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Lesson 6
Basic Laws of Electric Circuits
Nodal Analysis
Basic CircuitsNodal 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
Basic CircuitsNodal 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
Basic CircuitsNodal Analysis: Concept Illustration:
r e fe r e n c e n o d e
v 1v 2 v 3
R 2
R 1R 3
R 4
I
Figure 6.1: Partial circuit used to illustrate nodal analysis.
IR
VV
R
V
R
V
R
VV
4
31
3
1
1
1
2
21 Eq 6.1
3
Basic Circuits
Nodal Analysis: Concept Illustration:
Clearing the previous equation gives,
IVR
VR
VRRRR
3
42
21
4321
111111 Eq 6.2
We would need two additional equations, from theremaining circuit, in order to solve for V1, V2, and V3
4
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 6I1
v 1 v 2
+
_
v 6
Figure 6.2: Circuit for Example 6.1.
5
Basic Circuits
Nodal Analysis: Example 6.1, the nodal equations.
R 2 R 3
R 1 R 4
R 5
R 6I1
v 1 v 2
+
_
v 6
065
2
4
2
3
12
13
21
21
1
RR
V
R
V
R
VV
IR
VV
RR
VEq 6.3
Eq 6.46
Basic CircuitsNodal Analysis: Example 6.1: Set up for solution.
065
2
4
2
3
12
13
21
21
1
RR
V
R
V
R
VV
IR
VV
RR
V
01111
111
2
6543
1
3
12
3
1
321
VRRRR
VR
IVR
VRRR
Eq 6.3
Eq 6.4
Eq 6.5
Eq 6.6
7
Nodal Analysis: Example 6.2, using circuit values.
v1v2
1 0
5
2 0 4 A
2 A
Figure 6.3: Circuit for Example 6.2.
Find V1 and V2.
At v1:
2510
211
VVV
At v2:
6205
212 VVV
Eq 6.7
Eq 6.8
Basic Circuits
8
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
Basic CircuitsNodal Analysis: Example 6.3: With voltage source.
R 1
R 3
I
v2v1
+_ R 2 R 4E
Figure 6.4: Circuit for Example 6.3.
At V1:
IR
VV
R
V
R
EV
3
21
2
1
1
1
At V2:
IR
VV
R
V
3
12
4
2
Eq 6.11
Eq 6.12
10
Basic CircuitsNodal Analysis: Example 6.3: Continued.
Collecting terms in Equations (6.11) and (6.12) gives
12
3
11
3
1
2
1
1
1
R
EIV
RV
RRR
IVRR
VR
2
4
1
3
11
2
1
Eq 6.13
Eq 6.14
11
Basic CircuitsNodal Analysis: Example 6.4: Numerical example with voltagesource.
v2v1
6
4
1 0 5 A
+ _
1 0 V
Figure 6.5: Circuit for Example 6.4.
What do we do first?
12
Basic CircuitsNodal Analysis: Example 6.4: Continued
v2v1
6
4
1 0 5 A
+ _
1 0 V
At v1:
54
2101101
VVV
At v2:
04
110262
VVV
Eq 6.15
Eq 6.16
13
Basic CircuitsNodal Analysis: Example 6.4: Continued
Clearing Eq 6.15
4V1 + 10V1 + 100 – 10V2 = -200
or
14V1 – 10V2 = -300
Clearing Eq 6.16
4V2 + 6V2 – 60 – 6V1 = 0
or-6V1 + 10V2 = 60
Eq 6.17
Eq 6.18
V1 = -30 V, V2 = -12 V, I1 = -2 A14
Basic CircuitsNodal Analysis: Example 6.5: Voltage super node.
Given the following circuit. Solve for the indicated nodal voltages.
+_
6 A
5
4
2
1 0
v 1v 2 v 3
1 0 V
x
x
xx
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 thevoltage source and the tips of the branches connected to the nodes.
super node
15
Basic CircuitsNodal Analysis: Example 6.5: Continued.
+_
6 A
5
4
2
1 0
v 1v 2 v 3
1 0 V
At V1 62
315
21
VVVV
At supernode
02
1310
342
512
VVVVVV
Constraint Equation
V2 – V3 = -10 Eq 6.19
Eq 6.20
Eq 6.21
16
Basic CircuitsNodal Analysis: Example 6.5: Continued.
Clearing Eq 6.19, 6.20, and 6.21:
7V1 – 2V2 – 5V3 = 60
-14V1 + 9V2 + 12V3 = 0
V2 – V3 = -10
Eq 6.22
Eq 6.23
Eq 6.24
Solving gives:
V1 = 30 V, V2 = 14.29 V, V3 = 24.29 V
17
Basic CircuitsNodal Analysis: Example 6.6: With Dependent Sources.
Consider the circuit below. We desire to solve for the node voltagesV1 and V2.
1 0
2
4
5
2 A
+_1 0 V
5 V x
v1 v2
V x +_
Figure 6.7: Circuit for Example 6.6.
In this case we have a dependent source, 5Vx, that must be reckonedwith. Actually, there is a constraint equation of
012 VVV x Eq 6.25
18
Basic CircuitsNodal Analysis: Example 6.6: With Dependent Sources.
1 0
2
4
5
2 A
+_1 0 V
5 V x
v1 v2
V x +_
At node V1 22510
10 2111
VVVV
At node V22
4
5
2212
xVVVV
The constraint equation: 21 VVVx 19
Basic CircuitsNodal Analysis: Example 6.6: With Dependent Sources.
Clearing the previous equations and substituting the constraint VX = V1 - V2 gives,
887
3058
21
21
VV
VV Eq 6.26
Eq 6.27
which yields,
VVVV 03.5,9.6 21
20
circuits
End of Lesson 6Nodal Analysis