Q. A pure inductive coil of 30 mH is connected to an AC source of 220 V, 50 Hz. The rms current in the coil is
A) 50.35 A
B) 23.4 A
C) 30.5 A
D) 12.3 A
Explanation
(b) Given, $L = 30\ $mH
$V_{rms} = 220\ V$
$f = 50Hz$
Now, $X_{L} = \omega L = 2\pi fL$
$= 2 \times 3.14 \times 50 \times 30 \times 10^{- 3}$
$= 9.42\ \Omega$
The rms current in the coil is
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ i_{rms} = \frac{V_{rms}}{X_{L}} = \frac{220V}{9.42\ \Omega}$=23.4 A
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.