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?
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Correct Answer: Option A
Explanation:
Inductance (L) in series: 4 + 8 = 12
Therefore Energy stored =Â Â
W = \(\frac{1}{2} LI^2\)
= \(\frac{1}{2} \times 12 \times 10^2\)
= 600J
Inductance (L) in series: 4 + 8 = 12
Therefore Energy stored =Â Â
W = \(\frac{1}{2} LI^2\)
= \(\frac{1}{2} \times 12 \times 10^2\)
= 600J