A bulb of (220 V, 60 W) is operated on 110 V supply then power developed in it is

15 W

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30 W

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65 W

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60 W

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Solution

The correct option is A
15 W

$P = \frac{V^{2}}{R} \Rightarrow P α V^{2}$

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A bulb of (220 V, 60 W) is operated on 110 V supply then power developed in it is

The correct option is A 15 W P=V2R⇒PαV2

The correct option is A
15 W

$P = \frac{V^{2}}{R} \Rightarrow P α V^{2}$