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If an electric bulb of resistance \(400 \Omega\) is connected to an ac source of peak voltage 282.8 V, then the electric power of the bulb is
If an electric bulb of resistance \(400 \Omega\) is connected to an ac source of peak voltage 282.8 V, then the electric power of the bulb is
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\(V_{\text{rms}} = \frac{V_0}{\sqrt{2}}\) for sinusoidal AC.
Updated On: Apr 24, 2026
- \(200 \text{ W}\)
- \(150 \text{ W}\)
- \(300 \text{ W}\)
- \(100 \text{ W}\)
- \(250 \text{ W}\)
Show Solution
The Correct Option is D
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