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WB JEE Medical WB JEE Medical Solved Paper-2006

  • question_answer
    Three capacitors of capacitances \[1\,\mu F,2\mu F\]and 4uF 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, \[\frac{1}{{{C}_{1}}}=\frac{1}{1}+\frac{1}{2}+\frac{1}{4}=\frac{4+2+1}{4}=\frac{7}{4}\] \[\Rightarrow \]               \[{{C}_{1}}=\frac{4}{7}\mu F\] The parallel combination \[{{C}_{2}}=1+2+4=7\,\mu F\] \[\therefore \]  \[\frac{{{C}_{1}}}{{{C}_{2}}}=\frac{4/7}{7}=\frac{4}{49}\]


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