Three capacitors of capacitances 3,mu F,9,mu | vedclass

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(A) For series connection,the equivalent capacitance $C_s$ is given by $\frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$.Substituting th

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$1:15$

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(A) For series connection,the equivalent capacitance $C_s$ is given by $\frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$.Substituting the values: $\frac{1}{C_s} = \frac{1}{3} + \frac{1}{9} + \frac{1}{18} = \frac{6+2+1}{18} = \frac{9}{18} = \frac{1}{2}$.Thus,$C_s = 2\,\mu F$.For parallel connection,the equivalent capacitance $C_p$ is given by $C_p = C_1 + C_2 + C_3$.Substituting the values: $C_p = 3 + 9 + 18 = 30\,\mu F$.The ratio of equivalent capacitance in the two cases is $\frac{C_s}{C_p} = \frac{2}{30} = \frac{1}{15}$.

Three capacitors of capacitances $3\,\mu F$,$9\,\mu F$,and $18\,\mu F$ are connected once in series and another time in parallel. The ratio of equivalent capacitance in the two cases $\left( \frac{C_s}{C_p} \right)$ will be

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