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(B) $1$. Identify the capacitors in series: The $2\ \mu F$ capacitor at the top and the $2\ \mu F$ capacitor on the right are in series. Their equival

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(B) $1$. Identify the capacitors in series: The $2\ \mu F$ capacitor at the top and the $2\ \mu F$ capacitor on the right are in series. Their equivalent capacitance is $C_s = \frac{2 	imes 2}{2 + 2} = 1\ \mu F$.$2$. Simplify the circuit: Replace these with a single $1\ \mu F$ capacitor. This new capacitor is in parallel with the $1\ \mu F$ capacitor connected diagonally. Their equivalent capacitance is $C_p = 1 + 1 = 2\ \mu F$.$3$. Further simplification: Now,this $2\ \mu F$ capacitor is in series with the $2\ \mu F$ capacitor at the bottom. Their equivalent capacitance is $C_s = \frac{2 	imes 2}{2 + 2} = 1\ \mu F$.$4$. Final step: This $1\ \mu F$ capacitor is in parallel with the $1\ \mu F$ capacitor connected directly between $A$ and $B$. The total equivalent capacitance is $C_{eq} = 1 + 1 = 2\ \mu F$.

What is the equivalent capacitance between $A$ and $B$ in the given circuit (in $\mu F$)?

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