1. Topic 12
Mesh Analysis
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2. Mesh Currents R1 R2 R3 R4 R5 R6 R7 R8 vs
Remember our planar circuit example? R7 R1 R2 R3 R4 R5 R6 R8 vs M4
Mesh analysis can only be done on planar circuits M2
(the same restriction does not apply to node analysis) M1 M3
How many meshes?
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Mesh Method

3. Mesh Currents R7 R1 R2 R3 R4 R5 R6 R8 vs im1 im2 im3 im4 M1 M2 M3 M4
Imagine each of these meshes as having a circulating current
Here we have counterveiling currents
im1
im2
In this region we just have im1
In this region we just have im2
Net current down is im1-im2
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Mesh Method

4. Branch Currents R7 R1 R2 R3 R4 R5 R6 R8 vs i7 i1 im1 im2 im3 im4 i4 i8
Runs through vs, R1 & R2 i6 i5
Once we know all the mesh currents, we know all the branch currents (and we can find them by inspection)
Runs through R7
Runs through R4
Runs through R5
Runs through R6
Runs through R3
Runs through R8
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Mesh Method

5. Mesh Current Method ia ib 10 5 2k 5k 8k
Label the mesh currents including their direction
Write KVL around the meshes in terms of the mesh currents 1 2 1 2 1 2
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Mesh Method

6. Method Identify Law In terms of
Method Comparison
Method
Identify
Law
In terms of
Node-Voltage
critical nodes
KCL at crit. nodes
node voltages
Mesh-Current
meshes
KVL around meshes
mesh currents
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Mesh Method

7. Prelab 3, part 2.2b v1 + - v3 - + R1 R2 R3 R4 R5 vs1 vs2 i1 i3
Find the voltages and currents indicated using the mesh-current method ib ic ia v4 - + i4
Define the mesh currents
Write KVL around each mesh 1
vs1=12 v; vs2=6 v;
R1=820Ω; R2=330Ω;
R3=100Ω; R4=680Ω;
R5=120Ω; 2 3 1 2 3
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Mesh Method

8. 1 2 3 4
from 1 5
from 3 4 5
into 2
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Mesh Method

9. Find the voltages and currents indicated using the mesh-current method
Prelab 3, part 2.2b v1 + - v3 - + R1 R2 R3 R4 R5
vs1
vs2 i1 i3
Find the voltages and currents indicated using the mesh-current method
9.3 v4 - +
-3.8
10.7 i4
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Mesh Method

10. 5 26 90 8 80 30
Assessment 4.7 ib ia ic
Use the mesh method to find the power delivered by the source and the power absorbed by the 8Ω resistor
Define the mesh currents
Write KVL around each mesh
collecting terms 1 2 3
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Mesh Method

11. 1 2 3
25 x 1 + 3 4
25 x 2
+18 x 3 5
& 4 5 + 5
&
from 1
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Mesh Method

12. 5 26 90 8 80 30
Assessment 4.7 ib ia ic
Use the mesh method to find the power delivered by the source and the power absorbed by the 8Ω resistor ib ia
Current out of source is ia
Current through the 8Ω resistor is ib
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Mesh Method

13. Dependent Sources (Assessment 4.2)5 3 1 25 14 10
-3vφ vφ + -
Use mesh currents to find how much power is being delivered to the dependent source ib ia ic
Write KVL around each mesh in terms of the mesh currents 1 2 3 1 4 4 2
into 5 5 3
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Mesh Method

14. Dependent Sources (Assessment 4.2)5 3 1 25 14 10
-3vφ vφ + -
Dependent Sources (Assessment 4.2)
Use mesh currents to find how much power is being delivered to the dependent source ib ia ic -1 12 1 5 4 3 3
It is straightforward to solve the algebra, giving
ia = 4 A, ib = -1 A, ic = 3 A
& vφ = 12 v
triple negative!
ib is defined to go uphill but power absorbed is downhill power
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Mesh Method

15. 3 5 2 50
100 10 6 4 ic ia ib
The Supermesh
The current source gives us a problem (like a floating voltage source in the node method)
Notice that the emphasis is on being able to write KVL around the super mesh in terms of the mesh currents and voltage sources by inspection.
When we write KVL around meshes a and b we don’t know the voltage across the current source
But we do know
Since KVL works around any loop
And write KVL around the bigger loop (still using ia and ic)…
We mentally remove the current source
…well, actually just ia since ic =5+ia
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Mesh Method

16. 3 5 2 50
100 10 6 4 ic ia ib
The Supermesh
So now we have ic in terms of ia and one equation in ia and ib
We get our last equation from mesh b 1 2
Adding a resistor in series with the current source would not change the analysis at all
Gather terms… 3 2
equation 1 would be the same 3
KVL around the supermesh and mesh b would be the same 3 3x + 2 4
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Mesh Method

17. Assessment 4.10 3 2 8 16 30 6 4 5 ib ia ic
Use the mesh method to find the power dissipated in the 2Ω resistor
Identify the mesh currents
No. Since ic is the only mesh current through the branch with the current source, we know ic=16
Do we have a supermesh?
Leaves us only two to work with 1 1 2 2
These are readily solved as
&
The current through the 2Ω resistor is: So
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Mesh Method

18. Node vs Mesh Analysis How many essential nodes are there?
How many meshes are there?
Are there any supernodes which will reduce the no. of equations?
Are there any supermeshes which will reduce the no. of equations?
What exactly is it you need to solve? Will one of the methods give you that more directly or quickly?
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Mesh Method

19. Assesment 4.13: Node or Mesh?
Find the power delivered by the 2A current source
15Ω
10Ω
20v 2A
25v v1 20 25
two essential nodes
one node equation
two meshes
one supermesh
but
One node equation a little easier than one supermesh equation?
Current is uphill relative to the voltage, so…
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Mesh Method

20. Assesment 4.14: Node or Mesh?
Find the power delivered by the 4A current source 4 4Ω 3Ω
four essential nodes
three node equations ix
30ix 6Ω ib
128v ia
three meshes
one known so two equations 2Ω 5Ω
but
KVL around mesh a
around mesh b
Dependent current
Solving
Voltage across 4A
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Mesh Method