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(C) $1$. Observe the circuit carefully. The two $1\ \mu F$ capacitors connected to point $A$ are in parallel with each other.$2$. The equivalent capac

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

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(C) $1$. Observe the circuit carefully. The two $1\ \mu F$ capacitors connected to point $A$ are in parallel with each other.$2$. The equivalent capacitance of these two parallel capacitors is $C_p = 1\ \mu F + 1\ \mu F = 2\ \mu F$.$3$. Now,this $C_p$ is in series with the other $1\ \mu F$ capacitor. The equivalent capacitance of this series branch is $C_s = \frac{2 	imes 1}{2 + 1} = \frac{2}{3}\ \mu F$.$4$. Finally,this $C_s$ is in parallel with the $2\ \mu F$ capacitor connected directly between $A$ and $B$.$5$. The total equivalent capacitance is $C_{AB} = C_s + 2\ \mu F = \frac{2}{3} + 2 = \frac{2 + 6}{3} = \frac{8}{3}\ \mu F$.

What is the equivalent capacitance between points $A$ and $B$ in the given circuit?

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