MESH ANALYSIS: CONCEPTS, METHODS, EXAMPLES - PHYSICAL - 2024

The mesh analysis is a technique used to solve electrical circuits planes. This procedure may also appear in the literature as the method of circuit currents or the method of mesh (or loop) currents.

The foundation of this and other electrical circuit analysis methods is in Kirchhoff's laws and Ohm's law. Kirchhoff's laws, in turn, are expressions of two very important principles of conservation in Physics for isolated systems: both the electric charge and the energy are conserved.

Figure 1. Circuits are part of countless devices. Source: Pixabay.

On the one hand, electrical charge is related to current, which is charge in motion, while in a circuit energy is linked to voltage, which is the agent in charge of doing the work necessary to keep the charge moving.

These laws, applied to a flat circuit, generate a set of simultaneous equations that must be solved to obtain the current or voltage values.

The system of equations can be solved with already known analytical techniques, such as Cramer's rule, which requires the calculation of determinants to obtain the solution of the system.

Depending on the number of equations, they are solved using a scientific calculator or some mathematical software. There are also many options available online.

Important terms

Before explaining how it works, we will start by defining these terms:

Branch: section that contains an element of the circuit.

Node: point that connects two or more branches.

Loop: is any closed portion of a circuit, which begins and ends at the same node.

Mesh: loop that does not contain any other loop inside (essential mesh).

Methods

Mesh analysis is a general method used to solve circuits whose elements are connected in series, in parallel or in a mixed way, that is, when the type of connection is not clearly distinguished. The circuit must be flat, or at least it must be possible to redraw it as such.

Figure 2. Flat and non-flat circuits. Source: Alexander, C. 2006. Fundamentals of Electrical Circuits. 3rd. Edition. Mc Graw Hill.

An example of each type of circuit is shown in the figure above. Once the point is clear, to begin, we will apply the method to a simple circuit as an example in the next section, but first we will briefly review the laws of Ohm and Kirchhoff.

Ohm's Law: let V be the voltage, R the resistance and I the current of the ohmic resistive element, in which the voltage and the current are directly proportional, the resistance being the constant of proportionality:

Kirchhoff's Law of Voltage (LKV): In any closed path traveled in only one direction, the algebraic sum of the voltages is zero. This includes voltages due to sources, resistors, inductors, or capacitors: ∑ E = ∑ R _i. I

Kirchhoff's law of current (LKC): at any node, the algebraic sum of the currents is zero, taking into account that the incoming currents are assigned one sign and those that leave another. In this way: ∑ I = 0.

With the mesh current method it is not necessary to apply Kirchhoff's current law, resulting in fewer equations to solve.

- Steps to apply mesh analysis

We will start by explaining the method for a 2 mesh circuit. The procedure can then be extended for larger circuits.

Figure 3. Circuit with resistors and sources arranged in two meshes. Source: F. Zapata.

Step 1

Assign and draw independent currents to each mesh, in this example they are I ₁ and I ₂. They can be drawn either clockwise or counterclockwise.

Step 2

Apply Kirchhoff's Law of Tensions (LTK) and Ohm's law to each mesh. Potential falls are assigned a sign (-) while rises are assigned a sign (+).

Mesh abcda

Starting from point a and following the direction of the current, we find a potential rise in battery E1 (+), then a fall in R ₁ (-) and then another fall in R ₃ (-).

Simultaneously, the resistance R ₃ is also crossed by the current I ₂, but in the opposite direction, therefore it represents a rise (+). The first equation looks like this:

Then it is factored and terms are regrouped:

---------

-50 I ₁ + 10I ₂ = -12

Since it is a 2 x 2 system of equations, it can be easily solved by reduction, multiplying the second equation by 5 to eliminate the unknown I ₁:

-50 I ₁ + 10 I ₂ = -12

Immediately the current I ₁ is cleared from any of the original equations:

The negative sign in the current I ₂ means that the current in mesh 2 circulates in the opposite direction to that drawn.

The currents in each resistor are as follows:

The current I ₁ = 0.16 A flows through the resistance R ₁ in the direction drawn, through the resistance R ₂ the current I ₂ = 0.41 A flows in the opposite direction to the one drawn, and through the resistance R ₃ flows i ₃ = 0.16- (-0.41) A = 0.57 A down.

System solution by Cramer's method

In matrix form, the system can be solved as follows:

Step 1: Calculate Δ

The first column is replaced by the independent terms of the system of equations, maintaining the order in which the system was originally proposed:

Step 3: Calculate I

Step 4: Calculate Δ

Figure 4. 3-mesh circuit. Source: Boylestad, R. 2011. Introduction to Circuit Analysis.2da. Edition. Pearson.

Solution

The three mesh currents are drawn, as shown in the following figure, in arbitrary directions. Now the meshes are traversed starting from any point:

Figure 5. Mesh currents for exercise 2. Source: F. Zapata, modified from Boylestad.

Mesh 1

-9100.I ₁ + 18-2200.I ₁ + 9100.I ₂ = 0

Mesh 3

System of equations

Although the numbers are large, it can be solved quickly with the help of a scientific calculator. Remember that the equations must be ordered and add zeros in the places where the unknown does not appear, as it appears here.

The mesh currents are:

The currents I ₂ and I ₃ circulate in the opposite direction to that shown in the figure, since they turned out to be negative.

Table of currents and voltages in each resistance

Resistance (Ω)	Current (Amps)	Voltage = IR (Volts)
9100	I 1 –I 2 = 0.0012 - (- 0.00048) = 0.00168	15.3
3300	0.00062	2.05
2200	0.0012	2.64
7500	0.00048	3.60
6800	I 2 –I 3 = -0.00048 - (- 0.00062) = 0.00014	0.95

Cramer's rule solution

Since they are large numbers, it is convenient to use scientific notation to work with them directly.

Calculation of I ₁

The colored arrows in the 3 x 3 determinant indicate how to find the numerical values, multiplying the indicated values. Let's start by getting those of the first bracket in the determinant Δ:

(-11300) x (-23400) x (-10100) = -2.67 x 10 ¹²

9100 x 0 x 0 = 0

9100 x 6800 x 0 = 0

Immediately we obtain the second bracket in that same determinant, which is worked from left to right (for this bracket the colored arrows were not drawn in the figure). We invite the reader to verify it:

0 x (-23400) x 0 = 0

9100 x 9100 x (-10100) = -8.364 x 10 ¹¹

6800 x 6800 x (-11300) = -5.225 x 10 ¹¹

Similarly, the reader can also check the values ​​for the determinant Δ ₁.

Important: between both brackets there is always a negative sign.

Finally the current I ₁ is obtained through I ₁ = Δ ₁ / Δ

Calculation of I ₂

The procedure can be repeated to calculate I ₂, in this case, to calculate the determinant Δ ₂ the second column of the determinant Δ is replaced by the column of the independent terms and its value is found, according to the procedure explained.

However, as it is cumbersome due to large numbers, especially if you do not have a scientific calculator, the simplest thing is to substitute the already calculated value of I ₁ in the following equation and solve for:

Calculation of I3

Once with the values ​​of I ₁ and I ₂ in hand, that of I ₃ is found directly by substitution.

References

1. Alexander, C. 2006. Fundamentals of Electrical Circuits. 3rd. Edition. Mc Graw Hill.
2. Boylestad, R. 2011. Introduction to Circuit Analysis.2da. Edition. Pearson.
3. Figueroa, D. (2005). Series: Physics for Science and Engineering. Volume 5. Electrical Interaction. Edited by Douglas Figueroa (USB).
4. García, L. 2014. Electromagnetism. 2nd. Edition. Industrial University of Santander.
5. Sears, Zemansky. 2016. University Physics with Modern Physics. 14th. Ed. Volume 2.