Physics - Current Electricity Question with Solution | TestHub

Power through the resistor is given by, $P=I^{2}R$.

Now, $P_{\mathrm{int}}={I^2}_{\mathrm{int}}R$

and $P_{\mathrm{final}}={\left(0.8I_{\mathrm{int}}\right)}^{2}R$

Therefore, $\%$ change in power $=$

$\frac{P_{\mathrm{final}}-P_{\mathrm{int}}}{P_{\mathrm{int}}}\times 100=\left(0.64-1\right)\times 100=-36\%$