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 step by step, we will analyze the given circuit conditions and derive the necessary equations.

### Step 1: Analyze the first scenario (when L is removed)
When the inductor \( L \) is removed from the circuit, we are left with a resistor \( R \) and a capacitor \( C \). The phase difference \( \phi \) between the voltage and current is given as \( \frac{\pi}{3} \).

Using the formula for the phase difference in an RC circuit:
\[ \tan \phi = \frac{X_C}{R} \]
where \( X_C \) is the capacitive reactance.

Substituting \( \phi = \frac{\pi}{3} \):
\[ \tan\left(\frac{\pi}{3}\right) = \sqrt{3} \]
Thus, we have:
\[ \sqrt{3} = \frac{X_C}{R} \]
From this, we can express \( X_C \):
\[ X_C = \sqrt{3} R \quad \text{(Equation 1)} \]

### Step 2: Analyze the second scenario (when C is removed)
Now, when the capacitor \( C \) is removed, we have the resistor \( R \) and the inductor \( L \) in the circuit. The phase difference \( \phi \) is again given as \( \frac{\pi}{3} \).

Using the formula for the phase difference in an RL circuit:
\[ \tan \phi = \frac{X_L}{R} \]
where \( X_L \) is the inductive reactance.

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

### Step 3: Substitute \( X_C \) and \( X_L \) into the general equation
Now, we use the general equation for the phase difference in an LCR circuit:
\[ \tan \phi = \frac{X_L - X_C}{R} \]
Substituting the values from Equations 1 and 2:
\[ \tan \phi = \frac{\sqrt{3} R - \sqrt{3} R}{R} \]
This simplifies to:
\[ \tan \phi = \frac{0}{R} = 0 \]
Thus, we find:
\[ \phi = 0 \]

### Step 4: Calculate the power factor
The power factor \( \text{pf} \) is defined as:
\[ \text{pf} = \cos \phi \]
Since we found \( \phi = 0 \):
\[ \text{pf} = \cos(0) = 1 \]

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

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