## In This post you will get answer of Q.

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

Solution:

## 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 $100 ,V$ ).