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An electric bulb rated for 500 W at 100V 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 delivers 500W is … $Ω$

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detailed solution

Correct option is T

Resistance of the bulb is

$R = \frac{V_{R}^{2}}{P_{R}} = 20 Ω$

Bulb delivers a power of 500 W at 100 V

So $I^{2} R = 500 \Rightarrow I = 5 A$

$I = \frac{200}{20 + R} \Rightarrow R = 20 Ω$

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detailed solution

Correct answer is 20

Resistance of the bulb is

$R = \frac{V_{R}^{2}}{P_{R}} = 20 Ω$

Bulb delivers a power of 500 W at 100 V

So $I^{2} R = 500 \Rightarrow I = 5 A$

$I = \frac{200}{20 + R} \Rightarrow R = 20 Ω$

Talk to our academic expert!

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An electric bulb rated for 500 W at 100V 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 delivers 500W is …Ω