American engineer Norton explained his theorem in 1926 after 43 years publishing Thevenin’s theorem. Norton’s theorem is similar to Thevenin’s theorem.

Nortons theorem states that **“a linear two terminal circuit can be replaced by an equivalent current source I _{N} in parallel with a resistance R_{N} where R_{N} is the input or equivalent resistance at the terminals and I_{N} is the short circuit current through the terminals when all the independent sources are turned off”.**

Figure 1(a) circuit is a linear two terminal circuit. According to Norton’s theorem we replace this circuit by a current source and a equivalent resistance that is figure 1(b).

From source transformation The Thevenin’s and Norton’s resistances are equal

R_{N} = R_{Th}

As two terminal linear circuit is replaced to Norton’s equivalent circuit so they are equal. To find Norton’s current I_{N} from a to b terminal we have to short circuit and that it becomes figure 2.

Norton’s current I_{N} is equal to short circuit current, I_{N} = *i _{sc} *. Independent sources are turned off here as Thevenin’s theorem. As we know R

_{N}= R

_{Th}and

I_{N} = V_{Th} / R_{Th}

Norton theorem and Thevenin’s theorem are related to source transformation. So source transformation often known as Thevenin-Norton transformation.

As we know V_{Th} = V_{oc} , I_{N} = i_{sc}

R_{Th} = V_{oc} / i_{sc} = R_{N} .

To find Thevenin’s or Norton’s equivalent short circuit and open circuit tests are sufficient if one side of the circuit contains at least one independent source.

## Norton’s theorem problem**:**

Find Norton’s equivalent circuit from following circuit and also find I_{N}

## Norton theorem problem Answer**: **

we have to find R_{N} as the same as we find R_{Th} in Thevenin’s equivalent circuit. First we turn off all independent source equal to zero as we get figure 4(a).

Here 8, 4, 8Ω are in series and their equivalent resistance is in parallel with 5Ω resistance. Thus we get

R_{N} = 5 ││ ( 8 + 4 + 8 ) = 20 = (20×5) / 25 = 4Ω

We have to short circuit a and b terminals to find I_{N} then the circuit becomes figure 4(b).

Ignoring 5Ω resistance cause it is short circuited we apply mesh analysis and obtain,

i_{1 }= 2A, 20i_{2} – 4i_{1} – 12 = 0

From above equation we get_{, }i_{2} = 1A = i_{sc} = I_{N}

We can also determine I_{N} from the value of V_{Th }/ R_{Th }. Making open circuit terminals a and b we get V_{Th} from figure 4(c).

Applying mesh analysis,

i_{3} = 2A,

25i_{4} – 4i_{3} – 12 = 0

or i_{4} = 0.8A

and v_{oc }= V_{Th} = 5i_{4} = 4v

hence I_{N} = V_{Th} / R_{Th} = 4/4 = 1 A

as we got it previously. This is the confirmation of R_{Th} = V_{oc} / i_{sc} = 4/1 = 4Ω = R_{N}.

Thus we get Norton’s equivalent circuit as shown in figure 5.

nortons theorem |