ELL 100 - Introduction to Electrical Engineering
LECTURE 6: NODE ANALYSIS OF CIRCUITS
Outline





Introduction
Node Analysis
Super Node Concept
Nodal Analysis vs Mesh Analysis
Exercise/Numerical Analysis
2
INTRODUCTION
Applications of nodal circuit analysis:
Voltage summing using op-amps
 Used in DACs (Digital-to-Analog converters)
3
INTRODUCTION
Applications of nodal circuit analysis:
Wheatstone Bridge circuits
 Used to measure unknown resistances.
 Mechanical and civil engineers measure resistances of strain gauges
to find the stress and strain in machines and buildings.
4
INTRODUCTION
What is Nodal Analysis?
 Nodal analysis is a method that provides a general
procedure for analysing circuits using node voltages as
the circuit variables.
 Also called Node-Voltage Method.
5
INTRODUCTION
I1
Types of Nodes in Nodal Analysis:
 Non-Reference Node:
Node which has a definite
Node-Voltage.
 Reference Node: It is the
node which acts as reference
point to all the other node.
Also called
Datum Node/Ground.
1
I1
R2 2
R3
R1
0
Node 0 is Reference node
Nodes 1, 2 are Non-Reference node
6
INTRODUCTION
Reference node indication
(a)
(b)
(c)
Common symbols for
indicating reference node,
(a) common ground,
(b) (earth) ground,
(c) chassis ground.
Chassis ground: used in devices where the case, enclosure, or
chassis acts as reference point for all circuits.
Earth ground: used when the potential of the earth is used as
reference.
7
INTRODUCTION
Steps to find Node Voltages:
1. Select a node as the reference (ground) node. Assign
voltages v1, v2, v3,……., vn-1 to the remaining n-1 nodes.
2. Apply KCL to each of the n-1 non-reference nodes.
3. Use Ohm’s law to express the branch currents in terms
of node voltages.
4. Solve the resulting ‘n-1’ simultaneous equations to
obtain the unknown node voltages.
For ‘n’ nodes there will be ‘n-1’ simultaneous equations to
be solved to get all the node voltages.
8
INTRODUCTION
Example circuits depicting Nodes:
I1
1
I1
R4
R2 2
R3
R1
0
1
R1
2
R3 3
R2
2ix
ix
I1
0
9
NODE VOLTAGE ANALYSIS
• Dual of Mesh Analysis.
• Involves the application of KCL equations (KVL for mesh).
• One of the nodes is taken as reference or datum node
(node with maximum number of branches connected preferred).
• The reference node is assumed to be at ground or zero potential.
• Potentials of all the other nodes are defined with respect to
reference node.
• KCL equations are written for each node except for the
reference node. On solving, node voltages are obtained.
10
SUPER NODE ANALYSIS
Super node can be regarded as a
surface enclosing the voltage
source and its two nodes.
i4
5V 3
v3

i3
R 2 R
0
R i1 v2
i2

Definition of Super Node:
when a dependent or an
independent voltage source is
1
connected between two nonv1
reference nodes, then these
nodes can be combined to form 10V 

a generalised node which is
known as Super Node.
Supernode
R
Node 2 and 3 can be
combined into a Super-Node.
11
SUPER NODE ANALYSIS
Properties of super-node:
• A super-node cannot be assigned a single voltage value
• A super-node requires application of both KCL and KVL to solve it.
• Any element can be connected in parallel with the voltage source
forming the super-node without affecting the analysis.
• A super-node satisfies the KCL as like a simple node.
12
ADVANTAGES OF NODAL ANALYSIS
Nodal Analysis vs Mesh Analysis?
• Nodal analysis can be applied to both planar and nonplanar circuits.
But Mesh analysis can be applied to planar circuits only.
Planar circuit: a circuit that can be drawn on a plane with
no crossed wires.
• The choice between Node Voltage analysis and Mesh Current analysis
is usually unambiguous.
• To choose between methods, pick the one that involves solving the
fewest equations.
13
NODAL ANALYSIS VS MESH ANALYSIS
Nodal Analysis
Mesh analysis
•
The number of voltage variables,
and hence simultaneous
equations to solve, equals the
number of nodes minus one.
•
The number of current variables,
and hence simultaneous
equations to solve, equals the
number of meshes.
•
Every voltage source connected
to the reference node reduces the
number of unknowns by one.
•
Every current source in a mesh
reduces the number of
unknowns by one.
•
Nodal analysis is thus best for
circuits with voltage sources.
•
Mesh analysis is thus best for
circuits with current sources.
14
NUMERICAL
Q1. Calculate Node Voltages in the given circuit.
5A
1
v1
2Ω
i1
4Ω i2
i3
2
v2
i5
i4
10A
6Ω
0
15
NUMERICAL
Soln: There are 3 nodes:
0: ground, 1 & 2: non-reference
nodes.
5A
1
v1
Step1. Node voltages (v1 and v2) are
assigned. Also direction of branch
currents are marked.
2Ω
i1
4Ω i2
i3
2
v2
i5
i4
10A
6Ω
0
Step2. Apply KCL to nodes 1 and 2
16
CONTD.
5A
KCL at Node 1:
i1  i2  i3
KCL at Node 2:
i2  i4  i1  i5
1
v1
i1
4Ω i2
i3
2Ω
2
v2
i5
i4
10A
6Ω
0
Step3. Apply Ohm’s Law to KCL equations.
Ohm’s Law to KCL equation at Node 1:
v1  v2 v1  0
i1  i2  i3  5 

 3v1  v2  20
4
2
…(1)
17
CONTD.
Now, Ohm’s Law to KCL equation at Node 2:
v1  v2
v2  0
i2  i4  i1  i5 
 10  5 
4
6
i1
 3v1  5v2  60 …(2)
1
v1
On solving (1) and (2) we get,
v1  13.33V and v2  20V
2Ω
5A
4Ω i2
i3
2
v2
i5
i4
10A
6Ω
0
18
NUMERICAL
Q2. Calculate Node Voltages and Branch Currents in the given
circuit.
7Ω i2

60V 
v1
1
i1
i3
4Ω
12Ω
0
19
NUMERICAL
7Ω i2

Soln: There are 2 nodes:
0: ground, 1: non-reference node
60V 
v1
1
i1
i3
4Ω
12Ω
0
Step1. Node voltage (v1) is
assigned. Also direction of branch
currents are marked.
Step2. Apply KCL to node 1
20
CONTD.
KCL at Node 1:

i1  i2  i3  0
v1
7Ω i2
60V 
1
i1
4Ω
12Ω
Step3. Apply Ohm’s Law
i3
0
to KCL equations.
Ohm’s Law to KCL
v1  0 v1  60 v1  0



0
equation at Node 1:
12
7
4
 v1  18V
21
CONTD.
Now, from v1 we can find the branch currents (i1, i2 and i3)
v1  60 18  60
i2 

 6A
7
7

v1  0 18
i1 
  1.5A
12
12
7Ω i2
60V 
v1
1
i1
i3
4Ω
12Ω
0
v1  0 18
  4.5A
i13 
4
4
22
NUMERICAL
Q3. Calculate Node Voltages in the given circuit.
2Ω
5A
v2


v1
4V
6Ω
3Ω
2A
23
NUMERICAL
5A
v1
4V
v2


Soln: Here 4V voltage source is
connected between Node-1 and Node-2
and it forms a Super-node with a 2Ω
resistor in parallel.
2Ω
6Ω
2A
3Ω
v2 – v1 = 4V (always, whatever may be
the value of resistor in parallel). So 2Ω
resistor is irrelevant for applying KCL to
the super-node.
24
CONTD.
KCL in terms of nodes voltages:
v1  0 v2  0
5

2
6
3
 10
v2 2+1212
30 = 2νv11 + 2ν
i2
2A
(1)
4V
1
(2)
On solving (1) and (2) we get,
vν11 = 2V
6V
3.33and
V vand
2 ν
2 = 7.33 V
v1




v1  4  v2  0
 v2  v1  4
3Ω
6Ω

 vν21 + 2ν
2v21= 18
2
5A
v2
v1
i1
2
v2

25
NUMERICAL
Q4. Calculate current through the 4Ω resistor in the given circuit.
2A


1
3
4Ω
6V
3Ω
2
5Ω
7A
0
26
NUMERICAL
2A


1
4Ω 3
6V
3Ω
2
5Ω
7A
0
Soln. Here 6V voltage source is connected between Node-1 and Node-2
and it forms a Super node with a 4Ω resistor in series.
The two node voltages are related as:
v1 – v2 = 6
…(1)
27
CONTD.
KCL for this super-node 1 and 2:
And applying KCL at node 3:
v3 v3  v2

 7
5
4
 9v3  5v2  140
1
(2)
2A
4Ω 3
6V


v1  0 v2  v3

2
3
4
 4v1  3v2  3v3  24 …(2)
3Ω
2
7A
5Ω
0
…(3)
(3)
28
CONTD.
On solving (1), (2) and (3) we get,
v1  2.77 V, v2  8.77V and v3  20.43V
2A
Now, current through


1
4Ω 3
6V
3Ω
2
5Ω
7A
0
4Ω resistor is:
v2  v3 8.77  (20.42)

 2.9125A
4
4
29
NUMERICAL
Q5. Solve for voltage V across the 3Ω resistor using node voltage
analysis for the given circuit.
1 v1
2Ω

4A
2
1Ω
3Ω
5A
4


v
3
6V
30
NUMERICAL
1 v1
4A
1Ω
3Ω
5A

v
4

To find the number of independent
nodes, the circuit is redrawn as shown
below by turning off the sources.
2
2Ω

Soln: Here 6V voltage source is
connected between Node-3 and Node-4,
so node 3 and 4 are constrained to one
another.
3
6V
31
CONTD.
There are three nodes, in which two of them
are independent.
However, if we add the two series resistors
(2Ω and 3Ω) then we will have only one
independent node i.e. node 1 only.
And hence we will have to solve only one
equation.
1
2
2Ω
1Ω
3Ω
0
Then using voltage divider rule we can find
the unknown voltage across 3Ω resistor.
32
CONTD.
 v1  4.1667V
1 v1
4A
1Ω
3Ω
5A

Now, voltage across 3Ω resistor is
3
v  4.1667 
 2.5V
23
2
2Ω

v1  6 v1  0
4

5
1
5
 6v1  25
Node 4 is taken as ground (0 V)
v
4

Applying KCL ate node 1,
3
6V
33
NUMERICAL
Q6. Find Is for the circuit shown. Take V0 = 16V.
V
Is V1 6Ω 1
 0 4
2


1
4Ω
8Ω V0

34
NUMERICAL
Soln. Applying KCL at node 1:
KCL at node 2 gives:
V1 V1  V0
Is  
6
4
5V1
 Is 
4
12
V1 V0  V1 V0


4
4
8
V1 16  V1 16
 

4
4
8
2V1

8
4
 V1  12V
V
Is V1 6Ω 1
 0 4
2

4Ω

1
(1)
8Ω V0 =16 V

5  12
Is 
 4  5  4  1A
12
(2)
35
NUMERICAL
Q7. Solve the following network using the nodal analysis and determine
the current through 2S resistor. (S indicates Siemens i.e. conductance)
-3A
-8A
1
v1
3S
2
2S
v3 3
v2
4S
-8A
1S
-25A
5S
0
36
NUMERICAL
Soln: Here there are 4 nodes, node 0 is reference node. So we have to
write to 3 KCL equations.
-3A
-8A
1
v1
3S
2
2S
v3 3
v2
4S
-8A
1S
-25A
5S
0
37
CONTD.
We can write nodal voltage equation in matrix form, as below
3
4   v1  (3)  (8) 
(4  3)
 3
(3  2  1)
2  v2    (3) 

  

2
(4  2  5)   v3   ( 25) 
 4
-3A
-8A
 7 3 4   v1   11
  3 6 2  v2    3 

  

 4 2 11   v3   25 
1
v1
3S
2
2S
v3 3
v2
4S
-8A
1S
5S
-25A
0
38
CONTD.
-3A
-8A
On solving ,we get
v1  1V, v2  2V and v3  3V
1
v1
3S
2
2S
v3 3
v2
4S
-8A
1S
5S
-25A
0
Putting the value of v2 to find current through 2S resistor we have
I2S
v2  v3

 G (v2  v3 )  2(2  3)  2A
R
39
NUMERICAL
Q8. Determine the voltages at the nodes of the given circuit.
4Ω
1
2Ω
2
8Ω 3
4Ω
2ix
ix
3A
0
40
NUMERICAL
Soln: Here there are 4 nodes,
node-0 is reference node.
There are 3 non-reference nodes in
this network. So we have to write
three KCL equations.
4Ω
1
2Ω
8Ω 3
4Ω
2ix
ix
3A
Assign voltages to the three nodes
as shown below and label the
currents.
2
0
41
CONTD.
Applying KCL at node 1:
3  i1  ix
v1  v3 v1  v2
3

4
2
 3v1  2v2  v3  12
4Ω
i1
1 v1
(1)
ix
Applying KCL at node 2:
ix  i2  i3
v1  v2 v2  v3 v2  0



2
8
4
 4v1  7v2  v3  0
2
i2 8Ω 3
2Ω v2
v3
3A
i3
4Ω
2ix
0
(2)
42
CONTD.
Applying KCL at node 3:
i1  i2  2ix
4Ω
i1
v1  v3 v2  v3 2(v1  v2 )
1 v1



4
8
2
ix
 2v1  3v2  v3  0
(3)
2
i2 8Ω 3
2Ω v2
v3
3A
i3
4Ω
2ix
0
On solving (1), (2) and (3) we get,
v1  4.8V, v2  2.4V and v3  2.4V
43
UNSOLVED NUMERICAL.
Q1. Determine the node voltages and the current ix for the given circuit
having 4 nodes with the following values, R1 = 2Ω, R2 = 1Ω, R3 = 5Ω, R4 =
6Ω and I1 = 6A.
R4
1
R1
2
R3 3
R2
2ix
ix
I1
0
44
UNSOLVED NUMERICAL.
Q2. Determine the node voltages and the current through 8Ω resistor for the
given circuit having 4 nodes.
8Ω
1
4Ω
8A
2
6Ω
1Ω 3
9Ω
0
45
UNSOLVED NUMERICAL.
Q3. Determine the node voltages for the given circuit having 4 nodes.
-4A
1
8Ω
5Ω
-9A
2
2Ω
10Ω -25A
3
1Ω
0
46
UNSOLVED NUMERICAL.
Q4. Determine the node voltages and the current through 2Ω Resistor, for
the given circuit having 3 nodes.



4Ω
2
1 6V
 14V
3Ω
2Ω
6Ω
0
47
UNSOLVED NUMERICAL.
Q5. Determine the node voltages using supernode analysis, for the given
circuit having 5 nodes.

3Ω
vx 
1 6V 2 6Ω
2Ω
3vx



3
10A 4Ω

4
1Ω
0
48
UNSOLVED NUMERICAL.
Q5. Determine the node voltages using supernode analysis, for the given
circuit having 5 nodes.

3Ω
vx 
1 6V 2 6Ω
2Ω
3vx



3
10A 4Ω

4
1Ω
0
49
UNSOLVED NUMERICAL.
Q6. Using Nodal analysis, calculate the node voltages v1 and v2 in the given
circuit.
3Ω
1
v1
1A
5Ω
2
v2
7Ω
0
50
UNSOLVED NUMERICAL.
Q7. Using Nodal analysis, calculate the node voltages v1 and v2 in the given
circuit and the current in the 8Ω resistor.
5Ω
4V
1
v1
10Ω 2
12Ω
v2
15Ω
8Ω
6V
0
51
UNSOLVED NUMERICAL.
Q8. Using Nodal analysis, calculate the node voltages v1 and v2 in the given
circuit.
1Ω
1
v1
5V
2Ω
6Ω
2
v2
8Ω
0
52
REFERENCES
I. Edward Hughes; John Hiley, Keith Brown, Ian McKenzie
Smith,
“Electrical
and
Electronic
Technology”,
10th edition, Pearson Education Limited, Year: 2008.
II. Alexander, Charles K., and Sadiku, Matthew N. O.,
“Fundamentals of Electric Circuits”, 5th Ed, McGraw Hill,
Indian Edition, 2013.
53