The resistance of a 240V, 60 watt electric filament bulb is
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Correct Answer: Option E
Explanation:
power = IV
power = \(\frac{V^2}{R}\) = 60
= \(\frac{240^2}{R}\)
= 960\(\Omega\)
power = IV
power = \(\frac{V^2}{R}\) = 60
= \(\frac{240^2}{R}\)
= 960\(\Omega\)