From the circuit given below, the capacitance between terminals \(A\) and \(B\) shown in the circuit is ______ \(\mu\)F.
(take \(C_1=C_2=C_3=1\) \(\mu\)F and \(C_4=2\) \(\mu\)F.)

Let the nodes in the circuit be labeled to identify the connections:
- Let the input terminal be Node A.
- Let the junction between \(C_1\) and \(C_2\) be Node X.
- The top wire connects Node A to the junction between \(C_2\) and \(C_3\). Thus, this junction is also at the potential of Node A.
- The bottom wire connects Node X to the output terminal Node B, with capacitor \(C_4\) in this path.

Based on these nodes, the connections of each capacitor are as follows:
- \(C_1\) is connected between Node A and Node X.
- \(C_2\) is connected between Node X and Node A.
- \(C_3\) is connected between Node A and Node B.
- \(C_4\) is connected between Node X and Node B.

We can now simplify the circuit step-by-step:
Capacitors \(C_1\) and \(C_2\) are in parallel between nodes A and X. Their equivalent capacitance is:
\(C_{12} = C_1 + C_2 = 1\,\mu\text{F} + 1\,\mu\text{F} = 2\,\mu\text{F}\)

This equivalent capacitor \(C_{12}\) is in series with \(C_4\) (which is connected between X and B). The equivalent capacitance of this branch is:
\(C_{AXB} = \dfrac{C_{12} \times C_4}{C_{12} + C_4} = \dfrac{2 \times 2}{2 + 2} = 1\,\mu\text{F}\)

Finally, this branch is in parallel with \(C_3\) (which is connected directly between A and B). The total equivalent capacitance between A and B is:
\(C_{AB} = C_{AXB} + C_3 = 1\,\mu\text{F} + 1\,\mu\text{F} = 2\,\mu\text{F}\)

Answer: \(2\)