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?

By: |Published on: Nov 9|Categories: Physics JAMB| 0 comments

A. 600 J
B. 50 J
C. 133 J
D. 250 J

Correct Answer:

Option A – 600 J

Explanation

Inductance (L) in series: 4 + 8 = 12

Therefore Energy stored =

W = ½LI²
= ½ ×12×10²
= 600J

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