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

Options

A. `(sqrt(3))/(pi) Henry`
B. `pi Henry`
C. `(pi)/(sqrt(3)) Henry`
D. None of these

Explanation

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

### Step-by-Step Solution:

1. **Identify the Power Factor**:
The power factor (PF) is given as \( \frac{1}{2} \). The power factor is defined as:
\[ \text{PF} = \cos(\phi) \]
where \( \phi \) is the phase angle between the voltage and the current.

2. **Calculate the Phase Angle**:
To find \( \phi \):
\[ \phi = \cos^{-1}\left(\frac{1}{2}\right) \]
From trigonometric values, we know:
\[ \phi = 60^\circ \]

3. **Relate Inductive Reactance to Resistance**:
In a series \( RL \) circuit, the tangent of the phase angle is given by:
\[ \tan(\phi) = \frac{X_L}{R} \]
where \( X_L \) is the inductive reactance and \( R \) is the resistance.

4. **Express Inductive Reactance**:
The inductive reactance \( X_L \) can be expressed in terms of the inductance \( L \) and the angular frequency \( \omega \):
\[ X_L = \omega L \]
Thus, we can write:
\[ \tan(\phi) = \frac{\omega L}{R} \]

5. **Calculate \( \tan(60^\circ) \)**:
We know:
\[ \tan(60^\circ) = \sqrt{3} \]

6. **Substitute Values**:
Now substituting \( R = 100 \, \Omega \) and \( \tan(60^\circ) \):
\[ \sqrt{3} = \frac{\omega L}{100} \]
Rearranging gives:
\[ \omega L = 100 \sqrt{3} \]

7. **Determine Angular Frequency \( \omega \)**:
The angular frequency \( \omega \) is given by:
\[ \omega = 2 \pi f \]
Assuming a standard frequency \( f = 50 \, \text{Hz} \):
\[ \omega = 2 \pi \times 50 = 100 \pi \, \text{rad/s} \]

8. **Substitute \( \omega \) into the Equation**:
Now substituting \( \omega \) back into the equation:
\[ 100 \pi L = 100 \sqrt{3} \]
Dividing both sides by \( 100 \pi \):
\[ L = \frac{\sqrt{3}}{\pi} \]

9. **Final Result**:
Thus, the value of \( L \) is:
\[ L = \frac{\sqrt{3}}{\pi} \, \text{H} \]