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

  • question_answer
    Two identical capacitors, have the same capacitance C. One of them is charged to potential \[{{V}_{1}}\] and the other to \[{{V}_{2}}\]. The negative ends of the capacitors are connected together. When the positive ends are also connected, the decrease in energy of the combined system is                   [IIT-JEE (Screening) 2002]

    A)            \[\frac{1}{4}C(V_{1}^{2}-V_{2}^{2})\]

    B)                                      \[\frac{1}{4}C(V_{1}^{2}+V_{2}^{2})\]

    C)            \[\frac{1}{4}C{{\left( {{V}_{1}}-{{V}_{2}} \right)}^{2}}\]      

    D)            \[\frac{1}{4}C{{\left( {{V}_{1}}+{{V}_{2}} \right)}^{2}}\]

    Correct Answer: C

    Solution :

                       Initial energy of the system                         \[{{U}_{i}}=\frac{1}{2}C{{V}_{1}}^{2}+\frac{1}{2}C{{V}_{2}}^{2}\]                    When the capacitors are joined, common potential \[V=\frac{C{{V}_{1}}+C{{V}_{2}}}{2C}=\frac{{{V}_{1}}+{{V}_{2}}}{2}\]                    Final energy of the system                    \[{{U}_{f}}=\frac{1}{2}(2C){{V}^{2}}=\frac{1}{2}2C\,{{\left( \frac{{{V}_{1}}+{{V}_{2}}}{2} \right)}^{2}}=\frac{1}{4}C{{({{V}_{1}}+{{V}_{2}})}^{2}}\]                    Decrease in energy = \[{{U}_{i}}-{{U}_{f}}=\frac{1}{4}C{{({{V}_{1}}-{{V}_{2}})}^{2}}\]


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