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 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 given conditions and apply the relevant formulas for phase difference in an RLC circuit.

### Step-by-Step Solution:

1. **Understanding the Circuit**:
- We have a series RLC circuit connected to an AC voltage source. The components are resistance (R), inductance (L), and capacitance (C).

2. **Phase Difference with Inductor Removed**:
- When the inductor (L) is removed, the circuit consists of only R and C. The phase difference between voltage and current is given as \( \phi = \frac{\pi}{3} \).
- The formula for phase difference in an RC circuit is:
\[ \tan(\phi) = \frac{X_C}{R} \]
- Here, \( X_C \) is the capacitive reactance.

3. **Phase Difference with Capacitor Removed**:
- When the capacitor (C) is removed, the circuit consists of only R and L. The phase difference is again given as \( \phi = \frac{\pi}{3} \).
- The formula for phase difference in an RL circuit is:
\[ \tan(\phi) = \frac{X_L}{R} \]
- Here, \( X_L \) is the inductive reactance.

4. **Calculating Reactances**:
- From both conditions, we can write:
\[ \tan\left(\frac{\pi}{3}\right) = \sqrt{3} \]
- Therefore, we have two equations:
\[ \sqrt{3} = \frac{X_C}{R} \quad \text{(1)} \]
\[ \sqrt{3} = \frac{X_L}{R} \quad \text{(2)} \]

5. **Equating Reactances**:
- From equations (1) and (2), we can equate the two expressions:
\[ X_C = X_L \]
- This means the capacitive reactance is equal to the inductive reactance.

6. **Finding the Phase Difference in the Original Circuit**:
- In the original RLC circuit, the phase difference is given by:
\[ \tan(\phi) = \frac{X_L - X_C}{R} \]
- Since \( X_L = X_C \), we have:
\[ \tan(\phi) = \frac{X_L - X_L}{R} = 0 \]
- Thus, \( \phi = 0 \).

7. **Calculating the Power Factor**:
- The power factor (PF) is given by:
\[ \text{PF} = \cos(\phi) \]
- Since \( \phi = 0 \):
\[ \text{PF} = \cos(0) = 1 \]

### Conclusion:
The power factor of the circuit is **1**, which corresponds to a unity power factor.