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JEE Main & Advanced AIEEE Solved Paper-2010

  • question_answer
      Let C be the capacitance of a capacitor discharging through a resistor R. Suppose\[{{t}_{1}}\]is the time taken for 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 \[{{t}_{1}}/{{t}_{2}}\]will be ?       AIEEE  Solved  Paper-2010

    A) 2             

    B)                        1                             

    C)        \[\frac{1}{2}\]                   

    D)        \[\frac{1}{4}\]

    Correct Answer: D

    Solution :

     \[U=\frac{{{Q}^{2}}}{2C}=\frac{Q_{0}^{2}{{e}^{-\frac{2t}{RC}}}}{2C}\]   \[Q={{Q}_{0}}{{e}^{-t/RC}}\] \[U=\frac{{{U}_{0}}}{2}\] \[\frac{Q_{0}^{2}}{2\times 2C}=\frac{Q_{0}^{2}{{e}^{-\frac{2{{t}_{1}}}{RC}}}}{2C}\]                \[\frac{{{Q}_{0}}}{4}={{Q}_{0}}{{e}^{-\frac{{{t}_{2}}}{RC}}}\] \[\frac{1}{2}={{e}^{-\frac{2{{t}_{1}}}{RC}}}\]                       \[{{\log }_{e}}4=\frac{{{t}_{2}}}{RC}\] \[{{t}_{1}}=\frac{RC{{\log }_{e}}2}{2}\]                   \[{{t}_{2}}=RC\,{{\log }_{e}}4\] \[\frac{{{t}_{1}}}{{{t}_{2}}}=\frac{1}{4}\]


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