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

  • question_answer
    Two capacitors of capacitances \[3\mu \,F\] and \[6\mu F\] are charged to a potential of 12 V each. They are now connected to each other, with the positive plate of each joined to the negative plate of the other. The potential difference across each will be                                                                    [KCET 2002]

    A)            6 volt                                       

    B)            4 volt

    C)            3 volt                                       

    D)            Zero

    Correct Answer: B

    Solution :

               \[V=\frac{{{C}_{1}}{{V}_{1}}-{{C}_{2}}{{V}_{2}}}{{{C}_{1}}+{{C}_{2}}}=\frac{6\times 12-3\times 12}{3+6}=4\,volt\]


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