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

  • question_answer
    The capacities of two conductors are \[{{C}_{1}}\]and \[{{C}_{2}}\] and their respective potentials are \[{{V}_{1}}\]and\[{{V}_{2}}\]. If they are connected by a thin wire, then the loss of energy will be given by [MP PMT 1986]

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

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

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

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

    Correct Answer: C

    Solution :

               Initial energy \[{{U}_{i}}=\frac{1}{2}{{C}_{1}}V_{1}^{2}+\frac{1}{2}{{C}_{2}}V_{2}^{2}\], Final energy \[{{U}_{f}}=\frac{1}{2}({{C}_{1}}+{{C}_{2}}){{V}^{2}}\]  (where \[V=\frac{{{C}_{1}}{{V}_{1}}+{{C}_{2}}{{V}_{2}}}{{{C}_{1}}{{C}_{2}}})\] Hence energy loss \[\Delta U={{U}_{i}}-{{U}_{f}}=\frac{{{C}_{1}}{{C}_{2}}}{2({{C}_{1}}+{{C}_{2}})}{{({{V}_{1}}-{{V}_{2}})}^{2}}\]


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