An electric bulb marked 40, W and 200, V,is u | vedclass

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(D) The power rating of the bulb is $P_1 = 40\, W$ at a voltage $V_1 = 200\, V$.Since the resistance $R$ of the bulb remains constant,we use the formu

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$10$

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(D) The power rating of the bulb is $P_1 = 40\, W$ at a voltage $V_1 = 200\, V$.Since the resistance $R$ of the bulb remains constant,we use the formula $P = \frac{V^2}{R}$.Therefore,the ratio of the new power $P_2$ to the rated power $P_1$ is given by $\frac{P_2}{P_1} = \frac{V_2^2}{V_1^2}$.Given $V_2 = 100\, V$,we have $\frac{P_2}{40} = \left( \frac{100}{200} \right)^2$.$\frac{P_2}{40} = \left( \frac{1}{2} \right)^2 = \frac{1}{4}$.$P_2 = \frac{40}{4} = 10\, W$.

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

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