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JEE Main & Advanced Physics Electrostatics & Capacitance Question Bank Critical Thinking

  • question_answer
    The plates of a capacitor are charged to a potential difference of 320 volts and are then connected across a resistor. The potential difference across the capacitor decays exponentially with time. After 1 second  the potential difference between the plates of the capacitor is 240 volts, then after 2 and 3 seconds the potential difference between the plates will be                                                               [MP PET 1996]

    A)            200 and 180 V

    B)                                      180 and 135 V

    C)            160 and 80 V                         

    D)            140 and 20 V

    Correct Answer: B

    Solution :

               \[V={{V}_{0}}({{e}^{-\lambda t}})\] After 1 seconds \[{{V}_{1}}=320({{e}^{-\lambda }})\Rightarrow 240=320({{e}^{-\lambda }})\,\Rightarrow \,{{e}^{-\lambda }}=\frac{3}{4}\] After 2 seconds                               \[{{V}_{2}}=320\,{{({{e}^{-\lambda }})}^{2}}=320\times {{\left( \frac{3}{4} \right)}^{2}}=180\,\,volt\] After 3 seconds                               \[{{V}_{3}}=320\,{{({{e}^{-\lambda }})}^{3}}=320\times {{\left( \frac{3}{4} \right)}^{3}}=135\,\,volt\]


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