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(D) $1$. First,simplify the parallel combinations in the circuit. The two $1\,\mu F$ capacitors in parallel give $C_p = 1 + 1 = 2\,\mu F$. The $2\,\mu

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$\frac{2}{5}\,\mu F$

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(D) $1$. First,simplify the parallel combinations in the circuit. The two $1\,\mu F$ capacitors in parallel give $C_p = 1 + 1 = 2\,\mu F$. The $2\,\mu F$ and $2\,\mu F$ capacitors in series give $C_s = \frac{2 	imes 2}{2 + 2} = 1\,\mu F$.$2$. After these simplifications,the circuit reduces to three capacitors in series: $1\,\mu F$,$1\,\mu F$,and $1\,\mu F$ (from the $2\,\mu F$ and $2\,\mu F$ series combination and the remaining $1\,\mu F$ capacitor).$3$. Wait,looking at the circuit diagram,the final series combination is $1\,\mu F$,$2\,\mu F$,and $1\,\mu F$. $4$. The equivalent capacitance $C_{eq}$ is given by $\frac{1}{C_{eq}} = \frac{1}{1} + \frac{1}{2} + \frac{1}{1} = 1 + 0.5 + 1 = 2.5 = \frac{5}{2}$.$5$. Therefore,$C_{eq} = \frac{2}{5}\,\mu F$.

Find the equivalent capacitance between $A$ and $B$.

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