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(B) The circuit consists of two parts connected in series.First part (left): Two capacitors of capacitance $C$ are in series on the top branch,and two

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$2$

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(B) The circuit consists of two parts connected in series.First part (left): Two capacitors of capacitance $C$ are in series on the top branch,and two are in series on the bottom branch. The middle capacitor is in parallel with these branches.Equivalent capacitance of the top branch = $C/2$.Equivalent capacitance of the bottom branch = $C/2$.These two branches are in parallel with the middle capacitor $C$.So,$C_{eq1} = C/2 + C/2 + C = 2C$.Second part (right): Two capacitors of capacitance $C$ are in series on the top branch,and two are in series on the bottom branch.Equivalent capacitance of the top branch = $C/2$.Equivalent capacitance of the bottom branch = $C/2$.These two branches are in parallel.So,$C_{eq2} = C/2 + C/2 = C$.Now,$C_{eq1}$ and $C_{eq2}$ are in series.$C_{eq} = \frac{C_{eq1} 	imes C_{eq2}}{C_{eq1} + C_{eq2}} = \frac{2C 	imes C}{2C + C} = \frac{2C^2}{3C} = \frac{2C}{3}$.Given $C = 3 \mu F$,we have:$C_{eq} = \frac{2 	imes 3 \mu F}{3} = 2 \mu F$.

Find the equivalent capacitance between two points $A$ and $B$ for the given circuit. (Capacitance of each capacitor is $C = 3 \mu F$.) (in $\mu F$)

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