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An electric bulb of 500 W at 100 volt is used in a circuit having a 200 V supply. Calculate the resistance R to be connected in series with the bulb so that the power delivered by the bulb is 500 W.

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$20 Ω$

$30 Ω$

$5 Ω$

$10 Ω$

answer is A.

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Detailed Solution

$R_{bulb} = \frac{V^{2}}{P} = \frac{100 \times 100}{500} = 20 Ω$

$\Rightarrow$ The voltage drop must be 100V across bulb and the resistance, for equal p.d series resistance

$R_{b} = 20 Ω$

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