ICSE Class 10 Physics 2017 Solved Paper

Question : 18

Total: 65

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

Solution:

Given: R=500Ω,I=0.4A,P= ?, V= ?
We know,
Power, P‌=I2R
‌=(0.4)2×500
‌=80W
Potential difference,
V‌=R
‌=0.4×500
‌=200‌Volt.

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