An alternation current i=10cos⁡100πt A is flowing through a 100 mH inductor. Then average magnetic energy stored in it is

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2.5 Joule

2 Joule

4 Joule

5 Joule

answer is A.

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Detailed Solution

Irms=102A∴ U=12L Irms2=12×100×10−3×(102)2 JouleU=2.5 Joule

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