Question

An electric bulb marked $40W$ and $200V$, is used in a circuit of supply voltage $100V$. Now its power is

• A. 100 W
• B. 20 W
• C. 40 W
• D. 10 W

Answer 1

Answer: (D)

Actual power of bulb $\left(P_1\right)=40W$
Actual voltage of bulb $\left(V_1\right)=200V$
and supply voltage $\left(V_2\right)=100V$.
Power $(P)=\frac{V^2}{R}\propto V^2.$
Therefore $\frac{P_1}{P_2}=\frac{V_1^2}{V_2^2}$
or, $\frac{40}{P_2}=\frac{(200{)}^2}{(100{)}^2}=4$ or
$P_2=\frac{40}{4}=10W$
(where $P_2=$ power when voltage is $100V$ ).