Basic Electronics on the Go - Thevenin’s Theorem

From http://www.electronics-tutorials.ws/dccircuits/dcp_7.html

In this tutorial we will look at one of the more common circuit analysis theorems (next to Kirchhoff´s) that has been developed, Thevenin’s Theorem.

Thevenin’s Theorem states that “Any linear circuit containing several voltages and resistances can be replaced by just one single voltage in series with a single resistance connected across the load“. In other words, it is possible to simplify any electrical circuit, no matter how complex, to an equivalent two-terminal circuit with just a single constant voltage source in series with a resistance (or impedance) connected to a load as shown below.

Thevenin’s equivalent circuit.

  

Consider the circuit from the previous section.

 

Firstly, to analyse the circuit we have to remove the centre 40Ω load resistor connected across the terminals A-B, and remove any internal resistance associated with the voltage source(s). This is done by shorting out all the voltage sources connected to the circuit, that is v = 0, or open circuit any connected current sources making i = 0. The reason for this is that we want to have an ideal voltage source or an ideal current source for the circuit analysis.

The value of the equivalent resistance, Rs is found by calculating the total resistance looking back from the terminals A and B with all the voltage sources shorted. We then get the following circuit.

Find the Equivalent Resistance (Rs)

 

 

The voltage Vs is defined as the total voltage across the terminals A and B when there is an open circuit between them. That is without the load resistor R_L connected.

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:

This current of 0.33 amperes (330mA) is common to both resistors so the voltage drop across the 20Ω resistor or the 10Ω resistor can be calculated as:

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

or

V_AB  =  10  +  (10Ω x 0.33amps)  =   13.33 volts, the same.

Then the Thevenin’s Equivalent circuit would consist or a series resistance of 6.67Ω’s and a voltage source of 13.33v. With the 40Ω resistor connected back into the circuit we get:

and from this the current flowing around 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 circuit analysis tutorial.

Thevenin’s theorem can be used as another type of circuit analysis method and is particularly useful in the analysis of complicated circuits consisting of one or more voltage or current source and resistors that are arranged in the usual parallel and series connections.

While Thevenin’s circuit theorem can be described mathematically in terms of current and voltage, 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, Thevenin’s equivalent circuits of Transistors, Voltage Sources such as batteries etc, are very useful in circuit design.

Thevenin’s Theorem Summary

The basic procedure for solving a circuit using Thevenin’s 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.