Physics - Current Electricity Question with Solution | TestHub

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

Now, $P_{int}={\left(I^2\right)}_{int}R$

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

Therefore, $\%$ change in power $=$

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