Find the Norton equivalent for the circuit shown in Figure 2.53(a).

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Accepted Answer

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 [latex]v_{oc}[/latex] as a node-voltage variable. Writing a current equation at the top of the circuit, we have

[latex]\frac{v_x}{4}+\frac{v_{oc}-15}{R_1}+\frac{v_{oc}}{R_2+R_3}[/latex] = 0 (2.76)

Next, we use the voltage-divider principle to write an expression for [latex]v_{x}[/latex] in terms of resistances and [latex]v_{oc}[/latex]:

[latex]v_{x} = \frac{R_3}{R_2+R_3}v_{oc} = 0.25v_{oc}[/latex]

Substituting into Equation 2.76, we find that

[latex] \frac{0.25v_{oc}}{4}+\frac{v_{oc}-15}{R_1}+\frac{v_{oc}}{R_2+R_3} = 0[/latex]

Substituting resistance values and solving, we observe that [latex]v_{oc}[/latex] = 4.62 V.
Next, we consider short-circuit conditions as shown in Figure 2.53(b). In this case, the current through [latex]R_{2}[/latex] and [latex]R_{3}[/latex] is zero. Thus, [latex]v_{x}[/latex]= 0, and the controlled current source appears as an open circuit. The short-circuit current is given by

[latex]i_{sc} =\frac{v_s}{R_1} = \frac{15 V}{20 Ω}[/latex] = 0.75 A

Now, we can find the Thévenin resistance:

[latex]R_t=\frac{v_{oc}}{i_{sc}}=\frac{4.62}{0.75} [/latex] = 6.15 Ω

The Norton equivalent circuit is shown in Figure 2.53(c).

Question 2.19: Find the Norton equivalent for the circuit shown in Figure 2...

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