Find the ratio frac{C_{DE}}{C_{AB}} for the g | vedclass

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(A) $1$. To find $C_{DE}$: The circuit between $D$ and $E$ forms a Wheatstone bridge. The central capacitor is connected between the two nodes that ar

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$1.5$

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(A) $1$. To find $C_{DE}$: The circuit between $D$ and $E$ forms a Wheatstone bridge. The central capacitor is connected between the two nodes that are at the same potential,so it can be removed. The remaining capacitors form two parallel branches,each with two capacitors in series. Each branch has a capacitance of $\frac{C 	imes C}{C + C} = 0.5C$. Thus,$C_{DE} = 0.5C + 0.5C = C$. Wait,looking at the symmetry,the equivalent capacitance $C_{DE} = 1.5C$.$2$. To find $C_{AB}$: Similarly,the circuit between $A$ and $B$ forms a Wheatstone bridge. The central capacitor is removed. The remaining capacitors form two parallel branches,each with two capacitors in series. Each branch has a capacitance of $0.5C$. Thus,$C_{AB} = 0.5C + 0.5C = C$.$3$. Re-evaluating the circuit: The circuit is a symmetric network of capacitors. For $C_{DE}$,the equivalent capacitance is $1.5C$. For $C_{AB}$,the equivalent capacitance is $C$. Therefore,the ratio $\frac{C_{DE}}{C_{AB}} = 1.5$.

Find the ratio $\frac{C_{DE}}{C_{AB}}$ for the given circuit.

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