If the power factor is `1//2` in a series `RL` circuit with `R = 100 Omega`. If `AC` mains, `50 Hz` is used then `L` is

Options

A. `(sqrt(3))/(pi)` henry
B. `pi` henry
C. `sqrt(3)` henry
D. `sqrt(3) pi` henry

Explanation

To solve the problem step by step, we need to find the inductance \( L \) in a series \( RL \) circuit where the power factor is \( \frac{1}{2} \), the resistance \( R = 100 \, \Omega \), and the frequency \( f = 50 \, \text{Hz} \).

### Step 1: Understand the relationship between power factor, resistance, and impedance
The power factor \( \cos \phi \) is given by the formula:
\[ \cos \phi = \frac{R}{Z} \]
where \( Z \) is the impedance of the circuit.

### Step 2: Substitute the known values into the power factor equation
Given that \( \cos \phi = \frac{1}{2} \) and \( R = 100 \, \Omega \):
\[ \frac{1}{2} = \frac{100}{Z} \]

### Step 3: Solve for \( Z \)
Rearranging the equation gives:
\[ Z = 100 \times 2 = 200 \, \Omega \]

### Step 4: Relate impedance to resistance and reactance
In a series \( RL \) circuit, the impedance \( Z \) is given by:
\[ Z = \sqrt{R^2 + X_L^2} \]
where \( X_L \) is the inductive reactance.

### Step 5: Substitute the known values into the impedance equation
Substituting \( R = 100 \, \Omega \) and \( Z = 200 \, \Omega \):
\[ 200 = \sqrt{100^2 + X_L^2} \]

### Step 6: Square both sides to eliminate the square root
\[ 200^2 = 100^2 + X_L^2 \]
\[ 40000 = 10000 + X_L^2 \]

### Step 7: Solve for \( X_L^2 \)
\[ X_L^2 = 40000 - 10000 = 30000 \]

### Step 8: Take the square root to find \( X_L \)
\[ X_L = \sqrt{30000} = 100\sqrt{3} \, \Omega \]

### Step 9: Relate inductive reactance to inductance
The inductive reactance \( X_L \) is related to the inductance \( L \) and frequency \( f \) by:
\[ X_L = \omega L = 2\pi f L \]
where \( \omega = 2\pi f \).

### Step 10: Substitute \( X_L \) and solve for \( L \)
Substituting \( X_L = 100\sqrt{3} \) and \( f = 50 \, \text{Hz} \):
\[ 100\sqrt{3} = 2\pi (50) L \]
\[ L = \frac{100\sqrt{3}}{2\pi (50)} \]
\[ L = \frac{100\sqrt{3}}{100\pi} = \frac{\sqrt{3}}{\pi} \, \text{H} \]

### Final Answer
The inductance \( L \) is:
\[ L = \frac{\sqrt{3}}{\pi} \, \text{H} \]