Thévenin’s Theorem: What It Consists Of, Applications And Examples
The Thevenin ‘s theorem states that a circuit with terminals A and B may be substituted by one equivalent consisting of a source and a series resistance whose values give the same potential difference between A and B and the same impedance as the original circuit .
This theorem was made known in 1883 by the French engineer Léon Charles Thévenin, but it is claimed that it was enunciated thirty years earlier by the German physicist Hermann von Helmholtz.
Figure 1. Thévenin’s theorem. Source: self made
Its usefulness lies in the fact that, even when the original circuit is complex or unknown, for the purposes of a load or impedance that is placed between terminals A and B, the simple Thévenin equivalent circuit behaves in the same way as the original .
Article index

one
How is the equivalent voltage calculated step by step?

1.1
– Experimentally

1.2
– Solving the circuit


two
Applications of Thévenin’s theorem (part I)

2.1
Example 1a (calculation of equivalent stress step by step)

2.2
Example 1b (current in the load using the Thévenin equivalent)


3
Proof of Thévenin’s theorem

4
Application of Thévenin’s theorem (part II)

4.1
Example 2a (Thévenin equivalent resistance)

4.2
Example 2b

4.3
Example 2c


5
Application of Thévenin’s theorem (part III)

5.1
Example 3


6
References
How is the equivalent voltage calculated step by step?
How is the equivalent voltage calculated step by step?
The voltage or potential difference of the equivalent circuit can be obtained in the following ways:
– Experimentally
– Experimentally
Obtaining the Thévenin equivalent voltage
If it is a device or equipment that is in a “black box”, the potential difference between terminals A and B is measured with a voltmeter or an oscilloscope. It is very important that no load or impedance is placed between terminals A and B.
A voltmeter or an oscilloscope does not represent any load on the terminals, since both devices have a very large impedance (ideally infinite) and it would be as if terminals A and B were without load.
The voltage or voltage obtained in this way is the Thévenin equivalent voltage.
Obtaining the Thévenin equivalent impedance
To obtain the equivalent impedance from an experimental measurement, a known resistance is placed between terminals A and B and the voltage drop or voltage signal is measured with an oscilloscope.
From the voltage drop in the known resistance between the terminals, the current flowing through it can be obtained.
The product of the current obtained with the equivalent resistance plus the voltage drop measured in the known resistance is equal to the equivalent Thévenin voltage previously obtained. From this equality the equivalent Thévenin impedance is cleared.
– Solving the circuit
– Solving the circuit
Calculation of the Thévenin equivalent voltage
First, any load or impedance is disconnected from terminals A and B.
As the circuit is known, mesh theory or Kirchhoff’s laws are applied to find the voltage at the terminals. This tension will be the Thévenin equivalent.
Calculation of the Thévenin equivalent impedance
To obtain the equivalent impedance we proceed to:
– Replace the voltage sources of the original circuit with short circuits “zero impedance” and the current sources of the original circuit with open ones “infinite impedance”.
– Then the equivalent impedance is calculated following the rules of impedances in series and impedances in parallel.
Applications of Thévenin’s theorem (part I)
Applications of Thévenin’s theorem (part I)
We will apply Thévenin’s theorem to solve some circuits. In this first part we consider a circuit that only has voltage sources and resistors.
Example 1a (calculation of equivalent stress step by step)
Example 1a (calculation of equivalent stress step by step)
Figure 2 shows the circuit that is in a celestial box which has two electromotive force batteries V1 and V2 respectively and resistors R1 and R2, the circuit has terminals A and B in which a load can be connected.
Figure 2. Example 1 of Thévenin’s theorem. Source: self made
The goal is to find the Thévenin equivalent circuit, that is, to determine the Vt and Rt values of the equivalent circuit.
Apply the following values: V1 = 4V, V2 = 1V, R1 = 3Ω, R2 = 6Ω, and R = 1Ω.
Step by step solution
Step 1
We will determine the voltage across terminals A and B when no load is placed on them.
Step 2
The circuit to be solved consists of a single mesh through which a current I circulates, which we have taken to be positive in the clockwise direction.
Step 3
We go through the mesh starting with the lower left corner. The path leads to the following equation:
V1 – I * R1 – I * R2 – V2 = 0
Step 4
We solve for the mesh current I and obtain:
I = (V1 V2) / (R1 + R2) = (4V – 1V) / (3Ω + 6Ω) = ⅓ A
Step 5
With the mesh current we can determine the voltage difference between A and B, which is:
Vab = V1 – I * R1 = 4V – ⅓ A * 3Ω = 3V
That is
Thevenin equivalent voltage is: Vt = 3V
.
Step 6 (Thévenin equivalent resistance)
We now proceed to calculate the Thévenin equivalent resistance, for which and as previously mentioned, the voltage sources are replaced by a cable.
In that case, we only have two resistors in parallel, so the Thévenin equivalent resistance is:
Rt = (R1 * R2) / (R1 + R2) = (3Ω * 6Ω) / (3Ω + 6Ω) =
2Ω
Example 1b (current in the load using the Thévenin equivalent)
Example 1b (current in the load using the Thévenin equivalent)
Connect as a load to terminals A and B a resistance R = 1Ω to the equivalent circuit and find the current that flows through said load.
Solution
When resistance R is connected to the Thevenin equivalent circuit, we have a simple circuit consisting of a source Vt a resistance Rt in series with resistance R.
We will call the current flowing through the load R Ic, so that the mesh equation looks like this:
Vt – Ic * Rt – Ic * R = 0
from which it follows that Ic is given by:
Ic = Vt / (Rt + R) = 3V / (2Ω + 1Ω) = 1 A
Proof of Thévenin’s theorem
Proof of Thévenin’s theorem
To verify that Thévenin’s theorem holds, connect R to the original circuit and find the current flowing through R by applying the mesh law to the resulting circuit.
The resulting circuit remains and its mesh equations remain as shown in the following figure:
Figure 3. Mesh currents. (Own elaboration)
By adding the mesh equations, it is possible to find the mesh current I1 as a function of the current I2. Then it is substituted into the second mesh equation and an equation is left with I2 as the only unknown. The following table shows the operations.
Figure 4. Details of the operations. (Own elaboration)
Then the values of the resistance and voltages of the sources are substituted, obtaining the numerical value of the mesh current I2.
Figure 5. Detail of results. (Own elaboration)
The mesh current I2 is the current that flows through the load resistance R and the value found of 1 A fully coincides with that previously found with the Thévenin equivalent circuit.
Application of Thévenin’s theorem (part II)
Application of Thévenin’s theorem (part II)
In this second part, the Thévenin theorem will be applied in a circuit that has voltage sources, current sources and resistances.
Example 2a (Thévenin equivalent resistance)
Example 2a (Thévenin equivalent resistance)
The objective is to determine the Thévenin equivalent circuit corresponding to the circuit in the following figure, when the terminals are without the resistance of 1 ohm, then the resistance is placed and the current flowing through it is determined.
Figure 6. Circuit example 2. (Own elaboration)
Solution
To find the equivalent resistance, remove the load resistance (in this case the 1 ohm). Also, voltage sources are replaced by a short circuit and current sources by an open circuit.
In this way, the circuit for which the equivalent resistance will be calculated is the one shown below:
Figure 7. Detail for the calculation of the equivalent resistance (Own elaboration)
Rab = (12Ω * 4Ω) / (12Ω + 4Ω) = 3Ω which is the Thevenin equivalent resistance (Rth).
Example 2b
Example 2b
Calculate the Thévenin equivalent voltage.
Solution
To calculate the Thévenin equivalent voltage we consider the following circuit, in which we will place the currents in I1 and I2 in the branches indicated in the following figure:
Figure 8. Details for the Thévenin stress calculation. (Own elaboration)
The previous figure shows the equation of the current nodes and the equation of voltages when the external mesh is traversed. From the second of the equations the current I1 is cleared:
I1 = 2 – I2 * (5/3)
This equation is substituted in the equation of the nodes:
I2 = 2 – (5/3) I2 + 2 ===> I2 (8/3) = 4 ===> I2 = 12/8 = 1.5 A
This means that the voltage drop across the 4 ohm resistor is 6 volts.
In short, the Thévenin voltage is Vth = 6 V.
Example 2c
Example 2c
Find the Thevenin equivalent circuit and current in the load resistor.
Figure 9. Current in the load with Thévenin equivalent. (Own elaboration)
Solution
The previous figure shows the Thévenin equivalent circuit with the load resistance R. From the voltage equation in the mesh, the current I that flows through the load resistance R.
I = Vth / (Rth + R) = 6V / (3Ω + 1Ω) = 1.5 A
Application of Thévenin’s theorem (part III)
Application of Thévenin’s theorem (part III)
In this third part of the application of Thévenin’s theorem, an alternating current circuit is considered that contains an alternating voltage source, a capacitor, an inductance and a resistance.
Example 3
Example 3
The objective is to find the Thévenin Circuit equivalent to the following circuit:
Figure 10. Thévenin in an alternating current circuit. (Own elaboration)
Solution
The equivalent impedance corresponds to that of the capacitor in parallel with the series combination of the resistance and inductance.
The inverse of the equivalent impedance is given by:
Zeq ^ 1 = (5j) ^ – 1 + (5 + 5j) ^ – 1 = (1/5) j + ((1/10 + (1/10) j) = (1/10 + 3 / 10 j) Mho
And the equivalent impedance will then be:
Zeq = (1 – 3 j) Ohm
The complex current I can be derived from the mesh equation:
50V∠0 – I (5 j + 5 + 5j) = 50V∠0 – I * 5 = 0 ===> I = 10A ∠0
Now the voltage drop in the resistance plus the inductance is calculated, that is to say the voltage Vab which will be the equivalent Thévenin voltage:
Vab = I * (5 + 5 j) Ω = 10A ∠0 * 5Ω∠45º = 50V∠45º
In other words, the equivalent voltage has the same peak value of the original source but is 45 degrees out of phase: Vth = 50V∠45º
References
References

Electronics tutorials, Thevenin’s theorem. Recovered from: electronicstutorials.ws

Network theory questions and answers. Thevenin’s theorem. Recovered from: sanfoundry.com

Thevenin’s theorem. Step by step procedure. Recovered from: electricaltechnology.org

Thevenin’s theorem. Solved example step by step. Recovered from: electricalsimple.blogspot.com

Workshop on Thevenin’s and Norton’s theorems. Recovered from: web.iit.edu

Wikipedia. Thévenin’s theorem. Recovered from: wikipedia.com