CPE220 Electric Circuit
Analysis
Chapter 3:
Nodal and Mesh Analyses
Circuit -F2003: Nodal and Mesh Analysis
92
BYST
Chapter 3
3. Nodal Analysis
1
Nodal
analysis provides a general
procedure for analyzing electrical circuits
using node voltages as the circuit variables.
The given circuit commonly contains only
current sources and KCL equation for each
node is set up by expressing all the
unknown currents coming through or
leaving out of that node as a function of the
node voltage. Then, the node voltages in
the given circuit can be determined by
solving those KCL equations.
Nodal analysis can be used with either
planar or non-planar circuits. Hence, it is a
more general method compared with the
mesh method which will be discussed later.
Circuit -F2003: Nodal and Mesh Analysis
93
BYST
Recall that nodes are points at which two or
more elements have a common connection.
In the other words, nodes are the connected
segments of conductor that remain when we
remove the circuit elements.
16 W
9A
8W
12 W
3A
(a)
A single
node!
(b)
Figure 3.1 (a) An electrical circuit. (b) Redraw
the circuit with the circuit elements
removed.
Circuit -F2003: Nodal and Mesh Analysis
94
BYST
Fig. 3.1 illustrates nodes in an electrical
circuit after all circuit elements are
removed.
Node and Branch Voltages
A node voltage associated with a given node
is defined to be the voltage difference
between the given node and a reference
node, which has been chosen from among
the nodes. Usually, the reference node is
the node to which the highest number of
branches is connected. Hence, for a circuit
with “n” essential nodes, there are “n–1”
node voltages.
Once the set of node voltages is determined,
all the other voltages and currents can be
obtained in a straightforward manner.
Circuit -F2003: Nodal and Mesh Analysis
95
BYST
Example 3.1
Identify a reference node and
corresponding node voltages for the circuit
in Fig. 3.1(a).
Solution:
From the redraw circuit in Fig. 3.1(b), we
select the bottom node (node 3) as the
reference node.
16 W
v1
9A
node 1
8W
v2
12 W
node 2
3A
node 3
Reference node,
ground, earth, “sea
level”
Ans.
Circuit -F2003: Nodal and Mesh Analysis
96
BYST
A branch voltage is measured as the
difference between two node voltages. As
illustrated in Fig. 3.2, the voltage across the
16 W resistor is not a node voltage. It is the
branch voltage and is actually the
difference voltage between two node
voltages v1 and v2.
In general,
vj
vk
vjk = vj - vk
(3.1)
where vjk = the branch voltage in volts (V)
vj and vk = the node voltages in volts (V)
Circuit -F2003: Nodal and Mesh Analysis
97
BYST
Measurement of Branch and Node
Voltages
voltmeter
v1
16 V
+ –
The voltage
being measured
is a branch
voltage.
16 W
v2
1
9A
2
42 W
12 W
voltmeter
64 V
+ –
3A
voltmeter
48 V
+ –
Node
voltage
Node
voltage
Figure 3.2 Branch voltage v12 in term of the
node voltages v1 and v2.
Circuit -F2003: Nodal and Mesh Analysis
98
BYST
Example 3.2
Determine the relationship among the
branch voltages and the node voltages in
the following circuit.
v1
v12 -
+
v13
2
+
+
1
v2
v23
-
-
3
Ref. node
Solution:
There are three nodes. We choose node 3 as
the reference node. Hence, only two node
voltages which are v1 and v2.
For branch voltages, there are three branch
voltages: v12, v13, and v23. However, the
Circuit -F2003: Nodal and Mesh Analysis
99
BYST
branch voltages v13 and v23 are clearly
equal to the node voltage v1 and v2,
respectively. The branch voltage v12 is the
difference between v1 and v2.
That is,
v13 = v1
v23 = v2
v12 = v1 - v2
Ans.
Branch Currents
Currents flowing through any resistors are
defined as the branch currents. The value
of each branch current can be determined
by following the Ohm's law. For example,
the current flowing through the resistor R
in Fig. 3.3 can be determine as following:
Circuit -F2003: Nodal and Mesh Analysis
100
BYST
vj
R
Node j
vk
Node k
Figure 3.3 Branch currents flowing through
the resistor R.
 The branch current flowing from
node j to node k which can be calculated
as:
vj -vk
ijk =
R
(3.2)
 The branch current flowing from
node k to node j which can be calculated
as:
vk -vj
ikj =
R
Circuit -F2003: Nodal and Mesh Analysis
101
(3.3)
BYST
3.1.1 Nodal Analysis by Examples
The rationale for nodal analysis is that
once the node voltages are determined, all
the other voltages and currents can be
obtained in a simple manner.
The reference node is chosen by the circuit
analyst. In electronic circuits, we
frequently choose the node to which lots of
branches are connected. In power systems,
we usually choose “ground” or “earth.”
Basic Procedure for Nodal Analysis:
1. Select the node to which the highest
number of branches is connected as the
reference node.
2. Set up KCL equations for other nodes
by expressing the unknown currents as a
Circuit -F2003: Nodal and Mesh Analysis
102
BYST
function of the node voltages measured
with respect to (w.r.t.) the reference
node.
3. If the given circuit contains voltage
sources, KCL equations of those two
nodes connected by a voltage source are
combined to eliminate the redundancy of
KCL equations since the additional
information is available through the
node voltage. For example, if nodes j
and k are connected by a voltage source,
then vj - vk is already known.
4. Solve KCL equations to determine the
node voltages.
Circuit -F2003: Nodal and Mesh Analysis
103
BYST
Example 3.3
Determine the node voltages v1, v2, and v3
in the following circuit.
-3A
W
v2
v1
W
v3
3W
-8A
5W
1W
-25A
Reference node
Solution:
KCL @ node 1
v1 -v2
v1 -v3
-8 -3 =
+
3
4
Circuit -F2003: Nodal and Mesh Analysis
104
(3.4)
BYST
KCL @ node 2
v2 -v1
v2
+
3
1
v2 -v3
= -(-3) = 3
+
7
(3.4)
KCL @ node 3
v3 -v2
v3
+
7
5
v3 -v1
= -(-25) = 25
+
4
(3.5)
From Eq. 3.3, 3.4, and 3.5, we get
v1 = 5.412 V, v2 = 7.736 V, and v3 = 46.32 V
Ans.
Circuit -F2003: Nodal and Mesh Analysis
105
BYST
3.1.2 Nodal Analysis with Voltage
Sources
As mentioned earlier, nodal analysis
usually involves with current sources. For
electric circuits containing voltage sources,
some special treatment is needed to
eliminate the redundancy of KVL
equations, since additional information is
available on the node voltages. For
example, if node j and node k are
connected by the voltage source, the vj - vk
is known.
There are two possibilities of connecting
the voltage source:
 A voltage source is connected between
the reference node and a non-reference
node. In this case, we simply set the
Circuit -F2003: Nodal and Mesh Analysis
106
BYST
voltage at the non-reference node equal
to the voltage of the voltage source.
 A voltage source is connected between
the two non-reference nodes. In this
case, the two non-reference nodes form
a "supernode". Both KCL and KVL are
applied to determine the node voltages.
Supernodes
A supernode is a set of nodes connected to
each other by voltage source, but not to the
reference node by a path of voltage source.
vj
5V
vj = 5 V
Reference
node
(a)
Circuit -F2003: Nodal and Mesh Analysis
107
BYST
vj
5V
supernode
vk
vj - vk = 5 V
(b)
Figure 3.4 (a) A voltage source is connected between
the reference node and the non-reference
node. (b) A voltage source is connected
between two non-reference nodes. A
supernode is formed.
The voltage source in Fig. 3.4(a) is
connected between node j and the reference
node. Hence, the node voltage vj equal to 5
V. On the other hand, the voltage source in
Fig. 3.4(b) is connected between two nonreference nodes (node j and k). Hence,
node j and k form a supernode and the
branch voltage vjk (vj - vk) equal to 5 V.
Circuit -F2003: Nodal and Mesh Analysis
108
BYST
Example 3.4
Determine the node voltages v1 in the
following circuit.
-3A
W
v2
v1
22 V
v3
Supernode
3W
-8A
1W
5W
-25A
Reference node
Solution:
If we choose the bottom node as the
reference, then the voltage source will be
connected between two non-reference
nodes. Hence, node 2 and 3 form a
supernode.
Circuit -F2003: Nodal and Mesh Analysis
109
BYST
KCL @ node 1
v1 -v2
v1 -v3
-8 -3 =
+
3
4
(3.6)
KCL @ supernode 2 & 3
v2 -v1
v2
v3 -v1
v3
3+25 =
+
+
+
3
1
4
5
(3.7)
Since the branch voltage between node 2
and 3 must equal to 22 V. That is,
v3 - v2 = 22
V.
(3.8)
From Eq. 3.6, 3.7, and 3.8, the solution for
v1 is 1.071 V.
Ans.
Circuit -F2003: Nodal and Mesh Analysis
110
BYST
Example 3.5
Determine the values of the unknown nodeto-reference voltages in the following
circuit.
2W
+
0.
5W
v2
vx
14A
12 V
Ref.
v1
1W
0.5 vx
v3
y
v
2
0.
5W
2.
+
+
vy
Supernode
v4
Solution:
Only node 3 and 4 form a supernode since
the independent voltage source 12 V is
connected between node 1 and the reference
node.
Circuit -F2003: Nodal and Mesh Analysis
111
BYST
KCL @ node 2
v2 -v1
v2 -v3
+
0.5
2
= 14
(3.9)
KCL @ supernode 3&4
v3 -v2
v4
v4 -v1
= 0.5vx
+
+
2
1
2.5
(3.10)
Since
v3 - v4 = 0.2vy
(3.11)
v4 - v1 = v y
(3.12)
v2 - v1 = v x
(3.13)
and
v1 = -12
(3.14)
From all of above Eqs., we get
v1= -12 V., v2=-4 V., v3 = 0 V. and v4=-2 V.
Ans.
Circuit -F2003: Nodal and Mesh Analysis
112
BYST
3. Mesh Analysis
2
For planar networks the mesh analysis
provides another general procedure for
analyzing electric circuits using mesh
currents as the circuit variables. Basically,
mesh analysis initially assumes that the
currents flowing around the meshes are
independent. A mesh is a circuit loop that
does not enclose any other loop within it. In
the other words, a mesh is the smallest
circuit loop.
A planar network is a network that can be
drawn in a plane with no branches crossing
one another. Fig. 3.5 illustrates examples of
planar and nonplanar networks. Network
in Fig. 3.5(a) and (b) are clearly a planar
network and a nonplanar network,
respectively. The network in Fig. 3.5(c) is
Circuit -F2003: Nodal and Mesh Analysis
113
BYST
Figure 3.5 Examples of planar and nonplanar
networks. (a) and (c) are planar networks
whereas (b) is a nonplanar network.
drawn in such a way as to make it appear
nonplanar. However, it is the planar
network since it can be drawn in plan
without any crossing branches.
A mesh current is defined as a current that
flows only around the perimeter of a mesh.
It is traditionally labeled in the clockwise
direction.
Circuit -F2003: Nodal and Mesh Analysis
114
BYST
3.2.1 Mesh Analysis by Examples
Basic Procedure for Mesh Analysis:
1. Assign all mesh currents.
2. Set up KVL equation for each mesh.
Use Ohm's law to express the voltages in
terms of the mesh currents.
3. If the given circuit contains current
sources on the perimeter of any meshes.
That is, two meshes share current source
in common. Such meshes form a
supermesh to eliminate the redundancy
of KVL equations .
4. Solve KVL equations to determine the
mesh currents.
Circuit -F2003: Nodal and Mesh Analysis
115
BYST
Example 3.6
Determine the values of the mesh current i1
and i2 in the following circuit.
6W
42 V
i1
4W
i2
3W
10 V
Solution:
KVL in mesh 1:
6i1 + 3(i1 - i2) = 42
9i1 - 3i2 = 42
(3.14)
KVL in mesh 2:
4i2 + 3(i2 - i1) = 10
-3i1 + 7i2 = 10
Circuit -F2003: Nodal and Mesh Analysis
116
(3.15)
BYST
From Eq. 3.14 and 3.15, the mesh currents
can be determined as:
i1 = 6 A. and i2 = 4 A.
Ans.
Example 3.7
Determine the values of the mesh current i1,
i2 and i3 in the following circuit.
i2
1W
2W
3W
7V
i1
6V
i3
1W
W
Circuit -F2003: Nodal and Mesh Analysis
117
BYST
Solution:
KVL in mesh 1:
3i1 - i2 - 2i3 = 1
(3.17)
KVL in mesh 2:
-i1 + 6i2 - 3i3 = 0
(3.18)
KVL in mesh 3:
-2i1 - 3i2 + 6i3 = 6
(3.19)
From Eq. 3.17, 3.18 and 3.19, we get
i1 = 3 A. , i2 = 2 A. and i3 = 3 A.
Ans.
Circuit -F2003: Nodal and Mesh Analysis
118
BYST
3.2.2 Mesh Analysis with Current
Sources
Current source on the perimeter of meshes
can exist in two possible cases:
 A current source exists only in one
mesh. Hence, the mesh current is
known. For example, the mesh current
i2 in Fig. 3.6 equal to -5A.
6W
10 V
i1
4W
i2
3W
5A
Figure 3.6 A circuit with a current source existing
only in one mesh.
Circuit -F2003: Nodal and Mesh Analysis
119
BYST
 A current source exists between two
meshes. In this case, a supermesh is
formed. That is, a supermesh results
when two meshes have a dependent or
independent current source in common.
As shown in Fig. 3.7(a), two meshes
share the current source 6A in common.
A supermesh is formed by excluding the
current source as shown in Fig. 3.7(b).
6W
20 V
i1
10W
6W
W
i2
W
20 V
10W
i1
i2
W
6A
x
Exclude
(a) these
elements
(b)
Figure 3.7 (a) A circuit with a current source existing
between two meshes. (b) A supermesh is
formed by excluding the current source.
Circuit -F2003: Nodal and Mesh Analysis
120
BYST
Example 3.8
For the circuit in Fig. 3.7(a), determine the
values of the mesh current i1 and i2 using
mesh analysis.
Solution:
At node x in Fig. 3.7(a), we apply KCL and
get
(3.20)
i2 = i 1 + 6
A supermesh must, however, satisfy KVL
similar to any other mesh. Thus, applying
KVL to the supermesh in Fig. 3.7(b) yields
6i1 + 14i2 = 20
(3.21)
Solving Eq. 3.20 and 3.21, we get
i1 = -3.2 A. and i2 = 2.8 A.
Ans.
Circuit -F2003: Nodal and Mesh Analysis
121
BYST
Example 3.9
For the following circuit, determine the
values of the mesh current i1, i2, and i3 using
mesh analysis.
i2
1W
2W
3W
7V
i1
7A
i3
1W
W
Exclude these
elements
Solution:
Since mesh 1 and 3 share the current source
7A in common. They form a supermesh by
excluding the branch where the current
source is connected.
Circuit -F2003: Nodal and Mesh Analysis
122
BYST
KVL in supermesh:
i1 - 4i2 + 4i3 = 7
(3.22)
KVL in mesh 2:
-i1 + 6i2 - 3i3 = 0
(3.23)
@ the bottom node:
i1 = i 3 + 7
(3.24)
From Eq. 3.22, 3.23 and 3.24, we get
i1 = 9 A. , i2 = 2.5 A. and i3 = 2 A.
Ans.
Circuit -F2003: Nodal and Mesh Analysis
123
BYST
Example 3.10
For the following circuit, determine the
values of the mesh current i1, i2, and i3 using
mesh analysis.
1W
i2
2W
3W
15 A
i1
1
9
+ vx
vx
i3
1W
W
Solution:
KVL in mesh 2:
-i1 + 6i2 - 3i3 = 0
Circuit -F2003: Nodal and Mesh Analysis
124
(3.25)
BYST
From the given circuit, we get
i1 = 15
(3.26)
1
v x = i 3 - i1
9
(3.27)
and
Since
vx = 3(i3 - i2)
(3.28)
Substitute Eq. 3.28 into 3.27, we get
1
2
i2 +
- i1 +
3
3
i3 = 0
(3.29)
From Eq. 3.25, 3.26 and 3.29, we get
i1 = 15 A. , i2 = 11 A. and i3 = 17 A.
Ans.
Circuit -F2003: Nodal and Mesh Analysis
125
BYST