Question

An electric bulb is operating at 110 V. Power consumed by it, if it is rated 220 V - 110 W, will be

• A. 75 W
• B. 90 w
• C. 50 W
• D. 27.5 W

Answer 1

Answer: (D)

Power of an electric bulb is directly proportional to the square of voltage.
Hence, for given condition
$P_1\propto (V_1{)}^2$ and $P_2\propto (V_2{)}^2$

$\Rightarrow \frac{P_1}{P_2}=\frac{(V_1{)}^2}{(V_2{)}^2}$

$\therefore P_1=\frac{(V_1{)}^2}{(V_2{)}^2}{P}_2$

Here, an electric bulb is operating at 110 V and it is rated as 220 V - 110 W

hence, $P_1=\frac{(110{)}^2}{(220{)}^2}110$

$P_1=\frac{(1{)}^2}{(2{)}^2}110=\frac{110}{4}\approx 27.5W$