Question

The wattage 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?

• A. $50W$
• B. $100W$
• C. $22W$
• D. $11W$

Answer 1

Answer: (C)

$22W$

Using $P=\frac{V^2}{R}$ the resistance 50 W bulb is $R_1=\frac{{110}^2}{50}=242\Omega$

and resistance of 100 W bulb is $R_2=\frac{{110}^2}{100}=121\Omega$

As both bulb are in series so same current $I$ will flow through them.

So, $I=\frac{110}{R_1+R_2}=\frac{110}{242+121}=0.30A$

Power dissipation in 50 W bulb is $P_1=I^2{R}_1=(0.30{)}^2\times 242=21.78\approx 22W$