JEE Main 2026 — Apr 2, Shift 2 — Physics solution | Fundamenthol

JEE Main2026Apr 2, Shift 2Physics

Q.

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 ( $ϕ$ ) between the current and source voltage is $3 0^{\circ}$ and phase difference is $6 0^{\circ}$ when $S_{2}$ is closed keeping $S_{1}$ open. The value of $(3 L_{1} - L_{2})$ is___ H.

A
$\frac{9}{2}$
B
$\frac{2}{9}$
C
$\frac{1}{3}$
D
$3$

Solution

For $S_{1}$ is closed and $S_{2}$ open

$V = V_{0} sin (300 t)$

$\dfrac{X_{L_1}-X_C}{R} = \tan(30^\circ) = \dfrac{1}{\sqrt{3}}$

$\sqrt{3}\,(X_{L_1}-X_C)=R \qquad \cdots (1)$

For $S_{1}$ is open and $S_{2}$ is closed.

$V = V_{0} sin (300 t)$

$\dfrac{\left|X_{L_2}-X_C\right|}{R} = \tan(60^\circ) = \sqrt{3}$

$\dfrac{\left|X_{L_2}-X_C\right|}{\sqrt{3}} = R \qquad \cdots (2)$

From Eq. (1) and (2)

$\sqrt{3}\,(X_{L_1}-X_C) = \dfrac{(X_{L_2}-X_C)}{\sqrt{3}}$

$3\left(\omega L_1-\dfrac{1}{\omega C}\right) = \left(\omega L_2-\dfrac{1}{\omega C}\right)$

$\left|3L_1-L_2\right| = \dfrac{3}{\omega^2 C} - \dfrac{1}{\omega^2 C} = \dfrac{2}{9\times10^4\times100\times10^{-6}}$

$\left|3L_1-L_2\right| = \dfrac{2}{9}$

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