Q111 / 160 · JIPMER 2010 · Physics

Three capacitors of capacitances $1 \mu \mathrm{F}, 2 \mu \mathrm{F}$ and $4 \mu \mathrm{F}$ 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
  • Correct answer
    C
    4:49
  • D
    49:4

Solution

In series combination, $\frac{1}{C_1}=\frac{1}{1}+\frac{1}{2}+\frac{1}{4}=\frac{4+2+1}{4}=\frac{7}{4}$ $\Rightarrow \quad C_1=\frac{4}{7} \mu \mathrm{F}$ In parallel combination, $\begin{aligned} & C_1=1+2+4=7 \mu \mathrm{F} \\ \therefore \quad & \frac{C_1}{C_2}=\frac{4 / 7}{7}=\frac{4}{49} \end{aligned}$

Hence, the correct answer is option: 'C'.

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