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

Concept:
In an \(L-R\) series circuit: \[ \cos\phi=\frac{R}{Z} \] and \[ \tan\phi=\frac{X_L}{R}=\frac{\omega L}{R} \] ip

Step 1: Find phase angle.
Given power factor: \[ \cos\phi=\frac{\sqrt{3}}{2} \] So, \[ \phi=\frac{\pi}{6} \] Thus, \[ \tan\phi=\frac{1}{\sqrt{3}} \] ip

Step 2: Use \(\tan\phi=\frac{\omega L}{R}\).
\[ \frac{\omega L}{R}=\frac{1}{\sqrt{3}} \] \[ L=\frac{R}{\sqrt{3}\omega} \] ip

Step 3: Substitute \(R=50\Omega\).
To get a numerical value of \(L\), frequency is required because: \[ \omega=2\pi f \] But frequency is not stated in the pasted text. If the intended frequency is \(25\text{ Hz}\), then: \[ \omega=50\pi \] and \[ L=\frac{50}{\sqrt{3}\cdot 50\pi}=\frac{1}{\sqrt{3}\pi} \] which matches option (D). ip The keyed Option corresponds to:
\[ \boxed{(D)\ \frac{1}{\sqrt{3}\pi}} \]