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JAMB Physics 2011 Paper
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© africaexams.com
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
Validate
Solution:
Inductance (L) in series: 4 + 8 = 12
Therefore Energy stored =
W =
1
2
L
I
2
=
1
2
×
12
×
10
2
= 600J
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