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(B) The circuit consists of an infinite number of rows connected in parallel. Each row contains a certain number of capacitors connected in series. -

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(B) The circuit consists of an infinite number of rows connected in parallel. Each row contains a certain number of capacitors connected in series. - The first row has $1$ capacitor of $1\,\mu F$. Its equivalent capacitance is $C_1 = 1\,\mu F$. - The second row has $2$ capacitors in series,each of $1\,\mu F$. Its equivalent capacitance is $C_2 = \frac{1}{1+1} = \frac{1}{2}\,\mu F$. - The third row has $4$ capacitors in series,each of $1\,\mu F$. Its equivalent capacitance is $C_3 = \frac{1}{1+1+1+1} = \frac{1}{4}\,\mu F$. - The $n^{th}$ row has $2^{n-1}$ capacitors in series. Its equivalent capacitance is $C_n = \frac{1}{2^{n-1}}\,\mu F$. Since all these rows are connected in parallel,the total equivalent capacitance $C_{eq}$ is the sum of the capacitances of each row: $C_{eq} = C_1 + C_2 + C_3 + ... = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ...$ This is an infinite geometric progression with the first term $a = 1$ and common ratio $r = \frac{1}{2}$. The sum of an infinite geometric series is given by $S = \frac{a}{1-r}$. Substituting the values: $C_{eq} = \frac{1}{1 - 1/2} = \frac{1}{1/2} = 2\,\mu F$.

An infinite number of identical capacitors,each of capacitance $1\,\mu F$,are connected as shown in the adjoining figure. The equivalent capacitance between $A$ and $B$ is......$\mu F$.

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