What is Mesh Analysis : Procedure and Its Examples

In the domain of electronics, it is more crucial to analyze even simple circuits. For the analysis of simple circuits, principles such as Kirchhoff’s voltage and Kirchhoff’s current law are used. Whereas in the situation of complicated circuits which has multiple controlled voltage and current sources, there have to be additional tools along with KVL and KCL laws. Only with KVL and KCL principles, the analysis turns out to be inaccurate and not reliable too. So, to go with exact analysis and to know the variables in those circuits, approaches like mesh and nodal has to be implemented. With these methods, variables such as current and voltages can be known easily. Let us clearly know about Mesh Analysis, Super mesh analysis in this article.

What is Mesh Analysis?

Mesh is considered as a loop that has no other loops inside the circuit. Here, mesh currents are used as variables in the place of currents in order to find out the entire circuit analysis. Because of this, the technique needs a minimum number of equations to solve. Mesh analysis is implemented in the circuits using Kirchhoff’s voltage law to know the unknown current values.


This is also termed as the mesh current loop technique. After this, voltage values can also be known by the implementation of Ohm’s law. A branch is considered as the path where it connects two nodes and it is included with a circuit element. When a mesh consists of only one branch, then the branch current is termed as the mesh current. Whereas when a mesh consists of two branches, then the mesh current is considered either as the sum or difference of the two mesh loops when they are either in similar or opposite paths.

Steps

  • In knowing the variables of a circuit, there is a procedure to be followed for the implementation of mesh analysis and the steps can be explained as follows:
  • In the first phase, find out the meshes and mark out the mesh currents either in the anti-clockwise or clockwise directions.
  • Look into the amount of current flow that flows through each element corresponding to mesh currents.
  • Write down all the mesh equations for the observed meshes. The mesh equations are written by applying Kirchhoff’s law and then after by applying Ohm’s law
  • To find out the mesh currents, solve the observed mesh equations as per step 3.
  • With this, the flow of current and voltage values across every element in the circuit can be known by the application of mesh currents.

General Form to Set Up Equations in Mesh Analysis

Upon the identification of meshes in the circuit, every consists of one equation. The equations are the total of the voltage drop in the entire loop of the mesh current. In the cases of circuits, which have more than voltage and current, the voltage drop is considered as the impedance of the circuit which is multiplied with the specific loop mesh current.

When the voltage source exists internal to the loop, then the voltage present at the source can be either added up or subtracted based on the condition whether it is the loss of voltage or increase in the voltage for that mesh. But in the condition when the current source does not in between the meshes, then the mesh current will consider either a negative or positive value of the source based on the mesh current source direction.

Mesh Current Method

With the below circuit, mesh current method analysis can be known easily. In the circuit, loop currents I1 and I2 are applied in the clockwise direction


Depending on the direction of the loop current, polarities of the voltage drops takes place at the resistances R1, R2, and R3. Here, I1 and I2 currents will have opposite current flow paths because of the resistor R2 shares both the loops.

So, both the polarities of voltages can be known. Whereas in the practical scenarios, R2 can be categorized as two phases, but the loop currents are especially applicable for analysis applications. There is no impact on the polarities of the voltage sources because they are constant.

Upon the application of Kirchhoff’s voltage law, the below two equations can be written

R2(I1 – I2) + R1I1 = V1 – Derived from loop 1

R2(I2 – I1) + R1I2 = -V2 – Derived from loop 2

The similar terms in the above equations are combined and upon arrangement, the same terms appear at a similar position in every equation. When the loop currents are known, then branch currents can be evaluated. The rearranged equations are:

I1(R1 + R2) – I2R2 = V1 – For Loop 1

-I1R2 + (R2 – R3) I3 = -V2 – For Loop 2

Mesh Analysis Solved Problems

This section shows the solved examples of finding current in a circuit using the mesh current method.

In the below circuit, find out the amount of voltage that is through the 15 Amps current source by the method of mesh analysis. Provided all are current sources

Mesh Analysis To Find Voltage
Mesh Analysis To Find Voltage

As per the circuit, there is the chance of changing the voltage source to current using parallel resistance. To do this, a resistor is placed in series connection with the voltage source and the resistor should possess the same value as of voltage source and the voltage is

Vs = IsRs = 4 * 4 = 16V

Find out the branch currents (I1 and I2) for the loops and signify the current flow directions in both the loops.

Assigning Branch Currents
Assigning Branch Currents

Then, for every mesh (loop), apply KVL law

Mesh – 1

Vx – (I1 – I2) – 18 = 0

Here, I1 = 15

So, Vx +  (6 * I2) = 90

Mesh – 2

18 – 6 (I2 – I1) – 4 * I2 – 16 = 0

I2 = 78/10

= 7.8 Amps

As per the Mesh-1 equation

Vx = 90 – 44.4

Vx = 45.6 V

This is the solved example of Solving Two Meshes using Mesh Current Analysis

Here, we need to find out the voltage and branch currents. Consider the below circuit.

Mesh Current Method Across Three Meshes
Mesh Current Method Across Three Meshes

By the application of KVL law to the first loop, we get

V1 – R2 (I1 – I3) – R4 (I1 – I2) = 0

4 – 2(I1) – 2(I3) – 4 (I1) – 4(I2) = 0

-2(I3) – 6 (I1) = 4

On the application of KVL law to the second mesh, we get

-Vc – R4 (I2 – I1) – R3 (I2 – I3) = 0

-Vc = -4(I1) + 6(I2) – 2 (I3) = 0

As I2 = -2A, we get

-Vc = -4(I1) -12 – 2 (I3) = 0

On the application of KVL law to the third mesh, we get

-R1(I3) – R3(I3 – I2) – R2 (I3 – I1) = 0

Substituting I2 = -2A

2(I1) – 8(I3) = 0

Bu solving the first and third mesh equations, we get

I1 = 4.46 and I3 = -0.615

So, Vc = 28.61V

And branch current is

Iac = I1 – I3

Iac = 5.075 amps

This is the solved example of solving three meshes using Mesh Current Analysis

These are the sample examples solved through mesh analysis. A thorough analysis of this concept allows us to solve complex circuits too.

Super Mesh Analysis

For the analysis of huge and complex circuits, super mesh analysis serves as the best approach than that of Mesh analysis where because in the super mesh there will be two meshes sharing a common component as the current source.

The same technique is followed for supernode circuit analysis as an alternative for nodal circuit analysis because this method streamlines those complex circuits by closing the voltage element and minimizing the number of reference nodes for every voltage source.  In the super mesh analysis, the current source lies inner to the super mesh section, so that one can be able to minimize the meshes by one for every current source those are present.

When the current source is present on the circuit’s permitter, then a single mesh might not be considered. On the other hand, KVL is implemented only for those meshes in the modified electric circuit.

Let us consider an example of super mesh analysis for better understanding.

Using super mesh analysis, find out the values of V3, i1, i2, and i3 for the below circuit?

On the application of KVL to the Mesh-1, we get

10i1 + 80(i1 – i2) + 30 (i1 – i3) = 80

We get 60i1 – 20i2 – 30i3 = 80

By the application of the super mesh technique to Mesh 2 and Mesh 3, we get

30 =40i3+ 30 (i3 – i1) + 20(i2 – i1)

70i3 – 50i1+ 20i2 = 30

The individual current source that is in the super mesh corresponds to the expected mesh currents which are

15ix = i3 – i2

I3 = 15ix + i2

By solving all the above three equations, we get

i1 = 0.58 Amps, i2 = -6.16 Amps and i3 = 2.6 Amps

To find V3, we have v3 = i3 * R3, So

V3 = 2.6 * 40 = 104V

Mesh Analysis Uses

The foremost use of mesh analysis is for resolving planar circuits to know the current values at any position in both the simple and complicated electric circuits

The other usage is that normal computations to solve equations are tough and more mathematical formulas are needed, whereas through mesh analysis fewer computations are enough.

The other usage of Mesh Current analysis is an unbalanced wheat stone bridge. To know this, consider the below example

As the proportions of the resistors, R1/R4 and R2/R5 are not equal, we can understand that there will be some amount of voltage and current flow at R3. As we are aware that solving these types of circuits is complicated by the approach of the general series-parallel technique, we need another approach to solve this.

So, in regard to that, we can go with applying the branch current method, but this method needs six currents from Ia to If which leads to work on any number of equations. So, this complexity can be easily reduced through Mesh current method where this requires only a few variables.

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