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Manipal Engineering Manipal Engineering Solved Paper-2012

  • question_answer
    The current in a coil changes from\[0\]to\[2A\]in\[0.05\,\,s\]. If the induced emf is\[80\,\,V\], the self-inductance of the coil is,

    A) \[1\,\,H\]                                            

    B) \[0.5\,\,H\]

    C) \[1.5\,\,H\]        

    D)        \[2\,\,H\]

    Correct Answer: D

    Solution :

    From the formula,\[\varepsilon =-L\frac{dl}{dt}\] Take magnitude on both sides.                 \[|\varepsilon |\,\,=\left| -L\frac{dI}{dt} \right|\]     or     \[\varepsilon =L\frac{dI}{dt}\] \[\therefore \]  \[80=L\left( \frac{2}{0.05} \right)\]    \[\Rightarrow \]     \[80=L\times 40\] \[\Rightarrow \]               \[L=\frac{80}{40}\]    \[\Rightarrow \]    \[L=2H\]


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