In the given circuit, the equivalent resistance between the terminals A and B is _____ $$\Omega$$

The vertical wire connected at the far right of the circuit acts as an ideal short circuit. This wire provides a path of zero resistance, effectively bypassing the two $$4\text{ }\Omega$$ resistors and the rightmost vertical $$2\text{ }\Omega$$ resistor. Consequently, the equivalent resistance of this entire section is $$0\text{ }\Omega$$.

Moving left, the horizontal $$2\text{ }\Omega$$ resistor is connected in series with the previously identified shorted section ($$0\text{ }\Omega$$):

$$R_{\text{series}} = 2\text{ }\Omega + 0\text{ }\Omega = 2\text{ }\Omega$$

This $$2\text{ }\Omega$$ equivalent branch is in parallel with the first vertical $$2\text{ }\Omega$$ resistor. The equivalent resistance ($$R_{p}$$) for this part is: $$R_{p} = \frac{2\text{ }\Omega \times 2\text{ }\Omega}{2\text{ }\Omega + 2\text{ }\Omega} = 1\text{ }\Omega$$

The circuit now reduces to three components in series between terminals A and B: the $$3\text{ }\Omega$$ top resistor, the $$1\text{ }\Omega$$ equivalent middle network, and the $$6\text{ }\Omega$$ bottom resistor.

$$R_{AB} = 3\text{ }\Omega + 1\text{ }\Omega + 6\text{ }\Omega$$

$$R_{AB} = 10\text{ }\Omega$$