Question

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

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

Answer 1

Answer: (D)

10 W

Answer 2

Actual power of bulb $\left(P_{1}\right)=40\, W$
Actual voltage of bulb $\left( V _{1}\right)=200\, V$
and supply voltage $\left( V _{2}\right)=100\, V$.
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
$P_{2}=\frac{40}{4}=10\, W$
(where $P _{2}=$ power when voltage is $100 \,V$ ).