Find the net capacitance between A and B. | vedclass

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(C) The given circuit is a balanced Wheatstone bridge. Let the nodes be $A$,$B$,and the top and center junctions. Due to the symmetry of the circuit,t

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$2\,C$

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(C) The given circuit is a balanced Wheatstone bridge. Let the nodes be $A$,$B$,and the top and center junctions. Due to the symmetry of the circuit,the potential at the top junction and the center junction is the same. Therefore,no charge flows through the central capacitor connecting these two junctions,and it can be removed from the circuit. After removing the central capacitor,the circuit consists of two parallel branches connected between $A$ and $B$. Each branch consists of two capacitors of capacitance $C$ in series. The equivalent capacitance of each branch is $C_{eq1} = \frac{C 	imes C}{C + C} = \frac{C}{2}$. Since there are two such branches in parallel,the total net capacitance is $C_{net} = C_{eq1} + C_{eq1} = \frac{C}{2} + \frac{C}{2} = C$. Wait,looking at the diagram again,the capacitors are arranged such that the two upper capacitors are in series and the two lower capacitors are in series,and these two branches are in parallel with the bottom capacitor $C$. Actually,the circuit simplifies to two branches in parallel: one branch with two capacitors in series $(C/2)$ and another branch with two capacitors in series $(C/2)$,and the third capacitor $C$ is also in parallel. Total $C_{net} = C/2 + C/2 + C = 2C$.

Find the net capacitance between $A$ and $B$.

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