Physics - Current Electricity Question with Solution | TestHub

Electric power is given by $P=\frac{V^2}{R}$, here, $V$ is voltage and $R$ is resistance.

Now, for each bulb, $R_1=\frac{V^2}{P}=\frac{220^2}{100}=484\Omega$ and $R_2=\frac{V^2}{P}=\frac{220^2}{60}=484\left(\frac{10}{6}\right)\Omega$

Net current through the series combination is $I=\frac{220}{484+484\times \frac{10}{6}}\text{A}=0.171\text{A}$

Hence, the power consumed by the $100\text{W}$ bulb is $P_1=I^2{R}_1=14.06\text{W}$