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JEE Main 2026 · 2 Apr · Shift 2, Physics question 42

The figure given below shows an LCRLCR series circuit with two switches S1S_1 and S2S_2. When switch S1S_1 is closed keeping S2S_2 open, the phase difference (ϕ\phi) between the current and source voltage is 30° and phase difference is 60° when S2S_2 is closed keeping S1S_1 open. The value of (3L1−L2)(3L_1-L_2) is ________ H.

Series loop with source v = V₀ sin(300t), a capacitor C = 100 μF and a resistor R. Inductor L₂ sits on the left with switch S₁ connected across it; inductor L₁ sits on the right with switch S₂ connected across it. Closing a switch short-circuits the inductor beside it.
Choose one option
Answer and worked solution

Official answer: B

Which inductor is in circuit. Closing a switch shorts the inductor beside it:

  • S1S_1 closed: L₁, C, R in series, tan⁡30∘=ωL1−XCR\tan30^\circ=\dfrac{\omega L_1-X_C}R.
  • S2S_2 closed: L₂, C, R in series, tan⁡60∘=ωL2−XCR\tan60^\circ=\dfrac{\omega L_2-X_C}R.

Eliminate R. 3(ωL1−XC)=ωL2−XC3(\omega L_1-X_C)=\omega L_2-X_C, so ω(3L1−L2)=2XC\omega(3L_1-L_2)=2X_C.

Value. XC=1300×10−4=1003X_C=\dfrac1{300\times10^{-4}}=\dfrac{100}3 Ω, so 3L1−L2=2(100/3)300=293L_1-L_2=\dfrac{2(100/3)}{300}=\dfrac29 H.

(B)

Topics:LCR circuits and resonanceReactance and impedance
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