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WB JEE Medical WB JEE Medical Solved Paper-2006

  • question_answer
    A capacitor of capacitance \[\text{5 }\!\!\mu\!\!\text{ F}\]is connected as shown in the figure. The internal resistance of the cell is \[0.5\Omega .\] The amount of charge on the capacitor plates is :

    A) \[8\mu C\]                         

    B)  \[40\mu C\]

    C)  \[20\mu C\]                      

    D)  \[10\mu C\]

    Correct Answer: D

    Solution :

                     In steady state, there will be no current in the  capacitor branch. Net resistance of the circuit. \[R=1+1+0.5=2.5\,\Omega \] current drawn from the cell, \[i=\frac{V}{R}=\frac{2.5}{2.5}=1A\] Potential drop across two parallel branches \[V=E-ir=2.5-1\times 0.5\]                 \[=2.5-0.5\] \[=2.0\,\text{volt}\] So, charge on the capacitor plates \[q=CV\] \[=5\times 2\] \[=10\,\mu C\]


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