BCECE Medical BCECE Medical Solved Papers-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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