Electrical mesh analysis

1. Chapter Three
Basic Circuit Laws and Analysis
Mesh Analysis
1

2. Basic Circuits
Mesh Analysis: Basic Concepts:
 In formulating mesh analysis we assign a mesh current to
each mesh.
2
 A loop is a closed path with no node passed more
than once.
 A mesh is a loop which does not contain any
other loops within it.
 Mesh analysis is also known as loop analysis or
the mesh-current method.

3. Basic Circuits
Mesh Analysis: Basic Concepts:
I1 I2 I3
3
1. Assign mesh currents i1, i2, . . . , in to the n meshes.
2. Apply KVL to each of the n meshes. Use Ohm’s law to
express the voltages in terms of the mesh currents.
3. Solve the resulting n simultaneous equations to get the mesh
currents.
 S t e p s t o D e t e r m i n e Mesh C u r r e n t s :

4. Basic Circuits
Mesh Analysis: Basic Concepts:
R1
Rx
R2
+
_ I1 I2
+
_
VA VB
+ +
+
_
_
_
V1
VL1
V2
Figure 4.1: A circuit for illustrating mesh analysis.
 
A
X
X
X
L
A
L
V
I
R
I
R
R
so
R
I
I
V
R
I
V
where
V
V
V








2
1
1
2
1
1
1
1
1
1
1
)
(
,
;
Eq 4.1
Around mesh 1, we have:
4

5. Basic Circuits
Mesh Analysis: Basic Concepts:
R1
Rx
R2
+
_ I1 I2
+
_
VA VB
+ +
+
_
_
_
V1
VL1
V2
B
X
X
B
X
X
X
L
B
L
V
I
R
R
I
R
or
V
I
R
R
I
R
gives
in
Eq
ng
Substituti
R
I
V
R
I
I
V
with
V
V
V
have
we
mesh
Around














2
2
1
2
2
1
2
2
2
1
2
1
2
1
)
(
)
(
,
2
.
4
3
.
4
;
)
(
;
:
,
2
Eq 4.2
Eq 4.3
Eq 4.4 5

6. Basic Circuits
Mesh Analysis: Basic Concepts:
We are left with 2 equations: From (4.1) and (4.4) we have,
A
X
X V
I
R
I
R
R 

 2
1
1 )
(
B
X
X V
I
R
R
I
R 



 2
2
1 )
(
Eq 4.5
Eq 4.6
We can easily solve these equations for I1 and I2.
6

7. Basic Circuits
Mesh Analysis: Basic Concepts:
The previous equations can be written in matrix form as:

















































B
A
X
X
X
X
B
A
X
X
X
X
V
V
R
R
R
R
R
R
I
I
or
V
V
I
I
R
R
R
R
R
R
1
2
1
2
1
2
1
2
1
)
(
)
(
)
(
)
(
Eq (4.7)
Eq (4.8)
7

8. Basic Circuits
Mesh Analysis: Example 4.1.
Write the mesh equations and solve for the currents I1, and I2.
+
_
10V
4  2 
6 
7 
2V 20V
I1 I2
+
+
_
_
Figure 4.2: Circuit for Example 4.1.
Mesh 1: 4I1 + 6(I1 – I2) = 10 - 2
Mesh 2: 6(I2 – I1) + 2I2 + 7I2 = 2 + 20
Eq (4.9)
Eq (4.10)
8

9. Basic Circuits
Mesh Analysis: Example 4.1 continued…
Simplifying Equations (4.9) and (4.10) gives,
10I1 – 6I2 = 8
-6I1 + 15I2 = 22
Eq (4.11)
Eq (4.12)
I1 = 2.2105
I2 = 2.3509
9

10. Basic Circuits
Mesh Analysis: Example 4.2
Solve for the mesh currents in the circuit below.
+
_
6 
10 
9 
11 
3 
4 
20V 10V
8V
12V
I1 I2
I3
+
+
_
_
_
_
+
+
_
Figure 4.3: Circuit for Example 4.2.
The plan: Write KVL, clockwise, for each mesh. Look for a
pattern in the final equations.
10

11. Basic Circuits
Mesh Analysis: Example 4.2 continued…
+
_
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 (4.13)
Eq (4.14)
Eq (4.15)
11

12. Basic Circuits
Mesh Analysis: Example 4.2 continued…
Clearing Equations (4.13), (4.14) and (4.15) gives,
20I1 – 4I2 – 10I3 = 30
-4I1 + 18I2 – 11I3 = -18
-10I1 – 11I2 + 30I3 = 20
In matrix form:






































20
18
30
3
2
1
30
11
10
11
18
4
10
4
20
I
I
I
Standard Equation form
12

13. Basic Circuits
Mesh Analysis: Example 4.3 - 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 4.4: Circuit diagram for Example 4.3.



































15
25
10
3
2
1
30
10
0
10
50
10
0
10
30
I
I
I
Eq (4.16)
13

14. Basic Circuits
Mesh Analysis: With current sources (supermesh) in the circuit
14
 A supermesh results when two meshes have a (dependent or
independent) current source in common. The presence of the
current sources in between meshes reduces the number of
equations.
 The following are properties of a supermesh:
1. The current source in the supermesh is not completely ignored; it
provides the constraint equation necessary to solve for the mesh
currents.
2. A supermesh has no current of its own.
3. A supermesh requires the application of both KVL and KCL.

15. Basic Circuits
Mesh Analysis: With current sources (supermesh) in the circuit
Example 4.4: For the circuit in Figure below, find i1 to i4 using mesh analysis.
Figure 4.5: Circuit diagram for Example 4.4.
15
4 Ω 2 Ω
8 Ω
i2 i3 i4
10 V
i1
2 Ω
6 Ω
5 A
3i0
i3
i2
Q
i2
i1
P
i0

16. Basic Circuits
Mesh Analysis: With current sources in the circuit
16
Applying KVL on the supermesh, we can obtain:
For the independent current source, we apply KCL to node P:
For the dependent current source, we apply KCL to node Q:
Applying KVL in mesh 4:
From the equations above, we can solve to find:
Eq (4.19)
Eq (4.18)
Eq (4.17)
Eq (4.20)

17. 10V
20V
4A
10 
5 
20 
2 
+
_
15 
+
_
I1 I2
I3
Basic Circuits
Mesh Analysis:
Find the three mesh currents I1, I2 and I3 in the circuit below.
Figure 4.6. Circuit diagram with current sources
17
exercise