Physics - Current Electricity Question with Solution | TestHub

$R_1=\frac{V^2}{P}=\frac{220\times 220}{25}=44\times 44$

$R_2=\frac{220\times 220}{100}=22\times 22$

As resistance are in series current through them will be same.

$\frac{P_1}{P_2}=\frac{i^2{R}_1}{i^2{R}_2}=\frac{44\times 44}{22\times 22}=\frac{4}{1}$

Therefore, among the given options, the option matching is $P_1=16\text{W},P_2=4\text{W}$