In mesh analysis, when a current source is present between two mesh then supermesh analysis has to be performed. Mesh analysis stands to be one of the most universal method of solving electrical circuits or networks. It is used to determine currents flowing in individual mesh by forming system of linear equations with help of Kirchhoff’s Voltage Law.
In an electrical circuit mesh is the most fundamental or elementary form of a loop which cannot be further divided into other loops.
Illustration on Super Mesh Analysis
Let us consider an electrical circuit as shown below.
The circuit above consists of three mesh as indicated in the figure itself. It is clear from the figure that a current source IS comes between mesh 1 and 3. So in such cases we have to apply the supermesh analysis.
For super mesh analysis, as current source comes between mesh 1 and 3. Mesh 1 and 3 are considered to be the supermesh.
For solving above circuit using mesh analysis, we have three unknown currents and accordingly we should form three equations.
In case of supermesh analysis, the three equations can be formed in following ways.
First formulate equation of the mesh which is not a super mesh. Such as in above case, mesh 2 does not fall under super mesh and we can easily write KVL (Kirchhoff’s Voltage Law) for this mesh.
So, the KVL equation for mesh 2 is
Secondly, formulate current equation from the current source present between two mesh. Such as in above the equation can be formed from the current source IS present between the mesh 1 and 3.
The equation will be
Thirdly, write the KVL equation for the supermesh. While writing the KVL equation for supermesh do not consider the branch of the circuit which contains the current source. Such as in the above circuit, write KVL equation for super mesh 1 and 3 and while writing the equation do not consider the branch containing the current source IS.
The KVL equation for the super mesh will be
Finally the system of linear equations can be solved to find the unknown mesh currents.
What if the current source is present in the perimeter of any individual mesh?
As shown in the circuit below if the current source is present in the perimeter of any individual mesh then we do not apply the supermesh analysis. Supermesh analysis is to be performed only when the current source is present between two individual mesh.
So in the case of current source present in the perimeter, the mesh current of that particular mesh will be equal to the value of the current source.
Such as in the above circuit the mesh current of mesh 1 i.e. I1 = IS.
And other equations for the mesh 2 and mesh 3 can be formed using KVL.
Examples on Supermesh Analysis
Consider a given circuit where mesh analysis has to be performed.
Here, a current source of 5 Amps is present between mesh I and III. So we have to perform supermesh analysis.
First, the KVL equation for mesh II is
Now, the current of the common boundary of meshes I and III is given by
Then, applying KVL for supermesh I and II we have
Finally, solving equation (i), (ii) and (iii) we have