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(B) Let the equivalent capacitance between $A$ and $B$ be $C_{eq}$.Given that each capacitor has a capacitance $C = 1 \ \mu F$.The circuit consists of

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$2$

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(B) Let the equivalent capacitance between $A$ and $B$ be $C_{eq}$.Given that each capacitor has a capacitance $C = 1 \ \mu F$.The circuit consists of capacitors connected in a ladder network.The equivalent capacitance is given by the infinite series:$C_{eq} = C + \frac{C}{2} + \frac{C}{4} + \frac{C}{8} + \dots$This is a geometric progression with the first term $a = C$ and common ratio $r = 1/2$.The sum of an infinite geometric series is given by $S_{\infty} = \frac{a}{1 - r}$.Substituting the values:$C_{eq} = \frac{C}{1 - 1/2} = \frac{C}{1/2} = 2C$.Given $C = 1 \ \mu F$,we get:$C_{eq} = 2 	imes 1 \ \mu F = 2 \ \mu F$.

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

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