Q. A coil has resistance 30 ohm and inductive reactance 20 Ohm at 50 Hz frequency. If ac source, of 200 volt, 100 Hz, is connected across the coil, the current in the coil will be

A) $\frac{20}{\sqrt{13}}A$
B) 2.0$A$
C) 4.0$A$
D) 8.0$A$
Explanation

(c) If $\omega = 50 \times 2\pi$ then $\omega L = 20\Omega$
If $\omega^{'} = 100 \times 2\pi$ then $\omega^{'}L = 40\Omega$
$I = \frac{200}{Z} = \frac{200}{\sqrt{R^{2} + \left( \omega^{'}L \right)^{2}}} = \frac{200}{\sqrt{(30)^{2} + (40)^{2}}}$
$I = 4\ A$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.