Nodal Analysis Method

1. Week 12
Nodal Analysis Method

2. LEARNING OUTCOMES
After completing this module you will be able to:
1. Apply Ohm’s law and Kirchhoff’s current law to solve
circuits using nodal analysis.

3. 1. Node voltage analysis is the most general method for the
analysis of electric circuits. It can be used to solve 3-D
circuits that cannot be solved using mesh analysis.
2. The node voltage method is based on defining the voltage
at each node as an independent variable. One of the
nodes is selected as the reference node (usually – but not
necessarily – ground), and each of the other node
voltages is referenced to this node. Once each node
voltage is defined, Ohm’s law is applied between any
adjacent nodes to determine the current flowing in each
branch.
3. In the node voltage method, each branch current is
expressed in terms of one or more node voltages; thus
currents do not explicitly enter into the equations.
What is nodal analysis?

4. Terminologies
To apply nodal analysis we need to understand a number of
terms:
1. Branch
2. Node - major node and minor node
3. Reference node
4. Node voltage
1. Branch: elements connected end-to-end, nothing coming
off in between (in series)

5. 2. Node - major node and minor node
A minor node is one that just has two branches
connected to it.
A major node is one that has more than two branches
connected to it.
Minor node
Major nodes
Node - a point in a circuit where two or more circuit
components are joined

6. 3. Reference node
• the node to which node voltages are referenced.
• is normally the node with the most connections, or the
node at the bottom of the circuit diagram
• In fact, any node (minor or major) can be chosen to be
the reference node.
Note
1. The reference node is called the ground node where V = 0
Reference node
va vb vc

7. 3. Reference node (continued)
Common symbols for indicating a reference node,
(a) common ground, (b) ground, (c) chassis.
Ground symbol attached here indicates
that it has been chosen as the reference
node. Voltage here is taken as at 0 V.

8. 4. Node Voltages
The voltage drop from node X to a reference node
(ground) is called the node voltage Vx.
Example:
va vb vc
Example:

9. 1. Choose a reference (ground) node.
2. Assign node voltages to the other nodes.
3. Apply KCL to each node other than the reference node;
express currents in terms of node voltages.
4. Solve the resulting system of linear equations for the nodal
voltages.
Steps of Nodal Analysis
The rest of this module will how nodal analysis is applied in
different situations.

10. To use the nodal analysis method, we first need to know how
to define the branch current i flowing through a resistor in
terms of the associated node voltages.
Consider a resistor R attached between two nodes, A and B,
as shown in the figure below. Let VA and vB be the node
voltages at the two ends of the resistor, measured with respect
to a common reference point O, and let i be the corresponding
resistor current, whose reference direction is chosen according
to the passive sign convention.
vA vB
i
Reference node. O
R
A B
Using Ohm’s law, we can write
R
v
i AB

Branch Current Formulation in Nodal Analysis

11. Since vAB = VA – VB, then we obtain the required expression
for the branch current i in terms of the node voltages, VA and
VB, as follows:
R
v
v
i B
A 

The circuit given earlier is sometimes drawn in the simplified
form as shown in the figure below, without showing explicitly
the reference node nor the reference direction of the node
voltages.
Note
vA vB
i
R

12. To apply the nodal voltage method in circuit analysis, we also
need to know how to write current equations in terms of node
voltages.
To see how, we can write a current equation in terms of node
voltages, consider the circuit shown below. By KCL, we can
write
i1 – i2 – i3 = 0
vA vB vD
vC
R1 R2
R3
KCL Formulation in Nodal Analysis

13. Now, application of Ohm’s law to the three resistors gives us the
equations
0






BD
D
B
BC
C
B
AB
B
A
R
v
v
R
v
v
R
v
v
vA vB vD
vC
R1 R2
R3
KCL Formulation in Nodal Analysis (continued)
AB
B
A
R
v
v
i


1
BC
C
B
R
v
v
i


2 0
3 


BD
D
B
R
v
v
i
Thus,

14. As an illustration of the nodal analysis method, consider the circuit
shown in Figure x. What we want to do here is to solve for all branch
voltages and currents using the node voltage method.
3 Ω
6 Ω
22 V
6 V
2 Ω
Figure x
Worked Example 1

15. 3 Ω
6 Ω
22 V
6 V
2 Ω
A B C
D
To start the analysis, we first label all
the nodes as shown in Figure x. Next,
we need to select one node as our
reference node or datum node. By
inspection, we node that node D has
the most elements tied to it. So, we
select D as the reference node for this
circuit. The voltage of this node is then,
by definition, zero; that is, vD = 0.
Also, by inspection, we note that
vAD = 6 V
and
vCD = 22 V
Since
vAD = VA – VD
And vD = 0, then
vCD = VC – VD
vA = 6 V, and
vC = 22 V

16. 6
6 B
AB
B
A
AB
AB
AB
v
R
v
v
R
v
i





3
0 B
DB
B
D
DB
DB
DB
v
R
v
v
R
v
i





At node B where the voltage remains unknown we write a KCL equation
iAB + iCB + iDB = 0
By Ohm’s law we can write
2
22 B
CB
B
C
CB
CB
CB
v
R
v
v
R
v
i





3 Ω
6 Ω
22 V
6 V
2 Ω
A B C
D
iAB
iDB
iCB
Figure y. The circuit of Figure x
with element voltages and currents
defined.

17. 0
2
22
3
6
6




 B
B
B v
v
v
A
1
6
12
6
6
6





 B
AB
v
i
A
4
3
12
0
3
0





 B
DB
v
i
Hence, upon substitution of Eqs. (), into Eq. (), we obtain
or
vB = 12 V
We then find the currents:
A
6
2
12
22
2
22




 B
CB
v
i

18. Calculate the node voltage in the circuit shown in Fig. x
Worked Example 2

19. At node 1
3
2
1 i
i
i 

Solution
4
2
1
2
v
v
i


2
0
1
3


v
i
Ohm’s law gives
And, by inspection,
A
5
1 
i

20. At node 2, KCL gives
5
1
4
2 i
i
i
i 


Solution (continued)
4
2
1
2
v
v
i


6
0
2
5


v
i
Ohm’s law gives
And, by inspection, A
10
4 
i

21. In matrix form:



























5
5
4
1
6
1
4
1
4
1
4
1
2
1
2
1
v
v
Solution (continued)
After simplification, we get

























5
5
4
1
4
1
2
1
v
v
12
5
4
3

22. Solving using Cramer’s rule,





12
5
4
3
4
1
4
1




12
5
5
5
4
1
1




5
4
3
4
1
5
2
Therefore,



 1
1
v 


 2
2
v

23. Exercise

24. Exercise
Use nodal analysis to find vo in the circuit below.

25. 2A 3A

2

4 
8


V
I
Exercise
Find V and I for the following circuit.

26. Solution 2A 3A

2

4 
8


V
I
a b
c
.....[1]
..........
8
2V
-
3V
2
4
V
2
V
V
b
a
a
b
a




KCL at node a.
KCL at node b.
.....[2]
..........
4
2
5V
4V
-
3
8
V
2
V
V
b
a
b
a
b





Matrix equation
1.856A
8
14.85
8
V
I
12.57V
V
V
14.85V
V
12.57V
V
24
8
Vb
Va
5
4
2
3
b
a
b
a






























27. Example 3
VB IB
2
R
1
R 3
R
a b
c
[2]
.......
..........
R
V
IB
R
V
V
b,
Node
[1]
......
..........
VB
V
a,
Node
3
b
2
b
a
a




Applying nodal
analysis in the circuit
with current and
voltage sources.

28. Cont…
3
2
2
b
2
3
2
b
2
3
b
2
b
3
b
2
b
2
3
b
2
b
2
3
b
2
b
R
1
R
1
R
VB
IB
V
R
VB
IB
R
1
R
1
V
R
VB
IB
R
V
R
V
IB
R
V
R
V
R
VB
-
IB
R
V
R
V
R
VB
R
V
IB
R
V
VB
[2]
into
[1]
substitute


























29. Worked Example
Calculate the node voltages va and vb for the following circuit.
5A
10A

2

4

6
a
b

30. [1]
......
..........
20A
V
3V
5A
4
2V
V
V
5A
4
2V
4
V
4
V
5A
2(2)
(2)
V
4
V
-
V
0
5A
2
0
V
4
V
-
V
1,
Node
2
1
1
2
1
1
2
1
1
2
1
1
2
1














Solution
5A
10A

2

4

6
V1 V2

31. [2]
......
..........
60A
5V
3V
-
5A
6
V
4
V
4
V
5A
6
V
4
V
-
V
0
5A
6
V
4
V
-
V
0
5A
10A
6
0
V
4
V
-
V
2,
Node
2
1
2
2
1
2
1
2
2
1
2
2
1
2
















Cont…
5A
10A

2

4

6
V1 V2

32. 20V
V
13.33V
V
60
20
V2
V1
5
1
3
3
equation
matrix
the
Solving
2
1




















 

5A
10A

2

4

6
V1 V2
Cont…

33. We need to put Eqn. [1] and Eqn. [2] in matrix form
as
12
3
15
1))
(
3
(
5)
(3
5
1
3
3
Δ
is
matrix
the
of
t
determinan
The
60
20
V2
V1
5
1
3
3






























 


34. Cont…
20V
12
60
180
Δ
60
20
3
-
3
Δ
Δ
V
13.33V
12
60
100
Δ
5
1
60
20
Δ
Δ
V
as
V
and
V
obtain
we
Now,
2
2
1
1
2
1












35. When there exists a floating voltage source in a branch, we cannot
apply Ohm’s law to express the current flowing through it in terms of
its node voltages.
Analysing Circuits Containing Floating Voltage Sources
For example, in the circuit below, the current i flowing through the 7 V
source cannot be expressed in terms of its node voltages, va and vb.
2A
7V

4 
2 
4
1
V
a b
A floating voltage
source
Floating voltage sources have neither end connected to a known fixed
voltage. We have to change how we form the KCL equations slightly.

36. The way to overcome this problem is to form a supernode that
encloses the 7 V voltage source and the nodes a and b.
What is a Supernode?
A supernode is a closed surface used to enclose a part of the big
circuit and replace it with a node. In the example above, the
supernode encloses the 7-V floating voltage source and the two
non-reference nodes to which it is connected. The supernode
then becomes the common node to which all the four circuit
elements are connected.
2A
7V

4 
2 
4
1
V
Supernode
a b

37. How does creating a supernode helps us solve the problem?
1. The supernode hides the floating voltage source from
view, allowing us write a KCL equation for the supernode,
without having to consider its (the voltage source)
presence.
2A
7V

4 
2 
4
1
V
Supernode
Suppose the currents at the supernode in Figure x are
labelled as i1, i2, i3, and i4. Then, application of KCL to the
supernode gives us the current equation
i1
i2
i3
i4
0
4
3
2
1 


 i
i
i
i

38. 2. To generate a support equation that relates the voltage
source to its node voltages we apply KVL to the 7 V
source. The support equation in this case is
va - vb = 7 V
2A
7V

4 
2 
4
1
V
Supernode
a b
O V
va vb

39. 2A
7V

4 
2 
4
1
V
Worked Example
Find V1 for the following circuit.
The following worked examples show how we apply the supernode
concept to tackle the complication caused by the presence of a
floating voltage source in the circuit being analysed.
a b

40. Note: This circuit contains a floating voltage source that prevents
us from writing an Ohm’s law equation to relate the node
voltages va and vb to the branch current.
Step 1.To overcome this problem, we form a supermesh that
encloses the 7 V voltage source and the nodes a and b.
2A
7V

4 
2 
4
1
V
Supernode
a b
Solution

41. [1]
........
..........
7
b
a 
v
v
Step 2. Find the support equation.
For the 7-V source, the support equation is
2A
7V

4 
2 
4
1
V
a b
O V
va vb
Solution (continued)

42. Step 3. Find the supernode current equation from Figure x.
Let the reference directions of the currents at the supernode
be defined as shown in the figure below. Then, KCL gives
0
4
3
2
1 


 i
i
i
i
2A
7V

4 
2 
4
1
V
a b
i1
i2 i3
i4
Solution (continued)

43. 0




4
2
4
2 b
b
a v
v
v
Solution (continued)
Ohm’s law gives
Thus, for the supernode, we obtain the current equation
[2]
..........
8
3 b
a 
 v
v
The above equation can be simplified and written as
4
2
a
v
i 
2
3
b
v
i  and
4
4
b
v
i 

44. 

















 
8
7
3
1
1
1
b
a
v
v
Step 4. Solve for voltages va, vb.
Eqs.(1) and (2) can be written in matrix form as
Solving for va and vb using Cramer’s rule, we obtain
4
)
1
)(
1
(
)
3
)(
1
(
3
1
1
1







29
)
8
)(
1
(
)
3
)(
7
(
3
8
1
7






a
1
)
7
)(
1
(
)
8
)(
1
(
8
1
7
1




b
Solution (continued)

45. V
25
.
0
4
1




 b
b
v
V
25
.
7
4
29




 a
a
v
Therefore,
V
25
.
0
4
4
4
1 



 b
b
v
v
R
i
V
and

46. Exercise
In the circuit shown in Figure x, use node analysis to find vA, vB, and vC.
12 Ω
2 Ω
8 Ω
12 V
10 V
A
B
C
D
8 Ω
Figure x
Ans.: vA = 12 V
vB = 6 V
vC = 16 V

49. END