Mesh Analysis - Supermesh

Mesh analysis works by arbitrarily assigning mesh currents in the essential meshes (also referred to as independent meshes). An essential mesh is a loop in the circuit that does not contain any other loop. Figure 1 labels the essential meshes with one, two, and three.

VS=10V , IS=4A , R1=2Ω , R2=6Ω , R3=1Ω , R4=2Ω .

Solution:
There are three meshes in the circuit. So, we need to assign three mesh currents. It is better to have all the mesh currents loop in the same direction (usually clockwise) to prevent errors when writing out the equations.


A mesh current is the current passing through elements which are not shared by other loops. This is to say that for example the current of the voltage source is I1 , the current ofR2 is I2 and so on. But how about elements shared between two meshes? Current of such elements is the algebraic sum of both meshes. For example if we assume the current of R1is defined with direction from top to bottom, its current would be I1−I2 . If one assume the inverse direction, i.e. from bottom to top, it would be I2−I1 .
Now, lets write the equation for mesh of I1 (Mesh I). A mesh equation is in fact a KVL equation using mesh currents. We start from a point and calculate algebraic sum of voltage drops around the loop:
First the voltage source:
−VS+... 
Now we have reached R1 from its upper node. So its current is I1−I2 and we have:
−VS+R1×(I1−I2)+... 
One important point that you should remember and it always help you to validate the equation is that the current part for resistors is always equal to the current of the mesh that we are writing the equation for minus the current of the other mesh. Of course this is only valid if you respect the convention to

• define all the mesh currents in the same direction
• go around the loop at the same direction as the mesh current that you defined while writing the equation
• Let's continue writing the equation. The next element, which is also the last element, is 
• R3 . Without further thinking we can say that the term associated with this element isR3×(I1−I3) . (Why?)
  −VS+R1×(I1−I2)+R3×(I1−I3)=0 .
  The equation for Mesh I is done. The next mesh is Mesh II. But wait! what is the voltage across the current source to write in the mesh KVL equation? We don't know. There are two ways to resolve this issue:
  Assign a voltage to the current source ( VIS ). Write equations using VIS and later add equations of Mesh II and III to get rid of VIS .
  Write the equation for the Supermesh II & III.
  A supermesh is a larger loop which has both meshes inside.
  Let's try both methods.
  1) Using VIS 
  
• Supermesh
• A supermesh occurs when a current source is contained between two essential meshes. The circuit is first treated as if the current source is not there. This leads to one equation that incorporates two mesh currents. Once this equation is formed, an equation is needed that relates the two mesh currents with the current source. This will be an equation where the current source is equal to one of the mesh currents minus the other. The following is a simple example of dealing with a supermesh.
• Here is the supermesh:
  
  Around the loop clockwise:
  R1×(I2−I1)+R2×(I2)+R4×(I3)+R3×(I3−I1)=0 .
  As you can see, we were able to write the equation in one shot. That is why the supermesh method is preferred.
  
  Now, we have two equations: one for Mesh I and one for the supermesh. But there are three unknowns: I1 , I2 andI3 . So we need another equation. The third equation comes from the current source by writing KCL one of its nodes. We choose the node which is not shared by third loop which is the loop at the right hand side for this example. This way we minimize the number of terms in the equation. Note that the current ofR2 and R4 are I2 and I3 , respectively, but the terms for R2 and R3 are more complicated because of I1 involvement.
  Let's apply KCL for the right hand side node. I2 and IS are entering to the node and I3 is leaving.
  
  −I2−IS+I3=0
  Now we have all three equations:
  ⎧⎩⎨⎪⎪−VS+R1×(I1−I2)+R3×(I1−I3)=0R1×(I2−I1)+R2×(I2)+R4×(I3)+R3×(I3−I1)=0−I2−IS+I3=0
  Let's substitute values:
  VS=10V , IS=4A , R1=2Ω , R2=6Ω , R3=1Ω , R4=2Ω .
  ⎧⎩⎨⎪⎪2(I1−I2)+(I1−I3)=102(I2−I1)+6I2+2I3+I3−I1=0−I2+I3=4 
  ⎧⎩⎨⎪⎪3I1−2I2−I3=10−3I1+8I2+3I3=0−I2+I3=4 
  4.9166
  ⎧⎩⎨⎪⎪I1=4.92AI2=0.25AI3=4.25A 
• With DJ Matildo
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