2. Solved Papers
  3. BHU PMT

BHU PMT BHU PMT Solved Paper-2004

  • question_answer
    In 0.1 s, the current in a coil increases from 1 A to 1.5 A. If inductance of this coil is 60 mH, then induced current in an external resistance of \[3\,\,\Omega \] will be:                                                                                  [BHU PMT-2004]

    A)  1 A                                        

    B)  0.5 A

    C)  0.2 A                                    

    D)  0.1 A

    Correct Answer: D

    Solution :

                     The coefficient of self-induction of a coil is numerically equal to the emf (e) induced in the coil when the rate of change of current \[\left( \frac{\Delta \,i}{\Delta \,t} \right)\] in the coil is unity. \[\therefore \]                  \[L=-\frac{e}{\Delta \,i/\Delta \,t}\] \[\Rightarrow \]                               \[e=-L\frac{\Delta \,i}{\Delta \,t}\] Given, \[L=60\,mH=60\times {{10}^{-3}}\,H,\]                                 \[\Delta \,i=\left( 1.5-1 \right)A=0.5\,A\]                                 \[\Delta \,t=0.1\,s\] \[\therefore \]  \[e=-\frac{60\times {{10}^{-3}}\times 0.5}{0.1}=-0.3\,\,V\] Induced current \[i=\frac{e}{R}=\frac{0.3}{3}=0.1\,\,A\]


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