Q. A coil of $200\Omega$ resistance and $0.1\ H$ inductance is connected to an ac source of frequency $200/2\pi\ Hz$. Phase angle between potential and current will be
A) $30{^\circ}$
B) $90^{o}$
C) $45{^\circ}$
D) $0{^\circ}$
Explanation
(c) $\tan\phi = \frac{X_{L}}{R} = \frac{2\pi vL}{R} = \frac{2\pi \times \frac{200}{2\pi} \times 1}{200} = 1 \Rightarrow \phi = 45{^\circ}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.