EE2010 Fundamentals of Electric
Circuits
Lecture 7
Nodal Analysis
Nodal Analysis
• One of the systematic ways to
determine every
voltage
and
current in a circuit
• “Node Voltages” The voltages of each
node with respect to a pre-selected
reference node
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
• 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.
Nodal Analysis - Example
• Apply nodal analysis to the network in Fig.
• Steps 1 and 2: The network has two nodes, as
shown in Fig. The lower node is defined as the
reference node at ground potential (zero
volts), and the other node as V1, the voltage
from node 1 to ground.
Nodal Analysis - Example
• Step 3: I1 and I2 are defined as leaving the node in Fig, and
Kirchhoff’s current law is applied as follows:
Nodal Analysis - Example
• The currents I1 and I2 can then be determined
Nodal Analysis – Example -2
Applying KCL to each nodes Example
in node v1
V1 V1  V2

 I 2  I1  0
R1
R2
node v2
V2 V2  V1

 I2  0
R3
R2
7
7
Nodal Analysis – Example -3
At node 1, applying KCL,
v1
v  v2
 1
5  0
2
4
v1
v2
2 0  2 v1  v1  v 2
3 v1  v 2  2 0
At node 2, applying KCL,
v2
v  v1
 2
 5  10  0
6
4
- 3v1  5v2  60
8
v1
v2
Nodal Analysis – Example -3
Using Cramer’s Rule or Format Approach:
 3  1  v1  20
 3 5  v   60

 2   
3 1

 15  3  12
3 5
20  1
1 60 5 100 60
v1 


 13.33V


12
3 20
 2  3 60 180 60
v2 


 20V


12
9
Nodal Analysis – Example -2
Apply nodal analysis to the network in Fig
Steps 1 and 2: The network has
three nodes, as defined in Fig., with
the bottom node again defined as
the reference node (at ground
potential, or zero volts), and the
other nodes as V1 and V2.
Step 3: For node V1, the currents are
defined as shown in Fig. and
Kirchhoff’s current law is applied:
Nodal Analysis – Example -2
Nodal Analysis – Example -3
• For node V2 the currents are defined as shown in Fig., and
Kirchhoff’s current law is applied:
Nodal Analysis – Example -3
Nodal Analysis – Example -4
2A
v1 

v2
5
10 
20 
4A
Find V1 and V2.
At v1:
V1
10
At v2:

V2  V1
5
14
V1  V2
5

V2
20
2
Eq 3.1
 6
Eq 3.2
Nodal Analysis – Example -4
From Eq 3.1:
V1 + 2V1 – 2V2 = 20
or
Eq 3.3
3V1 – 2V2 = 20
From Eq 3.2:
4V2 – 4V1 + V2 = -120
or
-4V1 + 5V2 = -120
Solution: V1 = -20 V,
9
15
Eq 3.4
V2 = -40 V
Nodal Analysis – Example -5
Given the following circuit.
Set-up the equations to solve for V1 and V2.
Also solve for the voltage V6.
R2
v1
R3

v2
R5

+
R1
16
I1
R4
v6
_
R6
Nodal Analysis – Example -5
v1
R2
v2
R3

R5

+
R1
I1
V1
R1  R2
V2  V1
R3
17
R4

V1  V2

V2
R3
R4

 I1
V2
R5  R6
v6
_
R6
Eq 2.1
0
Eq 2.2
Nodal Analysis – Example -5
V1
R1  R2
V2  V1
R3
7
18

V1  V2

V2
R3
R4

 I1
V2
R5  R6
Eq 2.3
0
Eq 2.4
 1
 1 
1 

  V1    V2  I1
 R1  R2 R3 
 R3 
Eq 2.5
 1 
 1
1
1 
 V2  0
   V1   

 R3 
 R3 R4 R5  R6 
Eq 2.6
Supernode - Example
Determine the nodal voltages V1 and V2 in Fig. using the concept of a
supernode.
Replacing the independent voltage
Source of 12 V with a short-circuit
equivalent results in the network in Fig.
Relating the defined nodal voltages to the independent voltage source, we
have
V1 - V2 = E = 12 V
Supernode - Example