Question 2.11: Develop the Norton Equivalent for the circuit given below....
Develop the Norton Equivalent for the circuit given below.
The requirement “Norton Equivalent” implies the determination of I_{N} and R_{N}. Combination of these two key components, in parallel, as shown later in the solu-tion will represent the Norton Equivalent.
Using Ohm’s law, first replace the series combination of the 24 V source and the 4 Ω resistor with a parallel combination of a 6 A current source in shunt with the original 4 Ω resistor as shown below.
Combination of the two parallel current sources, 6 A and 3 A, results in a net, total, Norton current source of 9 A.
I_{SC}=I_{N}=(I_1+I_2)=6 A+3 A=9 A
Next, we focus on R_N. To determine the value of R_N, we view the circuit from the vantage point of A and B, or ground/common. As you look to the left from points A and B, you find that 12 Ω and the 4 Ω resistors are in parallel. Also, in accordance with the principles of Norton’s Theorem, when computing R_N, you neutralize the 6 A and 3 A current sources. This amounts to “open-circuiting” the two current sources. Then, the parallel combination of the 12 Ω and the 4 Ω resistors, as shown below, would result in R_Nof 3 Ω.
R_N=\frac{(12 Ω)(4 Ω)}{12 Ω+4 Ω}=3 ΩThe Norton Equivalent for the given circuit would then appear as follows:
