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Punjab Medical Punjab - MET Solved Paper-2008

  • question_answer
    The emf induced in a secondary coil is \[20000\,\,V\] when the current breaks in the primary coil. The mutual inductance is \[5H\] and the current reaches to zero in \[{{10}^{-4}}s\] in the primary. The maximum current in the primary before it breaks is

    A) \[0.1A\]                                               

    B) \[0.4A\]

    C) \[0.6A\]                               

    D) \[0.8A\]

    Correct Answer: B

    Solution :

    \[e=\frac{M{{i}_{\max }}}{t}\] or            \[20000=5\times \frac{{{i}_{\max }}}{{{10}^{-4}}}\] or            \[{{i}_{\max }}=\frac{20000\times {{10}^{-4}}}{5}=0.4\,\,A\]


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