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

  • question_answer
    A condenser of capacitance \[10\mu F\] has been charged to 100\[volts\]. It is now connected to another uncharged condenser in parallel. The common potential becomes 40\[volts\]. The capacitance of another condenser is                                                                                            [MP PET 1992]

    A)                    \[15\mu F\]

    B)                                      \[5\mu F\]

    C)                    \[10\mu F\]

    D)                                      \[16.6\mu F\]

    Correct Answer: A

    Solution :

                       By using \[V=\frac{{{C}_{1}}{{V}_{1}}+{{C}_{2}}{{V}_{2}}}{{{C}_{1}}+{{C}_{2}}}\]                    Þ \[40=\frac{10\times 100+{{C}_{2}}\times 0}{10+{{C}_{2}}}\] Þ \[{{C}_{2}}=15\mu F\]


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