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(C) To find the equivalent capacitance between points $A$ and $B$,let us analyze the circuit.Let the potential at point $A$ be $V_A$ and at point $B$

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$\frac{8}{3} \, \mu F$

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(C) To find the equivalent capacitance between points $A$ and $B$,let us analyze the circuit.Let the potential at point $A$ be $V_A$ and at point $B$ be $V_B$. Let the potential at the junction between the first two capacitors be $V_C$.The circuit consists of a $1 \, \mu F$ capacitor in parallel with another $1 \, \mu F$ capacitor,which is then in series with a $1 \, \mu F$ capacitor,and this whole combination is in parallel with a $2 \, \mu F$ capacitor.Wait,looking at the circuit diagram,the $1 \, \mu F$ capacitor (top) and the $1 \, \mu F$ capacitor (middle) are in parallel between point $A$ and the junction $C$. Their equivalent capacitance is $C_1 = 1 \, \mu F + 1 \, \mu F = 2 \, \mu F$.Now,this $C_1$ is in series with the $1 \, \mu F$ capacitor (right side). The equivalent capacitance of this branch is $C_{branch} = \frac{2 \, \mu F 	imes 1 \, \mu F}{2 \, \mu F + 1 \, \mu F} = \frac{2}{3} \, \mu F$.Finally,this branch is in parallel with the $2 \, \mu F$ capacitor connected between $A$ and $B$. Thus,the total equivalent capacitance is $C_{eq} = \frac{2}{3} \, \mu F + 2 \, \mu F = \frac{2 + 6}{3} \, \mu F = \frac{8}{3} \, \mu F$.

The equivalent capacitance between points $A$ and $B$ of the circuit shown will be

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