What is Norton Theorem with Example?

In this article, we will discuss Norton’s Theorem, its statement, diagram, formula, and examples.

Statement of Norton’s Theorem

The statement of Norton’s theorem is as under:

A linear bilateral electric circuit consisting of independent and/or dependent energy sources and linear bilateral circuit elements can be replaced by an equivalent electric circuit that consists of a current source in parallel with a resistor.

This equivalent electric circuit consisting of a current source and a parallel resistor is called Norton’s equivalent circuit.

In Norton’s equivalent circuit, the value of the current source is equal to the short-circuit current across the load terminals, and the resistance value of the parallel resistor is equal to the internal resistance of the source network looking through the open-circuited load terminals.

Explanation of Norton’s Theorem

The following steps explain the procedure of solving an electric circuit using Norton’s theorem.

Consider an electric circuit as shown in the figure-1. We have to find Norton’s equivalent circuit for this given electric circuit.

Step I – Open the load terminals by removing the load resistor R_L, and calculate the internal resistance R_N (Norton’s equivalent resistance) of the network by looking through the open-circuited load terminals.

`\R_N=R_2+(R_1 R_3)/(R_1+R_3)`

Step II – Short-circuit the load terminals. Determine the short-circuit current, which will be Norton’s current (I_N).

`\I=V/(R_1+((R_2 R_3)/(R_2+R_3 )))`

Therefore,

`\I_N=I×R_3/(R_2+R_3)`

Step III – Thus, Norton’s equivalent circuit and the load current will be.

`\I_L=I_N×R_N/(R_N+R_L)`

Hence, in this way, we obtain Norton’s circuit equivalent to a complex linear bilateral active electric circuit.

Now, let us solve a numerical example to understand the application of Norton’s theorem in circuit analysis.

Example – Determine the electric current through the 10 Ω resistor in the following circuit by using Norton’s theorem.

Solution – The given electric circuit can be solved using Norton’s theorem as follows.

Step I – Find R_N (Norton Resistance):

`\R_N=((2×3)/(2+3))+5`

`\R_N=(6/5)+5`

`\∴R_N=6.2 Ω`

Step II – Find I­_N (Norton Current):

`\R_(eq)=((5×2)/(5+2))+3`

`\⇒R_(eq)=(10/7)+3`

`\R_(eq)=4.43 Ω`

Therefore, the total current in the circuit from the voltage source is,

`\I=V/R_(eq) =10/4.43`

`\I=2.26 A`

Thus, using the current division rule, the Norton current will be,

`\I_N=2.26×(2/(2+5))`

`\∴I_N=0.645 A`

Step III – Equivalent Norton Circuit and the load current is,

`\I_L=0.645×(6.2/(6.2+10))`

`\∴I_L=0.246 A`

Hence, this is all about Norton’s theorem, its statement and examples.