The figure given below shows an $LCR$ series circuit with two switches S$_1$ and S$_2$. When switch S$_1$ is closed keeping S$_2$ open, the phase difference ($\varphi$) between the current and source voltage is $30^\circ$ and phase difference is $60^\circ$ when S$_2$ is closed keeping S$_1$ open. The value of $(3L_1-L_2)$ is ______ H.
Answer: (B) $\dfrac29$
Each switch is connected across one inductor. Closing S$_1$ shorts $L_2$, leaving $L_1$, $C$ and $R$; closing S$_2$ shorts $L_1$, leaving $L_2$, $C$ and $R$.
$\omega=300$ rad/s, so $X_C=\dfrac{1}{\omega C}=\dfrac{1}{300\times100\times10^{-6}}=\dfrac{100}{3}\ \Omega$.
$\tan30^\circ=\dfrac{\omega L_1-X_C}{R},\qquad\tan60^\circ=\dfrac{\omega L_2-X_C}{R}$
Since $\tan60^\circ=3\tan30^\circ$: $\ \omega L_2-X_C=3(\omega L_1-X_C)\Rightarrow\omega(3L_1-L_2)=2X_C$
$$3L_1-L_2=\frac{2X_C}{\omega}=\frac{2\times100/3}{300}=\frac29\ \text{H}$$
Solution by Sreeraj P, M.Sc Physics