Norton’s Theorem – Norton’s Current and Resistance

Definition of Norton’s Theorem

As per Norton’s theorem, we can replace a complex linear active two-port network with a single current source in parallel with a resistance.

The next question comes to our mind what would be the current and resistance of the source. First of all, we have to short circuit the ports of the network. As a result, the current flowing through the short circuit path between the ports is the current value of the source. Then we have to consider the value of the parallel resistance of the source. Because of that, we have to replace each connected source of the active network with its internal resistance. Then we calculate the equivalent resistance of the network between the said ports. This calculated value is nothing but the value of the internal resistance of the current source. Hence we connect this resistance in parallel with that imaginary source.

Edward Lawry Norton
Edward Lawry Norton

Explanation of Norton’s Theorem

To explain Norton’s Theorem in detail let us consider a common active linear network having one voltage source, one current source and resistances connected in certain manner.

Norton’s Current

Let us examine the value of current (IL) flowing through the resistance RL connected in the network between point A and B. This will be the current value of the source and we call it as Norton’s current.

To find out Norton’s Current, we first remove the resistance RL.

Now, we will short circuit the terminal A and B.

By applying any circuit analysis we can easily find out the current through the short circuit path between A and B. That current would be the source current of the equivalent current source.

Norton’s Resistance

Now we will open A and B and replace all the sources of the active network by their internal resistance. If the voltage sources of the network are ideal, we will replace them with short circuits. Since the internal resistance of an ideal voltage source is zero. If the current sources of the network are ideal, we will replace them with open circuits. Because the internal resistance of an ideal current source is infinite.

After doing that, we will calculate the equivalent resistance of the network across terminal A and B.

Now we can draw a current source in which source current is short circuit current between A and B.  Likewise the parallel resistance is the equivalent resistance between A and B.

Lastly, we will connect back the resistance RL across the source. Now we have to calculate the current through the resistance RL. So we have seen that we can very easily find out the current through the branch AB as the circuit or network is now in its simplest form.

This calculated current through the resistance RL is exactly equal to the current which would have been flowing through the resistance RL if the entire network was connected to RL. This is the simplest explanation of Norton’s Theorem.

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