BEF 12403 - Week 12 - Nodal Analysis Method.ppt

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 groundnode where V= 0
Reference node
v
a v
b v
c

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 voltageV
x.
Example:
v
a v
b v
c
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 iflowing 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 V
Aand v
Bbe the node
voltages at the two ends of the resistor, measured with respect
to a common reference point O, and let ibe the corresponding
resistor current, whose reference direction is chosen according
to the passive sign convention.
v
A
v
B
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 v
AB= V
A–V
B, then we obtain the required expression
for the branch current iin terms of the node voltages, V
Aand
V
B, as follows:R
vv
i
BA

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
v
A
v
B
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
i
1–i
2–i
3= 0
v
A
v
B
v
D
v
C
R
1
R
2
R
3
KCL Formulation in Nodal Analysis

13. Now, application of Ohm’s law to the three resistors gives us the
equations0





BD
DB
BC
CB
AB
BA
R
vv
R
vv
R
vv
v
A
v
B
v
D
v
C
R
1
R
2
R
3
KCL Formulation in Nodal Analysis (continued)AB
BA
R
vv
i


1 BC
CB
R
vv
i


2 0
3 


BD
DB
R
vv
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 V6 V
2 Ω
Figure x
Worked Example 1

15. 3 Ω
6 Ω
22 V6 V
2 ΩA 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, v
D= 0.
Also, by inspection, we note that
v
AD= 6 V
and
v
CD= 22 V
Since
v
AD= V
A–V
D
And v
D= 0, then
v
CD= V
C–V
D
v
A= 6 V, and
v
C= 22 V

16. 6
6
B
AB
BA
AB
AB
AB
v
R
vv
R
v
i



 3
0
B
DB
BD
DB
DB
DB
v
R
vv
R
v
i



 At node B where the voltage remains unknown we write a KCL equation
i
AB+ i
CB+ i
DB= 0
By Ohm’s law we can write2
22
B
CB
BC
CB
CB
CB
v
R
vv
R
v
i




3 Ω
6 Ω
22 V6 V
2 ΩA B C
D
i
AB
i
DB
i
CB
Figure y. The circuit of Figure x
with element voltages and currents
defined.

17. 0
2
22
36
6




BBB vvv A 1
6
126
6
6





B
AB
v
i A 4
3
120
3
0





B
DB
v
i Hence, upon substitution of Eqs. (), into Eq. (), we obtain
or
v
B= 12 V
We then find the currents:A 6
2
1222
2
22





B
CB
v
i

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

19. At node 1321 iii
Solution4
21
2
vv
i

 2
0
1
3


v
i
Ohm’s law gives
And, by inspection,A 5
1i

20. At node 2, KCL gives5142 iiii 
Solution (continued)4
21
2
vv
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
1v 



2
2v

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. Solution2A 3A
2
4 8


V
Ia b
c .....[1].......... 82V-3V
2
4
V
2
VV
ba
aba



KCL at node a.
KCL at node b......[2].......... 425V4V-
3
8
V
2
VV
ba
bab



Matrix equation1.856A
8
14.85
8
V
I
12.57VVV
14.85VV
12.57VV
24
8
Vb
Va
54
23
b
a
b
a



























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

28. Cont…32
2
b
232
b
23
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
VVB
[2] into [1] substitute

















29. Worked Example
Calculate the node voltages v
aand v
bfor the following circuit.5A
10A
2
4
6
a
b

30. [1]................20AV3V
5A
4
2VVV
5A
4
2V
4
V
4
V
5A
2(2)
(2)V
4
V-V
05A
2
0V
4
V-V
1, Node
21
121
121
121
121







 Solution5A
10A
2
4
6
V1 V2

31. [2]................60A5V3V-
5A
6
V
4
V
4
V
5A
6
V
4
V-V
05A
6
V
4
V-V
05A10A
6
0V
4
V-V
2, Node
21
221
212
212
212







 Cont…5A
10A
2
4
6
V1 V2

32. 20VV
13.33VV
60
20
V2
V1
5
1
3
3
equationmatrix theSolving
2
1





















 5A
10A
2
4
6
V1 V2 Cont…

33. We need to put Eqn. [1] and Eqn. [2] in matrix form
as123151))(3(5)(3
5
1
3
3
Δ
is matrix the of tdeterminan The
60
20
V2
V1
5
1
3
3

























34. Cont…20V
12
60180
Δ
60
20
3-
3
Δ
Δ
V
13.33V
12
60100
Δ
5
1
60
20
Δ
Δ
V
as Vand Vobtain we Now,
2
2
1
1
21









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 iflowing through the 7 V
source cannot be expressed in terms of its node voltages, v
aand v
b. 2A
7V
4 2 4
1V
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 aand 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 1V
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 1V
Supernode
Suppose the currents at the supernode in Figure x are
labelled as i
1, i
2, i
3, and i
4. Then, application of KCL to the
supernode gives us the current equation
i
1
i
2
i
3
i
40
4321  iiii

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
v
a-v
b= 7 V2A 7V 4 2 4 1V
Supernode
a b
O V
v
a
v
b

39. 2A
7V
4 2 4
1V Worked Example
Find V
1for 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 v
aand v
bto the branch current.
Step 1.To overcome this problem, we form a supermesh that
encloses the 7 V voltage source and the nodes aand b.2A 7V 4 2 4 1V
Supernode
a b
Solution

41. [1]..................7
bavv Step 2.Find the support equation.
For the 7-V source, the support equation is2A 7V 4 2 4 1V
a b
O V
v
a
v
b
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 gives0
4321  iiii 2A 7V 4 2 4 1V
a bi
1
i
2 i
3
i
4
Solution (continued)

43. 0
424
2
bba vvv Solution (continued)
Ohm’s law gives
Thus, for the supernode, we obtain the current equation[2] .......... 83
ba vv
The above equation can be simplified and written as4
2
av
i 2
3
bv
i
and4
4
bv
i

44. 


















8
7
3
1
1
1
b
a
v
v Step 4.Solve for voltages v
a, v
b.
Eqs.(1) and (2) can be written in matrix form as
Solving for v
aand v
busing Cramer’s rule, we obtain4)1)(1()3)(1(
31
11


 29)8)(1()3)(7(
38
17



a 1)7)(1()8)(1(
81
71

b
Solution (continued)

45. V 25.0
4
1




b
bv V 25.7
4
29




a
av Therefore,V 25.04
4
41 
b
b
v
v
RiV
and

46. Exercise
In the circuit shown in Figure x, use node analysis to find v
A, v
B, and v
C.
12 Ω
2 Ω
8 Ω12 V
10 V
A
B
C
D
8 Ω
Figure x
Ans.: v
A= 12 V
v
B= 6 V
v
C= 16 V

49. END