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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 its end. |
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Answer» Given : Resistance (R) = 500 Ω, Current (I) = 0.4 A Power (P) = I2R = (0.4)2 x 500 W = 80 W Potential difference at its ends (V) = IR = 0.4 x 500 V=200 V |
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