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

  • question_answer
    In the circuit shown here \[{{C}_{1}}=6\mu F,\ {{C}_{2}}=3\mu F\]and battery\[B=20V\]. The switch \[{{S}_{1}}\] is first closed. It is then opened and afterwards \[{{S}_{2}}\] is closed. What is the charge finally on \[{{C}_{2}}\]

    A)                    \[120\mu C\]

    B)                    \[80\mu C\]

    C)                    \[40\mu C\]

    D)                    \[20\mu C\]

    Correct Answer: C

    Solution :

               Common potential \[V=\frac{6\times 20+3\times 0}{(6+3)}=\frac{120}{9}Volt\]                    So, charge on \[3\,\mu F\] capacitor                    \[{{Q}_{2}}=3\times {{10}^{-6}}\times \frac{120}{9}=40\,\mu C\]


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