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JEE Main & Advanced Sample Paper JEE Main Sample Paper-25

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

    A)  1                    

    B)  \[\frac{1}{2}\]

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

    D)  2

    Correct Answer: C

    Solution :

    \[U=\frac{1}{2}\,\frac{{{q}^{2}}}{C}\,=\frac{1}{2}\,{{({{q}_{0}}{{e}^{-t/\tau }})}^{2}}\] \[=\frac{q_{0}^{2}}{2C}{{e}^{-2t/\tau }}\,(\tau =CR)\] \[U={{U}_{i}}{{e}^{-2t/\tau }}\] \[\Rightarrow \,\frac{1}{2}{{U}_{i}}={{U}_{i}}\,{{e}^{-2{{t}_{1}}/\tau }}\] \[\Rightarrow \,\frac{1}{2}\,={{e}^{-2{{t}_{1}}/\tau }}\] \[\Rightarrow \,\,{{t}_{i}}=\frac{\tau }{2}\,\ln 2\] Now, \[q={{q}_{0}}\,{{e}^{-t/\tau }}\] \[\Rightarrow \,\,\frac{1}{4}\,{{q}_{0}}={{q}_{0}}{{e}^{-{{t}_{2}}/\tau }}\] \[{{t}_{2}}=\tau \,\ln \,4=2\tau \ln \,2\] \[\therefore \,\,\frac{{{t}_{1}}}{{{t}_{2}}}=\frac{1}{4}\]


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