An electric bulb rated as 200 , W at 100 , V | vedclass

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(D) The resistance of the bulb $R_B$ is calculated using the formula $P = \frac{V^2}{R_B}$.Given $P = 200 \, W$ and $V = 100 \, V$, we have $R_B = \fr

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$50$

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(D) The resistance of the bulb $R_B$ is calculated using the formula $P = \frac{V^2}{R_B}$.Given $P = 200 \, W$ and $V = 100 \, V$, we have $R_B = \frac{100^2}{200} = \frac{10000}{200} = 50 \, \Omega$.To ensure the bulb delivers the same power, it must operate at its rated voltage of $100 \, V$ even when connected to a $200 \, V$ supply.Let $R$ be the resistance connected in series. The circuit current $I$ is given by $I = \frac{V_{supply}}{R + R_B} = \frac{200}{R + 50}$.The voltage across the bulb is $V_B = I 	imes R_B = 100 \, V$.Substituting $I$, we get $\frac{200}{R + 50} 	imes 50 = 100$.$\frac{10000}{R + 50} = 100 \implies 100 = R + 50 \implies R = 50 \, \Omega$.

An electric bulb rated as $200 \, W$ at $100 \, V$ is used in a circuit having $200 \, V$ supply. The resistance $R$ that must be put in series with the bulb so that the bulb delivers the same power is $..... \, \Omega$.

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