CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2010

  • question_answer
    A 50 Hz AC current of peak value 2 A flows through one of the pair of coils. If the mutual inductance between the pair of coils is 150 mH, then the peak value of voltage induced in the second coil is

    A)  \[30\,\pi V\]                     

    B)         \[60\,\pi V\]

    C)  \[15\,\pi V\]                     

    D)         \[300\,\pi V\]

    E)  \[3\,\pi V\]

    Correct Answer: A

    Solution :

    \[M=150\times {{10}^{-3}}H\] \[i=2A\] \[f=50\,Hz\]       Induced voltage,                 \[e=\frac{d\phi }{dt}\]                 \[=\frac{d}{dt}(Mi)=\frac{d}{dt}(M{{i}_{0}}\sin \omega t)\omega \cos \omega t\] e is maximum when \[\cos \omega t=1\] \[\therefore \]  \[e=M{{i}_{0}}\omega \] \[=150\times {{10}^{-3}}\times i\times 2\pi f\] \[=150\times {{10}^{-3}}\times 2\times 2\pi \times 50\] \[=30\text{ }\pi V\]


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