Question 2.18: Find the Thévenin equivalent for the circuit shown in Figure...

Because this circuit contains a dependent source, we cannot find the Thévenin resistance by zeroing the sources and combining resistances in series and parallel. Thus, we must analyze the circuit to find the open-circuit voltage and the short-circuit current.
We start with the open-circuit voltage. Consider Figure 2.50(b). We use node-voltage analysis, picking the reference node at the bottom of the circuit. Then, $v_{oc}$ is the unknown node-voltage variable. First, we write a current equation at node 1.

$i_x + 2i_x= \frac{v_{oc}}{10}$        (2.75)

Next, we write an expression for the controlling variable $i_{x}$ in terms of the node voltage $v_{oc}$:

Substituting this into Equation 2.75, we have

Solving, we find that $v_{oc}$ = 8.57 V.
Now, we consider short-circuit conditions as shown in Figure 2.50(c). In this case, the current through the 10-Ω resistance is zero. Furthermore, we get

$i_{x}=\frac{10  V}{5  Ω}$ = 2 A

and

$i_{sc} = 3i_x$ = 6 A
Next, we use Equation 2.74 to compute the Thévenin resistance:

$R_{t}=\frac{v_{oc}}{i_{sc}}=\frac{8.57  V}{6  A}$ = 1.43 Ω

Finally, the Thévenin equivalent circuit is shown in Figure 2.50(d).