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If voltage across a bulb rated 220 Volt-100 Watt drops by 2.5% of its rated value, the percentage of the rated value by which the power would decrease is
A
10%
B
20%
C
2.5%
D
5%
Detailed Solution
$P = \frac{V^2}{R}$, with R constant, so $P \propto V^2$
$\frac{\Delta P}{P}\times100 = 2\times\frac{\Delta V}{V}\times100$
$= 2\times2.5 = 5\%$
$\frac{\Delta P}{P}\times100 = 2\times\frac{\Delta V}{V}\times100$
$= 2\times2.5 = 5\%$