Circuits (MTE 120) (Spring 2010)
Akrem El-ghazal
http://pami.uwaterloo.ca/~akrem/
University of Waterloo,
Electrical and Computer Engineering Dep.
Nodal Analysis
Objective:
To analyze circuits using a systematic technique: the nodal analysis.
Node Voltages:
In the figure below, V means the voltage of the node a with respect to node r. If we
consider the node r as a reference node, then V is simply written as V , because the voltage
of the reference node is zero.
V means the voltage of the node b with respect to node r. If we consider the node r as
a reference node, then V is simply written as V , because the voltage of the reference bode
is zero.
Node voltages
Var
Vbr
RΩ
+
-
+
-
Vr=0, we can redraw the circuit
as follows
Vb
RΩ
Va
+
-
+
-
r
0V
Remark:
A current through a resistor can be calculated by using node-voltages as follows:
Node votlages
V1
i12
I12=
R
V1-V2
R
Node votlages
V2
V1
i21
I21=
R
V2-V1
R
V2
Circuits (MTE 120) (Spring 2010)
Akrem El-ghazal
A voltage across a resistor can be calculated by using node-voltages as follows:
Nodal Analysis
The nodal analysis is a systematic way of applying KCL at each essential node of a circuit and
represents the branch current in terms of the node voltages. This will give us a set of equations
that we solve together to find the node voltages. Once we find the node voltages we can use
them to calculate any other currents or voltages of interest.
Case #1: Circuits with Independent Current Sources.
Example #1:
Find the node voltages of the following circuit.
Solution
Step #1
Identify all of the essential nodes and choose one of them as a reference node.
Step #2
Assign voltages variable to all nodes except the reference node.
Step #3
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Circuits (MTE 120) (Spring 2010)
Akrem El-ghazal
Apply the KCL at each node except the reference node.In this step for each node we assume the
branch current is leaving from the node, and then we describe the branch current in term of
node voltages.
Apply KCL at Node # 1
Node#1 has three branches, so we apply the KCL to obtain an equation with three terms as
follows:
V 0 V V
0
1 6
12
1
3V 2V 12
Apply KCL at Node # 2
Node#2 has three branches, so we apply the KCL to obtain an equation with three terms as
follows:
V 0 V V
0
6
6
2
V 2V 24
4
Step #4
Solving question 1 and 2 for the unknown node voltages (V1 and V2):
V 6V
V 15 V
3
Circuits (MTE 120) (Spring 2010)
Akrem El-ghazal
Example #2:
Use the nodal analysis to find the ix , iy , iz , in , im , Va , Vb , Vc , Vd .
Solution:
First we apply the nodal analysis technique in order to find the node voltages, then we use the
node voltages to calculate: ix , iy , iz , in , im , Va , Vb , Vc , Vd .
Step #1
Identify all of the essential nodes and choose one of them as a reference node.
Step #2
Assign voltages variable to all nodes except the reference node.
Step #3
Apply KCL at each node except the reference node.In this step for each node we assume the
branch current is leaving from the node, and then we describe the branch current in term of
node voltages.
Apply KCL at Node # 1
Node#1 has three branches, so we apply the KCL to obtain an equation with three terms as
follows:
4
Circuits (MTE 120) (Spring 2010)
Akrem El-ghazal
V V V V V
0
10
5
20
0.35 V 0.2 V 0.05 V 0
1
Apply KCL at Node # 2
Node#2 has three branches, so we apply KCL to obtain an equation with three terms as follows:
V V V V
0
5
10
0.2 V 0.3 V 0.1 V 10
10 2
Apply KCL at Node # 2
Node#3has three branches, so we apply KCL to obtain an equation with three terms as follows:
V V V V V
0
10
20
5
0.05 V 0.1 V 0.35 V 0
3
Step #4
Solving the questions 1 ,2 and 3 for the unknown node voltages( V1 ,V2 and V3):
V 45.45 V, V 72.73 V , V 27.27 V
After we find the node voltages, we can use the node voltages to calculate any other
unknowns:
Now, let us use the node voltages to find the unknown currents ix , iy , iz , in , im :
V V 45.45 27.27
i 0.909A
20
20
i V V 45.45 72.73
1.364A
20
20
i V V 27.27 72.73
2.273 A
20
20
5
Circuits (MTE 120) (Spring 2010)
Akrem El-ghazal
V 27.27
5.454 A
5
5
V 45.45
i 4.545 A
10
10
i Now, let us use the node voltages to calculate the unknown voltages: , Va , Vb , Vc and Vd .
V V 27.27 V
V V 45.45V
V V V 45.45 27.27 18.18 V
V" V V 27.27 72.73 45.46 V
Case #2: Circuits with Independent Voltage Sources
(a) If the voltage sources are connected to the reference node
Example #3:
Find i0 by using the nodal analysis:
Solution:
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Circuits (MTE 120) (Spring 2010)
Akrem El-ghazal
We do not need to apply the KCL at Node 1, because the node voltage at this node is known,
V1=12 v.
Also, we do not need to apply the KCL at Node 3, because the node voltage at this node is
known V3=-6 v. Notice that the V3 is negative because its polarity is opposite to the polarity of
the voltage source. Notice that, for node voltages the negative sign of the polarity is at the
reference node.
Now, we have only one unknown node voltage (V2).
Apply KCL at node #2:
V V V V V
0
6k
12 k
12 k
Substitute the values of V1 and V3:
V V 12 V 6
0
6k
12 k
12 k
Solve for V2
3
V v
2
(b) If the Voltage Source Are Connected Between Two Unreferenced Nodes
Example #4:
Find the node voltages for the following circuit.
Solution:
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Circuits (MTE 120) (Spring 2010)
Akrem El-ghazal
Since we can not express the current through the voltage source in term of node voltages, we
combine the two nodes 2 and 3 as one supernode (as shown in the figure below) and relate
one of the node voltages (V2 or V3) in term of the other one as follows:
V V 22
Then
V V 22
Now, apply the KCL at the supernode:
We have six branches connected to the supernode (as indicated in the above figure):
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Circuits (MTE 120) (Spring 2010)
Akrem El-ghazal
V V V
V 22 V
V 22
3
25 0
1
1
1
1
3
4
5
7 V 13 V 170
1
Apply the KCL at node #1
1
We have four branches connected to node #1:
V V 22
V V
3
0
8
1
1
4
3
7 V 7 V 77
2
Solving equations 1 and 2 for V1 and V2:
V 4.5 V
V 15.5 V
Then
V V 22 15.5 22 6.5 V
Example #5:
In the pervious example, find the current through the 22-voltage source?
Solution:
We assume the direction of the current as show below (you can assume the opposite
direction), then we apply the KCL at node 2 or 3 to find the current i% :
Apply the KCL at node #2:
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Circuits (MTE 120) (Spring 2010)
Akrem El-ghazal
V V V
3 i% 0
1
1
3
V V V
15.5 15.5 4.5
i% 3
3 51.5 A
1
1
1
1
3
3
Also, we can find i% by applying the KCL at node #3
V 22 V
V 22
25 i% 0
1
1
4
5
V 22 V
V 22
i% 25 1
1
4
5
15.5 22 4.5
15.5 22
25 51.5 A
1
1
4
5
Example #6: (an extra problem)
Write the node-voltage equations for the following circuits.
Solution
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Circuits (MTE 120) (Spring 2010)
Akrem El-ghazal
1/4 Ω
1/3 Ω
V1
2v
V2
1/6 Ω
- +
1 Ω
8A
V3
-25 A
1/5 Ω
In order to find the current in the (1/6 Ω) resistor, we assume a node voltage Vx between the
voltage source and the (1/6 Ω) resistor (as show in the figure below) and relate Vx to the node
voltage V3 and the value of the voltage source (2 v) as follows:
V V 2
Then,
V V 2
Now, we can easily find the current through the voltage source and (1/6 Ω) resistor in term of
node voltages ((V 2 and V .
1/4 Ω
1/3 Ω
V1
V2 1/6 Ω
Vx=V3-2
2v
1 Ω
8A
- +
V3
-25 A
1/5 Ω
Apply the KCL at node #3
25 V
V V V 2 V
0
1
1
1
4
6
5
4 V 6 V 15 V 37
11
1
Circuits (MTE 120) (Spring 2010)
Akrem El-ghazal
Apply the KCL at node #2
V
V V V V 2
0
1
1
1
3
6
2
3 V 10V 6 V 12
Apply the KCL at node #1
V V V V
0
1
1
4
3
7 V 3 V 4 V 8
8
3
Case #3 Nodal Analysis for Circuits with Dependent Sources
For circuits that include dependent sources, first we ignore the fact that the dependent
source is a dependent source and we write the node-voltage equations as we would for a
circuit with independent sources. The node-voltage equations will have extra unknown
variables for the dependent sources beside to the normal unknown node voltages. All the extra
unknown variable of the dependent sources must be eliminated by writing an expression for
each one in term of the node voltages. There MUST be a relation between the unknown
variable of the dependent and the node voltages, because node voltages can be used to
describe any current or voltage of any branch in the circuit. The following examples will show
you how to apply the nodal analysis for circuits with dependent sources.
(a) Nodal Analysis for Circuits with Dependent Current Sources
Example #7:
Find the node voltages for the following:
Solution
12
Circuits (MTE 120) (Spring 2010)
Akrem El-ghazal
Apply the KCL at node # 1:
2 i* V
V V
0
12 k
6k
a
The above equation has an extra unknown variable i* . We should relate the extra
unknown variable of the dependent source to the node voltages. From the above figure it is
clear that:
V
i* b
3k
Substitute i* in equation a:
V
V
V V
2x
0
1
3 k 12 k
6k
V 2 V 0
Apply the KCL at node #2:
V V V
2 mA 0
3k
6k
V 3 V 12
Solving for the node voltages
24
V V
5
12
V V
5
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Circuits (MTE 120) (Spring 2010)
Akrem El-ghazal
(b) Nodal Analysis for Circuits with Dependent Voltage Sources
Example #8:
Find the current I0 by using the nodal analysis.
6Ω
2 Vx
12 kΩ
+
+
6Ω
12 Ω
I0
Vx
+
-
6v
-
Solution
Since there is a voltage source between two unreferenced nodes, we combine the nodes in one
supernode.
Then, we relate one of the node voltages (V1) to the other one (V2) using the value of the
voltage source between them. From the above figure:
V V 2V
Then,
V V 2V
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Circuits (MTE 120) (Spring 2010)
Akrem El-ghazal
Apply the KCL at the supernode:
The supernode has four branches so the node equation should have four terms:
V 2 V V 2 V V V V V
0 a
12 k
6k
6k
12 k
From the figure it is clear that V 6, so we do have to write the node-voltage equation for
node #3.
Substitute the value of V3, in equation a :
V 2 V V 2 V 6 V V 6
0 b
12 k
6k
6k
12 k
The above equation has an extra unknown variable ./ . We should relate the extra unknown
variable of the dependent source to the node voltages. From the above figure it is clear that:
V V
Substitute V in equation b:
V 2 V V 2 V 6 V V 6
0
12 k
6k
6k
12 k
V 2 V 2V 2 V 6 2 V V 6 0
Solving for V2
V 1.5 V
and
V V 1.5 .
Then,
V V 2V
V 1.5 2 0 1.5 4.5 V
I* V 4.5
0.375 A
12 12
15