# 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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