Discussion
If an electric bulb of resistance 400 Ω is connected to an ac source of peak voltage 282.8 V, then the electric power of the bulb is
Answer: D
The power is 100 W.
Power in an ac circuit uses the rms voltage:
\[ \begin{aligned} V_{rms} &= \frac{282.8}{\sqrt{2}} = 200\ \text{V} \\ P &= \frac{V_{rms}^{2}}{R} = \frac{40000}{400} = 100\ \text{W} \end{aligned} \]
Using the peak voltage gives 200 W, the usual slip; 150, 250 and 300 W do not follow.
For a sinusoidal supply the rms value is \(\frac{1}{\sqrt{2}}\) of the peak, and 282.8 V was chosen as \(200\sqrt{2}\) so that \(V_{rms}\) is exactly 200 V.
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