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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 20

  • question_answer
    Suppose C be the capacitance of a capacitor discharging through a resistor R. Suppose  is the time taken for the energy stored in the capacitor to reduce to half its initial value and \[{{t}_{2}}\]is taken for the charge to reduce to one- fourth its initial value. Then the ratio\[\frac{{{t}_{1}}}{{{t}_{2}}}\]will be

    A) 2                     

    B) 1

    C) \[\frac{1}{2}\]              

    D) \[\frac{1}{4}\]

    Correct Answer: D

    Solution :

    [d]: As\[U={{U}_{0}}{{e}^{-\frac{2t}{RC}}}\] \[\therefore \]\[\frac{{{U}_{0}}}{2}={{U}_{0}}{{e}^{-\frac{-2{{t}_{1}}}{RC}}}\]\[\Rightarrow \]\[\frac{2{{t}_{1}}}{RC}={{\log }_{e}}2\]                    ?(i) Also,\[q={{q}_{0}}{{e}^{-\frac{t}{RC}}}\] \[\therefore \]\[\frac{{{q}_{0}}}{4}={{q}_{0}}{{e}^{-\frac{{{t}_{2}}}{RC}}}\] \[\Rightarrow \]\[\frac{{{t}_{2}}}{RC}={{\log }_{e}}4=2{{\log }_{e}}2\]             ?(ii) From equation (i) and (ii) \[\frac{2{{t}_{1}}}{{{t}_{2}}}=\frac{{{\log }_{e}}2}{2{{\log }_{e}}2}\]or\[\frac{{{t}_{1}}}{{{t}_{2}}}=\frac{1}{4}\]


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