Q. In an electrical circuit $R,\ L,\ C$ and an a.c. voltage source are all connected in series. When $L$ is removed from the circuit, the phase difference between the voltage and the current in the circuit is $\pi/3$. If instead, $C$ is removed from the circuit, the phase difference is again $\pi/3$. The power factor of the circuit is
A) 1/2
B) $1/\sqrt{2}$
C) 1
D) $\sqrt{3}/2$
Explanation
(c) [image:7_image61.emf]
$\frac{X_{C}}{R} = \tan\frac{\pi}{3}$
[image:7_image62.emf]
$X_{C} = R\tan\frac{\pi}{3}\ \ \ \ \ \ \ \ \ \ldots(i)$
$\frac{X_{L}}{R} = \tan\frac{\pi}{3}$
[image:7_image63.emf]
$X_{L} = R\tan\frac{\pi}{3}\ \ \ \ \ \ \ \ \ldots(ii)$
Net impedance $Z = \sqrt{R^{2} + \left( X_{L} - X_{C} \right)^{2}} = R$
Power factor $\cos{\phi = \frac{R}{Z}} = 1$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.