The given circuit can be redrawn, as shown in Fig. with the 1 Ω resistor removed from terminals A and B. The current source has been converted into its equivalent voltage source as shown in Fig.. For finding Vth, we will find the currents x and y in Fig. .

Applying KVL to the first loop, we get
3 − (3 + 2) x − 1 = 0 or x = 0.4 A
∴ Vth = VAB = 3 − 3 × 0.4 = 1.8 V
The value of Rth can be found from Fig. by replacing the two voltage sources by short circuits. In this case Rth = 2 || 3 = 1.2 Ω.

Thevenin’s equivalent circuit is shown in Fig.. The current through the reconnected 1 Ω resistor is = 1.8/(12.1 + 1) = 0.82 A.