The figure given below shows an LCR series circuit with two switches $\mathrm{S}_{1}$ and $\mathrm{S}_{2}$. When switch $\mathrm{S}_{1}$ is closed keeping $\mathrm{S}_{2}$ open, the phase difference $(\phi)$ between the current and source voltage is $30^{\circ}$ and phase difference is $60^{\circ}$ when $S_{2}$ is closed keeping $\mathrm{S}_{1}$ open. The value of $\left(3 \mathrm{~L}_{1}-\mathrm{L}_{2}\right)$ is $\_\_\_\_$ H.

JEE Main · 2026 · 02 Apr 2026 · Shift 2 · Alternating Current

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})

____

Answer: (B)

With

S_{1}

closed and

S_{2}

open the inductor

L_{2}

is short-circuited, leaving

C

R

and

L_{1}

in series, for which

tan ϕ = \frac{X _{L_{1}} - X _{C}}{R}

, so

tan 3 0^{\circ} = \frac{1}{3} = \frac{X _{L_{1}} - X _{C}}{R}

. With

S_{2}

closed and

S_{1}

open,

L_{1}

is short-circuited and

C

R

L_{2}

remain, giving

tan 6 0^{\circ} = 3 = \frac{X _{L_{2}} - X _{C}}{R}

. Dividing,

X_{L_{2}} - X_{C} = 3 (X_{L_{1}} - X_{C})

, hence

3 X_{L_{1}} - X_{L_{2}} = 2 X_{C}

, i.e.

3 ω L_{1} - ω L_{2} = \frac{2}{ω C}

and

3 L_{1} - L_{2} = \frac{2}{ω ^{2} C}

. With

ω = 300 rad/s

and

C = 100 μ F

3 L_{1} - L_{2} = \frac{2}{( 300 ) ^{2} \times 100 \times 1 0 ^{- 6}} = \frac{2}{9} H

JEE Main · 2026 · 02 Apr 2026 · Shift 2 · Single correct option