An electric bulb has a rated power of 50 , W | vedclass

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(C) The resistance of the bulb is given by $R = \frac{V^2}{P} = \frac{100^2}{50} = 200 \, \Omega$.For the bulb to operate at its rated power, the curr

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$1.1$

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(C) The resistance of the bulb is given by $R = \frac{V^2}{P} = \frac{100^2}{50} = 200 \, \Omega$.For the bulb to operate at its rated power, the current $I$ must be $I = \frac{P}{V} = \frac{50}{100} = 0.5 \, A$.When connected to a $200 \, V$ $AC$ source, the total impedance $Z$ of the circuit is $Z = \frac{V_{source}}{I} = \frac{200}{0.5} = 400 \, \Omega$.The impedance of an $RL$ circuit is $Z = \sqrt{R^2 + X_L^2}$, where $X_L = 2\pi fL$.Substituting the values: $400 = \sqrt{200^2 + X_L^2}$.$160000 = 40000 + X_L^2 \implies X_L^2 = 120000$.$X_L = \sqrt{120000} = 200\sqrt{3} \, \Omega$.Since $X_L = 2\pi fL$, we have $200\sqrt{3} = 2 	imes \pi 	imes 50 	imes L$.$L = \frac{200\sqrt{3}}{100\pi} = \frac{2\sqrt{3}}{\pi} \approx \frac{2 	imes 1.732}{3.14} \approx 1.1 \, H$.

An electric bulb has a rated power of $50 \, W$ at $100 \, V$. If it is used on an $AC$ source of $200 \, V, 50 \, Hz$, a choke has to be used in series with it. This choke should have an inductance of (in $ \, H$)

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