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Two inductors of inductance 4 II and 8 II are arrange in series and a...

Physics
JAMB 2011

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
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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

There is an explanation video available below.


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Explanation Video

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