2. Basic Circuits
Nodal Analysis: The Concept.
• Every circuit has n nodes with one of the nodes being
designated as a reference node.
• We designate the remaining n – 1 nodes as voltage nodes
and give each node a unique name, vi.
• At each node we write Kirchhoff’s current law in terms
of the node voltages.
1
3. Basic Circuits
Nodal Analysis: The Concept.
• We form n-1 linear equations at the n-1 nodes
in terms of the node voltages.
• We solve the n-1 equations for the n-1 node voltages.
• From the node voltages we can calculate any branch
current or any voltage across any element.
2
4. Basic Circuits
Nodal Analysis: Concept Illustration:
reference node
v 1
v 2 v 3
R 2
R 1
R 3
R 4
I
Figure 6.1: Partial circuit used to illustrate nodal analysis.
I
R
V
V
R
V
R
V
R
V
V
4
3
1
3
1
1
1
2
2
1 Eq 6.1
3
5. Basic Circuits
Nodal Analysis: Concept Illustration:
Clearing the previous equation gives,
I
V
R
V
R
V
R
R
R
R
3
4
2
2
1
4
3
2
1
1
1
1
1
1
1
Eq 6.2
We would need two additional equations, from the
remaining circuit, in order to solve for V1, V2, and V3
4
6. Basic Circuits
Nodal Analysis: Example 6.1
Given the following circuit. Set-up the equations
to solve for V1 and V2. Also solve for the voltage V6.
R 2 R 3
R 1 R 4
R 5
R 6
I1
v1 v2
+
_
v6
Figure 6.2: Circuit for Example 6.1.
5
7. Basic Circuits
Nodal Analysis: Example 6.1, the nodal equations.
R 2 R 3
R 1 R 4
R 5
R 6
I1
v1 v2
+
_
v6
0
6
5
2
4
2
3
1
2
1
3
2
1
2
1
1
R
R
V
R
V
R
V
V
I
R
V
V
R
R
V
Eq 6.3
Eq 6.4
6
8. Basic Circuits
Nodal Analysis: Example 6.1: Set up for solution.
0
6
5
2
4
2
3
1
2
1
3
2
1
2
1
1
R
R
V
R
V
R
V
V
I
R
V
V
R
R
V
0
1
1
1
1
1
1
1
2
6
5
4
3
1
3
1
2
3
1
3
2
1
V
R
R
R
R
V
R
I
V
R
V
R
R
R
Eq 6.3
Eq 6.4
Eq 6.5
Eq 6.6
7
9. Nodal Analysis: Example 6.2, using circuit values.
v1
v2
10
5
20 4 A
2 A
Figure 6.3: Circuit for
Example 6.2.
Find V1 and V2.
At v1:
2
5
10
2
1
1
V
V
V
At v2:
6
20
5
2
1
2
V
V
V
Eq 6.7
Eq 6.8
Basic Circuits
8
10. Nodal Analysis: Example 6.2: Clearing Equations;
From Eq 6.7:
V1 + 2V1 – 2V2 = 20
or
3V1 – 2V2 = 20
From Eq 6.8:
4V2 – 4V1 + V2 = -120
or
-4V1 + 5V2 = -120
Eq 6.9
Eq 6.10
Solution: V1 = -20 V, V2 = -40 V
Basic Circuits
9
11. Basic Circuits
Nodal Analysis: Example 6.3: With voltage source.
R 1
R 3
I
v2
v1
+
_ R 2 R 4
E
Figure 6.4: Circuit for
Example 6.3.
At V1:
I
R
V
V
R
V
R
E
V
3
2
1
2
1
1
1
At V2:
I
R
V
V
R
V
3
1
2
4
2
Eq 6.11
Eq 6.12
10
12. Basic Circuits
Nodal Analysis: Example 6.3: Continued.
Collecting terms in Equations (6.11) and (6.12) gives
1
2
3
1
1
3
1
2
1
1
1
R
E
I
V
R
V
R
R
R
I
V
R
R
V
R
2
4
1
3
1
1
2
1
Eq 6.13
Eq 6.14
11
13. Basic Circuits
Nodal Analysis: Example 6.4: Numerical example with voltage
source.
v2
v1
6
4
10 5 A
+
_
10 V
Figure 6.5: Circuit for Example 6.4.
What do we do first?
12
14. Basic Circuits
Nodal Analysis: Example 6.4: Continued
v2
v1
6
4
10 5 A
+
_
10 V
At v1:
5
4
2
10
1
10
1
V
V
V
At v2:
0
4
1
10
2
6
2
V
V
V
Eq 6.15
Eq 6.16
13
16. Basic Circuits
Nodal Analysis: Example 6.5: Voltage super node.
Given the following circuit. Solve for the indicated nodal voltages.
+
_
6 A
5
4
2
10
v 1
v 2 v 3
1 0 V
x
x
x
x
Figure 6.6: Circuit for Example 6.5.
When a voltage source appears between two nodes, an easy way to
handle this is to form a super node. The super node encircles the
voltage source and the tips of the branches connected to the nodes.
super node
15
17. Basic Circuits
Nodal Analysis: Example 6.5: Continued.
+
_
6 A
5
4
2
10
v1
v2 v3
1 0 V
At V1 6
2
3
1
5
2
1
V
V
V
V
At super
node
0
2
1
3
10
3
4
2
5
1
2
V
V
V
V
V
V
Constraint Equation
V2 – V3 = -10 Eq 6.19
Eq 6.20
Eq 6.21
16
19. Basic Circuits
Nodal Analysis: Example 6.6: With Dependent Sources.
Consider the circuit below. We desire to solve for the node voltages
V1 and V2.
1 0
2
4
5
2 A
+
_
10 V
5 Vx
v1 v2
Vx
+
_
Figure 6.7: Circuit for
Example 6.6.
In this case we have a dependent source, 5Vx, that must be reckoned
with. Actually, there is a constraint equation of
0
1
2
V
V
V x Eq 6.25
18
20. Basic Circuits
Nodal Analysis: Example 6.6: With Dependent Sources.
1 0
2
4
5
2 A
+
_
1 0 V
5 Vx
v1 v2
Vx
+
_
At node V1 2
2
5
10
10 2
1
1
1
V
V
V
V
At node V2
2
4
5
2
2
1
2
x
V
V
V
V
The constraint equation: 2
1 V
V
Vx
19
21. Basic Circuits
Nodal Analysis: Example 6.6: With Dependent Sources.
Clearing the previous equations and substituting
the constraint VX = V1 - V2 gives,
8
8
7
30
5
8
2
1
2
1
V
V
V
V Eq 6.26
Eq 6.27
which yields,
V
V
V
V 03
.
5
,
9
.
6 2
1
20
