Question 2.18: Find the Thévenin equivalent for the circuit shown in Figure...
Find the Thévenin equivalent for the circuit shown in Figure 2.50(a).
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}:
i_x= \frac{10-v_{oc}}{5}Substituting this into Equation 2.75, we have
3 \frac{10-v_{oc}}{5}=\frac{v_{oc}}{10}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).