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

  • question_answer
    The capacitor of capacitance \[4\mu F\] and \[6\mu F\] are connected in series. A potential difference of \[500\ volts\] applied to the outer plates of the two capacitor system. Then the charge on each capacitor is numerically

    A)                    \[6000\,C\]

    B)                                      \[1200\ C\]

    C)                    \[1200\ \mu C\]          

    D)            \[6000\ \mu C\]

    Correct Answer: C

    Solution :

               \[{{C}_{eq}}=\frac{{{C}_{1}}{{C}_{2}}}{{{C}_{1}}+{{C}_{2}}}=2.4\mu F.\] Charge flown = 2.4 ´ 500 ´ 10?6 C =1200 mC.


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