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 Î© 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 Î©`

**Step
II** – Find I_{N} (Norton Current):

`\R_(eq)=((5×2)/(5+2))+3`

`\⇒R_(eq)=(10/7)+3`

`\R_(eq)=4.43 Î©`

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=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.