mesh analysis dc circuit
TRANSCRIPT
Module -1
EEE2001 Network Theory
Mesh Analysis
Dr. R.Sarvanakumar, Professor,SELECT,VIT University,Vellore
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 particularmesh current does not define the current in each branchof the mesh to which it is assigned.
I1 I2 I3
Mesh = A closed loop path which has no smaller loops inside
Dr. R.Sarvanakumar, Professor,SELECT,VIT University,Vellore
Mesh AnalysisProcedure
1. Count the number of meshes. Let the number equal N.
2. Define mesh current on each mesh. Let the values be I1, I2, I3, …
3. Use Kirchhoff's voltage law (KVL) on each mesh, generating N equations
4. Solve the equations
Dr. R.Sarvanakumar, Professor,SELECT,VIT University,Vellore
Mesh Analysis:R1
Rx
R2
+_ I1 I2
+_VA VB
+ ++
_
_
_V1
VL1
V2
Figure 1 : A circuit for illustrating mesh analysis.
Dr. R.Sarvanakumar, Professor,SELECT,VIT University,Vellore
Mesh Analysis:R1
Rx
R2
+_ I1 I2
+_VA VB
+ ++
_
_
_V1
VL1
V2
Figure 1 : A circuit for illustrating mesh analysis.
AXX
XL
AL
AL
VIRIRRso
RIIVRIVwhereVVVVVV
211
211111
11
11
)(,
;;;
;0
Eq.1.4
Around mesh 1:Eq.1.1
Eq.1.2
Eq.1.3
,2.13.1 givesinEqngSubstituti
Mesh Analysis:R1
Rx
R2
+_ I1 I2
+_VA VB
+ ++
_
_
_V1
VL1
V2
BXX
XL
BL
BL
VIRRIRgivesinEqngSubstituti
RIVRIIVwithVVV
VVV
221
222121
21
21
)(,2.23.2
;)(;
0 Eq 2.1
Eq 2.2
Eq. 2.4
Around mesh 2:
Eq 2.3
Dr. R.Sarvanakumar, Professor,SELECT,VIT University,Vellore
Mesh Analysis:
AXX VIRIRR 211 )(
BXX VIRRIR 221 )(
Eq 1.4
Eq 2.4
We can easily solve these equations for I1 and I2.
Dr. R.Sarvanakumar, Professor,SELECT,VIT University,Vellore
Mesh Analysis:
The previous equations can be written in matrix form as:
B
A
XX
XX
VV
II
RRRRRR
2
1
2
1
()(
R1
Rx
R2
+_ I1 I2
+_VA VB
+ ++
_
_
_V1
VL1
V2
Dr. R.Sarvanakumar, Professor,SELECT,VIT University,Vellore
Mesh Analysis: Example 1Write the mesh equations and solve for the currents I1, and I2.
+_
10V
4 2
6 7
2V 20V
I1 I2+
+_
_
Figure 1: Circuit for Example 1
Dr. R.Sarvanakumar, Professor,SELECT,VIT University,Vellore
Mesh Analysis: Example 1Write the mesh equations and solve for the currents I1, and I2.
+_
10V
4 2
6 7
2V 20V
I1 I2+
+_
_
Figure 2: Circuit for Example 2
Mesh 1 4I1 + 6(I1 – I2) = 10 - 2
Mesh 2 6(I2 – I1) + 2I2 + 7I2 = 2 + 20
Eq (1.1)
Eq (2.1)
Dr. R.Sarvanakumar, Professor,SELECT,VIT University,Vellore
Mesh Analysis: Basic Concepts:
The previous equations can be written in matrix form as:
228
156610
2
1
II
+_
10V
4 2
6 7
2V 20V
I1 I2+
+_
_
Dr. R.Sarvanakumar, Professor,SELECT,VIT
University,Vellore
Example 1, continued. With MATLAB
» % A MATLAB Solution» » R = [10 -6;-6 15];» » V = [8;22];» » I = inv(R)*V
I =
2.21052.3509
I1 = 2.2105
I2 = 2.3509
+_
10V
4 2
6 7
2V 20V
I1 I2+
+_
_
228
156610
2
1
II
Dr. R.Sarvanakumar, Professor,SELECT,VIT University,Vellore
Mesh Analysis: Standard form for mesh equationsConsider 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.
)2()1(
2
1
2221
1211
emfsemfs
II
RRRR
R12 = R21 = - resistance common between mesh 1 and 2when I1 and I2 are opposite through R1,R2.
)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.Dr. R.Sarvanakumar, Professor,SELECT,VIT University,Vellore
Basic CircuitsMesh Analysis: Example 2Solve for the mesh currents in the circuit below.
+
_
6
10
9
11
3
4
20V 10V
8V
12V
I1 I2
I3
+
+
___
_
++_
Figure 2: Circuit for Example 2.
The plan: Write KVL, clockwise, for each mesh. Look for apattern in the final equations.
Dr. R.Sarvanakumar, Professor,SELECT,VIT University,Vellore
Basic CircuitsMesh Analysis: Example 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 (1.1)
Eq (2.1)
Eq (2.3)Dr. R.Sarvanakumar, Professor,SELECT,VIT University,Vellore
Mesh Analysis: Example 2Clearing Equations (1.1), (2.1) and (3.1) gives,
20I1 – 4I2 – 10I3 = 30-4I1 + 18I2 – 11I3 = -18-10I1 – 11I2 + 30I3 = 20
In matrix form:
2018
30
321
3011101118410420
III
Standard Equation form
+
_
6
10
9
11
3
4
20V 10V
8V
12V
I1 I2
I3
+
+
___
_
++_
Dr. R.Sarvanakumar, Professor,SELECT,VIT University,Vellore
>> R=[20 -4 -10;-4 18 -11;-10 -11 30];>> V=[30;-18;20];>> I=inv(R)*V
I =
2.47320.59391.7088
2018
30
321
3011101118410420
III
+
_
6
10
9
11
3
4
20V 10V
8V
12V
I1 I2
I3
+
+
___
_
++_
Example 2 , continued. With MATLAB
Dr. R.Sarvanakumar, Professor,SELECT,VIT University,Vellore
Mesh Analysis: Standard form for mesh equationsConsider 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(
321
333231232221131211
emfsemfsemfs
III
RRRRRRRRR
Dr. R.Sarvanakumar, Professor,SELECT,VIT University,Vellore
Mesh Analysis: Standard form for mesh equations
R12 = R21 = - resistance common between mesh 1 and 2when I1 and I2 are opposite through R1,R2.
R13 = R31 = - resistance common between mesh 1 and 3when I1 and I3 are opposite through R1,R3.
R23 = R32 = - resistance common between mesh 2 and 3when 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.
Dr. R.Sarvanakumar, Professor,SELECT,VIT University,Vellore