Mesh Analysis - University of Babylon
Transcript of Mesh Analysis - University of Babylon
![Page 1: Mesh Analysis - University of Babylon](https://reader030.fdocuments.net/reader030/viewer/2022012211/61df217ec7efea684950312d/html5/thumbnails/1.jpg)
By
Hayder Hamzah
University of Babylon, Hillah, Iraq
+9647814471323
Mesh Analysis
![Page 2: Mesh Analysis - University of Babylon](https://reader030.fdocuments.net/reader030/viewer/2022012211/61df217ec7efea684950312d/html5/thumbnails/2.jpg)
1. Nodal method
2. Mesh method
3. Superposition method
4. Matrix method
![Page 3: Mesh Analysis - University of Babylon](https://reader030.fdocuments.net/reader030/viewer/2022012211/61df217ec7efea684950312d/html5/thumbnails/3.jpg)
1. Ohm's law
2. Kirchhoff's Voltage Law
![Page 4: Mesh Analysis - University of Babylon](https://reader030.fdocuments.net/reader030/viewer/2022012211/61df217ec7efea684950312d/html5/thumbnails/4.jpg)
Where
R: resistance
V: voltage
I: current
![Page 5: Mesh Analysis - University of Babylon](https://reader030.fdocuments.net/reader030/viewer/2022012211/61df217ec7efea684950312d/html5/thumbnails/5.jpg)
V4 = V1+ V2 + V3
V4 -V1- V2 - V3 = 0
![Page 6: Mesh Analysis - University of Babylon](https://reader030.fdocuments.net/reader030/viewer/2022012211/61df217ec7efea684950312d/html5/thumbnails/6.jpg)
Basic Circuits
Mesh Analysis: Basic Concepts:
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.
I1 I2 I3
![Page 7: Mesh Analysis - University of Babylon](https://reader030.fdocuments.net/reader030/viewer/2022012211/61df217ec7efea684950312d/html5/thumbnails/7.jpg)
Basic Circuits
Mesh Analysis: Basic Concepts:
R1
Rx
R2
+
_ I1 I2
+
_VA VB
+ +
+
_
_
_V1
VL1
V2
Figure 7.2: A circuit for illustrating mesh analysis.
AXX
XL
AL
VIRIRRso
RIIVRIVwhere
VVV
211
211111
11
)(,
;
Eq 7.1
Around mesh 1:
![Page 8: Mesh Analysis - University of Babylon](https://reader030.fdocuments.net/reader030/viewer/2022012211/61df217ec7efea684950312d/html5/thumbnails/8.jpg)
Basic Circuits Mesh Analysis: Basic Concepts:
R1
Rx
R2
+
_ I1 I2
+
_VA VB
+ +
+
_
_
_V1
VL1
V2
BVIRXRIXRor
BVIRXRIXR
givesinEqngSubstituti
RIVXRIILVwith
BVVLV
havewemeshAround
2)2(1
2)2(1
,2.73.7
222;)12(1;
21
2
Eq 7.2
Eq 7.3
Eq 7.4
![Page 9: Mesh Analysis - University of Babylon](https://reader030.fdocuments.net/reader030/viewer/2022012211/61df217ec7efea684950312d/html5/thumbnails/9.jpg)
Basic Circuits
Mesh Analysis: Basic Concepts:
We are left with 2 equations: From (7.1) and (7.4)
we have,
AXX VIRIRR 211 )(
BXX VIRRIR 221 )(
Eq 7.5
Eq 7.6
We can easily solve these equations for I1 and I2.
![Page 10: Mesh Analysis - University of Babylon](https://reader030.fdocuments.net/reader030/viewer/2022012211/61df217ec7efea684950312d/html5/thumbnails/10.jpg)
Basic Circuits
Mesh Analysis: Basic Concepts:
The previous equations can be written in matrix form as:
B
A
XX
XX
B
A
XX
XX
V
V
RRR
RRR
I
I
or
V
V
I
I
RRR
RRR
1
2
1
2
1
2
1
2
1
(
)(
(
)(Eq (7.7)
Eq (7.8)
![Page 11: Mesh Analysis - University of Babylon](https://reader030.fdocuments.net/reader030/viewer/2022012211/61df217ec7efea684950312d/html5/thumbnails/11.jpg)
Basic Circuits Mesh Analysis: Example 7.1.
Write the mesh equations and solve for the currents I1, and I2.
+
_10V
4 2
6 7
2V20V
I1 I2+
+_
_
Figure 7.2: Circuit for Example 7.1.
Mesh 1 4I1 + 6(I1 – I2) = 10 - 2
Mesh 2 6(I2 – I1) + 2I2 + 7I2 = 2 + 20
Eq (7.9)
Eq (7.10)
![Page 12: Mesh Analysis - University of Babylon](https://reader030.fdocuments.net/reader030/viewer/2022012211/61df217ec7efea684950312d/html5/thumbnails/12.jpg)
Basic Circuits
Mesh Analysis: Example 7.1, continued.
Simplifying Eq (7.9) and (7.10) gives,
10I1 – 6I2 = 8
-6I1 + 15I2 = 22
Eq (7.11)
Eq (7.12)
» % A MATLAB Solution
»
» R = [10 -6;-6 15];
»
» V = [8;22];
»
» I = inv(R)*V
I =
2.2105
2.3509
I1 = 2.2105
I2 = 2.3509
![Page 13: Mesh Analysis - University of Babylon](https://reader030.fdocuments.net/reader030/viewer/2022012211/61df217ec7efea684950312d/html5/thumbnails/13.jpg)
Basic Circuits Mesh Analysis: Example 7.2
Solve for the mesh currents in the circuit below.
+
_
6
10
9
11
3
4
20V 10V
8V
12V
I1 I2
I3
+
+
__
_
_
+
+_
Figure 7.3: Circuit for Example 7.2.
The plan: Write KVL, clockwise, for each mesh. Look for a
pattern in the final equations.
![Page 14: Mesh Analysis - University of Babylon](https://reader030.fdocuments.net/reader030/viewer/2022012211/61df217ec7efea684950312d/html5/thumbnails/14.jpg)
Basic Circuits Mesh Analysis: Example 7.2
+
_
6
10
9
11
3
4
20V 10V
8V
12V
I1 I2
I3
+
+
__
_
_
+
+_
Mesh 1: 6I1 + 10(I1 – I3) + 4(I1 – I2) = 20 + 10
Mesh 2: 4(I2 – I1) + 11(I2 – I3) + 3I2 = - 10 - 8
Mesh 3: 9I3 + 11(I3 – I2) + 10(I3 – I1) = 12 + 8
Eq (7.13)
Eq (7.14)
Eq (7.15)
![Page 15: Mesh Analysis - University of Babylon](https://reader030.fdocuments.net/reader030/viewer/2022012211/61df217ec7efea684950312d/html5/thumbnails/15.jpg)
Basic Circuits Mesh Analysis: Example 7.2
Clearing Equations (7.13), (7.14) and (7.15) gives,
20I1 – 4I2 – 10I3 = 30
-4I1 + 18I2 – 11I3 = -18
-10I1 – 11I2 + 30I3 = 20
In matrix form:
20
18
30
3
2
1
301110
11184
10420
I
I
I
WE NOW MAKE AN IMPORTANT
OBSERVATION!!
Standard Equation form
![Page 16: Mesh Analysis - University of Babylon](https://reader030.fdocuments.net/reader030/viewer/2022012211/61df217ec7efea684950312d/html5/thumbnails/16.jpg)
Basic Circuits Mesh Analysis: Standard form for mesh equations
Consider the following:
R11 =
of resistance around mesh 1, common to mesh 1 current I1.
R22 = of resistance around mesh 2, common to mesh 2 current I2.
R33 = of resistance around mesh 3, common to mesh 3 current I3.
)3(
)2(
)1(
3
2
1
333231
232221
131211
emfs
emfs
emfs
I
I
I
RRR
RRR
RRR
![Page 17: Mesh Analysis - University of Babylon](https://reader030.fdocuments.net/reader030/viewer/2022012211/61df217ec7efea684950312d/html5/thumbnails/17.jpg)
Basic Circuits Mesh Analysis: Standard form for mesh equations
R12 = R21 = - resistance common between mesh 1 and 2
when I1 and I2 are opposite through R1,R2.
R13 = R31 = - resistance common between mesh 1 and 3
when I1 and I3 are opposite through R1,R3.
R23 = R32 = - resistance common between mesh 2 and 3
when I2 and I3 are opposite through R2,R3.
)1(emfs = sum of emf around mesh 1 in the direction of I1.
)2(emfs = sum of emf around mesh 2 in the direction of I2.
)3(emfs = sum of emf around mesh 3 in the direction of I3.
![Page 18: Mesh Analysis - University of Babylon](https://reader030.fdocuments.net/reader030/viewer/2022012211/61df217ec7efea684950312d/html5/thumbnails/18.jpg)
Basic Circuits Mesh Analysis: Example 7.3 - Direct method.
20V
10V
15V
30V
20
10
30
10
12
8
+_
I1 I2 I3+
+
+
_
__
Use the direct method to write the mesh equations for the following.
Figure 7.4: Circuit diagram for Example 7.3.
15
25
10
3
2
1
30100
105010
01030
I
I
I
Eq (7.13)
![Page 19: Mesh Analysis - University of Babylon](https://reader030.fdocuments.net/reader030/viewer/2022012211/61df217ec7efea684950312d/html5/thumbnails/19.jpg)
Basic Circuits Mesh Analysis: With current sources in the circuit
Example 7.4: Consider the following:
10V
20V
4A
10
5
20
2
+
_
15
+
_
I1 I2
I3
Figure 7.5: Circuit diagram for Example 7.4.
Use the direct method to write the mesh equations.
![Page 20: Mesh Analysis - University of Babylon](https://reader030.fdocuments.net/reader030/viewer/2022012211/61df217ec7efea684950312d/html5/thumbnails/20.jpg)
10V
20V
4A
10
5
20
2
+
_
15
+
_
I1 I2
I3
Basic Circuits Mesh Analysis: With current sources in the circuit
This case is explained by using an example.
Example 7.4: Find the three mesh currents in the circuit below.
Figure 7.5: Circuit for Example 7.4.
When a current source is present, it will be directly related to
one or more of the mesh current. In this case I2 = -4A.
![Page 21: Mesh Analysis - University of Babylon](https://reader030.fdocuments.net/reader030/viewer/2022012211/61df217ec7efea684950312d/html5/thumbnails/21.jpg)
Basic Circuits Mesh Analysis: With current sources in the circuit
Example 7.4: Continued. An easy way to handle this case is to
remove the current source as shown below. Next, write the mesh
equations for the remaining meshes.
Note that I 2 is retained for writing the equations through the
5 and 20 resistors.
10V
20V
10
5
20
2
+
_
15
+
_
I1 I2
I3
![Page 22: Mesh Analysis - University of Babylon](https://reader030.fdocuments.net/reader030/viewer/2022012211/61df217ec7efea684950312d/html5/thumbnails/22.jpg)
Basic Circuits Mesh Analysis: With current sources in the circuit
Example 7.4: Continued.
10V
20V
10
5
20
2
+
_
15
+
_
I1 I2
I3 Equation for mesh 1:
10I1 + (I1-I2)5 = 10
or
15I1 – 5I2 = 10
Equations for mesh 2:
2I3 + (I3-I2)20 = 20
or
- 20I2 + 22I3 = 20
Constraint Equation
I2 = - 4A
![Page 23: Mesh Analysis - University of Babylon](https://reader030.fdocuments.net/reader030/viewer/2022012211/61df217ec7efea684950312d/html5/thumbnails/23.jpg)
Basic Circuits Mesh Analysis: With current sources in the circuit
Example 7.4: Continued. Express the previous equations in
Matrix form:
1
2
3
15 5 0 10
0 20 22 20
0 1 0 4
I
I
I
I1 = -0.667 A
I2 = - 4 A
I3 = - 2.73 A
![Page 24: Mesh Analysis - University of Babylon](https://reader030.fdocuments.net/reader030/viewer/2022012211/61df217ec7efea684950312d/html5/thumbnails/24.jpg)
End of Lesson 7
circuits
Mesh Analysis