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Manipal Medical Manipal Medical Solved Paper-2001

  • question_answer
    If a current of 1 A flows in 1 milli-second through a coil and the induced emf is 2 V, then the self inductance of the coil will be:

    A)  \[2\times {{10}^{-3}}H\]

    B)  \[1\times {{10}^{3}}H\]

    C)  \[4\times {{10}^{3}}H\]

    D)  \[1\times {{10}^{-3}}H\]

    Correct Answer: A

    Solution :

     Here: Current \[i=1\text{ }A\] Time t = 1 milli-sec \[=1\times {{10}^{-3}}sec\] The self inductance of the coil is given by \[L=\frac{2t}{i}=\frac{2\times {{10}^{-3}}}{1}\] \[=2\times {{10}^{-3}}H\]


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