Norton’s Theorem allows us to replace a complicated circuit with a simple equivalent circuit containing only a current source and a parallel connected resistor. This theorem is very important from both theoretical and practical viewpoints.
Concisely stated, Norton’s Theorem says:
Any two-terminal linear circuit can be replaced by an equivalent circuit consisting of a current source (IN) and a parallel resistor (RN).
It is important to note that the Norton equivalent circuit provides equivalence at the terminals only. Obviously, the internal structure and therefore the characteristics of the original circuit and its Norton equivalent are quite different.
Using Norton's theorem is especially advantageous when:
We can calculate the Norton equivalent in two steps:
To illustrate, let's find Norton’s equivalent circuit for the circuit below.
The TINA solution illustrates the steps needed for the calculation of the Norton parameters :
Of course, the parameters can be easily calculated by the rules of series-parallel circuits described in previous chapters:
RN = R2 + R2 = 4 ohm.
The short-circuit current (after restoring the source!) can be calculated using current division:
The resulting Norton equivalent circuit:
Find the Norton equivalent for the AB terminals of the circuit below
Find the current of the Norton equivalent using TINA by connecting a short circuit to the terminals, and then the equivalent resistance by disabling the generators.
Surprisingly, you can see that the Norton source might be zero current.
Therefore, the resulting Norton equivalent of the network is just a 0.75 Ohm resistor.
This example shows how the Norton equivalent simplifies calculations.
Find the current in the resistor R if its resistance is:
1.) 0 ohm; 2.) 1.8 ohm; 3.) 3.8 ohm 4.) 1.43 ohm
First, find the Norton equivalent of the circuit for the terminal pair connected to R by substituting for R an open circuit.
Finally, use the Norton equivalent to calculate the currents for the different loads: