An electric bulb marked as $50 W - 200 V$ is connected across a $100 V$ supply. The present power of the bulb is

37.5 W

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25 W

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12.5 W

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10 W

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Solution

The correct option is C $12.5 W$
We know that: $R = \frac{V^{2}}{P}$
Given: $V = 200 v o l t, P = 40 W$ and $V^{'} = 100 v o l t$
Hence, $R = \frac{V^{2}}{P} = \frac{200 \times 200}{50}$
Present power: $P^{'} = \frac{V^{' 2}}{R} = \frac{100 \times 100 \times 50}{200 \times 200}$
$P^{'} = 12.5 W$

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