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(A) By analyzing the circuit,we can see that the capacitors are arranged in two groups.In the first group,three capacitors are connected in parallel b

Vedclass · Accepted Answer

$1.5 C$

Vedclass · Answer

(A) By analyzing the circuit,we can see that the capacitors are arranged in two groups.In the first group,three capacitors are connected in parallel between the input and the intermediate node.Therefore,the equivalent capacitance of this group is $C_{123} = C + C + C = 3 C$.Similarly,in the second group,three capacitors are connected in parallel between the intermediate node and point $B$.Therefore,the equivalent capacitance of this group is $C_{456} = C + C + C = 3 C$.These two groups are connected in series with each other.Therefore,the total effective capacitance $C_{	ext{eq}}$ between points $A$ and $B$ is given by:$C_{	ext{eq}} = \frac{C_{123} 	imes C_{456}}{C_{123} + C_{456}} = \frac{(3 C)(3 C)}{3 C + 3 C} = \frac{9 C^2}{6 C} = 1.5 C$.

All capacitors used in the diagram are identical and each is of capacitance $C$. Then the effective capacitance between the points $A$ and $B$ is

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