Mesh Analysis
Dr. M. K. Uyguroglu
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EENG223 Mesh Analysıs
Mesh Analysis
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Nodal analysis was developed by applying
KCL at each non-reference node.
Mesh analysis is developed by applying KVL
around meshes/loops in the circuit.
Mesh analysis results in a system of linear
equations which must be solved for unknown
currents.
EENG223 Mesh Analysıs
Mesh Analysis
z
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quantity of interest is current
a mesh is a loop that does not contain
another loop within it
work for planar circuit only
planar circuit -> no branch passes over or
under other branch
M-meshes -> assign clockwise current for
each mesh
apply KVL around each mesh
EENG223 Mesh Analysıs
Planar Circuit
Nonplanar Circuit
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EENG223 Mesh Analysıs
Steps of Mesh Analysis
1. Identify meshes.
2. Assign a current to each mesh.
3. Apply KVL around each loop to get an
equation in terms of the loop currents.
4. Solve the resulting system of linear
equations.
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EENG223 Mesh Analysıs
Identifying the Meshes
1kΩ
V1
6
+
–
Mesh 1
1kΩ
1kΩ
Mesh 2
+
–
V2
EENG223 Mesh Analysıs
Steps of Mesh Analysis
1. Identify mesh (loops).
2. Assign a current to each mesh.
3. Apply KVL around each loop to get an
equation in terms of the loop currents.
4. Solve the resulting system of linear
equations.
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EENG223 Mesh Analysıs
Assigning Mesh Currents
1kΩ
V1
8
+
–
1kΩ
1kΩ
I1
I2
+
–
V2
EENG223 Mesh Analysıs
Steps of Mesh Analysis
1. Identify mesh (loops).
2. Assign a current to each mesh.
3. Apply KVL around each loop to get an
equation in terms of the loop currents.
4. Solve the resulting system of linear
equations.
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EENG223 Mesh Analysıs
Voltages from Mesh Currents
+ VR
R
–
I1
VR = I1 R
10
+
R
VR
I2
–
I1
VR = (I1 - I2 ) R
EENG223 Mesh Analysıs
KVL Around Mesh 1
1kΩ
+
V1
+
–
1kΩ
- +
I1
-
1kΩ
I2
+
–
V2
-V1 + I1 1kΩ + (I1 - I2) 1kΩ = 0
I1 1kΩ + (I1 - I2) 1kΩ = V1
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EENG223 Mesh Analysıs
KVL Around Mesh 2
1kΩ
V1
+
–
1kΩ
- +
1kΩ
I1
+
I2
+
–
V2
(I2 - I1) 1kΩ + I2 1kΩ + V2 = 0
(I2 - I1) 1kΩ + I2 1kΩ = -V2
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EENG223 Mesh Analysıs
Steps of Mesh Analysis
1. Identify mesh (loops).
2. Assign a current to each mesh.
3. Apply KVL around each loop to get an
equation in terms of the loop currents.
4. Solve the resulting system of linear
equations.
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EENG223 Mesh Analysıs
Matrix Notation
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The two equations can be combined into a
single matrix/vector equation.
− 1kΩ   I 1   V1 
1kΩ + 1kΩ
=



 − 1kΩ

−
I
V
1
k
1
k
Ω
+
Ω

 2   2 
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EENG223 Mesh Analysıs
Solving the Equations
Let:
Results:
V1 = 7V and V2 = 4V
I1 = 3.33 mA
I2 = -0.33 mA
Finally
Vout = (I1 - I2) 1kΩ = 3.66V
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EENG223 Mesh Analysıs
Another Example
2kΩ
2mA
12V
16
+
–
1kΩ
2kΩ
I0
4mA
EENG223 Mesh Analysıs
1. Identify Meshes
2kΩ
2mA Mesh 3
1kΩ
12V
17
+
–
Mesh 1
2kΩ
Mesh 2
I0
4mA
EENG223 Mesh Analysıs
2. Assign Mesh Currents
2kΩ
2mA
12V
18
+
–
I1
I3
1kΩ
2kΩ
I0
I2
4mA
EENG223 Mesh Analysıs
Current Sources
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The current sources in this circuit will have
whatever voltage is necessary to make the
current correct.
We can’t use KVL around the loop because
we don’t know the voltage.
What to do?
EENG223 Mesh Analysıs
Current Sources
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The 4mA current source sets I2:
I2 = -4 mA
The 2mA current source sets a constraint on
I1 and I3:
I1 - I3 = 2 mA
We have two equations and three unknowns.
Where is the third equation?
EEE 223 Mesh Analysıs
2kΩ
The
Supermesh
surrounds
this source!
12V
2mA
+
–
I3
2kΩ
I2
I1
1kΩ
The
Supermesh
does not
include this
source!
4mA
I0
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EENG223 Mesh Analysıs
KVL Around the Supermesh
-12V + I3 2kΩ + (I3 - I2)1kΩ + (I1 - I2)2kΩ = 0
I3 2kΩ + (I3 - I2)1kΩ + (I1 - I2)2kΩ = 12V
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EENG223 Mesh Analysıs
Matrix Notation
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The three equations can be combined into a
single matrix/vector equation.
1
0
  I 1  − 4mA 
 0
  I  =  2mA 
 1
0
−
1

 2  

2kΩ − 1kΩ − 2kΩ 2kΩ + 1kΩ  I 3   12V 
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EENG223 Mesh Analysı s
Solve Using MATLAB
>> A = [0 1 0; 1 0 -1;
2e3 -1e3-2e3 2e3+1e3];
>> v = [-4e-3; 2e-3; 12];
>> i = inv(A)*v
i = 0.0012
-0.0040
-0.0008
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EENG223 Mesh Analysıs
Solution
I1 = 1.2 mA
I2 = -4 mA
I3 = -0.8 mA
I0 = I1 - I2 = 5.2 mA
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EENG223 Mesh Analysıs
Class Example
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EENG223 Mesh Analysıs