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(A) The circuit consists of infinite rows connected in parallel between points $A$ and $B$.For the first row,the capacitors are in series: $C_1, 2C, 4

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$C$

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(A) The circuit consists of infinite rows connected in parallel between points $A$ and $B$.For the first row,the capacitors are in series: $C_1, 2C, 4C, 8C, \dots$The equivalent capacitance $C_1$ of the first row is given by:$\frac{1}{C_1} = \frac{1}{C} + \frac{1}{2C} + \frac{1}{4C} + \frac{1}{8C} + \dots$$\frac{1}{C_1} = \frac{1}{C} (1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots)$Using the sum of an infinite geometric series $S = \frac{a}{1-r}$ where $a=1$ and $r=1/2$:$\frac{1}{C_1} = \frac{1}{C} (\frac{1}{1 - 1/2}) = \frac{1}{C} (2) = \frac{2}{C} \Rightarrow C_1 = \frac{C}{2}$For the second row,the capacitors are: $C/2, C, 2C, 4C, \dots$$\frac{1}{C_2} = \frac{1}{C/2} + \frac{1}{C} + \frac{1}{2C} + \frac{1}{4C} + \dots = \frac{2}{C} + \frac{1}{C} + \frac{1}{2C} + \frac{1}{4C} + \dots$$\frac{1}{C_2} = \frac{2}{C} + \frac{1}{C} (1 + \frac{1}{2} + \frac{1}{4} + \dots) = \frac{2}{C} + \frac{1}{C} (2) = \frac{4}{C} \Rightarrow C_2 = \frac{C}{4}$Similarly,for the $n$-th row,$C_n = \frac{C}{2^n}$.The total equivalent capacitance $C_{eq}$ is the sum of these parallel rows:$C_{eq} = C_1 + C_2 + C_3 + \dots = \frac{C}{2} + \frac{C}{4} + \frac{C}{8} + \dots$$C_{eq} = \frac{C}{2} (1 + \frac{1}{2} + \frac{1}{4} + \dots) = \frac{C}{2} (2) = C$.

Consider an infinite matrix of capacitors as shown in the figure. The effective capacitance between point $A$ and $B$ will be

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