2. Solved Papers
  3. BCECE Engineering

BCECE Engineering BCECE Engineering Solved Paper-2001

  • question_answer
    When a current in coil changes from 4 A to 2 A in 0.05 s, emf of 8 V is induced in the coil. The coefficient of self-inductance in the coil is:

    A)  0.8 H                                    

    B)  0.4 H                    

    C)         0.2 H                    

    D)         0.1 H

    Correct Answer: C

    Solution :

    Key Idea: Emf induced in the coil is equal to negative rate of change of current multiplied by self-inductance. emf induced in the coil is given by \[e=-L\frac{di}{dt}\]                                       ?(i) Given,\[\frac{di}{dt}=\frac{2-4}{0.05}=-40A/s,\,e=8V\] Substituting the values in Eq. (i), we get \[8=-L\times (-40)\]                 or            \[L=\frac{8}{40}=0.2\,H\]


You need to login to perform this action.
You will be redirected in 3 sec spinner