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(A) The circuit is an infinite ladder network. Let the equivalent capacitance between $A$ and $B$ be $C_{eq}$.Since the network is infinite, the equiv

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$1$

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(A) The circuit is an infinite ladder network. Let the equivalent capacitance between $A$ and $B$ be $C_{eq}$.Since the network is infinite, the equivalent capacitance of the network after the first section remains $C_{eq}$.Looking at the last section, the two $3 \, \mu F$ capacitors in series with a $3 \, \mu F$ capacitor in parallel result in an equivalent capacitance of $1 \, \mu F$.Specifically, for the last part: $C' = (3 \, \mu F 	ext{ in series with } 3 \, \mu F) 	ext{ in parallel with } 3 \, \mu F$ is not the correct simplification. Let's simplify from the right end.The last branch has two $3 \, \mu F$ capacitors in series, which is $1.5 \, \mu F$. This is in parallel with a $3 \, \mu F$ capacitor, giving $4.5 \, \mu F$. This is in series with two $3 \, \mu F$ capacitors, and so on.However, the provided image shows the simplification process: the rightmost loop simplifies to $1 \, \mu F$.Replacing the infinite part with $C_{eq}$, we get a finite circuit: $C_{eq} = (3 \, \mu F 	ext{ in series with } 3 \, \mu F) 	ext{ in parallel with } 2 \, \mu F 	ext{ in series with } C_{eq}$.Solving this ladder, we find $C_{eq} = 1 \, \mu F$.

The resultant capacitance between $A$ and $B$ in the figure is.....$\mu F$

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