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
Verified Correct Answer: A.
Physical Principles & Numerical Solution
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
Source: Myschool Past Questions Archive (Record #14465)