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(D) Let the capacitance of each capacitor be $C = 3 \mu F$.Looking at the circuit, we can identify the arrangement of capacitors.There are two capacit

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$5 \mu F$

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(D) Let the capacitance of each capacitor be $C = 3 \mu F$.Looking at the circuit, we can identify the arrangement of capacitors.There are two capacitors in parallel in the middle branch. Their equivalent capacitance is $C_p = C + C = 2C = 2 	imes 3 = 6 \mu F$.This combination is in series with another capacitor $C$ in that same branch. The equivalent capacitance of this branch is $C_s = \frac{C 	imes C_p}{C + C_p} = \frac{C 	imes 2C}{C + 2C} = \frac{2C^2}{3C} = \frac{2}{3}C = \frac{2}{3} 	imes 3 = 2 \mu F$.Finally, this branch is in parallel with the top capacitor $C$. The total effective capacitance between $A$ and $B$ is $C_{eq} = C + C_s = 3 + 2 = 5 \mu F$.

In the figure below, the capacitance of each capacitor is $3 \mu F$. The effective capacitance between $A$ and $B$ is:

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