# MESH ANALYSIS THEORY

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```					MESH ANALYSIS THEORY
Loop Analysis
 Loop analysis is developed by applying KVL around
loops in the circuit.

 Loop (mesh) analysis results in a system of linear
equations which must be solved for unknown
currents.

EE2 Lect-02                               2
Mesh Analysis: Basic Concepts:

•In formulating mesh analysis we assign a mesh
current to each mesh.

•Mesh currents are sort of fictitious in that a particular
mesh current does not define the current in each branch
of the mesh to which it is assigned.

I1          I2      I3

EE2 Lect-02                                      3
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.

EE2 Lect-02                             4
Identifying the Meshes
1kW         1kW

1kW
V1   +                              +
Mesh 1    Mesh 2       V2
–                              –

EE2 Lect-02                         5
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.

EE2 Lect-02                             6
Assigning Mesh Currents
1kW           1kW

1kW
V1   +                                 +
I1         I2       V2
–                                 –

EE2 Lect-02                            7
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.

EE2 Lect-02                             8
Voltages from Mesh Currents
+    VR   –
+ VR        –              I2
R                    R

I1                     I1

VR = I1 R         VR = (I1 - I2 ) R

EE2 Lect-02                       9
KVL Around Mesh 1
1kW           1kW

1kW
V1   +                                   +
I1         I2         V2
–                                   –

I1 1kW + (I1 - I2) 1kW = V1
I1 2kW - I21kW = V1

EE2 Lect-02                              10
KVL Around Mesh 2
1kW           1kW

1kW
V1   +                                 +
I1         I2       V2
–                                 –

-(I1 ) 1kW + I2 2kW + = -V2
- I11kW + I2 2kW = -V2

EE2 Lect-02                            11
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.

EE2 Lect-02                             12
Matrix Notation
 The two equations can be combined into a single
matrix/vector equation.

 2kW  1kW  I1   V1 
 1kW 2kW  I    V 
           2   2 

EE2 Lect-02                           13
Solving the Equations
Let:           V1 = 7V and V2 = 4V
Results:
I1 = 3.33 mA
I2 = -0.33 mA

EE2 Lect-02                   14
Mesh Analysis: Example 1
Write the mesh equations and solve for the currents I 1, and I2.

4W                 2W

7W
6W
10V +
_         I1            I2
2V +               _
_                   20V
+

Mesh 1        4I1 + 6(I1 – I2) = 10 - 2               Eq (1)

Mesh 2     -6(I1 – I2) + 2I2 + 7I2 = 2 + 20           Eq (2)

EE2 Lect-02                                              15
Mesh Analysis: Example 1, continued.

Simplifying Eq (1) and (2) gives,

10I1 – 6I2 = 8

-6I1 + 15I2 = 22

I1 = 2.2105

I2 = 2.3509

EE2 Lect-02                 16
Mesh Analysis: Example 2
Solve for the mesh currents in the circuit below.
12V
9W
_
_+

+
I3                     8V
10 W                    11 W
+ _

6W                           4W
3W
I1       _                I2
20V +
_                         10V
_                    +

The plan: Write KVL, clockwise, for each mesh. Look for a
pattern in the final equations.

EE2 Lect-02                                         17
12V
9W
_
_+

+
I3                     8V
10 W                   11 W
+ _

6W                           4W
3W
I1      _                I2
20V +
_                         10V
_                    +

Mesh 1:      6I1 + 10(I1 – I3) + 4(I1 – I2) = 20 + 10          Eq (1)

Mesh 2:      -4(I1 – I2) + 11(I2 – I3) + 3I2 = - 10 - 8        Eq (2)

Mesh 3:      9I 3 - 11(I2 – I3) - 10(I1 – I2) = 12 + 8        Eq (3)

EE2 Lect-02                                               18
Clearing Equations (1), (2) and (3) gives,

Standard Equation form          In matrix form:

20I1 – 4I2 – 10I3 = 30        20  4  10  I1      30 
  4 18  11  I     18
-4I1 + 18I2 – 11I3 = -18                     2         
  10  11 30   I 3 
                      20 
    
-10I1 – 11I2 + 30I3 = 20

WE NOW MAKE AN IMPORTANT
OBSERVATION!!

EE2 Lect-02                                         19
Mesh Analysis: By Inspection
Consider the following:

 R11 R12         R13   I1       emfs(1)       20  4  10  I1      30 
  4 18  11  I     18
R                R23   I 2     emfs( 2)                    2         
 21 R22                                         10  11 30   I 3 
                      20 
    
 R31 R32
                 R33   I 3 
           emfs( 3) 
            
12V
9W
R11 =      of resistance around mesh 1,
_
_+

+
common to mesh 1 current I1.                       I3                     8V
10 W                  11 W
+ _
R22 =      of resistance around mesh 2,   6W                       4W
common to mesh 2 current I2.                                                   3W
I1       _                I2


20V +
_                    10V
_               +
R33 =       of resistance around mesh 3,
common to mesh 3 current I3.

EE2 Lect-02                                                                20
Mesh Analysis: By Inspection
R12 = R21 = - resistance common between mesh 1 and 2
when I1 and I2 are opposite through R1,R2.
R13 = R31 = - resistance common between mesh 1 and 3
when I1 and I3 are opposite through R1,R3.
R23 = R32 = - resistance common between mesh 2 and 3
when I2 and I3 are opposite through R2,R3.

 emfs(1) = sum of emf around mesh 1
in the direction of I1.
 emfs( 2) = sum of emf around mesh 2
in the direction of I2.
 emfs( 3) = sum of emf around mesh 3
in the direction of I3.

EE2 Lect-02                                      21
Mesh Analysis: Example 3 – By Inspection
Use the direct method to write the mesh equations for the following.

20 W               30 W              12 W

_                     8W
10 W
20V +
_           I1               I2     + 15V      I3
+                                   _
_    10V              10 W          + 30V

 30  10 0   I 1  10
  10 50  10  I    25
              2   
 0  10 30   I 3  15
                

EE2 Lect-02                                                22
Mesh Analysis: Example 4

2kW

2mA
1kW

+
12V                           2kW    4mA
–
I0

EE2 Lect-02                      23
1. Identify Meshes
2kW

2mA      Mesh 3
1kW

+                       2kW
12V              Mesh 1           Mesh 2   4mA
–
I0

EE2 Lect-02                            24
2. Assign Mesh Currents
2kW

2mA              I3
1kW

+                            2kW
12V                     I1               I2    4mA
–
I0

EE2 Lect-02                                25
Current Sources
 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?

EE2 Lect-02                                 26
Current Sources
 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?

EE2 Lect-02                              27
The
2kW                   Supermesh
The
Supermesh                                                 does not
surrounds             2mA           I3                 include this
this source!                                     1kW       source!

+                             2kW
12V                        I1             I2         4mA
–
I0

EE2 Lect-02                                            28
KVL Around the Supermesh
-12V + I3 2kW + (I3 – I2)1kW + (I1 - I2)2kW = 0

I3 2kW + (I3 - I2)1kW + (I1 - I2)2kW = 12V

I3 3kW - I23kW + I12kW = 12V

EE2 Lect-02                                 29
Matrix Notation
 The three equations can be combined into a single
matrix/vector equation.

 0      1         0   I 1    4mA
 1      0         1  I 2    2mA 
                                  
 2kW  3kW
                 3kW  I 3   12V 
             

EE2 Lect-02                             30
Solution
I1 = 1.2 mA
I2 = -4 mA
I3 = -0.8 mA

EE2 Lect-02                  31
Mesh Analysis: Example 5

20V
2W
_ +

10 W        I3
20 W

10V +
_        I1    5W        I2         4A

15 W

EE2 Lect-02                                    32
This case is explained by using an example.

20V
2W
_ +

10 W        I3
20 W

10V +
_           I1    5W        I2         4A

15 W

When a current source is present, it will be directly related to
one or more of the mesh current. In this case I 2 = -4A.

EE2 Lect-02                                               33
An easy way to handle this case is to remove the current source as shown
below. Next, write the mesh equations for the remaining meshes.

20V
2W
_ +

10 W         I3
20 W

10V +
_        I1     5W        I2

15 W

Note that I 2 is retained for writing the equations through the
5 W and 20 W resistors.

EE2 Lect-02                                                   34
20V
2W
_ +

10 W                 I3
20 W      Equation for mesh 1:

10V +
_       I1           5W          I2         10I1 + (I1-I2)5 = 10
15 W        or

15I1 – 5I2 = 10
Equations for mesh 2:

2I3 - (I2-I3)20 = 20                         Constraint Equation
or
I2 = - 4A
- 20I2 + 22I3 = 20

EE2 Lect-02                                          35
15 5 0   I1  10 
 0 20 22   I    20 
           2  
0
    1  0   I 3   4 
   

I1 = -0.667 A

I2 = - 4 A

I3 = - 2.73 A

EE2 Lect-02                        36

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