Alternating Current

JEE Main level, single correct, medium

56 of 190

Question 56 of 190

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°30° and phase difference is 60°60° when S2S_2 is closed keeping S1S_1 open. The value of (3L1−L2)(3L_1 - L_2) is _______ H.

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Answer: (B) 29\dfrac{2}{9}

From the source v=V0sin⁡(300t)v = V_0 \sin(300t), the angular frequency is ω=300 rad/s\omega = 300\text{ rad/s}.
Capacitance C=100 μF=10−4 FC = 100\ \mu\text{F} = 10^{-4}\text{ F}, so XC=1ωCX_C = \dfrac{1}{\omega C}.

Case 1 (S1S_1 closed, S2S_2 open):
Closing S1S_1 short-circuits L2L_2, leaving an LCR series circuit with CC, RR and L1L_1. With phase difference ϕ1=30∘\phi_1 = 30^\circ:
tan⁡30∘=ωL1−XCR\tan 30^\circ = \dfrac{\omega L_1 - X_C}{R}
13=ωL1−XCR⇒R=3 (ωL1−XC)\dfrac{1}{\sqrt{3}} = \dfrac{\omega L_1 - X_C}{R} \Rightarrow R = \sqrt{3}\,(\omega L_1 - X_C) ... (1)

Case 2 (S2S_2 closed, S1S_1 open):
Closing S2S_2 short-circuits L1L_1, leaving an LCR series circuit with CC, RR and L2L_2. With phase difference ϕ2=60∘\phi_2 = 60^\circ:
tan⁡60∘=ωL2−XCR\tan 60^\circ = \dfrac{\omega L_2 - X_C}{R}
3=ωL2−XCR⇒ωL2−XC=R3\sqrt{3} = \dfrac{\omega L_2 - X_C}{R} \Rightarrow \omega L_2 - X_C = R\sqrt{3} ... (2)

Substituting RR from (1) into (2):
ωL2−XC=3 (ωL1−XC)⋅3\omega L_2 - X_C = \sqrt{3}\,(\omega L_1 - X_C) \cdot \sqrt{3}
ωL2−XC=3ωL1−3XC\omega L_2 - X_C = 3\omega L_1 - 3X_C
2XC=3ωL1−ωL2=ω(3L1−L2)2X_C = 3\omega L_1 - \omega L_2 = \omega(3L_1 - L_2)

Therefore:
3L1−L2=2XCω=2ω2C3L_1 - L_2 = \dfrac{2X_C}{\omega} = \dfrac{2}{\omega^2 C}

Substituting ω=300 rad/s\omega = 300\text{ rad/s} and C=10−4 FC = 10^{-4}\text{ F}:
ω2C=(300)2⋅10−4=9×104⋅10−4=9\omega^2 C = (300)^2 \cdot 10^{-4} = 9 \times 10^4 \cdot 10^{-4} = 9

3L1−L2=29 H3L_1 - L_2 = \dfrac{2}{9}\text{ H}

Hence, the correct option is (2) 29(2)\ \dfrac{2}{9}.

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