Question

An electric bulb rated $220\, V,\, 100\, W$ is connected in series with another bulb rated $220\, V,\, 60\, W$. If the voltage across the combination is $220\, V$, the power consumed by the $100\, W$ 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
$R_{2}=\frac{(220 V )^{2}}{60 W }=\frac{4840}{6}=806.6\, \Omega$
$R_{eq}=R_{1}+R_{2}=(220)^{2}\left(\frac{1}{100}+\frac{1}{60}\right)$
Current flowing, $I=\frac{220 V }{R_{eq}}=\frac{220}{(220)^{2}}\left(\frac{100 \times 60}{160}\right)$
$I=\frac{1}{220}\left(\frac{75}{2}\right)=\frac{15}{88}\, A$
Power consumed by $100\, W$ bulb $=I^{2} R_{1}$
$=\left(\frac{15}{88}\right)^{2} \times \frac{(220)^{2}}{100}$
$=\frac{225}{16}=14\,W$