Question

An electric bulb rated $220V,100W$ is connected in series with another bulb rated $220V,60W$. If the voltage across the combination is $220V$, the power consumed by the $100W$ bulb will be about

• A. 25 W
• B. 14 W
• C. 60 W
• D. 100 W

Answer 1

Answer: (B)

Resistance of bulb 1
$R_1=\frac{V^2}{P}=\frac{(220{)}^2}{100}$
$R_1=484\Omega$
Resistance of bulb 2
$R_2=\frac{(220V{)}^2}{60W}=\frac{4840}{6}=806.6\Omega$
$R_{eq}=R_1+R_2=(220{)}^2\left(\frac{1}{100}+\frac{1}{60}\right)$
Current flowing, $I=\frac{220V}{R_{eq}}=\frac{220}{(220{)}^2}\left(\frac{100\times 60}{160}\right)$
$I=\frac{1}{220}\left(\frac{75}{2}\right)=\frac{15}{88}A$
Power consumed by $100W$ bulb $=I^2{R}_1$
$={\left(\frac{15}{88}\right)}^2\times \frac{(220{)}^2}{100}$
$=\frac{225}{16}=14W$