The figure shows a series LCR circuit with a 100 μF capacitor, a resistor R, two selectable inductors L₁ and L₂, and an AC source v = V₀ sin(300t). When switch $S_1$ is closed keeping $S_2$ open, the phase difference ( $\phi$ )
between the current and source voltage is 30° and phase difference is 60° when $S_2$
is closed keeping $S_1$ open. The value of ( $3L_1-L_2$ ) is___ H
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The figure shows a series LCR circuit with a 100 μF capacitor, a resistor R, two selectable inductors L₁ and L₂, and an AC source v = V₀ sin(300t). When switch S 1 S_1 S 1 is closed keeping S 2 S_2 S 2 open, the phase difference ( ϕ \phi ϕ ) between the current and source voltage is 30° and phase difference is 60° when S 2 S_2 S 2 is closed keeping S 1 S_1 S 1 open. The value of ( 3 L 1 − L 2 3L_1-L_2 3 L 1 − L 2 ) is___ H

Physics · Ac Cirucuits

JEE Mains 2026 2 April Evening Shift

G1 Physics · Accepted Answer

The angular frequency of the source is ω = 300 rad·s⁻¹. The capacitor is C = 100 μF = 100×10⁻⁶ F. For the switch condition giving phase angle 30°, the source solution gives (X(L₁)-X(C)) ÷ R = tan30° = 1 ÷ √3. Therefore, √3[X(L₁)-X(C)] = R. (1) For the switch condition giving phase angle 60°, (X(L₂)-X(C)) ÷ R = tan60° = √3. Therefore, [X(L₂)-X(C)] ÷ √3 = R. (2) Equate equations (1) and (2): √3[X(L₁)-X(C)] = [X(L₂)-X(C)] ÷ √3. Multiply by √3: 3[X(L₁)-X(C)] = X(L₂)-X(C). Thus, 3X(L₁)-X(L₂) = 2X(C)…