An alternation current i=10cos100πt A is flowing through a 100 mH inductor. Then average magnetic energy stored in it is
2.5 Joule
2 Joule
4 Joule
5 Joule
Irms=102A∴ U=12L Irms2=12×100×10−3×(102)2 JouleU=2.5 Joule