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

  • question_answer
    Three capacitors of \[2\mu F,\,3\mu F\] and \[6\mu F\] are joined in series and the combination is charged by means of a 24 volt battery. The potential difference between the plates of the \[6\mu F\] capacitor is                                  [MP PMT 2002]

    A)            4 volt

    B)            6 volt

    C)            8 volt                                       

    D)            10 volt

    Correct Answer: A

    Solution :

                       \[\frac{1}{{{C}_{eq}}}=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\Rightarrow {{C}_{eq}}=1\,\mu \,F\]                     Total charge Q = Ceq.V = 1 × 24 = 24 mC                    So p.d. across 6 mF capacitor = \[\frac{24}{6}=4\,volt\]                


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