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ETB LO1 Advanced Network Theorems (Part 1).pdf
1. 11/21/2017
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DEE1213
ELECTRICAL TECHNOLOGY B
Lecture #1
Advanced Network Theorems (Part 1)
1
Subject Learning Outcome (SLO)
• This Lecture partially contributing to the
fulfillment of the following SLO:
– Use circuit theory to analyze the three phase
power system.
2
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Lecture Learning Outcomes
Upon completion of this lecture, you will be able to:
• Apply Superposition Theorem to solve complex circuitry problems
3
Pre-Test
• Determine the current through the 6 Ω
resistor of the network below.
4
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Kirchhoff’s Voltage Law
• Consideration 1:
– We define a voltage drop as positive if we enter
the positive terminal and leave the negative
terminal.
7
+ _
v1
The drop moving from left to
right above is + v1.
+
_ v1
The drop moving from left to
right above is −v1.
Kirchhoff’s Voltage Law
• Consider the circuit below:
8
_ v2
+
+
+
+
_ _
_
v1 v4
v3
•
“a”
CW (from “a”):
- v1 – v2 + v4 + v3 = 0
CCW (from “a”):
- v3 – v4 + v2 + v1 = 0
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Nodes, Branches, Loops and
Current Division
• A branch is a single electrical element or device.
9
• A node can be defined as a connection point
between two or more branches.
Current Division
10
I
I2 I1
R2 R1
+
_
V
1 2
1 2
V V
I I I
R R
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Important!!
• All methods of analysis taught in ETA and ETB
are able to solve most of analytical problems
in circuit with DC sources.
• But, also circuits with AC sources!!
• Plus, circuits with mixture of DC and AC
sources!!
13
Mesh Analysis
• In formulating mesh analysis we assign a mesh
current to each mesh.
• Mesh currents are sort of fictitious in that a
particular mesh current does not define the
current in each branch of the mesh to which it
is assigned.
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I1 I2 I3
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Mesh Analysis
• Example:
15
+
_
10V
4 2
6
7
2V 20V
I1 I2
+
+
_
_
Loop 1: 4I1 + 6(I1 – I2) = 10 - 2
Loop 2: 6(I2 – I1) + 2I2 + 7I2 = 2 + 20
I1 = 2.2105 A
I2 = 2.3509 A
Nodal Analysis
16
R2 R3
R1 R4
R5
R6
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
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Electrical Technology B
Superposition Theorem
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Superposition
• Consider the circuit below that contains two
voltage sources.
• We assume that V1and V2 acting together
produce current I. 18
V1
V2
R1
R2
R3
+
+
_
_
I
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Superposition
• Superposition states that the current, I,
produced by both sources acting together is the
same as the sum of the currents, I1 + I2, where
I1 is produced by V1 and I2 is produced by V2.
19
V1
R1
R2
R3
+
_
I1
V2
R1
R2
R3
I2
+
_
V1 produces current I1 V2 produces current I2
Source Replacement
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Source Replacement: Example
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Superposition: Example 1
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2 3
VA
VB
VC
+
+
+
_
_
_
VA = 10 V, VB = 5 V, VC = 15 V
IT
With all sources acting: I`T = 6 A
With VA + VB acting, VC = 0: IA+B = 3 A
With VC acting, VA + VB = 0: IC = 3 A
We see that superposition holds.
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Superposition: Example 2
• Find the current I by using superposition.
1. First, deactivate the source IS and find I in the 6 resistor.
2. Second, deactivate the source VS and find I in the 6
resistor.
3. Sum the two currents for the total current. 23
I
6 VS = 54 V
IS = 3 A
12
+
_
Superposition: Example 2
• Step 1:
24
IVs
6 VS = 54 V
12
+
_
IVs = 3 A
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Superposition: Example 2
25
Is
6
IS = 3 A
12
Total current I: I = IS + Ivs = 5 A
• Step 2:
• Step 3:
A
I
I
s
s
2
)
12
6
(
12
3
Post-Test
26
• Determine the current through the 6 Ω
resistor of the network below.
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Homework
• Determine the current through the 2 Ω
resistor of the network below. (1 A)
29
Pre Lecture #2 Reading
• Thevenin’s & Norton’s Theorem (SLT: 2 hours,
Non F2F)
