SUPERPOSITION THEOREM: EXPLANATION, APPLICATIONS, SOLVED EXERCISES - PHYSICAL - 2024

The superposition theorem, in electrical circuits, states that the voltage between two points, or the current through them, is the algebraic sum of the voltages (or currents if it is the case), due to each source, as if each one will act independently.

This theorem allows us to analyze linear circuits that contain more than one independent source, since it is only necessary to calculate the contribution of each one separately.

Linear dependence is decisive for the theorem to apply. A linear circuit is one whose response is directly proportional to the input.

For example, Ohm's law applied to an electrical resistance states that V = iR, where V is the voltage, R is the resistance, and i is the current. It is then a linear dependence of voltage and current in a resistance.

In linear circuits, the superposition principle is applied taking into account the following:

-Each independent voltage source must be considered separately and for this it is necessary to turn off all the others. It is enough to put all those that are not under analysis to 0 V or to replace them in the scheme with a short circuit.

-If the source is current then the circuit must be opened.

-When considering the internal resistance of both current and voltage sources, they must remain in place, forming part of the rest of the circuit.

-If there are dependent sources, they must remain as they appear in the circuit.

Applications

The superposition theorem is used to obtain simpler and easier to handle circuits. But it should always be borne in mind that it only applies to those with linear responses, as stated at the beginning.

So it cannot be used directly to calculate power for example, since power is related to current by:

Since the current is squared, the response is not linear. Nor is it applicable to magnetic circuits in which transformers are involved.

On the other hand, the superposition theorem offers the opportunity to know the effect that each source has on the circuit. And of course, through its application it is possible to solve it completely, that is, to know currents and voltages through each resistance.

The superposition theorem can also be used in conjunction with other circuit theorems, for example Thévenin's, to solve more complex configurations.

In alternating current circuits the theorem is also useful. In this case, we work with impedances instead of resistances, as long as the total response of each frequency can be calculated independently.

Finally, in electronic systems the theorem is applicable for both direct current and alternating current analysis, separately.

Steps to apply the superposition theorem

-Deactivate all independent sources following the instructions given at the beginning, except the one to be analyzed.

-Determine the output, either voltage or current, produced by that single source.

-Repeat the two steps described for all other sources.

-Calculate the algebraic sum of all the contributions found in the previous steps.

Solved exercises

The worked examples below clarify the use of the theorem in some simple circuits.

- Example 1

In the circuit shown in the following figure, find the current through each resistor using the superposition theorem.

Solution

Voltage source contribution

To begin with, the current source is eliminated, which makes the circuit look like this:

The equivalent resistance is found by adding the value of each resistance, since they are all in series:

Applying Ohm's law V = IR and solving for the current:

This current is the same for all resistors.

Contribution of the current source

The voltage source is immediately eliminated, to work only with the current source. The resulting circuit is shown below:

The resistors in the right mesh are in series and can be replaced by a single one:

600 +400 + 1500 Ω = 2500 Ω

The resulting circuit looks like this:

The current of 2 mA = 0.002 A is divided between the two resistors in the figure, therefore the equation of the current divider is valid:

Where I _x is the current in the resistance R _x, R _eq symbolizes the equivalent resistance and I _T is the total current. It is necessary to find the equivalent resistance between both, knowing that:

Thus:

For this other circuit, the current that passes through the 7500 Ω resistor is found by substituting values ​​in the current divider equation:

While the one that passes through the 2500 Ω resistor is:

Application of the superposition theorem

Now the superposition theorem is applied for each resistance, starting with the 400 Ω:

I _{400 Ω} = 1.5 mA - 0.7 mA = 0.8 mA

Important: for this resistance, the currents are subtracted, as they circulate in the opposite direction, as can be seen from careful observation of the figures, in which the directions of the currents have different colors.

This same current flows equally through the 1500 Ω and 600 Ω resistors, since they are all in series.

The theorem is then applied to find the current through the 7500 Ω resistor:

I _{7500 Ω} = 0.7 mA + 0.5 mA = 1.2 mA

Important: in the case of the 7500 Ω resistor, note that the currents add up, because in both circuits they circulate in the same direction when passing through this resistor. Again it is necessary to carefully observe the directions of the currents.

- Exercise 2

Find the current and voltage across the 12 Ω resistor using the superposition theorem.

Solution

Source E ₁ is replaced with a short circuit:

The resulting circuit is drawn in the following way, to easily visualize the resistances that remain in parallel:

And now it is solved by applying series and parallel:

This resistance in turn is in series with the 2 Ω, therefore the total resistance is 5 Ω. The total current is:

This stream is divided as:

Therefore the voltage is:

Now source E ₁ is activated:

The resulting circuit can be drawn like this:

And in series with the 4 Ω there is an equivalent resistance of 40/7 Ω. In this case the total current is:

The voltage divider is applied again with these values:

The resulting current is: 0.5 - 0.4 A = 0.1 A. Note that they have been subtracted, since the current from each source has a different sense, as can be seen in the original circuit.

The voltage across the resistor is:

Finally, the total voltage is: 6V-4.8V = 1.2V

References

1. Alexander, C. 2006. Fundamentals of Electrical Circuits. 3rd. Edition. Mc Graw Hill.
2. Boylestad, R. 2011. Introduction to Circuit Analysis. 2nd. Edition. Pearson.
3. Dorf, R. 2006. Introduction to Electrical Circuits. 7th. Edition. John Wiley & Sons.
4. Edminister, J. 1996. Electrical Circuits. Schaum series. 3rd. Edition. Mc Graw Hill
5. Wikipedia. Current divider. Recovered from: es.wikipedia.org.