The total capacity of the system of capacitor | vedclass

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(A) Let the points be $A$ and $B$. The circuit consists of a bridge-like structure. By observing the circuit,the $2 \ \mu F$ capacitor (top) and the $

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$1$

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(A) Let the points be $A$ and $B$. The circuit consists of a bridge-like structure. By observing the circuit,the $2 \ \mu F$ capacitor (top) and the $2 \ \mu F$ capacitor (right) are in series. Their equivalent capacitance $C_1$ is given by $\frac{1}{C_1} = \frac{1}{2} + \frac{1}{2} = 1$,so $C_1 = 1 \ \mu F$. This $C_1$ is in parallel with the $1 \ \mu F$ capacitor (middle). Their equivalent capacitance $C_2 = 1 + 1 = 2 \ \mu F$. Finally,this $C_2$ is in series with the $2 \ \mu F$ capacitor (bottom). The total equivalent capacitance $C_{AB}$ is $\frac{1}{C_{AB}} = \frac{1}{C_2} + \frac{1}{2} = \frac{1}{2} + \frac{1}{2} = 1$. Thus,$C_{AB} = 1 \ \mu F$.

The total capacity of the system of capacitors shown in the adjoining figure between the points $A$ and $B$ is $..... \ \mu F$.

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