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(C) $1$. The two $2 \, \mu F$ capacitors are in parallel. Their equivalent capacitance is $C_p = 2 \, \mu F + 2 \, \mu F = 4 \, \mu F$.$2$. Now,the ci

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$4$

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(C) $1$. The two $2 \, \mu F$ capacitors are in parallel. Their equivalent capacitance is $C_p = 2 \, \mu F + 2 \, \mu F = 4 \, \mu F$.$2$. Now,the circuit consists of three $4 \, \mu F$ capacitors in series (the original top $4 \, \mu F$,the original left $4 \, \mu F$,and the original bottom $4 \, \mu F$ are not in series,let's re-examine the diagram).$3$. Looking at the diagram: The two $2 \, \mu F$ capacitors are in parallel,giving $4 \, \mu F$. This $4 \, \mu F$ is in series with the top $4 \, \mu F$ capacitor,giving $C_{top} = (4 	imes 4) / (4 + 4) = 2 \, \mu F$.$4$. Similarly,the bottom $4 \, \mu F$ capacitor is in series with the left $4 \, \mu F$ capacitor,giving $C_{bottom} = (4 	imes 4) / (4 + 4) = 2 \, \mu F$.$5$. Finally,these two branches ($C_{top}$ and $C_{bottom}$) are in parallel between $A$ and $B$. Thus,$C_{AB} = 2 \, \mu F + 2 \, \mu F = 4 \, \mu F$.

Find the equivalent capacitance between $A$ and $B$ as shown in the figure (in $\mu F$).

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