CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2001

  • question_answer
    Energy stored in a coil of self-inductance \[40\text{ }mH\]carrying a steady current of 2 A, is:

    A)  8 J                                         

    B)  0.8 J                     

    C)         0.08 J                   

    D)         80 J

    E)  4J

    Correct Answer: C

    Solution :

    \[{{U}_{L}}=\frac{1}{2}L{{I}^{2}}=\frac{1}{2}\times 40\times {{10}^{-3}}\times {{(2)}^{2}}=0.08\,J\]


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