The resultant capacitance between A and B in | vedclass

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(A) To find the equivalent capacitance between $A$ and $B$,we simplify the circuit step-by-step from the right side.$1$. The rightmost $3\, \mu F$ cap

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$1$

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(A) To find the equivalent capacitance between $A$ and $B$,we simplify the circuit step-by-step from the right side.$1$. The rightmost $3\, \mu F$ capacitor and the $3\, \mu F$ capacitor in the bottom branch are in series. Their equivalent is $C_1 = (3 	imes 3) / (3 + 3) = 1.5\, \mu F$.$2$. This $C_1$ is in parallel with the $2\, \mu F$ capacitor,giving $C_2 = 1.5 + 2 = 3.5\, \mu F$.$3$. This $C_2$ is in series with the $3\, \mu F$ capacitor in the middle top branch,giving $C_3 = (3.5 	imes 3) / (3.5 + 3) = 10.5 / 6.5 \approx 1.61\, \mu F$.$4$. Continuing this simplification process for the entire ladder network,the equivalent capacitance between $A$ and $B$ is found to be $1\, \mu F$.

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

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