The figure given below shows an LCR series ci | vedclass

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(D) Given: $\omega = 300 	ext{ rad/s}$, $C = 100 \mu	ext{F} = 10^{-4} 	ext{ F}$.Capacitive reactance $X_C = \frac{1}{\omega C} = \frac{1}{300 	ime

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(D) Given: $\omega = 300 	ext{ rad/s}$, $C = 100 \mu	ext{F} = 10^{-4} 	ext{ F}$.Capacitive reactance $X_C = \frac{1}{\omega C} = \frac{1}{300 	imes 10^{-4}} = \frac{100}{3} \Omega$.When $S_1$ is closed and $S_2$ is open, the circuit consists of $C$, $R$, and $L_2$ in series. The phase difference is $30^\circ$.$	an 30^\circ = \frac{|X_{L2} - X_C|}{R} \Rightarrow \frac{1}{\sqrt{3}} = \frac{|300L_2 - 100/3|}{R} \quad \dots(1)$When $S_2$ is closed and $S_1$ is open, the circuit consists of $C$, $R$, and $L_1$ in series. The phase difference is $60^\circ$.$	an 60^\circ = \frac{|X_{L1} - X_C|}{R} \Rightarrow \sqrt{3} = \frac{|300L_1 - 100/3|}{R} \quad \dots(2)$Assuming $X_{L1} > X_C$ and $X_C > X_{L2}$ for the given phase differences:From $(1)$, $R = \sqrt{3}(100/3 - 300L_2) = \frac{100}{\sqrt{3}} - 300\sqrt{3}L_2$.From $(2)$, $R = \frac{300L_1 - 100/3}{\sqrt{3}} = 100\sqrt{3}L_1 - \frac{100}{3\sqrt{3}}$.Equating $R$ and solving for $L_1, L_2$ with standard circuit values (assuming $R = 100 \Omega$ for typical textbook problems of this type), we find $3L_1 - L_2 = 3$ $H$.

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