1. Basic Laws of Electric Circuits
Nodal Analysis
Lesson 6

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: Eq 6.1
Figure 6.1: Partial circuit used to illustrate nodal analysis.
Eq 6.1 3

5. Basic Circuits Nodal Analysis: Concept Illustration:
Clearing the previous equation gives,
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.
Figure 6.2: Circuit for Example 6.1. 5

7. Basic Circuits Nodal Analysis: Example 6.1, the nodal equations.

8. Basic Circuits Nodal Analysis: Example 6.1: Set up for solution.
Eq 6.3
Eq 6.4
Eq 6.5
Eq 6.6 7

9. Basic Circuits Nodal Analysis: Example 6.2, using circuit values.
Figure 6.3: Circuit for
Example 6.2.
Find V1 and V2.
At v1:
Eq 6.7
At v2:
Eq 6.8 8

10. Basic Circuits Nodal Analysis: Example 6.2: Clearing Equations;
From Eq 6.7:
V1 + 2V1 – 2V2 = 20 or
Eq 6.9
3V1 – 2V2 = 20
From Eq 6.8:
4V2 – 4V1 + V2 = -120 or
-4V1 + 5V2 = -120
Eq 6.10
Solution: V1 = -20 V, V2 = -40 V 9

11. Basic Circuits Nodal Analysis: Example 6.3: With voltage source.
Figure 6.4: Circuit for
Example 6.3.
At V1:
Eq 6.11
At V2:
Eq 6.12 10

12. Basic Circuits Nodal Analysis: Example 6.3: Continued.
Collecting terms in Equations (6.11) and (6.12) gives
Eq 6.13
Eq 6.14 11

13. Basic Circuits
Nodal Analysis: Example 6.4: Numerical example with voltage
source.
Figure 6.5: Circuit for Example 6.4.
What do we do first? 12

14. Basic Circuits Nodal Analysis: Example 6.4: Continued At v1: Eq 6.1513

15. Basic Circuits Nodal Analysis: Example 6.4: Continued
Clearing Eq 6.15
4V1 + 10V – 10V2 = -200 or
14V1 – 10V2 = -300
Eq 6.17
Clearing Eq 6.16
4V2 + 6V2 – 60 – 6V1 = 0 or
-6V1 + 10V2 = 60
Eq 6.18
V1 = -30 V, V2 = -12 V, I1 = -2 A 14

16. Basic Circuits Nodal Analysis: Example 6.5: Voltage super node.
Given the following circuit. Solve for the indicated nodal voltages.
super node
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. 15

17. Basic Circuits Nodal Analysis: Example 6.5: Continued.
Constraint Equation
V2 – V3 = -10
Eq 6.19
At V1
Eq 6.20
At super
node
Eq 6.21 16

18. Basic Circuits Nodal Analysis: Example 6.5: Continued. Eq 6.22
Clearing Eq 6.19, 6.20, and 6.21:
Eq 6.22
7V1 – 2V2 – 5V3 = 60
-14V1 + 9V2 + 12V3 = 0
Eq 6.23
V2 – V3 = -10
Eq 6.24
Solving gives:
V1 = 30 V, V2 = V, V3 = V 17

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.
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
Eq 6.25 18

20. Basic Circuits Nodal Analysis: Example 6.6: With Dependent Sources.
At node V1
At node V2
The constraint equation: 19

21. Basic Circuits Nodal Analysis: Example 6.6: With Dependent Sources.
Clearing the previous equations and substituting
the constraint VX = V1 - V2 gives,
Eq 6.26
Eq 6.27
which yields, 20

22. circuits
End of Lesson 6
Nodal Analysis