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Calculate Inductance Difference in Dual Switch LCR Circuit

The figure given below shows an LCR 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^\circ and phase difference is 60∘60^\circ when S2S_2 is closed keeping S1S_1 open. The value of (3L1−L2)(3L_1 - L_2) is ______ H\text{H}.

Question Diagram 1

Options

A

92\frac{9}{2}

B

29\frac{2}{9}

Correct
C

13\frac{1}{3}

D

3

Topics & Concepts

Alternating CurrentAC Circuits

Step-by-Step Solution

To find the value of (3L1−L2)(3L_1 - L_2), we analyze the circuit under both switch configurations.

1. Circuit Parameters

From the given voltage equation v=V0sin⁡(300t)v = V_0 \sin(300t):

  • Angular frequency, ω=300 rad/s\omega = 300 \text{ rad/s}
  • Capacitance, C=100μF=10−4 FC = 100 \mu\text{F} = 10^{-4} \text{ F}

The capacitive reactance XCX_C is given by: XC=1ωC=1300×10−4=1003  ΩX_C = \frac{1}{\omega C} = \frac{1}{300 \times 10^{-4}} = \frac{100}{3} \;\Omega

2. Circuit Analysis for Switch Configurations

Case 1: S1S_1 closed and S2S_2 open

  • When switch S1S_1 is closed, inductor L2L_2 is short-circuited (bypassed).
  • When switch S2S_2 is open, inductor L1L_1 remains active in the series circuit.
  • The phase difference is ϕ1=30∘\phi_1 = 30^\circ.

Using the formula for the phase angle in an LCR series circuit: tan⁡ϕ1=∣XL1−XC∣R\tan \phi_1 = \frac{|X_{L1} - X_C|}{R} tan⁡30∘=13=∣XL1−XC∣R  ⟹  ∣XL1−XC∣=R3— (1)\tan 30^\circ = \frac{1}{\sqrt{3}} = \frac{|X_{L1} - X_C|}{R} \implies |X_{L1} - X_C| = \frac{R}{\sqrt{3}} \quad \text{--- (1)}

Case 2: S2S_2 closed and S1S_1 open

  • When switch S2S_2 is closed, inductor L1L_1 is short-circuited (bypassed).
  • When switch S1S_1 is open, inductor L2L_2 remains active in the series circuit.
  • The phase difference is ϕ2=60∘\phi_2 = 60^\circ.

tan⁡ϕ2=∣XL2−XC∣R\tan \phi_2 = \frac{|X_{L2} - X_C|}{R} tan⁡60∘=3=∣XL2−XC∣R  ⟹  ∣XL2−XC∣=R3— (2)\tan 60^\circ = \sqrt{3} = \frac{|X_{L2} - X_C|}{R} \implies |X_{L2} - X_C| = R\sqrt{3} \quad \text{--- (2)}

3. Relation Between L1L_1 and L2L_2

Multiplying equation (1) by 3: 3∣XL1−XC∣=3R3=R33 |X_{L1} - X_C| = \frac{3R}{\sqrt{3}} = R\sqrt{3}

Comparing this with equation (2): 3∣XL1−XC∣=∣XL2−XC∣3 |X_{L1} - X_C| = |X_{L2} - X_C|

Assuming (XL1−XC)(X_{L1} - X_C) and (XL2−XC)(X_{L2} - X_C) have the same sign (either both inductive or both capacitive): 3(XL1−XC)=XL2−XC3(X_{L1} - X_C) = X_{L2} - X_C 3XL1−3XC=XL2−XC3X_{L1} - 3X_C = X_{L2} - X_C 3XL1−XL2=2XC3X_{L1} - X_{L2} = 2X_C

Substitute XL1=ωL1X_{L1} = \omega L_1 and XL2=ωL2X_{L2} = \omega L_2: 3(ωL1)−ωL2=2XC3(\omega L_1) - \omega L_2 = 2X_C ω(3L1−L2)=2XC\omega (3L_1 - L_2) = 2X_C 3L1−L2=2XCω=2ω2C3L_1 - L_2 = \frac{2X_C}{\omega} = \frac{2}{\omega^2 C}

4. Numerical Calculation

Substituting the given values into the expression: 3L1−L2=2(300)2×10−43L_1 - L_2 = \frac{2}{(300)^2 \times 10^{-4}} 3L1−L2=290000×10−4=29 H3L_1 - L_2 = \frac{2}{90000 \times 10^{-4}} = \frac{2}{9} \text{ H}

Thus, the correct option is B.

Calculate Inductance Difference in Dual Switch LCR Circuit | Physics PYQ Solution - JEE Challenger