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JCECE Engineering JCECE Engineering Solved Paper-2013

  • question_answer
    In the adjoining figure,\[E=5\,\,V,\,\,r=1\Omega \]\[,\]\[{{R}_{2}}=4\Omega ,\,\,{{R}_{1}}={{R}_{3}}=1\Omega \] and \[C=3\mu F\]. The numerical value of the charge on each plate of the capacitor is

    A) \[3\mu C\]                                         

    B) \[6\mu C\]

    C) \[12\mu C\]                       

    D)  \[24\mu C\]

    Correct Answer: B

    Solution :

    From figure it is clear that, current will flow only through the branch containing\[{{R}_{2}}\]. \[\therefore \]  \[I=\frac{E}{{{R}_{2}}+r}=\frac{5}{4+1}=1A\] So,    potential   difference   across \[{{R}_{2}};\]\[V=I{{R}_{2}}=1\times 4=4V\] Let \[q\] be the charge on each plate of each capacitor, then                 \[\frac{q}{C}+\frac{q}{C}=4\] \[\Rightarrow \]               \[\frac{2q}{C}=4\] \[\Rightarrow \]               \[q=2\times C=2\times 3=6\mu C\]


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