- English
- Deutsch
- Português
- Español
- Electrical engineering
- Voltage Current Resistance
- Star Delta Transformation
- Practical Voltage and Current Sources, equivalent circuit diagram
- Capacitor to DC voltage
- Inductors in DC Circuits
- Alternating current
- AC Inductive Circuits
- Three-phase Current
- Transformer
- Complex numbers
- Locus Diagram in AC circuits
- Measurement error
- Videos electrical engineering
- Index electrical engineering

**Home ⇒ Overview Courses ⇒ Electrical engineering ⇒ Equivalent voltage source**

### Thévenin’s (voltage-source) equivalent circuit

Table of Contents

ToggleHere we show how to convert any circuit with a voltage source and several resistors into an equivalent voltage source with only one internal resistor.

Thevenin’s theorem states that any linear circuit, no matter how complex, can be simplified to an equivalent circuit consisting of a single voltage source with only one internal series resistance.

Learning by doing. Here is a simple example:

#### Thévenin's (voltage-source) equivalent circuit – how to use it - example

What we want: To simplify the circuit in pic a and get an equivalent voltage source with only one internal resistor R_{eq} as shown in pic b:

Step 1: Determining the new source voltage V_{eq}

The open-circuit voltage is the voltage that is tapped at terminals A and B when no load current is flowing. In this case, the open-circuit voltage corresponds to the voltage V_{R2}, as no current flows via R_{3}. The open-circuit voltage can therefore be easily determined using the voltage divider formula:

Step 2: Determine the equivalent resistance R_{eq}

To determine the equivalent internal resistance, we eliminate the voltage source by short-circuiting it.

The equivalent internal resistance is the resistance to be measured between terminals A and B. You can see that R_{1} and R_{2} are in parallel and also in series with R_{3}:

We have now determined all the parameters of the equivalent voltage source. If desired, the short-circuit current can be easily determined using Ohm's law:

**I _{max} = V_{eq}**

**. R**

_{eq}#### Exercises

Simplify the following real voltage source according to Thevenin's theorem.

Values: V_{0} = 30 V; R_{1} = 10 Ω; R_{1} = R_{2} = 20 Ω

The circuit would be easier to understand if we redraw it:

Now we proceed as usual:

**Step 1: Determining the new source voltage V _{eq}**

As you can see, the voltage V_{AB} in open circuit (i.e. without current flow) corresponds to the voltage at resistor R_{2}. We apply the voltage divider rule:

**V _{eq }=** V

_{R2}= V

_{R3}= V

_{o}/2 =

**15 V**

**Step 2: Determine the equivalent resistor R _{eq}**

To do this, we eliminate the voltage source by short-circuiting it:

**R _{eq} =** R

_{2}II R

_{3}+ R

_{1}= 20 Ω / 2 + 10 Ω =

**20 Ω**