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JAMB Physics 2011 Paper

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Question : 33

Total: 44

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?

600 J
50 J
133 J
250 J

Solution:

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

Therefore Energy stored =

W = $\frac{1}{2} L I^2$

= $\frac{1}{2} \times 12 \times 10^2$

= 600J

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