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(A) The circuit consists of eight capacitors,each of capacity $C = 2 \mu F$. By analyzing the symmetry of the circuit between points $A$ and $B$,we ca

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$10$

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(A) The circuit consists of eight capacitors,each of capacity $C = 2 \mu F$. By analyzing the symmetry of the circuit between points $A$ and $B$,we can identify the arrangement: $1$. The top branch has two capacitors in series: $C_{top} = C/2$. $2$. The bottom branch has two capacitors in series: $C_{bottom} = C/2$. $3$. The two middle branches each have two capacitors in parallel: $C_{mid1} = 2C$ and $C_{mid2} = 2C$. All these four branches are connected in parallel between points $A$ and $B$. Therefore,the equivalent capacitance $C_{AB}$ is: $C_{AB} = C_{top} + C_{bottom} + C_{mid1} + C_{mid2}$ $C_{AB} = \frac{C}{2} + \frac{C}{2} + 2C + 2C = C + 4C = 5C$ Given $C = 2 \mu F$,we get: $C_{AB} = 5 	imes 2 \mu F = 10 \mu F$.

Eight capacitors each of capacity $2 \mu F$ are arranged as shown in the figure. The effective capacitance between $A$ and $B$ is (in $\mu F$)

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