**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 & 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)

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**.

**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

♦

### Solved Example:

**Calculate and draw the Thévenine equivalent of the following circuit:**

#### 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**.

**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.

- It is an open circuit yes, but picture that we are measuring

**Rth**is the parallel combination of**R1**and**R2**

**Thus the Thévenin equivalent circuit:**

### 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.**

#### 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.

- Using
**Kirchhoff’s Voltage Law**, we go around the loop and write the**EMF**s 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.

**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:

- 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:

- 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**.

**optional reading:** Success in Electronics book by Tom Duncan

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