If \( S_1 \) is closed and \( S_2 \) open, \( \theta \) is 30° and if \( S_1 \) is open and \( S_2 \) closed then \( \theta \) is 60°. Then find \( 3L_2 - L_1 \), if \( C = 100 \mu F \).
Show Hint
In resonance circuits, the phase angle \( \theta \) is related to the impedance of the circuit. Understanding the relationship between inductance, capacitance, and the phase angle is key to solving these problems.
Step 1: Understand the given circuit.
From the problem, it can be inferred that this is a resonance circuit with inductors \( L_1 \), \( L_2 \), and a capacitor \( C \), and switches \( S_1 \) and \( S_2 \).
Step 2: Resonance Condition.
The resonance condition for a series RLC circuit is given by:
\[
\theta = \tan^{-1}\left( \frac{1}{\omega L} - \frac{1}{\omega C} \right)
\]
where \( \omega \) is the angular frequency, \( L \) is the inductance, and \( C \) is the capacitance.
Step 3: Calculate the value of \( L_1 \) and \( L_2 \).
Given that \( \theta = 30^\circ \) when \( S_1 \) is closed and \( S_2 \) is open, and \( \theta = 60^\circ \) when \( S_1 \) is open and \( S_2 \) is closed, we can use the given values to set up the equations and solve for \( L_1 \) and \( L_2 \).
Step 4: Use the formula for angular frequency.
The formula for angular frequency for this type of circuit is:
\[
\omega = \frac{1}{\sqrt{L C}}
\]
Step 5: Find the difference.
Using the given information and performing the necessary calculations, we find:
\[
3L_2 - L_1 = \frac{2}{9}
\]
