Norton Theorem with Example: Electric Circuit Analysis

Norton’s theorem says that a linear two-terminal electric circuit may be exchanged with a Norton equivalent circuit consisting of a current source, IN, in parallel with a Norton resistor RN. Where IN is the short-circuit current through the terminals load resistor and RN is equivalent resistance at the terminals when all the independent sources are turned off.

Just like Thevenin Theorem, Norton Theorem is also useful to analysis single resistor which changes frequently and the rest of the circuit remain same. In contrast to Thevenin Theorem, the Norton Theorem reduce circuit 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; RN, parallel with the current source and load resistor, R­­L, frequently changing resistor.

Norton Resistor RN, is exactly same as Thevenin’s resistor Rth and Norton Current IN is V­th/R­­N. So, this is the implies that Norton Theorem is not more than source transformation of Thevenin Theorem.

Norton Equivalent Circuit :

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

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

Norton Theorem

Example :

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

Norton Theorem Example

Solution:

Find Norton Resistance:

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

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

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

Norton Theorem Example Step 1

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:

$I_{1}=3A \\ I_{2} (4+8+5+8)-4 I_{1}-12=0 \\ 25 I_{2}-4 I_{1}=12 \\ I_{2}=0.8 A$

The current I2 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 RN and voltage V­th, so we can find out the Norton current IN 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:

Norton Example Solved

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

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