4. Equivalent Circuits
by George Sephton on October 3, 2014 at 1:44 pm | 0 comments.
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:
With this information we can calculate the current through or short circuit from earlier:
We can now calculate the resistance; R_{1} and R_{2} are in parallel so we’ll calculate this to form a new resistor in series with R_{3}, these can then be added together:
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:
This is the same as before; R_{1} and R_{2} are in parallel so we’ll calculate this to form a new resistor in series with R_{3}, these can then be added together:
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:
Norton | Thevenin | |
A |
Current | Voltage |
source in |
Parallel | Series |
with a |
Resistance | Resistance |
calculated with |
an Open Circuit | a Short Circuit |
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 R_{1} 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 R_{1} and R_{2} in order to simplify the circuit:
Move on to our next tutorial where we will be learning about complex numbers, reactive and resistive circuits.