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JEE Main & Advanced Physics Electrostatics & Capacitance Question Bank Mock Test - Capacitance

  • 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

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

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

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

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

    Correct Answer: C

    Solution :

    [c] Initial energy of the system \[{{U}_{i}}=\frac{1}{2}C{{V}^{2}}_{1}+\frac{1}{2}C{{V}^{2}}_{2}\] 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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