An electric bulb rated 100 ,W - 300 ,V is con | vedclass

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(C) The bulb is rated $P = 100 \,W$ and $V_b = 300 \,V$.The current through the bulb is $I = \frac{P}{V_b} = \frac{100}{300} = \frac{1}{3} \,A$.The re

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$4$

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(C) The bulb is rated $P = 100 \,W$ and $V_b = 300 \,V$.The current through the bulb is $I = \frac{P}{V_b} = \frac{100}{300} = \frac{1}{3} \,A$.The resistance of the bulb is $R = \frac{V_b^2}{P} = \frac{300^2}{100} = 900 \,\Omega$.When connected to a $V = 500 \,V$ supply,the total impedance $Z$ is given by $Z = \frac{V}{I} = \frac{500}{1/3} = 1500 \,\Omega$.In an $LR$ circuit,$Z^2 = R^2 + X_L^2$,so $X_L^2 = Z^2 - R^2$.$X_L^2 = 1500^2 - 900^2 = (1500 - 900)(1500 + 900) = 600 	imes 2400 = 1440000$.$X_L = \sqrt{1440000} = 1200 \,\Omega$.Since $X_L = 2 \pi f L$,we have $1200 = 2 \pi \left(\frac{150}{\pi}ight) L$.$1200 = 300 L$,which gives $L = 4 \,H$.

An electric bulb rated $100 \,W - 300 \,V$ is connected to an $AC$ supply of $500 \,V$ and $\frac{150}{\pi} \,Hz$. The required inductance to save the electric bulb is .......... $H$.

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