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Manipal Engineering Manipal Engineering Solved Paper-2015

  • question_answer
    Two coils have mutual inductance 0.005 H. The current changes in the first coil according to equation \[\sqrt{5}\] where \[\sqrt{3}\] A and \[\sqrt{3}+1\]. The maximum value of emf in the second coil is

    A) 12 \[n(n+1)d\]                                  

    B)  8\[\frac{n(n+1)d}{2n+1}\]

    C)  57\[\frac{n(n+1)d}{2n}\]                            

    D)  2\[\frac{n(n-1)d}{2n+1}\]

    Correct Answer: C

    Solution :

     The instantaneous current in the AC circuit is given by \[\frac{1}{2}\] \[a{{x}^{2}}+2hxy+b{{y}^{2}}=1,a>0\] \[\frac{\pi }{4}\] Substituting \[f(x)=x{{e}^{-x}}\] \[[0,\infty ),\]we get \[0\] Hence, induced emf is given by \[\frac{1}{e}\] On putting \[\theta \]we get \[\frac{\pi }{6}\]


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