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(D) The given circuit can be simplified by identifying the Wheatstone bridge structure. $1$. In the bridge part, all four capacitors have equal capaci

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$2$

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(D) The given circuit can be simplified by identifying the Wheatstone bridge structure. $1$. In the bridge part, all four capacitors have equal capacitance $C$. The ratio of capacitances in opposite arms is $C/C = C/C$, which satisfies the balanced Wheatstone bridge condition. $2$. Therefore, the central capacitor in the bridge has no potential difference across it and can be removed. $3$. The remaining four capacitors in the bridge form two series branches of $(C 	ext{ and } C)$ in parallel with each other. The equivalent capacitance of the bridge is $C_{eq1} = \frac{C}{2} + \frac{C}{2} = C$. $4$. This equivalent capacitance $C$ is in parallel with the remaining capacitor $C$ connected between $A$ and $B$. $5$. The total equivalent capacitance is $C_{eq} = C + C = 2\,C$.

Find the equivalent capacitance between $A$ and $B$. (in $C$)

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