The power factor of an ac circuit having resistance R and inductance L connected in series is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \displaystyle\frac{R}{(R^{2} + \omega^{2}L^{2})^{\frac{1}{2}}} \)
The power factor is \( \cos\phi \), where \( \phi \) is the phase difference between voltage and current. In a series circuit \( \cos\phi = R/Z \).
For a series LR circuit the voltages across \( R \) and \( L \) are \( 90^\circ \) apart, so the impedance adds as a right-angled triangle:
\( \displaystyle Z = \sqrt{R^{2} + X_L^{2}} = \sqrt{R^{2} + \omega^{2}L^{2}} \)
\( \displaystyle \cos\phi = \frac{R}{Z} = \frac{R}{(R^{2} + \omega^{2}L^{2})^{1/2}} \)
Note: A similar question was also asked in CUET (UG) 2023 on 27-05-2023 (Slot 3).