Question 2.19: Find the Norton equivalent for the circuit shown in Figure 2...
Find the Norton equivalent for the circuit shown in Figure 2.53(a).
Because the circuit contains a controlled source, we cannot zero the sources and combine resistances to find the Thévenin resistance. First, we consider the circuit with an open circuit as shown in Figure 2.53(a).We treat v_{oc} as a node-voltage variable. Writing a current equation at the top of the circuit, we have
\frac{v_x}{4}+\frac{v_{oc}-15}{R_1}+\frac{v_{oc}}{R_2+R_3} = 0 (2.76)
Next, we use the voltage-divider principle to write an expression for v_{x} in terms of resistances and v_{oc}:
v_{x} = \frac{R_3}{R_2+R_3}v_{oc} = 0.25v_{oc}Substituting into Equation 2.76, we find that
\frac{0.25v_{oc}}{4}+\frac{v_{oc}-15}{R_1}+\frac{v_{oc}}{R_2+R_3} = 0Substituting resistance values and solving, we observe that v_{oc} = 4.62 V.
Next, we consider short-circuit conditions as shown in Figure 2.53(b). In this case, the current through R_{2} and R_{3} is zero. Thus, v_{x}= 0, and the controlled current source appears as an open circuit. The short-circuit current is given by
i_{sc} =\frac{v_s}{R_1} = \frac{15 V}{20 Ω} = 0.75 A
Now, we can find the Thévenin resistance:
R_t=\frac{v_{oc}}{i_{sc}}=\frac{4.62}{0.75} = 6.15 Ω
The Norton equivalent circuit is shown in Figure 2.53(c).