An electric bulb rated for 500W at 100V is used in a circuit having a 200V supply. The resistance R that must be put in series with the bulb, so that the bulb draws 500W, is
A
18Ω
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B
20Ω
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C
40Ω
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D
700Ω
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Solution
The correct option is B20Ω Given that, as bulb draws 500W then voltage across it will be 100V.
For a bulb, we know that, P=VI,I=PVI=500W100V=5A
As bulb is connected in series to resistance, equal current flows through resistance and bulb.
Voltage across other resistor will be 200−100=100V