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MESH ANALYSIS THEORY

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MESH ANALYSIS THEORY Powered By Docstoc
					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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