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Question

Energy stored in a coil of self – inductance 40mH carrying a steady current of 2 A is

A
0.8 J
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B
8 J
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C
0.08 J
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D
80 J
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Solution

The correct option is C 0.08 J
U=12Li2
=1240×103×(2)2
=80×103
U=0.08 J

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