Question

The voltage rating of a light bulb indicates the power dissipated by the bulb, if it is connected across $110V$ DC potential difference. If a $50W$ and $100W$ bulb are connected in series to a $110V$ DC source, how much power will be dissipated in the $50W$ bulb (in $W$).

Answer 1

Using $R=\frac{V^2}{P}$

Resistance of $50\text{W},100\text{W}$ are $R_1=\frac{110\times 110}{50}$ $R_2=\frac{110\times 110}{100}$ respectively
Using Kirchhoffs law, current through the circuit is
$I=\frac{110}{R_1+R_2}=\frac{110\times 100}{3\times 110\times 110}=\frac{10}{33}\text{A}$

Power consumed by $50W$ bulb is $P_1=I^2\times R_1=\frac{10}{33}\times \frac{10}{33}\times \frac{110\times 110}{50}=\frac{200}{9}\approx 22.2\text{W}$