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JEE Main & Advanced Sample Paper JEE Main Sample Paper-46

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

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

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

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

    Correct Answer: C

    Solution :

     Initial energy of the system, \[{{U}_{i}}=\frac{1}{2}CV_{1}^{2}+\frac{1}{2}CV_{2}^{2}\] 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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