By what percentage will the illumination of the lamp decrease if the current drops by 20%?
Solution
<p>$$\begin{aligned}
& \mathrm{P}=\mathrm{i}^2 \mathrm{R} \\
& \mathrm{P}_{\text {int }}=\mathrm{I}_{\text {int }}^2 \mathrm{R} \\
& \mathrm{P}_{\text {final }}=\left(0.8 \mathrm{I}_{\text {int }}\right)^2 \mathrm{R}
\end{aligned}$$</p>
<p>% change in power $=$</p>
<p>$$\frac{P_{\text {final }}-P_{\text {int }}}{P_{\text {int }}} \times 100=(0.64-1) \times 100=-36 \%$$</p>
About this question
Subject: Physics · Chapter: Current Electricity · Topic: Ohm's Law and Resistance
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