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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