CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2004

  • question_answer
    A 50 mH coil carries a current of 2 A, the energy stored in joule is:

    A)  1                                            

    B)  0.05

    C)  10                         

    D)         0.5

    E)  0.1

    Correct Answer: E

    Solution :

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


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