An electric bulb of 300 draws a current of 0.4 A. Calculate the power of the bulb and the potential difference at its ends
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Solution
Current drawn by the bulb, I = 0.4 A
Resistance of the bulb, R = 300 Ω
Power of the bulb is P = I2R
∴ P= (0.4)2 x 300 = 48 W
Potential difference at the ends of the bulb is
V = IR
∴ V = 0.4 x 300 = 120 V
Concept: Electric Potential Difference
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