# Mesh Analysis

```Mesh Analysis
Discussion D2.4
Chapter 2
Section 2-8
1
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.
2
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.
3
Example
R2
R1
+
DC
Vs2
i1
+
R3
-
v3
+ v2 -
v1 -
v5 -
+
v7
-
R7
+ v6 -
R5
i3
+
i2
R6
DC
Vs1
i4
+
v4
-
R4
- v +
8
R8
Number of nodes, n = 7
Number of branches, b = 10
Number of loops, l = 4
l  b  n 1
4
R2
R1
Example
DC
Vs2
i1
v5 -
+
R3
Apply KVL to each mesh
-
v3
+ v2 -
v1 -
+
+
v7
-
R7
+ v6 -
R5
i3
+
Mesh 1:
Vs2  v1  v7  v5  0
Mesh 2:
v2  v6  v7  0
Mesh 3:
v5  vs1  v3  0
Mesh 4:
v4  v8  Vs1  v6  0
i2
R6
DC
Vs1
i4
+
R4
v4
-
- v +
8
R8
5
Mesh 1:
Vs2  v1  v7  v5  0
Mesh 2:
v2  v6  v7  0
Mesh 3:
Mesh 4:
+
DC
Vs2
v4  v8  Vs1  v6  0 R3
-
v3
+ v2 -
v1 i1
+
v5  vs1  v3  0
Express the voltage in terms of
the mesh currents:
R2
R1
v5 -
+
v7
-
R7
+ v6 -
R5
i3
R6
DC
+
Mesh 1: Vs  i1R1  (i1  i2 ) R7  (i1  i3 )R5  0
2
Mesh 2:
i2 R2  (i2  i4 ) R6  (i2  i1 ) R7  0
Mesh 3:
(i3  i1 )R5  Vs1  i3 R3  0
Mesh 4:
i4 R4  i4 R8 Vs1  (i4  i2 )R6  0
i2
Vs1
i4
+
v4
-
- v +
8
R8
6
R4
Mesh 1: Vs  i1R1  (i1  i2 ) R7  (i1  i3 )R5  0
2
Mesh 2:
i2 R2  (i2  i4 ) R6  (i2  i1 ) R7  0
Mesh 3:
(i3  i1 )R5  Vs1  i3 R3  0
Mesh 4:
i4 R4  i4 R8 Vs1  (i4  i2 )R6  0
Mesh 1:
(R1  R5  R7 )i1  R7i2  R5i3  Vs2
Mesh 2:
R7i1  ( R2  R6  R7 )i2  R6i4  0
Mesh 3:
R5i1  (R3  R5 )i3  Vs1
Mesh 4:
R6i2  (R4  R6  R8 )i4  Vs1
7
Mesh 1:
(R1  R5  R7 )i1  R7i2  R5i3  Vs2
Mesh 2:
R7i1  ( R2  R6  R7 )i2  R6i4  0
Mesh 3:
R5i1  (R3  R5 )i3  Vs1
Mesh 4:
 R1  R5  R7

 R7


 R5

0

R6i2  (R4  R6  R8 )i4  Vs1
 R7
R2  R6  R7
0
 R6
 R5
0
R3  R5
0
0
 i1   Vs2 
   0 
 R6

 i2   
 i3   Vs1 
0

  
R4  R6  R8   i4   Vs 
 1 
8
 R1  R5  R7

 R7


 R5

0

 R7
R2  R6  R7
0
 R6
 R5
0
R3  R5
0
0
 i1   Vs2 
   0 
 R6

 i2   
 i3   Vs1 
0

  
R4  R6  R8   i4   Vs 
 1 
Ri = v
R is an l x l symmetric resistance matrix
i is a 1 x l vector of mesh currents
v is a vector of voltages representing “known” voltages
9
Writing the Mesh Equations by Inspection
R2
R1
+
DC
Vs2
i1
+
R3
-
v3
+
+ v2 -
v1 -
v5 -
+
v7
-
R7
+ v6 -
R5
i3
 R1  R5  R7

 R7


 R5

0

i2
R6
DC
Vs1
i4
+
v4
-
R4
 R7
R2  R6  R7
0
 R6
 R5
0
R3  R5
0
0
 i1   Vs2 
   0 
 R6

 i2   
 i3   Vs1 
0

  


R4  R6  R8   i4 
 Vs1 
- v +
8
R8
•The matrix R is symmetric, rkj = rjk and all of the off-diagonal terms
are negative or zero.
The rkk terms are the sum of all resistances in mesh k.
The rkj terms are the negative sum of the resistances common to
BOTH mesh k and mesh j.
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.
10
MATLAB Solution of Mesh Equations
 R1  R5  R7

 R7


 R5

0

 R7
R2  R6  R7
0
 R6
 R5
0
R3  R5
0
0
 i1   Vs2 
   0 
 R6

 i2   
 i3   Vs1 
0

  
R4  R6  R8   i4   Vs 
 1 
Ri = v
1
iR v
11
Test with numbers
DC
R3
4V
3
R1
R2
2
3
i1
4
i2
R7
1
2
R5
R6
i3
DC
2V
i4
4
1
R8
4
1
0   i1   4 
 2  4 1

   

4
3

2

4
0

2

  i2    0 
 1
0
3 1
0   i3   2 

   
2
0
2  4  1  i4   2 
 0
12
R4
Test with numbers
4
1
0   i1   4 
 2  4 1

   

4
3

2

4
0

2

  i2    0 
 1
0
3 1
0   i3   2 

   
2
0
2  4  1  i4   2 
 0
 7 4 1 0   i1   4 

   

4
9
0

2

  i2    0 
 1 0 4 0   i3   2 

   
 0 2 0 7   i4   2 
Ri = v
13
MATLAB Run
DC
R3
4V
3
R1
R2
2
3
i1
4
i2
R7
1
2
R5
R6
i3
DC
2V
i4
4
R4
1
R8
4
1
0   i1   4 
 2  4 1

   

4
3

2

4
0

2

  i2    0 
 1
0
3 1
0   i3   2 

   
2
0
2  4  1  i4   2 
 0
14
PSpice Simulation
MATLAB:
15
What happens if we have independent
current sources in the circuit?
1. Assume the voltage across each current source is
known.
2. Write the mesh equations in the same way we
did for circuits with only independent or
dependent voltage sources.
3. Express the current of each independent current
source in terms of the mesh currents.
4. 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.
16
Example
3A
+
1
DC
10V
i1
va
-
i3
i3  3


i2

Write mesh equations by inspection.
3
1   i1   10 
1  3

  

 3 3  2  4 2   i2    0 
 1
  i   v 

2
2

1

 3   a 
17
 4 3 1   i1   10 

  


3
9

2
i

0

 2  

 1 2 3   3   v 

   a 
 4 3 0   i1   7 

   

3
9
0

  i2    6 
  1 2 1   v   9 

 a   
18
MATLAB Run
 4 3 0   i1   7 

   

3
9
0

  i2    6 
  1 2 1   v   9 

 a   
3A
+
1
DC
A
A
V
i1
i2
va
10V
i1
va
-
i3


i2

19
PSpice Simulation
+
-
i2
i1
MATLAB:
va
i1
i2
va
20
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?
21
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:
l  b  (n  1)
22
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.
23
Example
Transform voltage sources into current sources
4vx
+
vy
2S
-
V4
+

-
2vx
is

V1
is
V2
+
1/8

vx
-
S
V1
V3
+
-
3v y
vy  V1 V4  V1  (V2  2vx )
V2
+
8S
vx
S
12v y
-
b= 7
n= 3
24
Example
4vx
vy  V1  (V2  2vx )
2S
S
V1
is
V2
+
8S
vx
S
12v y
vx  V1
-
Write nodal equations
 11 3   V1   is  4vx 

    12v  4v 
x
 3 7  V2   y
25
Example
 11 3   V1   is  4vx 

    12v  4v 
x
 3 7  V2   y
vx  V1
vy  V1  (V2  2vx )
is  4V1

 11 3   V1  


   
 3 7  V2  12V1  12V2  24V1  4V1 
 7 3   V1   is 

 V    
 13 19   2   0 
26
Example
4vx
 7 3   V1   is 

 V    
 13 19   2   0  V
2S
S
1
7V1  3V2  is
13V1  19V2  0
is
V2
+
8S
vx
S
12v y
-
13
V2   V1
19
V1  0.1105is
3  13
7V1 
V1  is
19
V2  0.0756is
27
Example
Transform current source into a current source and write mesh equations
+
+
vy
-
V4
+


vx
V3
+
-
is
8
-
2vx

V2
+
-
+
V1
V2
+
1/8
1/8
V4
i2
2vx
V1
-

-

is
vy
V5
+
-
3v y
i2
vy 
2
vx
-

i1
V3
+
-
is i1
vx  
8 8
28
3v y
+
Example
vy
-
V4

i2
+
2vx

V1
V2
+
1/8
is
8
-
V5
+
-
vx
-

i1
V3
+
-
3v y
1   i1   is 8  3v y 
1 8  1  1 4

     2v 
1
1 2  1 i2  

x

29
1.375 1   i1   is 8  3v y 

     2v 
 1 1.5  i2  
x

is i1
vx  
8 8
Example
i2
vy 
2
1.375 1   i1   is 8  1.5i2 

     0.25i  0.25i 
 1 1.5  i2  
s
1
 1.375 0.5   i1   is 8 

     0.25i 
 1.25 1.5  i2  
s
30
Example
 1.375 0.5   i1   is 8 

     0.25i 
 1.25 1.5  i2  
s
+
vy
-
V4

i1  0.1163is
i2
is
8
-
2vx

V1
V2
+
1/8
i2  0.0698is
+
V5
+
-
vx
-

i1
V3
+
-
31
3v y
```
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