Q. There is a $5\Omega$ resistance in an ac, circuit. Inductance of $0.1H$ is connected with it in series. If equation of ac $e.$m.f. is $5\sin{50t}$, then the phase difference between current and $e$.m.f. is
A) $\frac{\pi}{2}$
B) $\frac{\pi}{6}$
C) $\frac{\pi}{4}$
D) 0
Explanation
(c) $\cos{\phi = \frac{R}{Z}} = \frac{R}{\sqrt{R^{2} + \omega^{2}L^{2}}} = \frac{5}{\sqrt{25 + (50)^{2} \times (0.1)^{2}}}$
$= \frac{5}{\sqrt{25 + 25}} = \frac{1}{\sqrt{2}} \Rightarrow \phi = \pi/4$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.