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

  • question_answer
    A condenser of capacity \[{{C}_{1}}\]is charged to a potential \[{{V}_{0}}\]. The electrostatic energy stored in it is \[{{U}_{0}}\]. It is connected to another uncharged condenser of capacity \[{{C}_{2}}\] in parallel. The energy dissipated in the process is        [MP PMT 1994]

    A)   \[\frac{{{C}_{2}}}{{{C}_{1}}+{{C}_{2}}}{{U}_{0}}\]

    B)                             \[\frac{{{C}_{1}}}{{{C}_{1}}+{{C}_{2}}}{{U}_{0}}\]

    C)              \[\left( \frac{{{C}_{1}}-{{C}_{2}}}{{{C}_{1}}+{{C}_{2}}} \right){{U}_{0}}\]   

    D)   \[\frac{{{C}_{1}}{{C}_{2}}}{2({{C}_{1}}+{{C}_{2}})}{{U}_{0}}\]

    Correct Answer: A

    Solution :

                       Loss of energy during sharing =\[\frac{{{C}_{1}}{{C}_{2}}{{({{V}_{1}}-{{V}_{2}})}^{2}}}{2({{C}_{1}}+{{C}_{2}})}\]                    In the equation, put \[{{V}_{2}}=0,\ {{V}_{1}}={{V}_{0}}\]                    \ Loss of energy \[=\frac{{{C}_{1}}{{C}_{2}}V_{0}^{2}}{2({{C}_{1}}+{{C}_{2}})}\] \[=\frac{{{C}_{2}}{{U}_{0}}}{{{C}_{1}}+{{C}_{2}}}\]  \[\left[ \because \ {{U}_{0}}=\frac{1}{2}{{C}_{1}}V_{0}^{2} \right]\]


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