An electric bulb is rated 60W, 220V. Calculate the resistance of its filament when it is operating normally
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Correct Answer: Option D
Explanation:
P = \(\frac{v^2}{P} = \frac{220^2}{100} = 806.7\Omega\)
P = \(\frac{v^2}{P} = \frac{220^2}{100} = 806.7\Omega\)