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

  • question_answer
    Two capacitances of capacity \[{{C}_{1}}\]and \[{{C}_{2}}\] are connected in series and potential difference \[V\] is applied across it. Then the potential difference across \[{{C}_{1}}\]will be             [MP PMT 1985]

    A)                    \[V\frac{{{C}_{2}}}{{{C}_{1}}}\]

    B)                                      \[V\frac{{{C}_{1}}+{{C}_{2}}}{{{C}_{1}}}\]

    C)                    \[V\frac{{{C}_{2}}}{{{C}_{1}}+{{C}_{2}}}\]            

    D)            \[V\frac{{{C}_{1}}}{{{C}_{1}}+{{C}_{2}}}\]

    Correct Answer: C

    Solution :

               Charge flowing \[=\frac{{{C}_{1}}{{C}_{2}}}{{{C}_{1}}+{{C}_{2}}}V\]. So potential difference across \[{{C}_{1}}=\frac{{{C}_{1}}{{C}_{2}}V}{{{C}_{1}}+{{C}_{2}}}\times \frac{1}{{{C}_{1}}}\]\[=\frac{{{C}_{2}}V}{{{C}_{1}}+{{C}_{2}}}\]


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