Thevenins Theorem

Thevenins Theorem

In the previous three tutorials we have looked at solving complex electrical circuits using Kirchoff’s Circuit Laws, Mesh Analysis and finally Nodal Analysis but there are many more “Circuit Analysis Theorems” available to choose from which can calculate the currents and voltages at any point in a DC circuit. In this tutorial we will look at one of the more common circuit analysis theorems (next to Kirchoff´s) that has been developed, Thevenins Theorem.

Thevenins Theorem states that “Any linear circuit containing several voltages and resistances can be replaced by just a Single Voltage in series with a Single Resistor“. In other words, it is possible to simplify any “Linear” circuit, no matter how complex, to an equivalent circuit with just a single voltage source in series with a resistance connected to a load as shown below.
Thevenins Theorem is especially useful in the Circuit Analysis of power or battery systems and other interconnected circuits were it will have an effect on the adjoining part of the circuit.

Thevenins equivalent circuit.

 

As far as the load resistor R_L is concerned, any “one-port” network consisting of resistive circuit elements and energy sources can be replaced by one single equivalent resistance Rs and equivalent voltage Vs, where Rs is the source resistance value looking back into the circuit and Vs is the open circuit voltage at the terminals.
For example, consider the circuit from the previous section.

 

Firstly, we have to remove the centre 40Ω resistor and short out (not physically as this would be dangerous) all the emf´s connected to the circuit, or open circuit any current sources. The value of resistor Rs is found by calculating the total resistance at the terminals A and B with all the emf´s removed, and the value of the voltage required Vs is the total voltage across terminals A and B with an open circuit and no load resistor Rs connected. Then, we get the following circuit.

Find the Equivalent Resistance (Rs)

Find the Equivalent Voltage (Vs)

 

We now need to reconnect the two voltages back into the circuit, and as V_S  =  V_AB the current flowing around the loop is calculated as:

so the voltage drop across the 20Ω resistor can be calculated as:

V_AB  =  20  –  (20Ω x 0.33amps)  =   13.33 volts.

Then the Thevenins Equivalent circuit is shown below with the 40Ω resistor connected.

 

and from this the current flowing in the circuit is given as:

which again, is the same value of 0.286 amps, we found using Kirchoff´s circuit law in the previous tutorial.
Thevenins theorem can be used as a Circuit Analysis method and is particularly useful if the load is to take a series of different values. It is not as powerful as Mesh or Nodal analysis in larger networks because the use of Mesh or Nodal analysis is usually necessary in any Thevenin exercise, so it might as well be used from the start. However, Thevenins equivalent circuits of Transistors, Voltage Sources such as batteries etc, are very useful in circuit design.

Thevenins Theorem Summary

The basic procedure for solving a circuit using Thevenins Theorem is as follows:

• 1. Remove the load resistor R_L or component concerned.
• 2. Find R_S by shorting all voltage sources or by open circuiting all the current sources.
• 3. Find V_S by the usual circuit analysis methods.
• 4. Find the current flowing through the load resistor R_L.

In the next tutorial we will look at Nortons Theorem which allows a network consisting of linear resistors and sources to be represented by an equivalent circuit with a single current source in parallel with a single source resistance.