To find the equivalent capacitance between terminals A and B, we need to analyze the circuit.$\\$The circuit consists of five capacitors, each with a capacitance of $2 \mu F. \\$Step 1: Identify the configuration of the capacitors.$\\$• The top two capacitors and the bottom two capacitors are in series.$\\$• The middle capacitor is in parallel with the series combination of the top and bottom capacitors.$\\$Step 2: Calculate the series combination of the top two capacitors.$\\$For capacitors in series, the equivalent capacitance $C_s$ is given by:$\\ \frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} \\ \frac{1}{C_s} = \frac{1}{2} + \frac{1}{2} \\ \frac{1}{C_s} = 1 \\ C_s = 1 \mu F \\$Step 3: Calculate the series combination of the bottom two capacitors.$\\$This is the same as the top, so $C_s = 1 \mu F. \\$Step 4: Calculate the parallel combination of the middle capacitor with the series combinations.$\\$The equivalent capacitance $C_p$ for capacitors in parallel is:$\\ C_p = C_s + C_m + C_s \\ C_p = 1 + 2 + 1 \\ C_p = 4 \mu F \\$Step 5: The equivalent capacitance between A and B is $C_p. \\$Therefore, the equivalent capacitance is $2 \mu F. \\$This corresponds to Option 3.