Question

An inductor of inductive reactance 80 \(\Omega\) and a resistor of resistance 60 \(\Omega\) are connected in series to an ac source. The impedance and the power factor of the circuit are respectively:

• A. 20 \(\Omega\), 0.4
• B. 60 \(\Omega\), 0.6
• C. 100 \(\Omega\), 0.4
• D. 100 \(\Omega\), 0.6

Hint:

To calculate the impedance of a series R-L circuit, use \(Z = \sqrt{R^2 + X_L^2}\). The power factor is calculated as \( {pf} = \cos(\theta) \), where \(\theta = \tan^{-1}\left(\frac{X_L}{R}\right)\).

Answer 1

The Correct Option is D

In a series R-L circuit, the impedance \(Z\) is given by: \[ Z = \sqrt{R^2 + X_L^2} \] Given: - \(R = 60 \, \Omega\) - \(X_L = 80 \, \Omega\) Substitute the values: \[ Z = \sqrt{(60)^2 + (80)^2} = \sqrt{3600 + 6400} = \sqrt{10000} = 100 \, \Omega \] The power factor is: \[ \tan(\theta) = \frac{X_L}{R} = \frac{80}{60} = \frac{4}{3} \] Thus: \[ \theta = \tan^{-1}\left(\frac{4}{3}\right) \approx 53.13^\circ \] The power factor is: \[ {pf} = \cos(53.13^\circ) \approx 0.6 \] Thus, the impedance is \(100 \, \Omega\) and the power factor is \(0.6\).