JEE Main 2026 — Alternating Current Question with Solution

From the source $v=V_0\sin (300t)$, the angular frequency is $\omega =300\text{ rad/s}$.
Capacitance $C=100\mu \text{F}=10^{-4}\text{ F}$, so $X_C=\frac{1}{\omega C}$.

Case 1 ($S_1$ closed, $S_2$ open):
Closing $S_1$ short-circuits $L_2$, leaving an LCR series circuit with $C$, $R$ and $L_1$. With phase difference ${\varphi }_1=30^{\circ }$:
$\tan 30^{\circ }=\frac{\omega L_1-X_C}{R}$
$\frac{1}{\sqrt{3}}=\frac{\omega L_1-X_C}{R}\Rightarrow R=\sqrt{3}(\omega L_1-X_C)$ ... (1)

Case 2 ($S_2$ closed, $S_1$ open):
Closing $S_2$ short-circuits $L_1$, leaving an LCR series circuit with $C$, $R$ and $L_2$. With phase difference ${\varphi }_2=60^{\circ }$:
$\tan 60^{\circ }=\frac{\omega L_2-X_C}{R}$
$\sqrt{3}=\frac{\omega L_2-X_C}{R}\Rightarrow \omega L_2-X_C=R\sqrt{3}$ ... (2)

Substituting $R$ from (1) into (2):
$\omega L_2-X_C=\sqrt{3}(\omega L_1-X_C)\cdot \sqrt{3}$
$\omega L_2-X_C=3\omega L_1-3X_C$
$2X_C=3\omega L_1-\omega L_2=\omega (3L_1-L_2)$

Therefore:
$3L_1-L_2=\frac{2X_C}{\omega }=\frac{2}{{\omega }^{2}C}$

Substituting $\omega =300\text{ rad/s}$ and $C=10^{-4}\text{ F}$:
${\omega }^{2}C=(300{)}^2\cdot 10^{-4}=9\times 10^4\cdot 10^{-4}=9$

$3L_1-L_2=\frac{2}{9}\text{ H}$

Hence, the correct option is $(2)\frac{2}{9}$.