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JIPMER Jipmer Medical Solved Paper-2010

  • question_answer
    Three capacitors of capacitances \[\text{1}\text{F,}\,\,\text{2}\text{F}\] and \[\text{4}\text{F}\] are connected first in a series combination, and then in a parallel combination. The ratio of their equivalent capacitances will be

    A)  2 : 49 

    B)                                         49 : 2                    

    C)  4 : 49                    

    D)         49 : 4

    Correct Answer: C

    Solution :

    In series combination, \[\frac{1}{{{C}_{1}}}=\frac{1}{1}+\frac{1}{2}+\frac{1}{4}=\frac{4+2+1}{4}=\frac{7}{4}\] \[\Rightarrow \]               \[{{C}_{1}}=\frac{4}{7}\mu F\] In parallel combination, \[{{C}_{1}}=1+2+4=7\mu F\] \[\therefore \]  \[\frac{{{C}_{1}}}{{{C}_{2}}}=\frac{4/7}{7}=\frac{4}{49}\]


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