Basic Laws of Electric Circuits
Mesh Analysis
Lesson 7
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
Basic Circuits
Mesh Analysis: Basic Concepts:
R1
+
VA
V1
R2
_
V2
_
+
+
_
+
I1
VL1
_
Rx
+
I2
_
VB
Figure 7.2: A circuit for illustrating mesh analysis.
Around mesh 1:
V1  VL1  V A
where V1  I1 R1 ; VL1  I1  I 2 RX
so, ( R1  RX ) I1  RX I 2  V A
Eq 7.1
Basic Circuits
Mesh Analysis: Basic Concepts:
R1
+
VA
V1
R2
_
+
V2
_
+
+
I1
_
VL1
_
Rx
+
I2
Around mesh 2 we have
V L1  V2  V B
with; V L1   ( I 2  I1 ) R X ; V2  I 2 R2
_
VB
Eq 7.2
Eq 7.3
Substituting Eq 7.3 in 7.2 gives,
R X I 1  ( R X  R2 ) I 2  V B
or
 R X I1  ( R X  R2 ) I 2   V B
Eq 7.4
Basic Circuits
Mesh Analysis: Basic Concepts:
We are left with 2 equations: From (7.1) and (7.4)
we have,
( R1  RX ) I1  RX I 2  VA
Eq 7.5
 RX I1  ( RX  R2 ) I 2  VB
Eq 7.6
We can easily solve these equations for I1 and I2.
Basic Circuits
Mesh Analysis: Basic Concepts:
The previous equations can be written in matrix form as:
 R X   I1   V A 
( R1  RX )

 R




( RX  R2   I 2   VB 
X

or
Eq (7.7)
1
 RX   V A 
 I1  ( R1  RX )
I     R



( RX  R2   VB 
X
 2 
Eq (7.8)
Basic Circuits
Mesh Analysis: Example 7.1.
Write the mesh equations and solve for the currents I1, and I2.
4
10V
+
_
2
7
6
I1
2V +_
I2
_
+
20V
Figure 7.2: Circuit for Example 7.1.
Mesh 1
Mesh 2
4I1 + 6(I1 – I2) = 10 - 2
Eq (7.9)
6(I2 – I1) + 2I2 + 7I2 = 2 + 20
Eq (7.10)
Basic Circuits
Mesh Analysis: Example 7.1, continued.
Simplifying Eq (7.9) and (7.10) gives,
10I1 – 6I2 = 8
Eq (7.11)
-6I1 + 15I2 = 22
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
Basic Circuits
Mesh Analysis: Example 7.2
Solve for the mesh currents in the circuit below.
9
+
12V
_
_+
10 
I3
6
20V
8V
+ _
11 
4
+
__
I1
_
+
10V
I2
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.
Basic Circuits
Mesh Analysis: Example 7.2
9
+
12V
_
_+
10 
I3
6
20V
8V
+ _
11 
4
+
__
I1
_
+
10V
I2
3
Mesh 1:
6I1 + 10(I1 – I3) + 4(I1 – I2) = 20 + 10
Eq (7.13)
Mesh 2:
4(I2 – I1) + 11(I2 – I3) + 3I2 = - 10 - 8
Eq (7.14)
Mesh 3:
9I3 + 11(I3 – I2) + 10(I3 – I1) = 12 + 8
Eq (7.15)
Basic Circuits
Mesh Analysis: Example 7.2
Clearing Equations (7.13), (7.14) and (7.15) gives,
Standard Equation form
20I1 – 4I2 – 10I3 = 30
-4I1 + 18I2 – 11I3 = -18
-10I1 – 11I2 + 30I3 = 20
In matrix form:
 20  4  10  I1 
 30 
  4 18  11  I     18

 2 


  10  11 30   I 3 
 20 
WE NOW MAKE AN IMPORTANT
OBSERVATION!!
Basic Circuits
Mesh Analysis: Standard form for mesh equations
Consider the following:
 R11
R
 21
 R31
R12
R22
R32
R13   I1 
  emfs (1) 
R23   I 2     emfs ( 2) 
  emfs ( 3) 
R33   I 3 
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.
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.
 emfs (1) = sum of emf around mesh 1 in the direction of I1.
 emfs ( 2) = sum of emf around mesh 2 in the direction of I2.
 emfs ( 3) = sum of emf around mesh 3 in the direction of I3.
Basic Circuits
Mesh Analysis: Example 7.3 - Direct method.
Use the direct method to write the mesh equations for the following.
20 
30 
I1
I2
+
_
10V
8
_
10 
20V +_
12 
+
15V
10 
I3
_
+ 30V
Figure 7.4: Circuit diagram for Example 7.3.
0   I1  10 
 30  10
 10 50  10  I   25

 2   
 0
 10 30   I 3  15 
Eq (7.13)
Basic Circuits
Mesh Analysis: With current sources in the circuit
Example 7.4: Consider the following:
20V
_ +
2
I3
10 
10V +_
I1
5
20 
I2
4A
15 
Figure 7.5: Circuit diagram for Example 7.4.
Use the direct method to write the mesh equations.
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.
20V
_ +
2
I3
10 
10V +_
I1
5
20 
I2
4A
15 
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.
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.
20V
_ +
2
I3
10 
10V +_
I1
5
20 
I2
15 
Note that I 2 is retained for writing the equations through the
5  and 20  resistors.
Basic Circuits
Mesh Analysis: With current sources in the circuit
Example 7.4: Continued.
20V
_ +
2
I3
10 
10V +_
20 
Equation for mesh 1:
10I1 + (I1-I2)5 = 10
I1
5
I2
15 
Equations for mesh 2:
2I3 + (I3-I2)20 = 20
or
- 20I2 + 22I3 = 20
or
15I1 – 5I2 = 10
Constraint Equation
I2 = - 4A
Basic Circuits
Mesh Analysis: With current sources in the circuit
Example 7.4: Continued. Express the previous equations in
Matrix form:
15 5 0   I1  10 
 0 20 22   I    20 

 2  
 0
1
0   I 3   4
I1 = -0.667 A
I2 = - 4 A
I3 = - 2.73 A
circuits
End of Lesson 7
Mesh Analysis