The capacities of three capacitors are in the ratio 1:2:3. Their equivalent capacity when formed in parallel is greater than that when formed in series by $\frac{120}{11} μ F,$ their individual capacities are ___

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$3 μ F, 6 μ F, 9 μ F$

$1 μ F, 2 μ F, 3 μ F$

$4 μ F, 8 μ F, 12 μ F$

$2 μ F, 4 μ F, 6 μ F$

answer is A.

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Detailed Solution

Let each capacity = C

$C_{p} = C + 2 C + 3 C = 6 C$

$\frac{1}{C_{s}} = \frac{1}{C} + \frac{1}{2 C} + \frac{1}{3 C} = \frac{6 + 3 + 2}{6 c} = \frac{11}{6 C} \Rightarrow C_{3} = \frac{6 C}{11}$

Given $C_{p} = C_{3} + \frac{120}{11} \Rightarrow 6 C = \frac{6 C}{11} + \frac{120}{11} \Rightarrow \frac{120}{11} = \frac{(66 - 6) C}{11}$

$C = \frac{120}{60} = 2 μ F {,C}_{1} = 2 μ F, C_{2} = 4 μ F . C_{3} = 6 μ F$

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