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 Concept:
The circuit has inductors, a capacitor, and a resistor in series, but switches can short out certain components.
Closing \(S_1\) shorts out \(L_2\), leaving \(L_1, C, R\) in the circuit.
Closing \(S_2\) shorts out \(L_1\), leaving \(L_2, C, R\) in the circuit.
We use the phase angle formula for series LCR circuits to establish equations for both cases.
Step 2: Key Formula or Approach:
The phase difference \(\phi\) in a series LCR circuit is given by \(\tan \phi = \frac{|X_L - X_C|}{R}\).
Assuming the circuit remains capacitive (\(X_C>X_L\)), we use \(\tan \phi = \frac{X_C - X_L}{R}\).
(Note: Assuming it is inductive yields the exact same final result for the required expression).
Step 3: Detailed Explanation:
Case 1: \(S_1\) closed, \(S_2\) open. \(L_2\) is shorted.
The phase difference is \(30^\circ\).
\[ \tan 30^\circ = \frac{X_C - X_{L1}}{R} \implies \frac{1}{\sqrt{3}} = \frac{X_C - X_{L1}}{R} \] \[ X_{L1} = X_C - \frac{R}{\sqrt{3}} \quad \text{--- (Eq 1)} \] Case 2: \(S_2\) closed, \(S_1\) open. \(L_1\) is shorted.
The phase difference is \(60^\circ\).
\[ \tan 60^\circ = \frac{X_C - X_{L2}}{R} \implies \sqrt{3} = \frac{X_C - X_{L2}}{R} \] \[ X_{L2} = X_C - R\sqrt{3} \quad \text{--- (Eq 2)} \] We are asked to find the value of \(3L_1 - L_2\).
Since \(X_L = \omega L\), we can write \(3L_1 - L_2 = \frac{3X_{L1} - X_{L2}}{\omega}\).
Let's compute \(3X_{L1} - X_{L2}\) using Eq 1 and Eq 2:
\[ 3X_{L1} - X_{L2} = 3\left(X_C - \frac{R}{\sqrt{3}}\right) - \left(X_C - R\sqrt{3}\right) \] \[ 3X_{L1} - X_{L2} = 3X_C - \sqrt{3}R - X_C + \sqrt{3}R = 2X_C \] The unknown resistance \(R\) cancels out perfectly!
Now, calculate \(X_C\):
Given \(\omega = 300 \text{ rad/s}\) and \(C = 100 \mu\text{F} = 10^{-4} \text{ F}\).
\[ X_C = \frac{1}{\omega C} = \frac{1}{300 \times 10^{-4}} = \frac{10^4}{300} = \frac{100}{3}\ \Omega \] Finally, evaluate the required expression:
\[ 3L_1 - L_2 = \frac{2X_C}{\omega} = \frac{2 \times (100/3)}{300} = \frac{200}{900} = \frac{2}{9} \text{ H} \] Step 4: Final Answer:
The value of \((3L_1 - L_2)\) is \(\frac{2}{9}\).