Thévenin's Theorem

Léon Charles Thévenin, born in Meaux France, was a telegraph engineer and a talented violinist that was very interested in solving the problems of measurement in electrical circuits. He deeply studied KCL, KVL and Ohm's Law eventually developing the famous Thévenin's Theorem in 1883 which elegantly states that "any linear circuit can be represented by a single voltage source, Vth, in series with resistance Rth."

What's also fascinating to note is that the theorem was independently derived in 1853 by the German scientist Hermann von Helmholtz.

By unknown photographer - Hermann von Helmholtz in 1948 by Leo Königsberger, Public Domain, https://en.wikipedia.org/w/index.php?curid=41453873

Let's decompose this theorem now, but first, what did Thévenin mean by a linear circuit? By linear circuit he simply meant that the circuit obeys the superposition principle. So what does this then mean? It means, we can simplify any network of voltage, current sources and resistances connected together to an equivalent voltage source and resistor connected to the outside world! It is really quite profound when you pause to think, I strongly encourage you to stop to think about this, any horrid rats nest of resistors and sources can be simplified to a voltage source in series with a resistor!

Note that a linear circuit contains all of the elements we have learned so far, namely, resistors, voltage sources and current sources. Later on we will encounter non-linear elements for which Thevenin's Theorem does not apply.

The circuit shown above on the left can be simplified to the circuit on the right "Thevenin Equivalent Circuit" using Thevenin's Theorem. If we attach a circuit to terminals X and Y, a load, then both circuits will drive the load identically.

Determining the Thevenin Equivalent Circuit

We can derive the Thevenin Equivalent Circuit using a process which can be broken down into two steps, calculating the Thevenin Voltage and the Thevenin Series Resistance.

Thevenin Voltage

The Thevenin Voltage is the open circuit voltage observed across terminals X and Y which can be connected to an external load.

Thevenin Resistance

The Thevenin Resistance is the resistance measured across terminals X and Y when all of the sources are "zeroed".

We can think of this as being the resistance looking back into terminals X and Y.

To "zero" a voltage source we replace it with a short circuit and to "zero" a current source we replace it with an open circuit. Recall that the voltage across a short is zero and no current will flow through an open circuit hence effectively zeroing their effect.

Examples:

1. a) Determine the Thevenin Voltage of the circuit below:

The calculate the Thevenin Voltage, we need to find the open circuit voltage across terminals X and Y. As an initial simplification we know that no current will flow through X and Y and hence no current will flow through R3 which means that the voltage across R3 is zero and that the voltage across R2 is then equal to Vxy. Knowing this we then have a series circuit composed of R1 and R2 and we can apply the voltage divider rule to find Vxy:

b) Determine the Thevenin Resistance.

To determine the Thevenin Resistance we need to "zero" sources and then redraw the circuit. In this case we simply replace the voltage source with a short circuit.

The Thevenin Resistance is then equal to the resistance seen across terminals X and Y which is R3 plus R1 in parallel with R2 which we can write as follows:

c) Draw the Thevenin Equivalent Circuit

2. a) Determine the thevenin voltage of the circuit below:

On a first look this circuit looks considerably more complex however, by recalling first principles we can determine the Thevenin Voltage MUCH easier than you think :) Firstly notice that Vs2 presents a voltage of 2V at the node which connects to R3 and R4 which means that we have a voltage difference of 5V across R4 and R5 as 0A flows through the open terminals X and Y. This means that R4 and R5 acts as a simple voltage divider and as R4 and R5 are the same value, it follows that half of the voltage will be dropped across R4 which means that Vth is simply 2.5V. It often pays to analyse a circuit from first principles first before going into a systematic technique!

b) Determine the Thevenin Resistance.

As before we need to "zero" the sources and then redraw the circuit.

By inspection this resistance looks tricky to solve but if we think of this from the perspective of current entering from X and returning to Y then we can make further simplifications. If we were to inject a current into node X from the outside world then current would flow through R5 and return to Y or flow through R4 and then where? Why through the short circuit of course! Which means that we can eliminate all of the circuit elements to the left hand side of R4 for a considerably simpler circuit!

From here Rth is simply R4 in parallel with R5 which is 1kΩ!

c) Draw the Thevenin Equivalent Circuit.

3. a) Determine the thevenin voltage of the circuit below:

This circuit has no obvious simplifications by inspection hence a systematic approach is warranted here. The Thevenin voltage is equal to the voltage across X Y which appears as the second principal node in the circuit. Let's employ Nodal Analysis here and begin by labelling the nodes:

At node Va:

At node Vb:

With two equations and two unknowns we can use the process of elimination to solve for Va and Vb. We can multiply [1] by 9 and then add to [2] to eliminate Vb and solve for Va:

Substiting back into [1] to solve for Vb:

Now Vb represents the open circuit voltage for the load and hence is equal to the Thevenin Voltage resulting in Vth = 14.78V.

b) Determine the Thevenin Resistance.

As before we need to "zero" the sources and then redraw the circuit. We need to replace Vs1 with a short circuit and Is1 with an open circuit:

The Thevenin Resistance is then equal to the resistance seen across terminals X and Y which is R5 in parallel with R3 and R4 plus R1 in parallel with R2 we can write as follows:

Which may look slightly intimidating/confusing on first glance but we can break this down to confirm that it is the case. It is clear that we have R5 in parallel with R3 and R4 but why does the equation show it equal to the parallel combination including R1 and R2? The reason is that the circuit fragment containing R1, R2, R3 and R4 can be simplified to a parallel resistance with R5 by first combining R3 in parallel with R4 and then this is in series with the parallel combination of R1 in parallel with R2 hence we simply add R1||R2 to the above.