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

  • question_answer
    Two capacitors of 3pF and 6pF are connected in series and a potential difference of 5000\[V\] is applied across the combination. They are then disconnected and reconnected in parallel. The potential between the plates is [MP PMT 1992]

    A)                    2250\[V\]

    B)                                      2222\[V\]

    C)                    \[2.25\times {{10}^{6}}V\]                                         

    D)            \[1.1\times {{10}^{6}}V\]

    Correct Answer: B

    Solution :

               \[\frac{1}{C}=\frac{1}{3}+\frac{1}{6}\] Þ \[C=2\,pF\]                    Total charge\[=2\times {{10}^{-12}}\times 5000={{10}^{-8}}\,C\]                    The new potential when the capacitors are connected in parallel is                    \[V=\frac{2\times {{10}^{-8}}}{(3+6)\times {{10}^{-12}}}=2222\,V\]


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