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ELECTRICAL AND ELECTRONIC ELECTRICAL AND ELECTRONIC TECHNOLOGYTECHNOLOGY
((BEX17003BEX17003) )
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Chapter 3(b): Direct Current Circuit
Analysis (II)
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Lecture Contents
Nodal Analysis Nodal Analysis
Mesh/Loop AnalysisMesh/Loop Analysis
Methods of AnalysisMethods of Analysis
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Nodal and Mesh/Loop
Analysis
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Nodal Analysis Nodal analysis – based on the systematic
application of KCL. [important technique – pay attention]
We can analyze any linear circuit by:obtaining a set of simultaneous equationssolved to obtain the required values (voltage or current)we can solve the simultaneous equation either using Cramer’s Rule or any other software such as MATLAB or MathCAD.
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Cont… Nodal analysis provides a general procedure
for analyzing circuit using node voltages as a circuit variables.
Important key idea resistance is a passive element, by the passive sign convention, current must always flow from a higher potential to a lower potential
Rvv lowerhigheri …..( Eq. 3.21)
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Example 3.12
I1
I2
R2
R1 R3v1v2
Consider the figure below using nodal analysis.
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Solution 3.12 1. Label the nodes (including the reference node)
I1
R2
R1R3v1
v2
v = 0
I2
+ +
- -
1 2
i1
i2
i3
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Cont…
Rvlowervhigheri
I1R2
R1 R3
v1 v2
v = 0
I21 2
i1
i2
i3
1
11
0Rvi
2
212 R
vvi
3
23
0Rvi
# Current flows from a higher potential to a lower potential in a resistor
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2. Write the KCL equation for node 1 and 2.
KCL at node 1.I1 = I2 + i1 +i2
KCL at node 2.I2 +i2 = i3
2
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1
121 R
vvRvII
3
2
2
212 R
vR
vvI
R2
R1 R3v1v2
v = 0
I2
+ +
- -
1
i1
i2
i3
2
I1
I2
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Example 3.13 Find the node voltages for the following circuit.
10A
5 A
R1=2 Ω R3=6 Ωv1
v2
1 2R2=4 Ω
v=0V
v3
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Solution 3.13
20v3vvv2v20:(x4)
4vv
20v5
ii5
21
211
211
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KCL at node 1.
KCL at node 2.
v1
605v3v2v601203v3v:(x12)
60v5104
vvi510i
21
221
221
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V1
5 A
10A
R1=2 Ω R3=6 Ωv2
V2R2=4 Ω
5A
i2
i1
10A
i3
5A
v=0V
v3(1)
(2)
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Using the elimination technique
Substituting
20Vv804v :(2) (1)
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20v2
13.33V340v20203v 11
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To use Cramer’s rule, we put the equation in matrix form
Vv
Vv
baba
abab
aaaa
vv
vvvv
2012240
33.1312
160
24060180)20)(3(603603203
16060100)1)(60(520560120
12315)1)(3(535313
6020
5313
6053203
22
11
221
1112
222
2111
2221
1211
2
1
21
21
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Consider the following circuit with current and voltage source. Look for Vb.
[2].................R
0-VbIR
VbVaIII
b, Node[1]................VVa
a, Node
3B
2
3B2
B
∑I entering the node = ∑I leaving the node
I1
I2
I3
IB
VB IB
2R1
R3R
a b
c
Va Vb
Vc=0
Example 3.14
1632
2
BB
b
2
BB
32b
2
BB
3
b
2
b
B3
b
2
b
2
B
B3
b
2
b
2
B
3
bB
2
bB
R1
R1
RVI
V
RVI
R1
R1V
RVI
RV
RV
IRV
RV
RV-
IRV
RV
RV
RVI
RVV
[2] into [1] substitute
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Example 3.15Determine the voltages for the nodes in the following figure.
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2 8
43A 2ix
1 2 3
ix
0
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Solution 3.15
-(3)------------- 032:)3/8(2
)(284
v-v 2
3, nodeAt -(2)--------------- 0v-7v-4v:8)(
40-v
8v-v
2v-v
2, nodeAt (1)------------- 1223:)4(
243 3
1, nodeAt
321
21323121
321
2322132
321
21311
vvv
vvvviii
iii
vvv
vvvvii
x
x
x
4
843A
2ixV1 V2 V3
1
i1
I=3A
ix 2 i2
i3
3i1
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(5)-------2vv 04v2v-
(3) and Eq.(2) Adding(4)2.45
12v-v
or125v-5v
2121
21
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:Eq(3) and (1) Eq Addingtechnique, neliminatio the Using
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-2.4Vv 2.4V,v 4.8V, vThus,
-2.4V-v4v-3v2v-3vvget, we (3) Eq. From
4.8V2vv 2.4,v 2.4v-2vyield (4) into Eq.(5) ngSubstituti
321
222123
21222
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Nodal Analysis with Voltage Sources
Important Supernode – formed by enclosing a
(dependent or independent) voltage source connected between two non-reference nodes and any elements connected in parallel with it.
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Mesh/Loop Analysis
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Mesh/Loop Analysis Mesh analysis provides another general
procedure for analyzing circuits. Recall that a loop is a closed path with
no node passed more than once. A mesh is a loop that does not contain
any other loop within it. Mesh analysis apply KVL to find
unknown currents
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Cont… Mesh analysis is not quite as general as
nodal analysis, because it is only applicable for a circuit that is planar.
A planar circuit is one that can be drawn in a plane with no branches crossing one another; otherwise it is nonplanar.
For better understanding, refer to Sadiku and Alexander, 2nd Edition page 90-91.
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Example 3.17
Write the mesh equation for the following circuit.
1VB
2R1R
5R 2VB4R
3R
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Solution 3.17
)1(V)(-RI)R(RI
0)RI-(IRIV-0VVV-
: 1 loopfor KVL0 clockwise V
loop, allFor
B142411
42111B1
R4R1B1
1VB
R31R
R42VB
R2
R5
1I
2I 3I
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1VB
R31R
R42VB
R2
R5
1I
2I 3I
-(2)---------0)(-RI)RR(RI)(-RI0)RI-(IRI)RI-(I
0VVV: 2 loopfor KVL
53542241
53222412
R5R2R4
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1VB
R31
R
R42
VB
R2
R5
1I
2I 3
I
)3(V)R(RI)(-RIVRI)RI-(I
0VVV: 3 loopfor KVL
B253352
B233523
B2R3R5
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2
1
3
2
1
535
55424
441
B253352
53542241
B142411
00
0matrix,In
)3(V)R(RI)(-RI-(2)---------0)(-RI)RR(RI)(-RI
)1(V)(-RI)R(RI
B
B
V
V
III
RRRRRRRR
RRR
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Example 3.18Find Vo for the following circuit.
4A
2 6 5V
873A Vo
313.04V(8)IV I
8 (21)I 0 8I4)-6(I5-3)7(I08I)I-6(I5-)I-7(I
0 VV5-V 0 clockwise V
:3 Loop forKVL 3AI
:Loop2 For4A I
:Loop1 For
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3
3
3 33
31323
867
2
1
A38.0
4A
2 6 5V
873A Vo
I1
I2 I3
Solution 3.18
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Mesh Analysis with Current SourcesCase 1: When a current source exists only in one mesh
Loop 2: I2= -5A
10V
4Ω 3Ω
6ΩI1 I2 5AI1= -2A
-10+4i1+6(i1-i2) = 0
Loop 1:
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Nodal vs. Mesh AnalysisHow do we know which method is
better or more efficient?
Nodal analysis is normally used when a circuit has fewer node equations than mesh equations.
Mesh analysis is normally used when a circuit has fewer mesh equations than node equations.
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Nodal vs. Mesh Analysis(Nature of network)
Nodal AnalysisNetworks that contain many :series-connected elementsvoltage sourcessupernodes
Mesh AnalysisNetworks that contain
many : parallel-connected elements current sources supermeshes
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Nodal vs. Mesh Analysis(Information required)
If node voltages are required, it maybe expedient to apply nodal analysis
If branch or mesh currents are required, it maybe expedient to apply mesh analysis