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An electric bulb rated for 500 W at 100 V is used in a circuit having a 200 V supply. The resistance R that must be put in series with the bulb, so that the bulb draws 500 W, is

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$18 Ω$

$20 Ω$

$40 Ω$

$700 Ω$

answer is B.

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Detailed Solution

$P = V I o r I = \frac{P}{V} o r I = \frac{500 W}{100 V} = 5 A$

Now, $5 R = 100 o r R = 20 Ω$

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