If voltage across a bulb rated 220 ,V, 100 ,W | vedclass

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(B) The power consumed by a bulb is given by $P = \frac{V^2}{R}$, where $V$ is the voltage and $R$ is the constant resistance of the bulb.Taking the l

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$5$

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(B) The power consumed by a bulb is given by $P = \frac{V^2}{R}$, where $V$ is the voltage and $R$ is the constant resistance of the bulb.Taking the logarithmic derivative, we get $\frac{dP}{P} = 2 \frac{dV}{V}$.Here, the percentage change in voltage is $\frac{dV}{V} = -2.5 \% = -0.025$.Therefore, the percentage change in power is $\frac{dP}{P} = 2 	imes (-2.5 \%) = -5 \%$.The negative sign indicates a decrease in power.Thus, the power decreases by $5 \%$ of its rated value.

If voltage across a bulb rated $220 \,V, 100 \,W$ drops by $2.5 \%$ of its rated value, then the percentage of the rated value by which the power would decrease is (in $\%$)

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