Norton’s theorem is suitable for a linear bilateral network where we desire to find the values of the current flowing through the resistor for its different values. The solution of the complicated electrical networks is simplified by the use of Norton’s theorem.

Norton’s theorem is an alternative to Thevenin’s theorem. Norton’s theorem reduces a complex linear and bilateral circuit into a simple circuit consisting of a current source in parallel with a resistor thereby feeding a load resistance. The figure below shows a circuit equivalent to Norton’s theorem.

Here, I_{N} is Norton’s equivalent current, R_{N} is Norton’s equivalent resistance and R_{L} is the load resistance.

Then, from the equivalent circuit, the current flowing in the load resistance is

Meanwhile, the above expression of current flowing through the load comes from the current division rule.

Further, the I_{N} is short-circuit current flowing through the load terminal when we remove the load and replace it with short-circuit. Secondly, R_{N }is the equivalent resistance as seen inward from the load terminal when the source has been removed and replaced by its internal resistance.

## Statement of Norton’s Theorem

Any linear, bilateral network having two terminal X and Y (or load terminal) can be replaced by a current source with current I_{N} in parallel with a resistance R_{N} feeding the load terminal. Here, I_{N} is Norton’s equivalent current which is short-circuited current flowing through the terminal X and Y when the load resistance R_{L} is removed and replaced by a short-circuit. Finally, R_{N} being Norton’s equivalent resistance which is measured between the terminals X and Y the load removed, and the sources being replaced by their internal resistances.

For instance, in the case of the idea source. Firstly, for an ideal voltage source, we replace it with short-circuit. Secondly, for an ideal current source, we replace it with an open-circuit.

## Application of Norton’s Theorem

In the above circuit, R_{L} is the load resistance where we want to determine the current flowing using Norton’s theorem.

For the application of Norton’s theorem, we need to determine the I_{N} and R_{N}.

**To determine the I**_{N}:

_{N}:

Firstly, to determine the I_{N} we will remove the load resistance R_{L} from the terminal XY and replace it with short-circuit. Then we will determine the current flowing in the short-circuit path. Finally, this current will be equal to the I_{N}.

To calculate the I_{N}

**To determine R**_{N}:

_{N}:

Secondly, to determine R_{N} we will remove the source from the circuit, leaving behind its internal resistance only. Then we will view the circuit inwards from the open terminal XY and determine the equivalent resistance. Finally, this equivalent resistance will be equal to the R_{N}.

For instance, if the internal resistance is negligible or the source is an ideal source, then replace the voltage source with short-circuit and the current source be an open-circuit.

For instance, to determine the R_{N} in the above illustration. We replace the voltage source with short-circuit and determine the equivalent resistance as seen inward from the terminal XY.

Finally, we will form Norton’s equivalent circuit.

Where load current I_{L} is

## Examples of Norton’s Theorem

Here in the following circuit, we will determine the current flowing 15 Ω resistor using Norton’s theorem.

Firstly, to determine the value of I_{N.} We will remove the 15 Ω resistor from the circuit and replace it with short-circuit.

Then we can determine the current flowing through the short circuit path will be

In the above circuit, all the current which is delivered by the source will flow through the 10 Ω resistor and short circuit terminal. As current always flows through the low resistance or impedance path. The short-circuit portion of the circuit will provide a low resistance path for the current, thereby all the current flows through 10 Ω resistors and then through terminal AB and back into the source. In short, both the 20 Ω resistance has no significance in the determination of I_{N}.

Secondly, to determine the R_{N.} We will replace the voltage source with a short-circuit and determine the equivalent resistance across the terminal AB.

Finally, we will draw Norton’s equivalent circuit.

Hence, we can calculate the current flowing in the 15 Ω resistor by using the current division rule.

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