In the given AC circuit,when switch S is at p | vedclass

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(A) In position $1$,the circuit is an $LR$ series circuit. The phase angle $\phi$ is given by $	an \phi = \frac{X_L}{R} = \frac{\omega L}{R}$.Given $

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current leads source $emf$ by $\frac{\pi}{4}$

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(A) In position $1$,the circuit is an $LR$ series circuit. The phase angle $\phi$ is given by $	an \phi = \frac{X_L}{R} = \frac{\omega L}{R}$.Given $\phi = \pi / 6$,$\omega = 1000 	ext{ rad/s}$,and $L = \sqrt{3} 	ext{ mH} = \sqrt{3} 	imes 10^{-3} 	ext{ H}$.$	an(\pi / 6) = \frac{1000 	imes \sqrt{3} 	imes 10^{-3}}{R} \implies \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{R} \implies R = 3 \, \Omega$.In position $2$,the circuit is an $RC$ series circuit. The phase angle $\phi$ is given by $	an \phi = \frac{X_C}{R} = \frac{1}{\omega C R}$.Given $C = \frac{1000}{3} \, \mu	ext{F} = \frac{1000}{3} 	imes 10^{-6} 	ext{ F}$.$	an \phi = \frac{1}{1000 	imes (\frac{1000}{3} 	imes 10^{-6}) 	imes 3} = \frac{1}{1000 	imes \frac{1}{3} 	imes 10^{-3} 	imes 3} = \frac{1}{1} = 1$.Since $	an \phi = 1$,$\phi = \pi / 4$. In an $RC$ circuit,current leads the source $emf$ by $\phi$. Therefore,current leads source $emf$ by $\pi / 4$.

In the given $AC$ circuit,when switch $S$ is at position $1$,the source $emf$ leads the current by $\pi / 6$. Now,if the switch is at position $2$,then

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