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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 - Physics

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

 

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