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(C) For capacitors connected in series,the equivalent capacitance $C_{eq}$ is given by the formula:$\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} +

Vedclass · Accepted Answer

$0.75$

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(C) For capacitors connected in series,the equivalent capacitance $C_{eq}$ is given by the formula:$\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + ... + \infty$Substituting the given values:$\frac{1}{C_{eq}} = \frac{1}{C} + \frac{1}{4C} + \frac{1}{16C} + ... + \infty$$\frac{1}{C_{eq}} = \frac{1}{C} \left( 1 + \frac{1}{4} + \frac{1}{16} + ... + \infty ight)$The term in the bracket is an infinite geometric progression ($G$.$P$.) with the first term $a = 1$ and common ratio $r = \frac{1}{4}$.The sum of an infinite $G$.$P$. is given by $S_{\infty} = \frac{a}{1 - r}$.$S_{\infty} = \frac{1}{1 - 1/4} = \frac{1}{3/4} = \frac{4}{3}$.Therefore,$\frac{1}{C_{eq}} = \frac{1}{C} 	imes \frac{4}{3} = \frac{4}{3C}$.Thus,$C_{eq} = \frac{3}{4}C = 0.75C$.

An infinite number of capacitors with capacitances $C, 4C, 16C, ... \infty$ are connected in series. What is the equivalent capacitance of the combination in terms of $C$?

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