) = 2 P_{ave} .$
20.50

We know the average power is 60.0 W, and so

$P_{0} = 2 (60 . 0 W) = 120 W .$
20.51
Discussão
So the power swings from zero to 120 W one hundred twenty times per second (twice each cycle), and the power averages 60 W.

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