Around the 1870’s there was a French gentleman called Léon Charles Thévenin. Before inventing his famous Thevenin’s Theorem, he worked on the development of long distance underground telegraph lines. Over time he became more and more interested in electrical circuits and one day he came up with a method to reduce a complex circuit to a more digestible equivalent circuit. Of course theoretically.

According to Thevenin, a complex circuit can be dumbed down to an equivalent circuit with only 2 components in series: a voltage source and a resistor.

Source and load example using an amplifier and a speaker

Source & Load

To understand more about Thevenin’s Theorem, first let’s divide circuits into to parts: source and load.

  • source – the output part of a circuit (terminals of amp below)
  • load – the input part of another circuit (terminals of speaker)
Source and load example using an amplifier and a speaker

The source comprises of numerous voltage sources, current sources, thousands of resistors. So it is complicated.

Thevenin came up with the idea to substitute the source with an equivalent circuit when we want to do some calculation on the load, or something else connected to the source. This equivalent circuit is a much simplified version of the original source with the same characteristics. Note, that this is all theoretical.

Let’s say we want to do some calculation on the load in the circuit. We really don’t want to have all the complex circuitry of the amp for this. After calculating the equivalent circuit of the amp – which is only a resistor and a voltage source – we can replace it with this circuit and connect the load to it.

Below there is a complex circuit on the left with a load attached (RL). On the right we have the Thevenine equivalent circuit comprising of only two components in series: Vth and Rth.

Demonstrating how using Thévenin’s Theorem simplifies a complex circuit

Any linear circuit consisting of voltage sources, current sources and resistors can be replaced with an equivalent circuit consisting only of a single voltage source and a single resistor in series.

We refer to the single voltage source as Thevenin voltage and the single resistor as Thevenin resistance.

How to find Vth and Rth:

  • Thevenin Voltage (Vth) – to get Vth we disconnect the load (if connected) to have an open circuit and calculate the voltage across the terminals.
  • Thevenin Resitacne (Rth) – the get Rth we short all voltage sources and open circuit all current sources inside the source.

The Thevenin Equivalent Circuit is a Voltage Divider:

  • The complex circuit the load is connected to is now only a voltage source and a resistor.
  • Thus, t is easier to calculate the voltages across the load using the voltage divider equation
Demonstrating how using Thévenin’s Theorem simplifies a complex circuit

Solved Example:

Calculate and draw the Thévenine equivalent of the following circuit:

Example circuit for finding the equivalent circuit using Thévenin’s Theorem

Solution:

For the Thévenin equivalent circuit we need to find the Thévenin voltage Eth and Thévenin resistance Rth.

Finding Eth:

  • To find Eth, we disconnect the load to have an open circuit and calculate the voltage across the terminals.
  • The circuit forms a potential divider, thus, the terminal voltage is the same as the voltage drop across R2. We can use the voltage divider equation to get Vth.
Calculating the Thevenin Voltage using the potetntial divider equation

Finding Rth:

  • To do this, we short circuit the voltage sources and open circuit the current sources if we had any.
  • R1 and R2 are in parallel
    • It is an open circuit yes, but picture that we are measuring Rth from the open terminals. If we consider the terminal branch, R1 and R2 are in parallel.
  • Rth is the parallel combination of R1 and R2
Calculating the Thevenine resistance

Thus the Thévenin equivalent circuit:

Final eqivalent circuit using Thévenin’s Theorem

Solved Example 2:

In an earlier topic, we solved this exercise by using both Kirchhoff’s Law and the Superposition Theorem. Now using Thévenin’s Theorem, find the current in the 4Ohm branch. You should get the same answer.

Example 2 circuit for calculation current in a branch using Thévenin’s Theorem

Solution:

To solve such a task, first we need to consider the 4Ohm resistor as the load.

Finding Rth:

  • To find Eth, we disconnect the load to have an open circuit and calculate the voltage across the terminals A and B (see below).
  • Now we have a loop. To find Eth we will use Kirchhoff’s Voltage Law. For this we need to know the voltage drops across the resistors. And for this we need to find the current in the loop.
Example circuit without load
  • Using Kirchhoff’s Voltage Law, we go around the loop and write the EMFs to the left of the equation and the voltage drops across the resistors to the right.
  • Note, the red arrows are pointing toward the higher potential
  • Starting from E1, we see that E2 is pointing agains the current flow, thus, E2 will be negative and will be subtracted from E1.
  • All voltage drops across the resistors are pointing against the current, thus, they will be added together on the right of the equation.
Finding the current around the loop using Kirchhoff's Voltage Law
  • Vth is the voltage measured across A and B
  • This voltage is also the same as E1 + the voltage drop across r1
  • This voltage is also the same as E2 + the voltage drop across r2
  • We calculate one of them:
Finding the Thevenin voltage
  • Note, using the left branch we would have to subtract the voltage drop across r1 from E1 as they are in the opposite direction.

 

Finding Rth:

  • To find Rth, we short the voltage sources.
  • As before, we end up with r1 and r2 in parallel:
Finding the Tevenin Resistance
  • Now that we have all the pieces of the puzzle, draw the equivalent circuit and put back the load.
  • We have a loop with one voltage source and 2 resistors in series.
  • Finding the current in the loop will give us the current through the 4 Ohm resistor.
Thevenin Equivalent circuit and finding the current flowing through the load

optional reading: Success in Electronics book by Tom Duncan
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