Tutorial 8 - Circuit Theorems

Learning Objective

To learn about constant voltage and current sources.

To apply Thévenin’s Theorem;

To apply Norton's Theorem

Key Question

What are constant voltage and constant current sources?

What is Thévenin’s Theorem?

What is Norton's Theorem?

Constant Voltage and Constant Current Sources

When we buy a 9 V battery, we want it to supply 9 V to the device throughout its life.  In effect we want it to be a constant voltage supply.  We want it to have a voltage of 9 V, whatever current is drawn.


An ideal voltage source is a source that provides a constant voltage, whatever current is drawn, or whatever current is supplied to the terminals.


The circuit symbol for a constant voltage source is shown below:

We can plot a graph of the voltage against the current:

A perfect battery behaves as a constant voltage source.  The nearest thing to a perfect battery is a lead-acid car battery.  In the lab, we can assume that most sources are ideal constant voltage sources, provided that the current taken by the circuit is low; the voltage does not change measurably.


A constant current source is one that provides the same current, whatever the voltage.  An ideal constant current circuit is one that has infinite internal resistance.

We can plot a graph of the voltage against the current:

Constant current sources can be made using electronic circuits, for example the circuit below:


Op amp circuits can also provide constant current sources.


Real sources do not behave like this.  Norton’s and Thévenin's Theorems help us to analyse circuits that do not have perfect constant voltage or constant current sources.




Thévenin’s Theorem

This was discovered by a French electrical engineer, Leon Charles Thévenin (1857 – 1926).  He came up with his rule after extensive study of Ohm’s work and Kirchhoff’s work.  Thévenin was also an excellent violinist and enjoyed coarse fishing.


The same ideas also occurred to Hermann von Helmholtz (1821 – 1894), a German physicist and physician, but Thévenin’s name is attached to the theorem.



Thévenin’s Theorem states:


The current in any branch of a network is that which would result if an EMF equal to the potential difference across a break made in the branch, were introduced to the branch, all other EMFs removes and represented by the internal resistances of the sources.


In other words we treat any circuit, however complicated, as a

Perfect battery in series with an internal resistor.


You could use a battery symbol to replace the constant voltage source.


Thévenin's Theorem allows us to simplify more complex circuits, and can be easier to use than Kirchhoff’s Laws.   There are certain steps that we have to take to simplify a network:

  • We need to remove a resistor and determine the open circuit voltage across the break.

  • Remove any other sources of EMF, and replace them with resistors of the same internal resistance.

  • Replace the external resistor and determine the value of the current.


The internal resistance equation is useful:

Thévenin Voltage

What are the Thévenin voltage (Vth) and the Thévenin resistance (Rth) of this circuit?

So the first thing to do is to remove the resistor R1 next to A.  We can do this by disconnecting wire AB.  This means that there is no current flowing through R1.


Now we work out the voltage across the gap AB.  We can do this with the voltage divider equation:


We can say that the Thévenin voltage is now 7.5 V

Thévenin Resistance

Now we need to remove the source of EMF and “look back” into the circuit.  The EMF is shorted out.

Let's make this a bit more "user friendly". 


Work out the resistance of this network:


R2 + R3 = 1000 + 1000 = 2000 W


Parallel combination = 1000 W


Total resistance = 1000 + 1000 = 2000 W


Therefore the Thévenin resistance is 2000 W



Worked example

Use Thévenin’s Theorem to find the current flowing in the 10 ohm resistor for the circuit below.





Remove the 10 ohm resistor to work out the Thévenin voltage:


As R3 is in open circuit, no current is flowing.  There is a current flowing through R1 and R2.

Since the total resistance in this loop is 10 W and the voltage is 10 V, it doesn’t take a genius to see that the current = 1 A.


So the voltage across R2 = 8 × 1 = 8 V


Now we need to get rid of the battery:



Work out the parallel resistors R1 and R2



Total resistance = 1.6 + 5 = 6.6 W


So our circuit is now:


Use the internal resistance equation:


Current = 8 ÷ 16.6 = 0.482 A



Now you have a go…

Question 1

          Look at this circuit.

Use Thévenin’s Theorem to calculate:

a.       the Thévenin voltage;

b.      the Thévenin resistance;

c.       the current in the 0.8 ohm resistor.


Norton’s Theorem

Along with Thévenin’s Theorem to model non-ideal constant voltage sources, there is Norton’s Theorem that models non-ideal constant current sources.  It was devised by an American electrical engineer, Edward Lawry Norton (1898 – 1983).


A very similar finding was derived by Hans Ferdinand Mayer (1895 – 1980), a German electrical engineer.


Norton’s Theorem states:


The current that flows in any branch of a network is the same as that which would flow in the branch is it were connected across a source of electrical energy, the short-circuit current of which is equal to the current that would flow in a short circuit across the branch, and the internal resistance of which is equal to the resistance which appears across the open-circuited branch terminals.


Got that?


In other words, any collection of batteries and resistors can be treated as a single ideal current source in parallel with a resistor, r (or in the diagram below, RNO).


So how do we go about using Norton’s Theorem?

Consider this network of a constant voltage and several resistors.

It’s actually the identical circuit to the one we used with Thévenin.

The first thing we do is short-circuit AB.  So let’s do that:

Question 2

Show that the current between A and B is 3.75 mA



Then we add an ammeter to determine the current in AB.  Actually a perfect ammeter has zero resistance, so acts as a short-circuit.



Now remove the source of EMF and replace it with a resistor of the same value as its internal resistance.  Then remove the connection AB and “look in”.


Then work out the equivalent resistance.  We have already shown that it’s 2 kW.

So our Norton circuit becomes:

If we connect a Norton equivalent source to a load, R, we get:

We can write an expression for the Norton current:


Worked Example


Use Norton’s Theorem to find the current flowing in the 10 W resistor.



Short circuit the branch with the 10 ohm resistor (which gets rid of both the 5 ohm and the 10 ohm resistor):


In a short circuit, there will be zero voltage across the 8 ohm resistor, so no current will go through it.  So we assume that all the current goes through the 2 ohm resistor.


 Now remove the EMF source and look in:

 It’s two parallel resistors:


Now use the Norton equivalent equation:


This is our Norton Equivalent network:


Question 3

Look at this circuit:

Find the Norton equivalent of this circuit between A and B.

Hence find the voltage and current across the 80 ohm resistor




Self Test