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. `pi`H
B. `sqrt(3)/piH`
C. `pi/sqrt(3)H`
D. `sqrt(2)/pi`H

Explanation

To solve the problem, we need to find the inductance \( L \) in a series \( R-L \) circuit given the power factor and resistance. Here's a step-by-step solution:

### Step 1: Understand the Power Factor
The power factor \( \cos \phi \) is given as \( \frac{1}{2} \). This implies that:
\[ \phi = \cos^{-1}\left(\frac{1}{2}\right) = 60^\circ \]

### Step 2: Relate Power Factor to Resistance and Reactance
In a series \( R-L \) circuit, the power factor is defined as:
\[ \cos \phi = \frac{R}{Z} \]
where \( Z \) is the impedance of the circuit. The impedance \( Z \) can also be expressed in terms of resistance \( R \) and inductive reactance \( X_L \):
\[ Z = \sqrt{R^2 + X_L^2} \]
where \( X_L = \omega L \) and \( \omega = 2\pi f \).

### Step 3: Calculate the Inductive Reactance
From the power factor relation, we can express \( Z \):
\[ \frac{R}{Z} = \frac{1}{2} \implies Z = 2R \]
Substituting \( R = 100 \, \Omega \):
\[ Z = 2 \times 100 = 200 \, \Omega \]

### Step 4: Set Up the Impedance Equation
Now, substituting \( Z \) back into the impedance equation:
\[ 200 = \sqrt{100^2 + X_L^2} \]
Squaring both sides:
\[ 200^2 = 100^2 + X_L^2 \]
\[ 40000 = 10000 + X_L^2 \]
\[ X_L^2 = 40000 - 10000 = 30000 \]
\[ X_L = \sqrt{30000} = 100\sqrt{3} \, \Omega \]

### Step 5: Relate Reactance to Inductance
Now, we can relate the inductive reactance to the inductance:
\[ X_L = \omega L \implies 100\sqrt{3} = (2\pi f)L \]
Given \( f = 50 \, \text{Hz} \):
\[ \omega = 2\pi \times 50 = 100\pi \]
Substituting this into the equation:
\[ 100\sqrt{3} = (100\pi)L \]
Dividing both sides by \( 100\pi \):
\[ L = \frac{\sqrt{3}}{\pi} \, \text{H} \]

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

### Conclusion
The correct option is \( \frac{\sqrt{3}}{\pi} \) Henry.

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