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BCECE Engineering BCECE Engineering Solved Paper-2003

  • question_answer
    The current in self-inductance L = 40 mH is increased uniformly from 1 A to 11 A in 4 milliseconds. The induced emf produced in L during this process will be:

    A)  100 V

    B)                                         0.2V

    C)         440 V                                   

    D)         40 V

    Correct Answer: A

    Solution :

                    emf induced in coil is given by \[e=-L\frac{\Delta i}{\Delta t}\] ?(i) Given, \[L=40\,MH=40\times {{10}^{-3}}H,\]                 \[\Delta i=11-1=10A\]                 \[\Delta t=4ms=4\times {{10}^{-3}}s\] Putting the values in Eq. (i), we obtain                 \[|e|=40\times {{10}^{-3}}\times \frac{10}{4\times {{10}^{-3}}}=100\,V\] Note: The minus sign in the Eq. (i) is the reflection of Lenzs law.


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