Question

A resistor of \(450 \, \Omega\) and an inductor are connected in series to an ac source of frequency \( \frac{75}{\pi} \, \text{Hz} \). If the power factor of the circuit is 0.6, then the inductance connected in the circuit is

• A. 6 mH
• B. 4 H
• C. 4 mH
• D. 6 H

Hint:

For an LR series AC circuit: - Impedance \( Z = \sqrt{R^2 + X_L^2} \), where \( X_L = \omega L = 2\pi f L \). - Power factor \( \cos\phi = \frac{R}{Z} \). - Phase angle \( \tan\phi = \frac{X_L}{R} \). If \( \cos\phi \) is given (e.g., 0.6 = 3/5), form a right triangle (3-4-5) to find \( \sin\phi \) (4/5) and \( \tan\phi \) (4/3).

Answer 1

The Correct Option is B

Resistance \( R = 450 \, \Omega \).
Frequency \( f = \frac{75}{\pi} \, \text{Hz} \).
Power factor \( \cos\phi = 0.
6 \).
The circuit is an LR series circuit.
The power factor is given by \( \cos\phi = \frac{R}{Z} \), where \(Z\) is the impedance of the circuit.
\( Z = \sqrt{R^2 + X_L^2} \), where \( X_L \) is the inductive reactance.
Given \( \cos\phi = 0.
6 \): \[ 0.
6 = \frac{450}{Z} \implies Z = \frac{450}{0.
6} = \frac{4500}{6} = 750 \, \Omega \] Now use \( Z^2 = R^2 + X_L^2 \): \[ (750)^2 = (450)^2 + X_L^2 \] \[ X_L^2 = (750)^2 - (450)^2 = (750-450)(750+450) \] \[ X_L^2 = (300)(1200) = 360000 \] \[ X_L = \sqrt{360000} = \sqrt{36 \times 10^4} = 6 \times 10^2 = 600 \, \Omega \] Alternatively, if \( \cos\phi = 0.
6 = 3/5 \), then \( \sin\phi = \sqrt{1-(3/5)^2} = \sqrt{1-9/25} = \sqrt{16/25} = 4/5 \).
Also, \( \tan\phi = \frac{\sin\phi}{\cos\phi} = \frac{4/5}{3/5} = \frac{4}{3} \).
For an LR circuit, \( \tan\phi = \frac{X_L}{R} \).
\[ \frac{X_L}{450} = \frac{4}{3} \implies X_L = 450 \times \frac{4}{3} = 150 \times 4 = 600 \, \Omega \).
This is consistent.
Inductive reactance \( X_L = \omega L = 2\pi f L \).
\[ 600 = 2\pi \left(\frac{75}{\pi}\right) L \] \[ 600 = 2 \times 75 L = 150 L \] \[ L = \frac{600}{150} = \frac{60}{15} = 4 \, \text{H} \] The inductance is 4 H.
This matches option (2).