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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2003

  • question_answer
    The magnetic flux linked with a coil at any   instant t is given by \[\phi =5{{t}^{3}}-100t+300,\] the emf induced in the coil at \[t=2\] second is:

    A)  40 V                                     

    B)  - 40 V

    C)  300 V                                   

    D)  140 V

    Correct Answer: A

    Solution :

    The induced e.m.f                 \[e=-\frac{d\phi }{dt}\] So,          \[e=-\frac{d}{dt}\,(5{{t}^{3}}-100\,t+300)\] or            \[e=-(15\,\,{{t}^{2}}-100)\] Induced e.m.f at \[t=2\,\,\sec \] \[e=-(15\times 4-100)=+40\] volt


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