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Three capacitors of capacitances 9 $μ F$ ,3 µF and 18µF are connected once in series and another time in parallel. The ratio of equivalent capacitance in the two cases $(\frac{C_{s}}{C_{p}})$ will be:

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$1 : 15$

$15 : 1$

$1 : 1$

$1 : 3$

answer is A.

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Detailed Solution

$\frac{1}{C_{s}} = \frac{1}{3} + \frac{1}{9} + \frac{1}{18} = \frac{1}{2} \Rightarrow C_{s} = 2 μ F$

$\Rightarrow C_{p} = 3 + 9 + 18 = 30 μ F \Rightarrow \frac{C_{s}}{C_{p}} = \frac{2}{30} = \frac{1}{15}$

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