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Basic Laws of Electric Circuits

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Basic Laws of Electric Circuits Nodal Analysis Lesson 6 Basic Circuits Nodal Analysis: The Concept. Every circuit has n nodes with one of the nodes being designated ... – PowerPoint PPT presentation

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Title: Basic Laws of Electric Circuits


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 Kirchhoffs 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
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.
Eq 6.3
Eq 6.4
6
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 6.15
At v2
Eq 6.16
13
15
Basic Circuits
Nodal Analysis Example 6.4 Continued
Clearing Eq 6.15
4V1 10V1 100 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.
Clearing Eq 6.19, 6.20, and 6.21
Eq 6.22
7V1 2V2 5V3 60
-14V1 9V2 12V3 0
Eq 6.23
Eq 6.24
V2 V3 -10
Solving gives
V1 30 V, V2 14.29 V, V3 24.29 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
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