Question

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

• A. $50\text{W}$
• B. $100\text{W}$
• C. $22\text{W}$
• D. $11\text{W}$

Answer 1

The correct option is C $22\text{W}$

$R=\frac{V^2}{P}$
$\therefore R_1=\frac{110\times 110}{50}=242\Omega$
and
$R_2=\frac{110\times 110}{100}=121\Omega$


$I=\frac{110}{242+121}=\frac{10}{33}\text{A}$

So, Power dissipated in $50\text{W}$ bulb$P_1=I^2\times R_1=\frac{10}{33}\times \frac{10}{33}\times 242$
or $P_1=\frac{2200}{99}\approx 22\text{W}$