Find the equivalent capacitance of the system | vedclass

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(B) $1$. Identify the circuit structure: The circuit consists of five capacitors,each of capacitance $C$. $2$. Simplify the rightmost part: The two ca

Vedclass · Accepted Answer

$1.6\,C$

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(B) $1$. Identify the circuit structure: The circuit consists of five capacitors,each of capacitance $C$. $2$. Simplify the rightmost part: The two capacitors on the right are in series. Their equivalent capacitance is $C_{s1} = \frac{C 	imes C}{C + C} = \frac{C}{2}$. $3$. Simplify the middle part: This $C_{s1} = \frac{C}{2}$ is in parallel with the middle vertical capacitor $C$. Their equivalent is $C_{p1} = C + \frac{C}{2} = \frac{3C}{2}$. $4$. Simplify the remaining series part: This $C_{p1} = \frac{3C}{2}$ is in series with the top horizontal capacitor $C$. Their equivalent is $C_{s2} = \frac{C 	imes (3C/2)}{C + (3C/2)} = \frac{3C^2/2}{5C/2} = \frac{3C}{5}$. $5$. Final parallel combination: This $C_{s2} = \frac{3C}{5}$ is in parallel with the leftmost vertical capacitor $C$. The total equivalent capacitance is $C_{eq} = C + \frac{3C}{5} = \frac{8C}{5} = 1.6\,C$.

Find the equivalent capacitance of the system of capacitors between points $A$ and $B$ as shown in the figure.

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