Nodal Analysis Theory by roshan.iesl

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									    LECT 06


Nodal Analysis Theory




  EE2 Lec 03
           Nodal Analysis: The Concept.



• Every circuit has n nodes with one of the nodes being
  designated as a reference node.

• We designate the remaining n – 1 nodes as voltage nodes
  and give each node a unique name, vi.

• At each node we write Kirchhoff’s current law in terms
  of the node voltages.



            EE2 Lec 03                                    2
          Nodal Analysis: The Concept.



• We form n-1 linear equations at the n-1 nodes
  in terms of the node voltages.

• We solve the n-1 equations for the n-1 node voltages.

• From the node voltages we can calculate any branch
  current or any voltage across any element.



           EE2 Lec 03                                     3
Steps of Nodal Analysis
1. Choose a reference node.
2. Assign node voltages to the other nodes.
3. Apply KCL to each node other than the reference-express
   currents in terms of node voltages.
4. Solve the resulting system of linear equations.




               EE2 Lec 03                                    4
             REFERENCE
             NODE



EE2 Lec 03               5
Steps of Nodal Analysis
1. Choose a reference node.
2. Assign node voltages to the other nodes.
3. Apply KCL to each node other than the reference-express
   currents in terms of node voltages.
4. Solve the resulting system of linear equations.




               EE2 Lec 03                                    6
EE2 Lec 03   7
Steps of Nodal Analysis
1. Choose a reference node.
2. Assign node voltages to the other nodes.
3. Apply KCL to each node other than the reference-express
   currents in terms of node voltages.
4. Solve the resulting system of linear equations.




               EE2 Lec 03                                    8
Current flows from a higher potential to a lower
  potential in a resistor.
        vhigher  vlower
   i
                R
     v1  0
i1            or i1  G1v1
       R1
     v1  v2
i2             or i2  G2 (v1  v2 )
       R2
     v2  0
i3            or i3  G3v2
       R3


                EE2 Lec 03                         9
NODE 01




             v1 v1  v2
        I1             I2  0
             R1   R2


    EE2 Lec 03                     10
NODE 02




        v1  v 2   v2
                      I2  0
          R2       R3

    EE2 Lec 03                   11
Steps of Nodal Analysis
1. Choose a reference node.
2. Assign node voltages to the other nodes.
3. Apply KCL to each node other than the reference-express
   currents in terms of node voltages.
4. Solve the resulting system of linear equations.




               EE2 Lec 03                                12
In terms of the conductance, we get
             I1  I 2  v1 (G1  G 2 )  v 2G 2
             I 2   v1G 2  v 2 (G 2  G 3 )

  In matrix form
     G1  G2         G2   v1   I1  I 2 
      G                    v    I 
                    G2  G3   2   2 
          2




     EE2 Lec 03                                   13
• Calculate the node voltages for the following circuit.


              V1         2kW       V2    2kW   V3
             1                 2                3
I1                                      1kW    2kW     I2
                        2kW




           EE2 Lec 03                                  14
KCL at Node 1                2kW
                       V1          V2


        I1
                            2kW



                     V1  V2    V1
                I1               0
                      2kW      2kW



         EE2 Lec 03                     15
KCL at Node 2
  V1   2kW       V2        2kW    V3


                          1kW
                                 V1  V2    V2   V2  V3
                                                       0
                                  2kW      1kW    2kW




             EE2 Lec 03                                   16
KCL at Node 3 2kW
                  V2        V3


                             2kW          I2




                 V2  V3    V3
                               I2  0
                  2kW      2kW


         EE2 Lec 03                            17
System of Equations
 Node 1:
               V1 0.5  0.5  0.5V2  I1
 Node 2:
             0.5V1  V2 0.5  1  0.5  0.5V3  0
 Node 3:


             0.5V2  V3 0.5  0.5  I 2


               EE2 Lec 03                              18
Matrix Notation
 The three equations can be combined into a single
 matrix/vector equation.

            1      0.5   0  V1   I 1 
             0.5   2         V    0 
                          0.5  2   
           
            0
                   0.5   1  V3   I 2 
                                 


               EE2 Lec 03                             19
 Nodal Analysis: Standard form for Node equations
     Consider the following:

G11 G12        G13  V1      current  sources (1) 
G              G23  V2    current  sources (2) 
 21 G22                                            
G31 G32
               G33  V3 
                             current  sources (3) 
                                                       
G11 =        of conductance connected to node 1
G22 =        of conductance connected to node 2
G33 =        of conductance connected to node 3

                   V1           2kW       V2    2kW   V3
                   1                  2                 3
I1                                             1kW     2kW   I2
                            2kW


                   EE2 Lec 03                                     20
G12 = G21 = - Conductance common between node 1 and 2

G13 = G31 = - Conductance common between node 1 and 3

G23 = G32 = - Conductance common between node 2 and 3


   current  sources(1) = sum of currents from sources
                              entering the node 1.
   current  sources(2) = sum of currents from sources
                              entering the node 2.
   current  sources(3)    = sum of currents from sources
                              entering the node 3.




            EE2 Lec 03                                       21
 Calculate the node voltages using inspection




           EE2 Lec 03                            22
 3 / 4  1 / 4  v1  5      3  1  v1  20
                             
 1 / 4 10 / 24 v   5   3 5 v   60
                 2                2  
             20  1
         1 60    5 100  60
                                              3 1
   v1                      13.33V
                    12                             15  3  12
                                             3   5

                3 20
           2  3 60 180  60
   v2                       20V
                     12



               EE2 Lec 03                                             23
 Determine the node voltages for the following circuit.




                EE2 Lec 03                                 24
                                 v1  v3 v1  v2           3v1  2v2  v3  12 .......... .......... .(1)
    Node1 : 3  i1  ix  3            
                                    4       2

                            v1  v2 v2  v3 v2  0        4v1  7v2  v3  0.......... .......... .(2)
  Node 2 : ix  i2  i3                  
                               2       8      4

                           v1  v3 v2  v3 2(v1  v2 )    2v1  3v2  v3  0.......... .......... .(3)
Node 3 : i1  i2  2ix                  
                              4       8        2




             EE2 Lec 03                                                                                     25
      3  2  1  v1  12
      4  7  1 v2    0 
                  
      2 3
              1 v3   0 
                   

                   1 48
             v1    4.8 V
                    10
                    2 24
             v 2    2 .4 V
                     10
                     3  24
               v3           2 . 4 V
                      10


EE2 Lec 03                                26
Nodal Analysis: With voltage source.
                                        I

                    R1   v1                 v2
                                   R3

   E     +                    R2             R4
         _




       At V1:       I  G1(V1  E)  G2V1  G3(V1  V2 )  0


       At V2:         I  G4V2  G3 (V2  V1)  0

       EE2 Lec 03                                              27
(G  G  G )V  G3V2  I  G E                  Eq 1
  1   2   3 1               1

 G3V1  (G3  G4 )V2   I                     Eq 2




   G1  G2  G3     G3   v1   I  G1E 
                         v     I 
             G3 G3  G4   2  
                                           



       EE2 Lec 03                                      28
Given the following circuit. Solve for the indicated nodal voltages.




                                       2W
                                                         x
                         v1             v2       _           v3
                                   x                 +
                              5W                 10 V
                                       x                 x
                   6A                       4W               10 W




            EE2 Lec 03                                                 29
                              2W
                                                           super node
                                                x
        v1                     v2       _           v3
                          x                 +
                  5W                    10 V
                              x                 x
      6A                           4W               10 W




When a voltage source appears between two nodes, an easy way to
handle this is to form a super node. The super node encircles the
voltage source and the tips of the branches connected to the nodes.


             EE2 Lec 03                                                 30
                  2W
                    v2                v3
   v1                         _
                                  +
           5W                 10 V

 6A                      4W            10 W   Constraint Equation

                                               V2 – V3 = -10        Eq 1


At V1      6  0.2( V1  V2 )  0.5( V1  V3 )  0                    Eq 2



                                                                           Eq 3
At super     0.2(V2  V1 )  0.25 V2  0.1V3  0.5 (V3  V1 )  0
node



             EE2 Lec 03                                                      31
      V2 – V3 = -10              Eq 1

    7V1 – 2V2 – 5V3 = 60         Eq 2

   -14V1 + 9V2 + 12V3 = 0        Eq 3

Solving gives:
  V1 = 30 V, V2 = 14.29 V, V3 = 24.29 V



         EE2 Lec 03                       32

								
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