Basic Electrical Engineering Series

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Mesh Analysis

Simple mesh circuits restricted to two meshes can be solved by applying Kirchoff’s laws to obtain two simultaneous equations. The two equations are then solved using standard elimination and back substitution.

For more complex circuits it may be possible to reduce the complexity using for example Thevenins or Nortons theorem. Also it may be possible to simplify a circuit using the general Star-Mesh transform (Rosen’s Theorem). In attempting a circuit analysis problem, the amount of work involved should always be considered and the best method adopted accordingly.

A more general method of solving multi-mesh problems is the subject of this tutorial – mesh analysis. An N mesh circuit will involve deriving a set of N simultaneous equations, the solution of which can be tackled in a veriaty of ways using Matrix algebra. Two methods will be considered: the use of determinants and Cramers Rule and the matrix inversion method.

But first we must define a general method of deriving the mesh equations.

Maxwell’s Circulating Currents

Consider the three-mesh circuit below. We can label each mesh with a current that has to be solved for the circuit. These are known as Maxwell’s circulating currents and can be drawn clockwise or anti-clockwise – however we shall use the convention always draw the currents clockwise. Recall that a current flowing into a component will cause a potential difference across that component with the sense arrow being at the end were the current enters (i.e. a voltage rise). By convention this is the most positive end.

We can now apply Kirchoffs Voltage Law (KVL) to each mesh and so derive a set of equations from these circulating currents. Note that it doesn’t matter if the currents are labeled in the wrong sense. As long as we follow the sign convention then the algebra will take care of itself and all will be correct.

However to be consistent, apply the currents in a clockwise direction. Note that for N meshes we will obtain N simultaneous equations. So lets do an example and derive the equations using KVL in each case.

Mesh 1

25I₁ + 60I₁ – 60I₂ - 25I₃ – 10 = 0

85 I₁ – 60 I₂ - 25 I₃ = 10 ………. 1

Mesh 2

60 I₂ + 40 I₂ - 60 I₁ - 40 I₃ + 20 = 0

- 60 I₁ + 100 I₂ - 40 I₃ = - 20 ……….. 2

Mesh 3

40 I₃ + 25 I₃ + 10 I₃ – 25 I₁ – 40 I₂ = 0

-25 I₁ - 40 I₂ + 75 I₃ = 0 ……….. 3

So we have our three mesh equations from the circulating currents in the three meshes. Note we write them down in a certain order

A I₁ + B I₂ + C I₃ = D

We can represent these simultaneous equations as a matrix equation as follows:

[Z]*[I] = [E] Ohms Law

That is, an impedance matrix multiplied by a column vector of unknown currents is equal to a column vector of applied emf’s, as follows:

Recall the numbering convention for a matrix, Z_i,j that is the i_th row and j_th column as follows:

The subscripts represent the meshes, therefore Z11 represents the self impedance of mesh 1, that is the sum of the impedances in mesh 1. Whereas Z21 represents the impedance on the boundary of mesh 2 and mesh 1.

Note well the following rules:

The numbers on the diagonal of the matrix are positive. These are the mesh self-impedance and are just the sum of the impedences in each mesh.

The numbers off the diagonal represent the total impedences from one mesh with respect to another i.e. Z₁₂, Z₁₃ etc. Note there are always two i.e. the impedance from mesh one to mesh two is the same as from mesh 2 to mesh 1. Hence for example Z₃₁ is the same as Z₁₃.

The numbers from one mesh with respect to another are always negative.

Note that the emf are positive when aiding a circulating current i.e. on the LHS of the mesh and negative when opposing a circulating current i.e. on the RHS of a mesh.

These are the rules that must be applied to any mesh and will always result in a correct solution.

The next step is that we need to solve this matrix equation for the three unknown currents.

There are many ways to do these all-involving varying amounts of work. But note this general technique of mesh analysis is amenable to solution by computer and is the basis of most circuit analysis packages.

We will look at two techniques. The first uses determinants and Cramers rule.

Solution using Cramers Rule

The next step is to form the determinant of the three equations. A determinant is a number, which is determined from the array of the coefficients of the unknown currents, as follows:

The first thing to note is that the determinant can be arrived at by expanding along any row or column.

Let’s choose column 1 to illustrate the procedure.

First we choose the first number of the column (85). We then obtain the ‘minor’ by covering up the row and the column that the number lies upon. The array that is left is the minor, as follows

The minor is evaluated by multiplying the diagonals and subtracting:

i.e. (100 x 75) – (-40 x -40).

This result is called the co-factor and this then multiplied by the number 85.

The procedure is repeated for each number in the column or row chosen.

Another thing to note is that the sign of the number in the determinant has a pattern as follows:

So to complete the exercise lets evaluate the determinant of the above along column 1.

This is:

D_R = (+) (85) [(100)(75) – (-40)(-40)]

(-) (-60) [(-60)(75) – (-25)(-40)]

(+) (-25) [(-60)(-40) – (-25)(100)]

Note the signs (+ - +) !!

D_R = 49000

As an exercise repeat the procedure along column 2 and row 3, just to check it’s the same.

The next step is to replace the coefficients of I₁ with the numbers on the right hand side of the equation, that is the column vector of applied emf's, as follows:

And then find this determinant D₁

Choosing Column 1 then, we obtain

D₁ = (+) (10) [(100).(75) – (-40).(-40)]

(-) (-20) [(-60).(75) – (-25).(-40)]

(+) (0) [ ]

D₁ = - 51000

Note the zero in column 1 helps us in reducing the amount of arithmetic – so always choose the one with a zero in it if one exists!!

The next step is to replace the coefficients of I₂ with the numbers on the right hand side of the equation, as follows:

And then find this determinant D₂

Choosing Column 2 then, we obtain

D₂ = ( -) (10) [(-60).(75) – (-40).(-25)]

(+) (-20) [(85).(75) – (-25).(-25)]

(-) (0) [ ]

D₂ = -60000

Again the zero in column 2 helps us in reducing the amount of arithmetic.

The next step is to replace the coefficients of I₃ with the numbers on the right hand side of the equation, as follows:

And then find this determinant D₃

Choosing Column 3 then, we obtain

D₃ = (+) (10) [(-60)(-40) – (100)(-25)]

( -) (-20) [(85)(-40) – (-60)(-25)]

(+) (0) [ ]

D₃ = - 49000

Finally we now apply Cramers Rule, as follows:

I₁ = D1/D_R = -1.04 A

I₂ = D2/D_R = -1.22 A

I₃ = D3/D_R = -1 A

And of course the minus signs just mean that the current is going in the opposite direction to that labeled.

Solution by Matrix Inversion

We have seen that the system of mesh equations results in the following:

[Z]*[I] = [E] Ohms Law

This can also be solved as follows:

[I] = [Z]^-1 [E]

Which involves finding the inverse matrix [Z]^-1

There are several methods of doing this. One method is as follows:

Where C^T is the adjoint of the matrix Z and ¦Z¦ is the determinant of the matrix Z.

The steps are as follows:

The determinant of Z

We have already done this in the other method as follows:

D_R = (+) (85) [(100).(75) – (-40).(-40)]

(-) (-60) [(-60).(75) – (-25).(-40)]

(+) (-25) [(-60)(-40) – (-25)(100)]

D_R = 49000 = ¦Z¦

Form a new matrix C consisting of the co-factors of each of the elements of Z i.e. evaluate the minor of each element.

Form the transpose of C, that is, swap rows and columns

Which in this case is just the same!

Find Z^-1 as follows

Finally [I] = [Z]^-1 [E]

As before

Note that as the number of meshes grows, so does the amount of work. There are many other methods of solving sets of equations, e.g. row transformation method, Gaussian elimination, Gauss-Sidel, Eigen vector method etc.

Using kirchoff’s current law, we can now label the circuit with all the currents

Exercise Example

Solve the above Circuit for each branch current using both Cramers Rule and the Matrix Inversion Method.