Q. Energy required to establish a current of 4 A in a coil of self-inductance L =200mH is
A) 0.16 J
B) 0.18 J
C) 0.40 J
D) 1.6 J
Explanation
(d) Energy stored in a self-inductor, $E = \frac{1}{2}Li^{2}$
$= \frac{1}{2} \times 200 \times 10^{- 3}\lbrack 4\rbrack^{2} = 1.6\ J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.