In this tutorial we will be looking at how components within a circuit can be interchanged in order to make the analysis either. We will start by looking at Norton’s Theorem which states that:

“Any network of sources and resistors with output terminals, A and B, can be replaced with an ideal current source in parallel with resistors at the terminals”

Norton’s Theorem

As stated above, Norton’s Theorem can be used to convert a circuit to a current source and resistor in parallel, as shown below:

In order to understand Norton’s Theorem, we will be looking at an example and an analysing it. We’ll start with a network of voltage sources and resistors:

We first need to find the short circuit between A and B (ie when we connect the two together with a zero-resistance wire).

Let’s find the current along the wire creating the short circuit; since we know that the voltage across this branch is the difference between b and a, we can now split the circuit into two parts:

The current across the short circuit is now easy to calculate:

Using the circuit on the left, we can calculate the voltage at a and b, however, before we do this, we need to combine R₂ and R₃ to find our new value, R_n:

R₂ and R₃ are in parallel so we used our resistor circuit laws from tutorial 2. Now to find the value of V_ab we use the potential divider equation (also in tutorial 2), since the two resistors, R₁ and R_n form a potential divider.

The voltage drop between a and b is 4.62V and we know that point b is at 0V as it connects to the negative terminal of the voltage source, therefore the voltage at point a is 4.62V.

With this information we can calculate the current through or short circuit from earlier:

Our next step is to find the resistance between the output terminals, A and B. This is a more simplified process and involves turning the sources off; having done this we can then find the equivalent resistance throughout the network:

We can now calculate the resistance; R₁ and R₂ are in parallel so we’ll calculate this to form a new resistor in series with R₃, these can then be added together:

We now have the short circuit current and the equivalent resistance and using these two we can create our Norton equivalent circuit. Remember that Norton’s circuit is a resistor and a current source in parallel:

We now have our equivalent circuit which has made the network a lot simpler.

Thevenin’s Theorem

Thevenin’s Theorem states that:

“Any network of voltage and current generator and resistors with output terminals, A and B is equivalent at those terminals to a single voltage source in series with a single resistance”

As we did with Norton’s Theorem, we will be using an example to help explain Thevenin’s Theorem. We’ll start with the same circuit from before:

The first step is to find the voltage output at A and B when there is nothing connected (an open circuit). An open circuit has a resistance of infinity and so current will not flow, using this information we can visualise where the current in this circuit will flow:

Using this we can find the voltage and a and b. As before the voltage at b will zero since it flows into the negative end of the voltage source and the voltage at a can be calculated using our current divider circuit:

Now that we know the voltage, we can now calculate the resistance that is in series. As with Norton’s Theorem, this is done by turning all of the sources in the network off:

This is the same as before; R₁ and R₂ are in parallel so we’ll calculate this to form a new resistor in series with R₃, these can then be added together:

Using these two, we now have our Thevenin equivalent Circuit:

Once again, we have managed to make our network a lot simpler.

Norton and Thevenin Duality

Although Norton and Thevenin’s Theorems are very similar, it can be easy to remember the difference:

Source Conversion

Sometimes it can be easier to analyse a circuit by converting a voltage source into a current source – or vice versa. It sounds quite complicated but it is just ohm’s law applied to a circuit. Again we will use the example from Norton’s and Thevenin’s Theorem:

First move the R₁ so that it is in parallel with the source and then change that voltage source to a current source with the new current value (which is simply ohm’s law):

Finally combine R₁ and R₂ in order to simplify the circuit:

We can then re-draw our circuit:

Move on to our next tutorial where we will be learning about complex numbers, reactive and resistive circuits.