2. Questions Bank
  3. JEE Main & Advanced
  4. Physics
  5. Electrostatics & Capacitance

JEE Main & Advanced Physics Electrostatics & Capacitance Question Bank Grouping of Capacitors

  • question_answer
    \[n\] identical condensers are joined in parallel and are charged to potential\[V\]. Now they are separated and joined in series. Then the total energy and potential difference of the combination will be                                         [MP PET 1993]

    A)                    Energy and potential difference remain same

    B)                    Energy remains same and potential difference is \[nV\]

    C)                    Energy increases \[n\] times and potential difference is \[nV\]

    D)                    Energy increases \[n\] times and potential difference remains same

    Correct Answer: B

    Solution :

               According to energy conservation, energy remains the same Þ \[{{U}_{parallel}}={{U}_{series}}\]Þ\[\frac{1}{2}(nC){{V}^{2}}=\frac{1}{2}\left( \frac{C}{n} \right)\,V{{'}^{2}}\]Þ V' = nV  (\[V'\]= potential difference across series combination)


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