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Q. A $ 50 \,mH $ coil carries a current of $ 2\,A $ , the energy stored in joule is:

KEAMKEAM 2004

Solution:

$ U=\frac{1}{2}L{{i}^{2}}=\frac{1}{2}\times 50\times {{10}^{-3}}\times 2\times 2=0.1\,J $