Question

When alternating current is passed through an \( L-R \) series circuit, the power factor is \( \frac{\sqrt{3}}{2} \) and \( R = 50\Omega \), then the value of \( L \) is: 
 
\[ \left[ \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}, \quad \sin \frac{\pi}{6} = \frac{1}{2}, \quad \tan \frac{\pi}{6} = \frac{1}{\sqrt{3}} \right] \]

• A. $\frac{1}{2}\pi$
• B. $\frac{\sqrt{3}}{2}\pi$
• C. $\frac{1}{2\sqrt{3}\pi}$
• D. $\frac{1}{\sqrt{3}\pi}$

Hint:

For an \(L-R\) AC circuit: \[ \tan\phi=\frac{\omega L}{R} \] So you need \(R\), phase angle, and frequency to calculate \(L\).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The power factor in an AC circuit is given by $\cos \phi$, where $\phi$ is the phase angle between voltage and current. In an L-R series circuit, the power factor is the ratio of resistance to impedance.
Step 2: Key Formula or Approach:
1. Power Factor: $\cos \phi = \frac{R}{Z}$
2. Phase Angle Relation: $\tan \phi = \frac{X_L}{R} = \frac{\omega L}{R}$
3. Angular Frequency: $\omega = 2\pi f$. (Assuming standard $f = 50 \text{ Hz}$ for this exam)
Step 3: Detailed Explanation:
Given $\cos \phi = \frac{\sqrt{3}}{2} \implies \phi = 30^\circ$ or $\frac{\pi}{6} \text{ radians}$.
From the given data, $\tan \left( \frac{\pi}{6} \right) = \frac{1}{\sqrt{3}}$.
Using the relation $\tan \phi = \frac{\omega L}{R}$:
\[ \frac{1}{\sqrt{3}} = \frac{2\pi(50)L}{50} \]
\[ \frac{1}{\sqrt{3}} = 2\pi L \implies L = \frac{1}{2\sqrt{3}\pi} \]
Step 4: Final Answer:
The value of inductance $L$ is $\frac{1}{2\sqrt{3}\pi}$.