NORTON THEOREM: DESCRIPTION, APPLICATIONS, EXAMPLES AND EXERCISES - PHYSICAL - 2025

The theorem Norton, applied to electrical circuits, sets a linear circuit with two terminals a and b, can be replaced by another fully equivalent, consisting of a current source I call _not connected in parallel with a resistance R _No.

Said current I _No or I _N is the one that would flow between points a and b, if they were short-circuited. The resistance R _N is the equivalent resistance between the terminals, when all independent sources turn off. All that has been said is outlined in Figure 1.

Figure 1. Norton equivalent circuit. Source: Wikimedia Commons. Drumkid

The black box in the figure contains the linear circuit to be replaced by its Norton equivalent. A linear circuit is one in which the input and the output have a linear dependence, such as the relationship between the voltage V and the direct current I in an ohmic element: V = IR

This expression corresponds to Ohm's law, where R is the resistance, which can also be an impedance, if it is an alternating current circuit.

Norton's theorem was developed by the electrical engineer and inventor Edward L. Norton (1898-1983), who worked for a long time for Bell Labs.

Applications of Norton's theorem

When you have very complicated networks, with many resistances or impedances and you want to calculate the voltage between any of them, or the current that flows through it, Norton's theorem simplifies the calculations, since as we have seen, the network can be replaced by a smaller and more manageable circuit.

In this way, Norton's theorem is very important when designing circuits with multiple elements, as well as to study the response of them.

Relationship between Norton and Thevenin theorems

Norton's theorem is the dual of Thevenin's theorem, which means that they are equivalent. Thevenin's theorem states that the black box in Figure 1 can be replaced by a voltage source in series with a resistor, called the Thevenin resistor R _Th. This is expressed in the following figure:

Figure 2. Original circuit on the left, and its Thévenin and Norton equivalents. Source: F. Zapata.

The circuit on the left is the original circuit, the linear network in the black box, circuit A at the top right is the Thevenin equivalent, and circuit B is the Norton equivalent, as described. Viewed from terminals a and b, the three circuits are equivalent.

Now note that:

-In the original circuit the voltage between terminals is V _ab.

-V _ab = V _Th in circuit A

-Finally, V _ab = I _N.R _N in circuit B

If terminals a and b are short-circuited in all three circuits, it must be observed that the voltage and current between these points must be the same for all three, since they are equivalent. So:

-In the original circuit the current is i.

-For circuit A, the current is i = V _Th / R _Th, according to Ohm's law.

-Finally in circuit B, the current is I _N

Therefore it is concluded that the Norton and Thevenin resistances have the same value, and that the current is given by:

i = I _N = V _Th / R _Th = V _Th / R _N

Example

To correctly apply Norton's theorem, the following steps are followed:

-Isolate from the network the section of the circuit for which the Norton equivalent is to be found.

-In the remaining circuit, indicate terminals a and b.

-Substitute the voltage sources for short circuits and the current sources for open circuits, to find the equivalent resistance between terminals a and b. This is R _N.

-Return all the sources to their original positions, short-circuit the terminals and find the current that circulates between them. This is I _N.

-Draw the Norton equivalent circuit according to what is indicated in figure 1. Both current source and equivalent resistance are in parallel.

Thevenin's theorem can also be applied to find R _Th, which we already know is equal to R _N, then by Ohm's law we can find I _N and proceed to draw the resulting circuit.

And now let's see an example:

Find the Norton equivalent between points A and B of the following circuit:

Figure 3. Example circuit. Source: F. Zapata.

The part of the circuit whose equivalent is to be found is already isolated. And points A and B are clearly determined. The following is to short-circuit the 10 V source and find the equivalent resistance of the obtained circuit:

Figure 4. Short-circuited source. Source: F. Zapata.

Viewed from terminals A and B, both resistors R ₁ and R ₂ are in parallel, therefore:

1 / R _eq = 1 / R ₁₂ = (1/4) + (1/6) Ω ^-1 = 5/12 Ω ^-1 → R _eq = 12/5 Ω = 2.4 Ω

Then the source is back in place and the points A and B are shorted to find the current flowing there, this will I _N. In that case:

Figure 5. Circuit to calculate Norton current. Source: F. Zapata.

I _N = 10 V / 4 Ω = 2.5 A

Norton equivalent

Finally the Norton equivalent is drawn with the found values:

Figure 6. Norton equivalent of the circuit in figure 3. Source: F. Zapata.

Exercise resolved

In the circuit of the following figure:

Figure 7. Circuit for the resolved exercise. Source: Alexander, C. 2006. Fundamentals of Electrical Circuits. 3rd. Edition. Mc Graw Hill.

a) Find the Norton equivalent circuit of the external network to the blue resistor.

b) Also find the Thévenin equivalent.

Solution to

Following the steps indicated above, the source must be short-circuited:

Figure 8. Source short-circuited in the circuit of figure 7. Source: F. Zapata.

RN calculation

Viewed from terminals A and B, the resistor R ₃ is in series with the parallel formed by the resistors R ₁ and R ₂, let's first calculate the equivalent resistance of this parallel:

And then this parallel is in series with R _3, so the equivalent resistance is:

This is the value of both R _N and R _Th, as explained earlier.

IN calculation

Terminals A and B are then short-circuited, returning the source to its place:

Figure 9. Circuits to find the Norton current. Source: F. Zapata.

The current through I ₃ is the current I _N sought, which can be determined with the mesh method or using series and parallel. In this circuit R ₂ and R ₃ are in parallel:

Resistor R ₁ is in series with this parallel, then:

The current coming out of the source (blue color) is calculated using Ohm's law:

This current is divided into two parts: one that passes through R ₂ and another that passes through R ₃. However, the current that passes through parallel R ₂₃ is the same that passes through R ₁, as can be seen in the intermediate circuit in the figure. The voltage there is:

Both resistors R ₂ and R ₃ are at that voltage, since they are in parallel, therefore:

We already have the Norton current sought, since as previously said I ₃ = I _N, then:

Norton equivalent

Everything is ready to draw the Norton equivalent of this circuit between points A and B:

Figure 10. Norton equivalent of the circuit in figure 7. Source: F. Zapata.

Solution b

Finding the Thévenin equivalent is very simple, since R _Th = R _N = 6 Ω and as explained in the preceding sections:

V _Th = I _N. R _N = 1 A. 6 Ω = 6 V

The Thévenin equivalent circuit is:

Figure 11. Thevenin equivalent of the circuit in figure 7. Source: F. Zapata.

References

1. Alexander, C. 2006. Fundamentals of Electrical Circuits. 3rd. Edition. Mc Graw Hill.
2. Boylestad, R. 2011. Introduction to Circuit Analysis. 2nd. Edition. Pearson.
3. Dorf, R. 2006. Introduction to Electrical Circuits. 7th. Edition. John Wiley & Sons.
4. Edminister, J. 1996. Electrical Circuits. Schaum series. 3rd. Edition. Mc Graw Hill.
5. Wikipedia. Norton's theorem. Recovered from: es.wikipedia.org.