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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 10

  • question_answer
    In the circuit shown, A is joined to B for a long time, and then A is joined to C. The total heat produced in resistance R is

    A) \[\frac{L{{E}^{2}}}{6{{R}^{2}}}\]   

    B) \[\frac{L{{E}^{2}}}{4{{R}^{2}}}\]

    C) \[\frac{4{{R}^{2}}}{L{{E}^{2}}}\]   

    D) \[\frac{L{{E}^{2}}}{{{R}^{2}}}\]

    Correct Answer: B

    Solution :

    [b] -Max. energy stored in \[2L\] inductance, \[U=\frac{1}{2}.2L{{\left( \frac{E}{2R} \right)}^{2}}=\frac{L{{E}^{2}}}{4{{R}^{2}}}\] This stored magnetic energy is completely converted into heat.


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