Physics

An electric bulb of resistance 500 Ω, draws a current of 0.4 A. Calculate the power of the bulb and the potential difference at it's end.

Current Electricity

ICSE 2017

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Answer

Given,

R = 500 Ω,

I = 0.4 A

Power of bulb P = I2 R
= (0.4)2 x 500 = 80 W

Potential difference at the ends of the bulb V = IR = 0.4 x 500 = 200 V

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