Question

In an electrical circuit \( R, L, C \) and an a.c. voltage source are all connected in series. When \( L' \) is removed from the circuit, the phase difference between the voltage and the current in the circuit is \( \frac{3}{4} \). If instead \( C' \) is removed from the circuit, the phase difference is again \( \frac{3}{4} \). The power factor of the circuit is

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

Hint:

The power factor in an a.c. circuit is determined by the phase difference between voltage and current. It gives the efficiency of energy transfer in the circuit.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We need to find the power factor of the complete LCR series circuit. The information provided about removing L and C helps determine the relative values of the reactances.
Step 2: Key Formula or Approach:
1. Phase difference \( \tan \phi = \frac{|X_L - X_C|}{R} \).
2. Power factor \( \cos \phi = \frac{R}{Z} \).
Step 3: Detailed Explanation:
When \( L \) is removed:
\[ \tan(\pi / 3) = \frac{X_C}{R} \implies \sqrt{3} = \frac{X_C}{R} \implies X_C = \sqrt{3}R \]
When \( C \) is removed:
\[ \tan(\pi / 3) = \frac{X_L}{R} \implies \sqrt{3} = \frac{X_L}{R} \implies X_L = \sqrt{3}R \]
Since \( X_L = X_C \), the circuit is in resonance.
In resonance, the total impedance \( Z = R \).
Power factor \( \cos \phi = \frac{R}{Z} = \frac{R}{R} = 1 \).
Step 4: Final Answer:
The power factor of the circuit is 1.