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

  • question_answer
    \[2\mu F\]capacitance has potential difference across its two terminals \[200\ volts\]. It is disconnected with battery and then another uncharged capacitance is connected in parallel to it, then P.D. becomes\[20\ volts\]. Then the capacity of another capacitance will be [CPMT 1991; DPMT 2001]

    A)                    \[2\mu F\]

    B)                                      \[4\mu F\]

    C)                    \[18\mu F\]                  

    D)            \[10\mu F\]

    Correct Answer: C

    Solution :

                       By using, common potential \[V=\frac{{{C}_{1}}{{V}_{1}}+{{C}_{2}}{{V}_{2}}}{{{C}_{1}}+{{C}_{2}}}\] Þ \[20=\frac{2\times 200+{{C}_{2}}\times 0}{2+{{C}_{2}}}\] Þ \[{{C}_{2}}=18\mu F\]


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