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JEE Main & Advanced JEE Main Paper (Held On 10-Jan-2019 Evening)

  • question_answer
    The self-induced emf of a coil is 25 volts. When the current in it is changed at uniform rate from 10 A to 25 A in 1s, the change in the energy of the inductance is- [JEE Main Online Paper (Held On 10-Jan-2019 Evening]

    A) 740 J                           

    B) 637.5 J

    C) 540 J   

    D)                  437.5 J

    Correct Answer: D

    Solution :

    Induced emf \[=\,\,L\frac{di}{dt}\,=\,\frac{L\times (25-10)}{1}\] \[=L\times 15\] \[\,L\times 15=25\] \[L=\frac{25}{15}\] Change in energy of inductor \[\,=\,\,\frac{1}{2}L\left[ {{25}^{2}}-{{10}^{2}} \right]\] \[=\,\,\frac{1}{2}\times \frac{25}{15}\times 15\times 35\] \[=\,\,\frac{25\times 35}{2}\] = 437.5 Joule


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