The magnetic potential energy stored in a certain inductor is 25 mJ, when the current in the inductor is 60 mA. This inductor is of inductance
- 0.138 H
- 138.88 H
- 1.389 H
- 13.89 H
Answer
(4) 13.89 H
Solution
U = 12LI2
L = 2UI2
= 2 25 10-3(60 10-3)2
= 50 10-33.6 10-3
= 13.89 H
Answer: (4)