# A Brief on Norton’s Theorem with Examples

- Frank
- Jun 08, 2022
- read:

Electrical and Electronics Engineering stream is involved of numerous engineering subjects that includes basic subjects such as network theorems, electrical circuit analysis, electronic devices and circuits, and so on. These network theorems are used to solve electrical circuits and also to calculate different parameters such as voltage, current, etc., of the circuits. Different types of theorems include Nortons theorem, Substitution theorem, Thvenins theorem, and so on. Here, in this article let us discuss in detail about a brief on Nortorn’s theorem with examples.

### Norton’s Theorem

Any linear electrical complex circuit can be simplified into simple circuit that consists of a single current source and parallel equivalent resistance connected across the load. Let us consider a few simple Norton theorem examples to understand in detail about Norton theory. The Norton’s equivalent circuit can be represented as shown in the figure below.

### Norton’s Theorem Statement

Norton’s theorem states that any linear complex electrical circuit can be reduced into a simple electric circuit with one current and resistance connected in parallel. For understanding in depth regarding norton theory, let us consider Norton’s theorem examples as follows.

#### Nortons Theorem Examples

Primarily, let us consider a simple electrical circuit that consists of two voltage sources and three resistors which are connected as shown in the above figure. The above circuit consists of three resistors among which R2 resistor is considered as load. Then, the circuit can be represented as shown below.

We know that, if the load changes, then the calculation of various parameters of electric circuits is difficult. So, network theorems are used for calculating the network parameters easily.

In this Norton’s theorem also we follow the procedure similar to thevenins theorem (up to some extent). Here, primarily remove the load (consider resistor R2=2 Ohms as load in the circuit) as shown in the above figure. Then, short circuit the load terminals with a wire (exactly opposite to the procedure that we follow in thevenins theorem, i.e., open circuit of load terminals) as shown in the below figure. Now, calculate the resultant current (current through resistors R1, R3, and short circuit line after removing R2) as shown in the figure below.

From the above figure, the Nortons source current is equal to 14A which is used in the Norton’s equivalent circuit as shown in the below figure. Norton’s theorem equivalent circuit consists of the Norton current source (INorton) in parallel with Norton’s equivalent resistance (RNorton) and load (here R2=2Ohms).

This Nortorn’s theorem equivalent circuit is a simple parallel circuit as shown in the figure. Now, for calculating Norton’s equivalent resistance we have to follow two procedures such as Thevenins theorem and Superposition theorem.

Primarily, remove the load resistance (similar to thevenins theorem step of calculating thevenins resistance). Then, replace voltage sources with short circuit (wires in case of ideal voltage sources and in case of practical voltage sources their internal resistances are used). Similarly, current sources with open circuit (breaks in case of ideal current sources and in case of practical current sources their internal resistances are used). Now, the circuit becomes as shown in the figure below and it is a simple parallel circuit with resistors.

As the resistors R1 and R3 are parallel to each other, the value of Norton’s resistance is equal to parallel resistance value of R1 and R3. Then, the total Norton’s theorem equivalent circuit can be represented as shown in the circuit below.

The formula for calculating the load current, Iload can be calculated using various basic laws such as Ohm’s law, Krichhoff’s voltage law, and Krichhoff’s current law.

Thus, the current passing through the load resistor Rload (R2) is given by

Where,

I N = Norton’s current (14A)

R N = Norton’s resistance (0.8 Ohms)

R L = Load resistance (2 Ohms)

Therefore, I load = current passing through load resistance = 4A.

Similarly, the large, complex, linear networks with several numbers of sources (current or voltage sources) and resistors can be reduced to simple parallel circuits with single current source in parallel with Norton’s resistance and load.

Thus, the Norton’s equivalent circuit with Rn and In can be determined and a simple parallel circuit can be formed (from a complex network circuit). The calculations of the circuit parameters can be easily analyzed. If one resistance in the circuit is changed rapidly (load), then the Norton’s theorem can be used to perform calculations easily.

Do you know any network theorems other than Norton’s theorem which are usually used in practical electrical circuits? Then, share your views, comments, ideas, and suggestions in the comments section below.