Question

In an electrical circuit $ R, L, C $ and AC 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{\pi}{3} $. If instead $ C $ is removed from the circuit, the phase difference is again $ \frac{\pi}{3} $. The power factor of the circuit is

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

Hint:

The power factor of a purely resistive circuit is always 1, as the voltage and current are in phase.

Answer 1

The Correct Option is C

In this case, the circuit contains resistance \( R \), inductance \( L \), and capacitance \( C \), connected in series with an AC voltage source. 
The phase difference \( \phi \) between the voltage and the current is influenced by the presence of the inductor and capacitor. 
The phase difference in an RLC circuit is given by the formula: \[ \tan(\phi) = \frac{X_L - X_C}{R} \] where: 
- \( X_L = \omega L \) is the inductive reactance, 
- \( X_C = \frac{1}{\omega C} \) is the capacitive reactance, 
- \( \omega = 2\pi f \) is the angular frequency of the AC voltage source. When the inductor \( L \) is removed, the phase difference becomes: \[ \tan\left(\frac{\pi}{3}\right) = \frac{-X_C}{R} \] When the capacitor \( C \) is removed, the phase difference becomes: \[ \tan\left(\frac{\pi}{3}\right) = \frac{X_L}{R} \] This implies that both reactances, inductive and capacitive, are equal, and the circuit behaves as a purely resistive circuit when either component is removed. 
Thus, the power factor \( \text{PF} \) of the circuit is: \[ \text{PF} = \cos(\phi) = \cos(0) = 1 \] Therefore, the correct answer is: \[ \text{(3) } 1 \]