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An electric bulb of resistance 500Omega,...

An electric bulb of resistance `500Omega`, draws a current of 0.4 A. Calculate the power of the bulb and the potential difference at its end.

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Given : Resistance (R ) = `500Omega`, Current (I) = 0.4A
Power (P) = `I^(2)R`
= `(0.4)^(2)xx500W=80W`
Potential difference at its ends (V) = IR
= `0.4xx500V` = 200V
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