Two inductors of inductance 4 II and 8 II are arrange in ...
Two inductors of inductance 4 II and 8 II are arrange in series and a current of 10 A is passed through them. What is the energy stored in them?
600 J
50 J
133 J
250 J
Correct answer is A
Inductance (L) in series: 4 + 8 = 12
Therefore Energy stored =
W = 12LI2
= 12×12×102
= 600J