Thank you for reporting, we will resolve it shortly
Q.
An electric bulb, marked $40\, W$ and $200\, V$, is used in a circuit of supply voltage $100\, V$. Now its power is
Current Electricity
Solution:
Power of bulb is given by, $P=\frac{V^{2}}{R}$
$\Rightarrow \frac{P_{2}}{P_{1}}=\frac{V_{2}^{2}}{V_{1}^{2}} (\because R$ is constant $)$
$\Rightarrow \frac{P_{2}}{P_{1}}=\left(\frac{100}{200}\right)^{2}=\frac{1}{4}$
$ \Rightarrow P_{2}=\frac{P_{1}}{4}=\frac{40}{4}=10 \,W$