Knowing how to analyze the various types of electrical circuits, using different methods is essential for the professional of the Electrical World! Thinking to help professionals, students and lovers of electricity to better understand the methods and concepts for analysis of electrical circuits, we created this article explaining in detail what is Kirchhoff’s second law and where we apply Kirchhoff’s second law, in addition to explaining step by step how to use Kirchhoff’s second law. Come on!

Kirchhoff’s laws are generally applied to more complex electrical circuits, such as circuits with more than one source and several components, whether they are in series or in parallel. Created and developed by the German physicist Gustav Robert Kirchhoff, Kirchhoff’s laws have fundamental concepts for the analysis of electrical circuits, from the simplest circuits to the most complex circuits.

To better understand what Kirchhoff’s second law is, it is important to understand what meshes are in electrical circuits. We can define that the grid is a closed path of electrical conductors, where the circulation of the electric current is the same in all points of this grid. Disregarding the association of resistors, the image below shows a parallel series circuit that has three loops, respectively indicated.

## Kirchhoff’s Second Law – Definition

Kirchhoff’s second law, which is also known as mesh law or Kirchhoff’s law for voltages (LKT) basically shows how electrical voltage is distributed across the elements of electrical circuits.

The second Kirchhoff’s Law states that when traversing a loop in a certain direction, starting and reaching the same point, the algebraic sum of the DDP is zero, that is, the sum of the stresses in a loop must be equal to zero.

To facilitate understanding, in the example below we have a series circuit , where the sum of all these voltages must be zero, or the sum of all voltage drops on the resistors must equal the voltage of the source.

## Kirchhoff’s Second Law – Application

Before applying the second law of Kirchhoff in the circuit of the previous image, it is important to highlight that it is necessary to arbitrate the direction that the electric current will travel in the circuit and establish the polarity of all the components of this same circuit.

### Determining the polarity of components

The first step to start the analysis of the circuit is to determine the direction of the electric current in the loop, so that the direction of the electric current is arbitrated equally for all the loops of the circuit to be analyzed, however when determining the direction of the electric current try do this in the most coherent way possible, as in this example, which was chosen to perform the analysis clockwise, as the current tends to flow from the highest electrical potential to the lowest potential.

Performing the analysis of this circuit is relatively simple, but the great tip for determining what we will add or subtract in the expression of the mesh analysis is to make the following considerations:

If the direction of the electrical current that was arbitrated is the same direction as the direction of the electrical current, considering the source of greatest value, we must consider the signal of the resistors as positive in the equation.

However, if the direction of the electric current that was arbitrated is different from the direction of the electric current, considering the source with the highest value, we must consider the signal of the resistors in the load as negative in the equation.

When analyzing the electrical circuit and following the flow of electrical current that was arbitrated, we must consider the signal that the arrow is entering the source, that is, if the arrow is coming from the most positive (positive) end of the source, we will consider adding this source in the equation, however if the arrow is entering the less positive (negative) end we must subtract the source in the equation, as we can see in the image below.

To better understand mesh analysis we will determine the electrical voltage over the resistor R5 of the circuit shown in the image below and check if the value found will be correct, come on!

As seen previously, the first step is to determine the direction of the current arbitrarily, so we will make two examples, arbitrating the direction of the electric current in a clockwise and counterclockwise direction.

### Application – Clockwise

In this example we will arbitrate the direction of the electric current to clockwise, which is the same direction as the electric current, so we will consider the signal of the voltage drops on the resistors as positive in the equation.

When analyzing the circuit, we will have 4V of R1, plus 6V of R2, plus 3V of R3, plus 5V of R4, plus the voltage of R5, which we don’t know. Following the flow of the electric current that was arbitrated, the arrow is entering the less positive (negative) end of the source, so that the source in the equation has the negative sign and this whole sum must be equal to zero, as we can see in the image below.

Now to know the value of the electrical voltage on resistor R5 we must isolate it in the equation, passing all other values to the other side of the equality, exchanging their respective signals. Solving the equation, the voltage drop on resistor R5 is 2V.

To verify that it is correct, just do the same analysis as we did previously, but replacing R5 with 2V and adding all the grid voltages. When performing the calculation we have that the result is equal to zero, as Kirchhoff’s second law states.

### Application – Counterclockwise

We will arbitrate the direction of the electric current to the counterclockwise direction, which in this case is the opposite direction to that of the electric current, so we will consider the sign of the voltage drops on the resistors as negative in the equation.

When analyzing the circuit, we will have the voltage drop on resistor R5, a voltage that we do not know, minus the voltage drops on resistors R4, R3, R2 and R1, with voltages -5V, -3V, -6V and -4V respectively.

Following the flow of the electric current that was arbitrated, the arrow is entering the most positive end of the source, so that the source in the equation has the positive sign and this whole sum must be equal to zero.

In order to know the value of the electrical voltage on resistor R5, we must isolate it in the equation, passing all other values to the other side of the equality, exchanging their respective signals, as was done in the previous resolution. Solving the equation, the voltage drop on resistor R5 is -2V.

To verify that it is correct, just do the same analysis as we did previously, but replacing R5 with -2V and adding all the grid voltages. When performing the calculation we have that the result is equal to zero, as Kirchhoff’s second law states.

To facilitate understanding of Kirchhoff’s second law, the video below the Mundo da Elétrica channel shows a little more about the mesh law.