**Norton’s theorem** says that a linear two-terminal electric circuit may be exchanged with a Norton equivalent circuit. Norton Thoerm circuit consisting of a current source, I_{N}, in parallel with a Norton resistor R_{N}. Where I_{N} is the short-circuit current through the terminals load resistor and R_{N} is equivalent resistance at the terminals when all the independent sources are turned off. In this post, we will learn how to use the Norton theorem with examples.

Just like Thevenin Theorem, Norton Theorem is also useful to analyze a single resistor that changes frequently and the rest of the circuit remains the same. In contrast to Thevenin Theorem, the Norton Theorem reduce it to the single current source instead of the voltage source. An American Engineer E. L. Norton, in 1926, proposed that a circuit can be reduced to the single current source, two resistors; R_{N}, parallel with the current source and load resistor, R_{L}, frequently changing a resistor.

Norton Resistor R_{N} is exactly the same as Thevenin’s resistor R_{th} and Norton Current I_{N} is Vth/R_{N}. So, this implies that Norton Theorem is not more than the source transformation of Thevenin Theorem.

## Norton Equivalent Circuit

According to the Norton theorem, any linear circuit can be reduced to a single current source in parallel with R_{N}, which is calculated from the load resistor end keeping all dependent sources.

And the current of the current source can be calculated from the load resistor terminal voltage keeping it open. The generic Norton Equivalent circuit is shown in the diagram below.

## Use Norton Theorem with Examples

Suppose we want to connect some variable resistor at terminal A-B and find Norton equivalent circuit to the following example for the terminals.

### Solution

#### Find Norton Equivalent Resistance

First of all, we have to replace all independent voltage sources with the closed- and current sources with an open-circuit, as shown in the figure below.

Now we will find equivalent resistance from the terminal A and B point of view, the direction shown by the arrow in the diagram above. So, Norton resistor R_{N} will be:

$R_{N}=5||( 8+4+8)=\frac{ 20\times 5}{20+5}$

$R_{N}=4\Omega $

### Find Norton Voltage:

To find I_{N}, first, we find out the voltage at terminal A-B, while it is open-circuited. So the circuit becomes something like this:

You can use any method you want, here I am using mesh analysis and writing equations as below:

$I_{1}=2A $

$ I_{2} (4+8+5+8)-4 I_{1}-12=0 $

$25 I_{2}-4 I_{1}=12 $

$I_{2}=0.8 A$

The current I_{2} is flowing through 5Ω, so the voltage across the resistor will be:

$V_{th}=5\times 0.8 $

$ V_{th}=4 v$

### Find Norton Current

Now we have both resistance R_{N} and voltage V_{th}, so we can find out the Norton current I_{N} as follow:

$I_{N}=\frac{V_{th}}{R_{N}} $

$=\frac{4}{4}=1A$

The Norton equivalent circuit for the terminal A-B will be as follow:

Whereat terminal A-B we can put any resistor we need. We can analyze the resistor current through the current divider rule instead of going into complex circuit calculations.

Find Norton Voltage:

To find IN, first we find out the voltage at terminal A-B, while it is open circuited. So the circuit becomes something like this:

Norton Theorem

You can use any method you want, here I am using mesh analysis and writing equations as below:

I1=3A is wrong. it is 2A, please correct it.

Thanks Dita. The corrections are made.