Question 2.17: Find the Thévenin resistance for the circuit shown in Figure...

To zero the sources, we replace the voltage source by a short circuit and replace the current source by an open circuit. The resulting circuit is shown in Figure 2.48(b).

The Thévenin resistance is the equivalent resistance between the terminals. This is the parallel combination of R_1 and R_2, which is given by

R_t= R_{eq}=\frac{1}{1/R_1+1/R_2}= \frac{1}{1/5+1/20} = 4 Ω

Next, we find the short-circuit current for the circuit. The circuit is shown in Figure 2.48(c). In this circuit, the voltage across R_2 is zero because of the short circuit. Thus, the current through R_2 is zero:

i_2 = 0

Furthermore, the voltage across R_1 is equal to 20 V. Thus, the current is

i_1= \frac{v_s}{R_1}=\frac{20}{5} = 4 A

Finally, we write a current equation for the node joining the top ends of R_2 and the 2-A source. Setting the sum of the currents entering equal to the sum of the currents leaving, we have

This yields i_{sc} = 6 A.
Now, the Thévenin voltage can be found. Applying Equation 2.74, we get

R_t=\frac{v_{oc}}{i_{sc}}            (2.74)

V_t = R_t i_{sc} = 4 × 6 = 24 V

The Thévenin equivalent circuit is shown in Figure 2.48(d).