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Manipal Medical Manipal Medical Solved Paper-2008

  • question_answer
    A 50 Hz alternating current of peak value 1A flows through the primary coil of a transformer. If the mutual inductance between the primary and secondary be 1.5 H, then the mean value of the induced voltage is

    A)  75 V           

    B)  150 V

    C)  225 V          

    D)  300 V

    Correct Answer: D

    Solution :

     Time period of AC \[T=\frac{1}{n}=\frac{1}{50}s\] Time interval\[\Delta t\]for current to decrease from 1 A to zero \[=\frac{T}{4}\] \[\therefore \] \[\Delta t=\frac{T}{4}=\frac{1}{50}\times \frac{1}{4}=\frac{1}{200}s\] Change in current, \[\Delta I=0-1=-1A\] Mean induced emf, \[e=-M\left( \frac{\Delta I}{\Delta t} \right)\] \[=1.5\times \frac{-1}{1/200}=300\,V\]


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