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JEE Main & Advanced JEE Main Paper (Held On 10 April 2015)

  • question_answer
    When current in a coil changes from 5 A to 2 A in 0.1 s, an average voltage of 50 V is produced. The self-inductance of the coils is: JEE Main Online Paper (Held On 10 April 2015)

    A)  3H                                         

    B)  1.67 H

    C)  6 H                                        

    D)  0.67 H

    Correct Answer: B

    Solution :

    \[{{I}_{1}}=5A\] \[{{I}_{2}}=2A\] \[\Delta I=2-5=-3A\] \[\Delta t=0.1\sec \] \[Emf=50v\] \[emf=\varepsilon =-L\frac{di}{dt}\] \[50=-L\left( \frac{-3}{0.1} \right)\] \[50=30L\] \[L=\frac{5}{3}=1.67H\]


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