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(C) In the given circuit,there are four capacitors,each of $2\,\mu F$. Let the capacitors be $C_1, C_2, C_3,$ and $C_4$. The capacitor connected direc

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

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(C) In the given circuit,there are four capacitors,each of $2\,\mu F$. Let the capacitors be $C_1, C_2, C_3,$ and $C_4$. The capacitor connected directly between $A$ and $B$ is in parallel with the combination of the other three capacitors connected in series.Let the top capacitor be $C_1 = 2\,\mu F$. The other three capacitors $(C_2, C_3, C_4)$ are connected in series along the bottom and sides. The equivalent capacitance of these three in series is given by $\frac{1}{C_s} = \frac{1}{2} + \frac{1}{2} + \frac{1}{2} = \frac{3}{2}$. Thus,$C_s = \frac{2}{3}\,\mu F$.Now,this series combination $C_s$ is in parallel with $C_1$. Therefore,the equivalent capacitance $C_{eq} = C_1 + C_s = 2 + \frac{2}{3} = \frac{6+2}{3} = \frac{8}{3}\,\mu F$.

The equivalent capacitance between the points $A$ and $B$ in the given diagram is

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