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JCECE Engineering JCECE Engineering Solved Paper-2005

  • question_answer
    The coefficient of mutual inductance between the primary and secondary of the coil is \[5\,\,H\]. A current of \[10\,\,A\] is cut-off in \[0.5\,\,s\]. The induced emf is:

    A) \[1\,\,V\]                                            

    B) \[10\,\,V\]

    C) \[5\,\,V\]                                            

    D) \[100\,\,V\]

    Correct Answer: D

    Solution :

    From Faraday's law of electromagnetic induction, the induced \[emf(e)\] is given by                 \[e=-M\frac{di}{dt}\] Given,\[M=5\,\,H,\,\,di=10\,\,A,\,\,dt=0.5\,\,s\] \[\therefore \]  \[e=-5\times \frac{10}{0.5}=-100\,\,V\] Note: Minus sign in the relation is reflection of Lenz's law.


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