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?
-
A.
600 J -
B.
50 J -
C.
133 J -
D.
250 J
Correct Answer: Option A
Explanation
Inductance (L) in series: 4 + 8 = 12
Therefore Energy stored =
W = (frac{1}{2} LI^2)
= (frac{1}{2} times 12 times 10^2)
= 600J