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: Relationship between phase difference and power factor.
The phase difference \( \phi \) between the voltage and the current in an a.c. circuit is related to the power factor \( \text{pf} \) by:
\[ \text{pf} = \cos(\phi). \]
Given that the phase difference is \( \frac{3}{4} \), we can find the power factor using:
\[ \text{pf} = \cos\left(\frac{3}{4}\right). \]

Step 2: Analyze the given condition for removing \( L' \) or \( C' \).
The fact that the phase difference remains \( \frac{3}{4} \) whether \( L' \) or \( C' \) is removed from the circuit suggests that the circuit is predominantly resistive, with the power factor being influenced by the overall impedance of the circuit.

Step 3: Applying the condition.
The given phase difference \( \frac{3}{4} \) corresponds to a power factor of \( \frac{\sqrt{3}}{2} \) (from standard values of cosine function for various angles).
\[ \text{pf} = \cos\left(\frac{3}{4}\right) = \frac{\sqrt{3}}{2}. \]
Final Answer:
Thus, the power factor of the circuit is:
\[ \boxed{\frac{\sqrt{3}}{2}}. \]