Question

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 \( \phi \) 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.}

• A. \( \frac{9}{2} \)
• B. \( \frac{2}{9} \)
• C. \( \frac{1}{3} \)
• D. 3

Answer 1

The Correct Option is A

Step 1: Understanding the given circuit.
In the LCR circuit, the phase difference \( \phi \) between the current and the voltage is given by: \[ \tan \phi = \frac{X_L - X_C}{R} \] where \( X_L \) is the inductive reactance, \( X_C \) is the capacitive reactance, and \( R \) is the resistance.
Step 2: Phase difference with different conditions.
- When \( S_1 \) is closed and \( S_2 \) is open, the inductance \( L_1 \) is in the circuit. The phase difference is 30°, so: \[ \tan 30^\circ = \frac{X_{L_1} - X_C}{R} \] - When \( S_2 \) is closed and \( S_1 \) is open, the inductance \( L_2 \) is in the circuit. The phase difference is 60°, so: \[ \tan 60^\circ = \frac{X_{L_2} - X_C}{R} \]
Step 3: Solving for \( (3L_1 - L_2) \).
By using the relationships for \( X_L = \omega L \) and simplifying the given equations, we find that the value of \( (3L_1 - L_2) \) is \( \frac{9}{2} \) H.
Final Answer: (A) \( \frac{9}{2} \)