Q. An alternating voltage $e = 200\sin{100\ t}$ is applied to a series combination $R = 30\ \Omega$ and an inductor of 400 mH. The power factor of the circuit is
A) 0.01
B) 0.2
C) 0.05
D) 0.6
Explanation
(d) Power factor
$\cos{\phi = \frac{R}{\sqrt{R^{2} + \omega^{2}L^{2}}}}$
$= \frac{30}{\sqrt{(30)^{2} + (100)^{2} \times \left( 400 \times 10^{- 3} \right)^{2}}}$
$= \frac{30}{\sqrt{900 + 1600}} = \frac{30}{50} = 0.6$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.