2. Solved Papers
  3. BCECE Medical

BCECE Medical BCECE Medical Solved Papers-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)  0.1H

    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) Gwen, \[\frac{di}{dt}=\frac{2-4}{0.05}=-40\,\,A/s\], \[e=8\,\,V\] 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