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

  • question_answer
    Two condensers \[{{C}_{1}}\]and \[{{C}_{2}}\] in a circuit are joined as shown in figure. The potential of point \[A\] is \[{{V}_{1}}\] and that of \[B\] is \[{{V}_{2}}\]. The potential of point \[D\] will be              [MP PMT 1997]

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

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

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

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

    Correct Answer: C

    Solution :

               Charge on \[{{C}_{1}}\] = charge on \[{{C}_{2}}\]                    Þ \[{{C}_{1}}({{V}_{A}}-{{V}_{D}})={{C}_{2}}({{V}_{D}}-{{V}_{B}})\]                    Þ \[{{C}_{1}}({{V}_{1}}-{{V}_{D}})={{C}_{2}}({{V}_{D}}-{{V}_{2}})\]Þ \[{{V}_{D}}=\frac{{{C}_{1}}{{V}_{1}}+{{C}_{2}}{{V}_{2}}}{{{C}_{1}}+{{C}_{2}}}\]


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