To light a 50,W, 100,V lamp,it is connected i | vedclass

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(B) The resistance of the lamp is $R = \frac{V_L^2}{P} = \frac{100^2}{50} = 200\,\Omega$.In an $RC$ series circuit,the total voltage $V$ is given by $

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$3$

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(B) The resistance of the lamp is $R = \frac{V_L^2}{P} = \frac{100^2}{50} = 200\,\Omega$.In an $RC$ series circuit,the total voltage $V$ is given by $V^2 = V_R^2 + V_C^2$,where $V_R = 100\,V$ is the voltage across the lamp.$200^2 = 100^2 + V_C^2 \Rightarrow V_C^2 = 40000 - 10000 = 30000 \Rightarrow V_C = 100\sqrt{3}\,V$.The current in the circuit is $I = \frac{V_R}{R} = \frac{100}{200} = 0.5\,A$.The capacitive reactance is $X_C = \frac{V_C}{I} = \frac{100\sqrt{3}}{0.5} = 200\sqrt{3}\,\Omega$.Since $X_C = \frac{1}{2\pi f C}$,we have $200\sqrt{3} = \frac{1}{2 	imes \pi 	imes 50 	imes C} \Rightarrow C = \frac{1}{200\sqrt{3} 	imes 100\pi} = \frac{1}{20000\pi\sqrt{3}}\,F$.Converting to $\mu F$: $C = \frac{10^6}{20000\pi\sqrt{3}} = \frac{50}{\pi\sqrt{3}}\,\mu F$.Comparing this with $C = \frac{50}{\pi\sqrt{x}}\,\mu F$,we get $x = 3$.

To light a $50\,W, 100\,V$ lamp,it is connected in series with a capacitor of capacitance $C = \frac{50}{\pi \sqrt{x}} \mu F$ to a $200\,V, 50\,Hz$ $AC$ source. The value of $x$ is........

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