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: \[ \text{(3) } 1 \]

Answer 2

In a series circuit with resistance \( R \), inductance \( L \), capacitance \( C \), and an AC voltage source, the phase difference \( \phi \) between the voltage and the current depends on the reactance of \( L \) and \( C \). The impedance \( Z \) of the circuit can be expressed as \( Z = R + j(X_L - X_C) \), where \( X_L = \omega L \) and \( X_C = \frac{1}{\omega C} \) are the inductive and capacitive reactances, respectively. The phase difference is given by \( \tan\phi = \frac{X_L - X_C}{R} \).

1. When \( L \) is removed, the circuit behaves as an RC circuit, and the phase difference is \( \frac{\pi}{3} \):

\[\tan\left(\frac{\pi}{3}\right) = \sqrt{3} = -\frac{X_C}{R}\] \[X_C = R\sqrt{3}\]

2. When \( C \) is removed, the circuit behaves as an RL circuit, and the phase difference is also \( \frac{\pi}{3} \):

\[\tan\left(\frac{\pi}{3}\right) = \sqrt{3} = \frac{X_L}{R}\] \[X_L = R\sqrt{3}\]

Since \( X_L = X_C = R\sqrt{3} \) in their respective scenarios, this symmetry implies that the circuit is under resonance when both \( L \) and \( C \) are present, i.e., \( X_L = X_C \). Under resonance:

\[X_L - X_C = 0\] \[\tan\phi = 0\] \[\phi = 0\]

The power factor \( \cos\phi \) at resonance is:

\[\cos(0) = 1\]

Thus, the power factor of the circuit is 1.