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 circuit when the inductor (L) and capacitor (C) are removed separately and how that affects the phase difference and the power factor.

### Step 1: Understand the Circuit Configuration
In the given circuit, we have a resistor (R), an inductor (L), a capacitor (C), and an AC voltage source all connected in series. The phase difference between the voltage and current is affected by the presence or absence of L or C.

### Step 2: Analyze the Circuit with Capacitor Removed
When the capacitor (C) is removed, we have an R-L circuit. The phase difference (φ) is given as π/6. The relationship between the phase difference and the reactance in an R-L circuit is given by:

\[ \tan(\phi) = \frac{X_L}{R} \]

Where \(X_L\) is the inductive reactance. For φ = π/6, we have:

\[ \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}} \]

Thus,

\[ \frac{X_L}{R} = \frac{1}{\sqrt{3}} \implies X_L = \frac{R}{\sqrt{3}} \]

### Step 3: Analyze the Circuit with Inductor Removed
When the inductor (L) is removed, we have an R-C circuit. The phase difference (φ) is again π/6. The relationship between the phase difference and the reactance in an R-C circuit is given by:

\[ \tan(\phi) = \frac{X_C}{R} \]

Where \(X_C\) is the capacitive reactance. For φ = π/6, we have:

\[ \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}} \]

Thus,

\[ \frac{X_C}{R} = \frac{1}{\sqrt{3}} \implies X_C = \frac{R}{\sqrt{3}} \]

### Step 4: Relate Inductive and Capacitive Reactance
Since we have both \(X_L\) and \(X_C\) equal to \(\frac{R}{\sqrt{3}}\), we can conclude:

\[ X_L = X_C \implies \frac{R}{\sqrt{3}} = \frac{R}{\sqrt{3}} \]

This means that the circuit is at resonance when both L and C are present.

### Step 5: Calculate the Impedance (Z)
The total impedance (Z) of the RLC series circuit at resonance is given by:

\[ Z = R \]

### Step 6: Calculate the Power Factor
The power factor (PF) is defined as:

\[ \text{Power Factor} = \cos(\phi) \]

Since φ = π/6:

\[ \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \]

### Final Answer
Thus, the power factor of the circuit is:

\[ \text{Power Factor} = \frac{\sqrt{3}}{2} \]