2. Solved Papers
  3. JEE Main & Advanced
  4. Physics
  5. Electrostatics & Capacitance

JEE Main & Advanced Physics Electrostatics & Capacitance JEE PYQ-Electrostatics and Capacitance

  • 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 2010]

    A) 2

    B) 1

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

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

    Correct Answer: D

    Solution :

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


You need to login to perform this action.
You will be redirected in 3 sec spinner