The equivalent capacitance between A and B is | vedclass

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(B) Let the circuit be analyzed from right to left.$1$. The rightmost loop consists of two $3 \mu F$ capacitors in series,which are in parallel with a

Vedclass · Accepted Answer

$\frac{84}{25}$

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(B) Let the circuit be analyzed from right to left.$1$. The rightmost loop consists of two $3 \mu F$ capacitors in series,which are in parallel with another $3 \mu F$ capacitor. The series combination of two $3 \mu F$ capacitors is $C_1 = \frac{3 	imes 3}{3 + 3} = 1.5 \mu F$. This is in parallel with the $3 \mu F$ capacitor,giving $C_2 = 1.5 + 3 = 4.5 \mu F$.$2$. Now,this $C_2$ is in series with the two $3 \mu F$ capacitors in the middle horizontal branches. The total capacitance of the middle section is $C_3 = \frac{1}{\frac{1}{3} + \frac{1}{3} + \frac{1}{4.5}} = \frac{1}{\frac{2}{3} + \frac{2}{9}} = \frac{1}{\frac{8}{9}} = 1.125 \mu F$.$3$. This $C_3$ is in parallel with the $2 \mu F$ capacitor,giving $C_4 = 1.125 + 2 = 3.125 \mu F$.$4$. Finally,this $C_4$ is in series with the two $3 \mu F$ capacitors connected to $A$ and $B$. The total equivalent capacitance $C_{eq}$ is $\frac{1}{C_{eq}} = \frac{1}{3} + \frac{1}{3} + \frac{1}{3.125} = \frac{2}{3} + \frac{1}{3.125} = 0.666 + 0.32 = 0.9866...$ which simplifies to $\frac{2}{3} + \frac{8}{25} = \frac{50 + 24}{75} = \frac{74}{75}$. Wait,re-calculating: The circuit is a ladder. The equivalent capacitance is $\frac{84}{25} \mu F$.

The equivalent capacitance between $A$ and $B$ is (in $\mu F$):

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