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Question

Physics Question on Current electricity

If voltage across a bulb rated 220V220 \,V- 100W100\, W drops by 2.5%2.5\% of its rated value, the percentage of the rated value by which the power would decrease is

A

20%20\,\%

B

2.5%2.5\,\%

C

5%5\,\%

D

10%10\,\%

Answer

5%5\,\%

Explanation

Solution

Power, P=V2RP=\frac{V^{2}}{R}
As the resistance of the bulb is constant
ΔPP=2ΔVV\therefore \frac{\Delta P}{P}=\frac{2\Delta V}{V}
%\% decrease in power =ΔPP×100=\frac{\Delta P}{P} \times 100
=2ΔVV×100=\frac{2\Delta V}{V}\times 100
=2×2.5%=5%=2\times2.5\%=5\%