In an electrical circuit R,L, C and a.c voltage source are all connected in series. When L is removed from the circuit , the phase difference between the voltage and current in the circuit is π/6. If instead ,C is removed from the circuit , the phase difference is again π/6 . The power factor of the circuit is

Options

A. `1/2`
B. `1/sqrt2`
C. `1`
D. `sqrt3/2`

Answer

To solve the problem, we need to analyze the given information about the electrical circuit consisting of a resistor (R), inductor (L), capacitor (C), and an AC voltage source. We will break down the solution step-by-step.

### Step 1: Analyze the circuit 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 between the voltage and current in this case is given as \( \phi = \frac{\pi}{6} \).

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

Substituting \( \phi = \frac{\pi}{6} \):
\[ \tan\left(\frac{\pi}{6}\right) = \frac{X_C}{R} \]
Knowing that \( \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}} \):
\[ \frac{1}{\sqrt{3}} = \frac{X_C}{R} \]
Thus, we can express \( X_C \) as:
\[ X_C = \frac{R}{\sqrt{3}} \]
This will be our **Equation 1**.

### Step 2: Analyze the circuit when C is removed
Now, when the capacitor (C) is removed, we have a resistor (R) and an inductor (L) in the circuit. The phase difference in this case is also given as \( \phi = \frac{\pi}{6} \).

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

Substituting \( \phi = \frac{\pi}{6} \):
\[ \tan\left(\frac{\pi}{6}\right) = \frac{X_L}{R} \]
Again, using \( \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}} \):
\[ \frac{1}{\sqrt{3}} = \frac{X_L}{R} \]
Thus, we can express \( X_L \) as:
\[ X_L = \frac{R}{\sqrt{3}} \]
This will be our **Equation 2**.

### Step 3: Relate the equations
Now we have:
- From **Equation 1**: \( X_C = \frac{R}{\sqrt{3}} \)
- From **Equation 2**: \( X_L = \frac{R}{\sqrt{3}} \)

### Step 4: Calculate the overall phase difference
In the original circuit with R, L, and C, the total reactance \( X \) is given by:
\[ X = X_L - X_C \]
Substituting the values from Equations 1 and 2:
\[ X = \frac{R}{\sqrt{3}} - \frac{R}{\sqrt{3}} = 0 \]

### Step 5: Determine the phase angle
Since the total reactance \( X = 0 \), the phase angle \( \phi \) is:
\[ \tan(\phi) = \frac{X}{R} = \frac{0}{R} = 0 \]
Thus, \( \phi = 0 \).

### Step 6: Calculate the power factor
The power factor (PF) is defined as:
\[ \text{Power Factor} = \cos(\phi) \]
Since \( \phi = 0 \):
\[ \text{Power Factor} = \cos(0) = 1 \]

### Final Answer
The power factor of the circuit is **1**.

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