In an electrical circuit `R,L,C` and an `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 `pi//3`. If instead, `C` is removed from the circuit, difference the phase difference is again `pi//3`. The power factor of the circuit is

Options

A. `1//2`
B. `1//sqrt(2)`
C. 1
D. `sqrt(3)//2`

Explanation

To solve the problem, we will analyze the circuit step by step, using the given information about the phase differences when either the inductor (L) or the capacitor (C) is removed.

### Step 1: Analyze the first case (L is removed)

When the inductor (L) is removed, the circuit consists of the resistor (R) and the capacitor (C). The phase difference between the voltage and current is given as \( \phi_1 = \frac{\pi}{3} \).

Using the relationship for the phase difference in an R-C circuit:
\[ \tan \phi_1 = \frac{X_C}{R} \]
where \( X_C \) is the capacitive reactance.

Substituting \( \phi_1 \):
\[ \tan\left(\frac{\pi}{3}\right) = \frac{X_C}{R} \]
We know that \( \tan\left(\frac{\pi}{3}\right) = \sqrt{3} \), so:
\[ \sqrt{3} = \frac{X_C}{R} \]
From this, we can express \( X_C \):
\[ X_C = \sqrt{3} R \quad \text{(Equation 1)} \]

### Step 2: Analyze the second case (C is removed)

When the capacitor (C) is removed, the circuit consists of the resistor (R) and the inductor (L). The phase difference is again given as \( \phi_2 = \frac{\pi}{3} \).

Using the relationship for the phase difference in an R-L circuit:
\[ \tan \phi_2 = \frac{X_L}{R} \]
where \( X_L \) is the inductive reactance.

Substituting \( \phi_2 \):
\[ \tan\left(\frac{\pi}{3}\right) = \frac{X_L}{R} \]
Again, since \( \tan\left(\frac{\pi}{3}\right) = \sqrt{3} \):
\[ \sqrt{3} = \frac{X_L}{R} \]
From this, we can express \( X_L \):
\[ X_L = \sqrt{3} R \quad \text{(Equation 2)} \]

### Step 3: Relate the reactances

From Equations 1 and 2, we have:
\[ X_C = X_L \]
This implies:
\[ \sqrt{3} R = \sqrt{3} R \]
This indicates that the circuit is in resonance, where the inductive reactance equals the capacitive reactance.

### Step 4: Calculate the power factor

In a resonant circuit, the impedance \( Z \) is equal to the resistance \( R \):
\[ Z = R \]
The power factor \( \text{PF} \) is given by:
\[ \text{PF} = \cos \phi = \frac{R}{Z} \]
Since \( Z = R \):
\[ \text{PF} = \frac{R}{R} = 1 \]

### Conclusion

The power factor of the circuit is \( 1 \).