BCECE Engineering BCECE Engineering Solved Paper-2001

  • question_answer
    Energy stored in a coil of self-inductance 40 mH carrying a steady current of 2 A is:

    A)  80 J                                       

    B)  0.08 J                   

    C)  0.8 J                     

    D)         8 J    

    Correct Answer: B

    Solution :

    An inductor has the capability of storing energy in its magnetic field. Hence, stored energy \[U=\frac{1}{2}L{{i}^{2}}\]                 Here, \[i=2A,L=40\,mH=40\times {{10}^{-3}}H\]                 \[\therefore \]  \[U=\frac{1}{2}\times 40\times {{10}^{-3}}\times {{(2)}^{2}}\]                                 \[=0.08\,J\] Note: Energy stored in inductor is similar to energy of a capacitor stored in the electric field between its plates.


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