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JEE Main & Advanced JEE Main Paper (Held on 08-4-2019 Morning)

  • question_answer
    A 20 Henry inductor coil is connected to a 10 ohm resistance in series as shown in figure. The time at which rate of dissipation of energy (joule's heat) across resistance is equal to the rate at which magnetic energy is stored in the inductor is:                                [JEE Main 8-4-2019 Morning]

    A) \[\frac{2}{\ell n2}\]                            

    B) \[\ell n2\]

    C) \[2\ell n2\]

    D) \[\frac{1}{2}\ell n2\]

    Correct Answer: C

    Solution :

    \[LIdI={{I}^{2}}R\]           \[L\times \frac{E}{10}\left( -{{e}^{-t/2}} \right)\times \frac{-1}{2}=\frac{E}{10}\left( 1-{{e}^{-t/2}} \right)\times 10\]           \[{{e}^{-t/2}}=1-{{e}^{-t/2}}\]           \[t=2\ell n2\]            


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