4/3/2015 Mesh Analysis - Supermesh | Solved Problems

http://www.solved-problems.com/circuits/electrical-circuits-problems/resistive-circuits/1226/mesh-analysis-supermesh/ 1/4

solved-problems.com

http://www.solved-problems.com/circuits/electrical-circuits-problems/resistive-circuits/1226/mesh-analysis-supermesh/

Mesh Analysis - Supermesh

Solve the circuit and find the power of sources:

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

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 of R2 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 R1 is defined with direction from

top to bottom, its current would be I1I2 . If one assume

the inverse direction, i.e. from bottom to top, it would be

I2I1 .

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 I1I2 and we have:

VS+R1(I1I2)+...

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 is R3(I1I3) . (Why?)

VS+R1(I1I2)+R3(I1I3)=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:

4/3/2015 Mesh Analysis - Supermesh | Solved Problems

http://www.solved-problems.com/circuits/electrical-circuits-problems/resistive-circuits/1226/mesh-analysis-supermesh/ 2/4

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

Mesh II:

R1(I2I1)+R2(I2)VIS=0

Note that for R1 , unlike the equation for Mesh I, the

current is I2I1 . This is because we are walking around

the loop with the direction of I2 , or briefly it is because

we are writing the equation for mesh of I2 .

Mesh III:

R3(I3I1)+VIS+R4(I3)=0 .

Let's add two equations:

R1(I2I1)+R2(I2)VIS+R3(I3I1)+VIS+R4(I3)=0+0

Simplifying:

R1(I2I1)+R2(I2)+R3(I3I1)+R4(I3)=0

2) Supermesh

Here is the supermesh:

Around the loop clockwise:

R1(I2I1)+R2(I2)+R4(I3)+R3(I3I1)=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

and I3 . 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 of R2 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.

I2IS+I3=0

Now we have all three equations:

VS+R1(I1I2)+R3(I1I3)=0R1(I2I1)+R2(I2)+R4(I3)+R3(I3I1)=0I2IS+I3=0

Let's substitute values:

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

4/3/2015 Mesh Analysis - Supermesh | Solved Problems

http://www.solved-problems.com/circuits/electrical-circuits-problems/resistive-circuits/1226/mesh-analysis-supermesh/ 3/4

2(I1I2)+(I1I3)=102(I2I1)+6I2+2I3+I3I1=0I2+I3=4

3I12I2I3=103I1+8I2+3I3=0I2+I3=4

4.9166

I1=4.92AI2=0.25AI3=4.25A

The circuit is solved. Any other voltage or current in the circuit can be easily found using mesh currents.

To find power of sources, we need current of the voltage source and voltage across the current source.

For the voltage source, current is equal to I1 as it is located at the unshared part of Mesh I. The current is

entering from the negative terminal. Therefore, the active sign convention should be used to find the sign

of power:

PVS=VSI1=49.2W

4/3/2015 Mesh Analysis - Supermesh | Solved Problems

http://www.solved-problems.com/circuits/electrical-circuits-problems/resistive-circuits/1226/mesh-analysis-supermesh/ 4/4