1 Mesh Analysis Discussion D2.4 Chapter 2 Section 2-8.
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Transcript of 1 Mesh Analysis Discussion D2.4 Chapter 2 Section 2-8.
![Page 1: 1 Mesh Analysis Discussion D2.4 Chapter 2 Section 2-8.](https://reader030.fdocuments.net/reader030/viewer/2022032523/56649d775503460f94a59bfd/html5/thumbnails/1.jpg)
1
Mesh Analysis
Discussion D2.4
Chapter 2
Section 2-8
![Page 2: 1 Mesh Analysis Discussion D2.4 Chapter 2 Section 2-8.](https://reader030.fdocuments.net/reader030/viewer/2022032523/56649d775503460f94a59bfd/html5/thumbnails/2.jpg)
2
Mesh Analysis• Mesh analysis applies KVL to find unknown
currents. • It is only applicable to planar circuits (a circuit that
can be drawn on a plane with no branches crossing each other).
• A mesh is a loop that does not contain any other loops.
• The current through a mesh is known as the mesh current.
• Assume for simplicity that the circuit contains only voltage sources.
![Page 3: 1 Mesh Analysis Discussion D2.4 Chapter 2 Section 2-8.](https://reader030.fdocuments.net/reader030/viewer/2022032523/56649d775503460f94a59bfd/html5/thumbnails/3.jpg)
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Mesh Analysis Steps
1. Assign mesh currents i1, i2, i3, … il, to the l meshes,
2. Apply KVL to each of the l meshes and use Ohm’s law to express the voltages in terms of the mesh currents,
3. Solve the l resulting simultaneous equations to find the mesh currents.
![Page 4: 1 Mesh Analysis Discussion D2.4 Chapter 2 Section 2-8.](https://reader030.fdocuments.net/reader030/viewer/2022032523/56649d775503460f94a59bfd/html5/thumbnails/4.jpg)
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Example
DC
DC
1R
3R5R
7R
2R
6R
8R
4R
1v 2v
3v 4v
5v6v
7v
8v
+ +
+ +
++
+
+
-
-- -
-
-
-
-
1sV
2sV 1i 2i
3i 4i
Number of nodes, n =
Number of branches, b =
Number of loops, l =
1l b n
7
10
4
![Page 5: 1 Mesh Analysis Discussion D2.4 Chapter 2 Section 2-8.](https://reader030.fdocuments.net/reader030/viewer/2022032523/56649d775503460f94a59bfd/html5/thumbnails/5.jpg)
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Example
Apply KVL to each mesh
2 1 7 5 0sV v v v
2 6 7 0v v v
15 3 0sv v v
Mesh 1:
Mesh 2:
Mesh 3:
14 8 6 0sv v V v Mesh 4:
DC
DC
1R
3R5R
7R
2R
6R
8R
4R
1v 2v
3v 4v
5v6v
7v
8v
+ +
+ +
++
+
+
-
-- -
-
-
-
-
1sV
2sV 1i 2i
3i 4i
![Page 6: 1 Mesh Analysis Discussion D2.4 Chapter 2 Section 2-8.](https://reader030.fdocuments.net/reader030/viewer/2022032523/56649d775503460f94a59bfd/html5/thumbnails/6.jpg)
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DC
DC
1R
3R5R
7R
2R
6R
8R
4R
1v 2v
3v 4v
5v6v
7v
8v
+ +
+ +
++
+
+
-
-- -
-
-
-
-
1sV
2sV 1i 2i
3i 4i
2 1 7 5 0sV v v v
2 6 7 0v v v
15 3 0sv v v
Mesh 1:
Mesh 2:
Mesh 3:
14 8 6 0sv v V v Mesh 4:
2 1 1 1 2 7 1 3 5( ) ( ) 0sV i R i i R i i R
2 2 2 4 6 2 1 7( ) ( ) 0i R i i R i i R
13 1 5 3 3( ) 0si i R V i R
Mesh 1:
Mesh 2:
Mesh 3:
14 4 4 8 4 2 6( ) 0si R i R V i i R Mesh 4:
Express the voltage in terms of the mesh currents:
![Page 7: 1 Mesh Analysis Discussion D2.4 Chapter 2 Section 2-8.](https://reader030.fdocuments.net/reader030/viewer/2022032523/56649d775503460f94a59bfd/html5/thumbnails/7.jpg)
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2 1 1 1 2 7 1 3 5( ) ( ) 0sV i R i i R i i R
2 2 2 4 6 2 1 7( ) ( ) 0i R i i R i i R
13 1 5 3 3( ) 0si i R V i R
Mesh 1:
Mesh 2:
Mesh 3:
14 4 4 8 4 2 6( ) 0si R i R V i i R Mesh 4:
Mesh 1:
Mesh 2:
Mesh 3:
Mesh 4:
21 5 7 1 7 2 5 3( ) sR R R i R i R i V
7 1 2 6 7 2 6 4( ) 0R i R R R i R i
15 1 3 5 3( ) sR i R R i V
16 2 4 6 8 4( ) sR i R R R i V
![Page 8: 1 Mesh Analysis Discussion D2.4 Chapter 2 Section 2-8.](https://reader030.fdocuments.net/reader030/viewer/2022032523/56649d775503460f94a59bfd/html5/thumbnails/8.jpg)
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Mesh 1:
Mesh 2:
Mesh 3:
Mesh 4:
21 5 7 1 7 2 5 3( ) sR R R i R i R i V
7 1 2 6 7 2 6 4( ) 0R i R R R i R i
15 1 3 5 3( ) sR i R R i V
16 2 4 6 8 4( ) sR i R R R i V
2
1
1
1 5 7 7 5 1
7 2 6 7 6 2
5 3 5 3
6 4 6 8 4
0
00
0 0
0 0
s
s
s
VR R R R R i
R R R R R i
VR R R i
R R R R i V
![Page 9: 1 Mesh Analysis Discussion D2.4 Chapter 2 Section 2-8.](https://reader030.fdocuments.net/reader030/viewer/2022032523/56649d775503460f94a59bfd/html5/thumbnails/9.jpg)
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Ri = v
R
v
iis an l x l symmetric resistance matrix
is a 1 x l vector of mesh currents
is a vector of voltages representing “known” voltages
2
1
1
1 5 7 7 5 1
7 2 6 7 6 2
5 3 5 3
6 4 6 8 4
0
00
0 0
0 0
s
s
s
VR R R R R i
R R R R R i
VR R R i
R R R R i V
![Page 10: 1 Mesh Analysis Discussion D2.4 Chapter 2 Section 2-8.](https://reader030.fdocuments.net/reader030/viewer/2022032523/56649d775503460f94a59bfd/html5/thumbnails/10.jpg)
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•The matrix R is symmetric, rkj = rjk and all of the off-diagonal terms are negative or zero.
Writing the Mesh Equations by Inspection
2
1
1
1 5 7 7 5 1
7 2 6 7 6 2
5 3 5 3
6 4 6 8 4
0
00
0 0
0 0
s
s
s
VR R R R R i
R R R R R i
VR R R i
R R R R i V
The vk (the kth component of the vector v) = the algebraic sum of the independent voltages in mesh k, with voltage rises taken as positive.
The rkj terms are the negative sum of the resistances common to BOTH mesh k and mesh j.
The rkk terms are the sum of all resistances in mesh k.
DC
DC
1R
3R5R
7R
2R
6R
8R
4R
1v 2v
3v 4v
5v6v
7v
8v
+ +
+ +
++
+
+
-
-- -
-
-
-
-
1sV
2sV 1i 2i
3i 4i
![Page 11: 1 Mesh Analysis Discussion D2.4 Chapter 2 Section 2-8.](https://reader030.fdocuments.net/reader030/viewer/2022032523/56649d775503460f94a59bfd/html5/thumbnails/11.jpg)
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MATLAB Solution of Mesh Equations
1i R v
2
1
1
1 5 7 7 5 1
7 2 6 7 6 2
5 3 5 3
6 4 6 8 4
0
00
0 0
0 0
s
s
s
VR R R R R i
R R R R R i
VR R R i
R R R R i V
Ri = v
![Page 12: 1 Mesh Analysis Discussion D2.4 Chapter 2 Section 2-8.](https://reader030.fdocuments.net/reader030/viewer/2022032523/56649d775503460f94a59bfd/html5/thumbnails/12.jpg)
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Test with numbers
DC
DC
1R
3R5R
7R
2R
6R
8R
4R
1
4
2
4
1
3
32
2V
4V 1i2i
3i4i
1
2
3
4
2 4 1 4 1 0 4
4 3 2 4 0 2 0
1 0 3 1 0 2
0 2 0 2 4 1 2
i
i
i
i
![Page 13: 1 Mesh Analysis Discussion D2.4 Chapter 2 Section 2-8.](https://reader030.fdocuments.net/reader030/viewer/2022032523/56649d775503460f94a59bfd/html5/thumbnails/13.jpg)
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Test with numbers
1
2
3
4
2 4 1 4 1 0 4
4 3 2 4 0 2 0
1 0 3 1 0 2
0 2 0 2 4 1 2
i
i
i
i
1
2
3
4
7 4 1 0 4
4 9 0 2 0
1 0 4 0 2
0 2 0 7 2
i
i
i
i
Ri = v
![Page 14: 1 Mesh Analysis Discussion D2.4 Chapter 2 Section 2-8.](https://reader030.fdocuments.net/reader030/viewer/2022032523/56649d775503460f94a59bfd/html5/thumbnails/14.jpg)
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MATLAB Run
DC
DC
1R
3R5R
7R
2R
6R
8R
4R
1
4
2
4
1
3
32
2V
4V 1i2i
3i4i
1
2
3
4
2 4 1 4 1 0 4
4 3 2 4 0 2 0
1 0 3 1 0 2
0 2 0 2 4 1 2
i
i
i
i
![Page 15: 1 Mesh Analysis Discussion D2.4 Chapter 2 Section 2-8.](https://reader030.fdocuments.net/reader030/viewer/2022032523/56649d775503460f94a59bfd/html5/thumbnails/15.jpg)
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PSpice Simulation
MATLAB:
![Page 16: 1 Mesh Analysis Discussion D2.4 Chapter 2 Section 2-8.](https://reader030.fdocuments.net/reader030/viewer/2022032523/56649d775503460f94a59bfd/html5/thumbnails/16.jpg)
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What happens if we have independent or dependent current sources in the circuit?
1. Write the mesh equations in the same way we did for circuits with only independent or dependent voltage sources.
2. Express the current of each independent or dependent current source in terms of the mesh currents.
3. Rewrite the equations with all unknown mesh currents on the left hand side of the equality and all known voltages on the r.h.s of the equality.
![Page 17: 1 Mesh Analysis Discussion D2.4 Chapter 2 Section 2-8.](https://reader030.fdocuments.net/reader030/viewer/2022032523/56649d775503460f94a59bfd/html5/thumbnails/17.jpg)
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Example
Write mesh equations by inspection.
1
2
3
1 3 3 1 10
3 3 2 4 2 0
1 2 2 1 a
i
i
i v
DC 10V
1
3A
+ v -a
1i 2i
3i3 3i
![Page 18: 1 Mesh Analysis Discussion D2.4 Chapter 2 Section 2-8.](https://reader030.fdocuments.net/reader030/viewer/2022032523/56649d775503460f94a59bfd/html5/thumbnails/18.jpg)
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1
2
4 3 0 7
3 9 0 6
1 2 1 9a
i
i
v
1
2
4 3 1 10
3 9 2 0
1 2 3 3 a
i
i
v
![Page 19: 1 Mesh Analysis Discussion D2.4 Chapter 2 Section 2-8.](https://reader030.fdocuments.net/reader030/viewer/2022032523/56649d775503460f94a59bfd/html5/thumbnails/19.jpg)
19
MATLAB Run
AAV
i1i2va
1
2
4 3 0 7
3 9 0 6
1 2 1 9a
i
i
v
DC 10V
1
3A
+ v -a
1i 2i
3i
![Page 20: 1 Mesh Analysis Discussion D2.4 Chapter 2 Section 2-8.](https://reader030.fdocuments.net/reader030/viewer/2022032523/56649d775503460f94a59bfd/html5/thumbnails/20.jpg)
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PSpice Simulation
MATLAB:i1
va
i2
i1
va
i2
+ -
![Page 21: 1 Mesh Analysis Discussion D2.4 Chapter 2 Section 2-8.](https://reader030.fdocuments.net/reader030/viewer/2022032523/56649d775503460f94a59bfd/html5/thumbnails/21.jpg)
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Nodal Analysis vs Mesh Analysis
• If the circuit is nonplanar we must use nodal analysis.
• If the circuit is planar, there are two principle considerations:
– The number of equations we need to write.– The information we need in the circuit. For
example, do we want to find a current or a voltage?
![Page 22: 1 Mesh Analysis Discussion D2.4 Chapter 2 Section 2-8.](https://reader030.fdocuments.net/reader030/viewer/2022032523/56649d775503460f94a59bfd/html5/thumbnails/22.jpg)
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The Number of Mesh Equations
If the circuit has only voltage sources or current sources in parallel with resistances (that is, current sources that can be exchanged for voltage sources) then after we have exchanged the current sources, we have to write l mesh equations and the fundamental theorem of network topology tells us that:
( 1)l b n
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The Number of Nodal Equations
We also know that if the circuit has only current sources or voltage sources in series with resistances (that is, sources that can be transformed to current sources) then after we have transformed the voltage sources, the number of nodal equations is (n - 1).
If we are not interested in the currents through voltage sources or voltages across current sources we may be able to reduce the number of equations we have to write.
![Page 24: 1 Mesh Analysis Discussion D2.4 Chapter 2 Section 2-8.](https://reader030.fdocuments.net/reader030/viewer/2022032523/56649d775503460f94a59bfd/html5/thumbnails/24.jpg)
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1/8
si
3 yv
+ -
+-
yv
xv+
+ -
-
2 xv
1V2V
3V
4V
Example
Transform voltage sources into current sources
8S
S
S
2S
si
12 yv
1V
xv+
-
4 xv
2V
b =
n =
7
3
1 4 1 2( 2 )y xv V V V V v
![Page 25: 1 Mesh Analysis Discussion D2.4 Chapter 2 Section 2-8.](https://reader030.fdocuments.net/reader030/viewer/2022032523/56649d775503460f94a59bfd/html5/thumbnails/25.jpg)
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Example
8S
S
S
2S
si
12 yv
1V
xv+
-
4 xv
2V
1
2
411 3
12 43 7s x
y x
i vV
v vV
1 2( 2 )y xv V V v
1xv V
Write nodal equations
![Page 26: 1 Mesh Analysis Discussion D2.4 Chapter 2 Section 2-8.](https://reader030.fdocuments.net/reader030/viewer/2022032523/56649d775503460f94a59bfd/html5/thumbnails/26.jpg)
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Example
1
2
411 3
12 43 7s x
y x
i vV
v vV
1 2( 2 )y xv V V v
1xv V
1 1
2 1 2 1 1
411 3
12 12 24 43 7sV i V
V V V V V
1
2
7 3
13 19 0sV i
V
![Page 27: 1 Mesh Analysis Discussion D2.4 Chapter 2 Section 2-8.](https://reader030.fdocuments.net/reader030/viewer/2022032523/56649d775503460f94a59bfd/html5/thumbnails/27.jpg)
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1
2
7 3
13 19 0sV i
V
Example
1 27 3 sV V i
1 213 19 0V V
2 1
13
19V V
1 1
3 137
19 sV V i
2 0.0756 sV i
1 0.1105 sV i
8S
S
S
2S
si
12 yv
1V
xv+
-
4 xv
2V
![Page 28: 1 Mesh Analysis Discussion D2.4 Chapter 2 Section 2-8.](https://reader030.fdocuments.net/reader030/viewer/2022032523/56649d775503460f94a59bfd/html5/thumbnails/28.jpg)
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1/8
si
3 yv
+ -
+-
yv
xv+
+ -
-
2 xv
1V2V
3V
4V
Example
Transform current source into a current source and write mesh equations
1/8
8si
3 yv
+ -
+-
yv
xv+
+ -
-
2 xv
1V2V
3V
4V
+-
5V
2i
1i
1
8 8s
x
i iv 2
2y
iv
![Page 29: 1 Mesh Analysis Discussion D2.4 Chapter 2 Section 2-8.](https://reader030.fdocuments.net/reader030/viewer/2022032523/56649d775503460f94a59bfd/html5/thumbnails/29.jpg)
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1/8
8si
3 yv
+ -
+-
yv
xv+
+ -
-
2 xv
1V2V
3V
4V
+-
5V
2i
1i
1
2
8 31 8 1 1 4 1
21 1 2 1s y
x
i vi
vi
Example
![Page 30: 1 Mesh Analysis Discussion D2.4 Chapter 2 Section 2-8.](https://reader030.fdocuments.net/reader030/viewer/2022032523/56649d775503460f94a59bfd/html5/thumbnails/30.jpg)
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1
2
8 31.375 1
21 1.5s y
x
i vi
vi
1
8 8s
x
i iv
2
2y
iv
21
12
8 1.51.375 1
0.25 0.251 1.5s
s
i ii
i ii
1
2
81.375 0.5
0.251.25 1.5s
s
ii
ii
Example
![Page 31: 1 Mesh Analysis Discussion D2.4 Chapter 2 Section 2-8.](https://reader030.fdocuments.net/reader030/viewer/2022032523/56649d775503460f94a59bfd/html5/thumbnails/31.jpg)
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1
2
81.375 0.5
0.251.25 1.5s
s
ii
ii
1/8
8si
3 yv
+ -
+-
yv
xv+
+ -
-
2 xv
1V2V
3V
4V
+-
5V
2i
1i2 0.0698 si i
1 0.1163 si i
Example