Mesh analysis is a technique for analyzing electrical circuits using loops or meshes. It involves assigning a mesh current to each loop and writing Kirchhoff's voltage law equations for each mesh. The mesh equations take the form of a matrix equation that can be solved for the unknown mesh currents. Key aspects of mesh analysis include defining mesh currents, writing KVL equations clockwise around each mesh, identifying common resistance terms between meshes, and handling circuits with current sources by removing the source and including it as a constraint.

1. Lesson 7
Basic Laws of Electric Circuits
Mesh Analysis

2. 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.
I 1 I 2 I 3

3. Basic Circuits
Mesh Analysis: Basic Concepts:
R 1
R x
R 2
+
_ I 1 I 2
+
_V A V B
+ +
+
_
_
_V 1
V L 1
V 2
Figure 7.2: A circuit for illustrating mesh analysis.
( )
AXX
XL
AL
VIRIRRso
RIIVRIVwhere
VVV
=−+
−==
=+
211
211111
11
)(,
;
Eq 7.1
Around mesh 1:

4. Basic Circuits
Mesh Analysis: Basic Concepts:
R 1
R x
R 2
+
_ I 1 I 2
+
_V A V B
+ +
+
_
_
_V 1
V L 1
V 2
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

5. 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.

6. 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)

7. Basic Circuits
Mesh Analysis: Example 7.1.
Write the mesh equations and solve for the currents I1, and I2.
+
_
1 0 V
4 Ω 2 Ω
6 Ω
7 Ω
2 V 2 0 V
I 1 I 2
+
+
_
_
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)

8. 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

9. Basic Circuits
Mesh Analysis: Example 7.2
Solve for the mesh currents in the circuit below.
+
_
6 Ω
1 0 Ω
9 Ω
1 1 Ω
3 Ω
4 Ω
2 0 V 1 0 V
8 V
1 2 V
I 1 I 2
I 3
+
+
__
_
_
+
+
_
Figure 7.3: Circuit for Example 7.2.
The plan: Write KVL, clockwise, for each mesh. Look for a
pattern in the final equations.

10. Basic Circuits
Mesh Analysis: Example 7.2
+
_
6 Ω
1 0 Ω
9 Ω
1 1 Ω
3 Ω
4 Ω
2 0 V 1 0 V
8 V
1 2 V
I 1 I 2
I 3
+
+
__
_
_
+
+
_
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)

11. 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

12. 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

13. 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.

14. Basic Circuits
Mesh Analysis: Example 7.3 - Direct method.
2 0 V
1 0 V
1 5 V
3 0 V
2 0 Ω
1 0 Ω
3 0 Ω
1 0 Ω
1 2 Ω
8 Ω
+
_
I 1 I 2 I 3
+
+
+
_
_
_
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)

15. Basic Circuits
Mesh Analysis: With current sources in the circuit
Example 7.4: Consider the following:
1 0 V
2 0 V
4 A
1 0 Ω
5 Ω
2 0 Ω
2 Ω
+
_
1 5 Ω
+
_
I 1 I 2
I 3
Figure 7.5: Circuit diagram for Example 7.4.
Use the direct method to write the mesh equations.

16. 1 0 V
2 0 V
4 A
1 0 Ω
5 Ω
2 0 Ω
2 Ω
+
_
1 5 Ω
+
_
I 1 I 2
I 3
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.

17. 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 I2 is retained for writing the equations through the
5 Ω and 20 Ω resistors.
1 0 V
2 0 V
1 0 Ω
5 Ω
2 0 Ω
2 Ω
+
_
1 5 Ω
+
_
I 1 I 2
I 3

18. Basic Circuits
Mesh Analysis: With current sources in the circuit
Example 7.4: Continued.
1 0 V
2 0 V
1 0 Ω
5 Ω
2 0 Ω
2 Ω
+
_
1 5 Ω
+
_
I 1 I 2
I 3 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

19. 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

20. End of Lesson 7
circuits
Mesh Analysis