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