Q. 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

Solution:

Power, $P = \frac{V^2}{R}$
As the resistance of the bulb is constant
$\therefore \, \, \frac{\Delta V^2}{P}=\frac{2\Delta V}{V}$
$\%$ decrease in power $ =\frac{\Delta P}{P} \times 100=\frac{2\Delta V}{V} \times 100$
$\hspace40mm = 2 \times \, 2.5 \% = 5\% $

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