In the given figure,the equivalent capacitanc | vedclass

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(A) The circuit consists of four parallel branches connected between points $A$ and $B$.$1$. The top branch has two capacitors of capacitance $C$ in s

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$2$

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(A) The circuit consists of four parallel branches connected between points $A$ and $B$.$1$. The top branch has two capacitors of capacitance $C$ in series,so their equivalent capacitance is $C_{1} = \frac{C 	imes C}{C + C} = \frac{C}{2}$.$2$. The bottom branch also has two capacitors of capacitance $C$ in series,so its equivalent capacitance is $C_{4} = \frac{C}{2}$.$3$. The middle two branches form a balanced Wheatstone bridge structure. The capacitors in the middle branch (between nodes $D$ and $E$) do not carry any charge because the potential at $D$ and $E$ is the same. Thus,the central capacitor can be ignored.$4$. For the second branch,the series combination of two $C$ capacitors gives $\frac{C}{2}$. Similarly,for the third branch,the series combination gives $\frac{C}{2}$.$5$. Since all four branches are in parallel,the total equivalent capacitance is $C_{AB} = \frac{C}{2} + \frac{C}{2} + \frac{C}{2} + \frac{C}{2} = 2C$.Therefore,the value is $2$.

In the given figure,the equivalent capacitance $C_{AB}$ is equal to ...... $C$.

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