MESH ANALYSIS
In mesh analysis, all the current sources are preferably transformed into their equivalent
voltage sources. The clockwise referenced mesh currents are assumed and finally KVL is
applied to each mesh. It involves systematic application of Ohm’s Law and KVL. A mesh
is defined as a loop which does not have any other loop in it. However, a loop may have
several meshes in it. Thus a mesh is necessarily a loop but a loop is not necessarily a
mesh. Figure 2.1 shown below gives the clear illustration of a mesh and a loop. As shown
in figure, ABCD, BCEF and EGHF are meshes whereas ABEGHFCD, ABEFCD and
BEGHFC are loops.
Fig. 2.1
While applying mesh analysis, the number of meshes are initially identified. The
directions of mesh currents and hence sign conventions at each circuit element are
assumed arbitrarily.
Apply KVL in Mesh1
V = I1 Z1 + (I1 – I2) Z4
V = (Z1 + Z4) I1 + (Z4) I2
or
(Z1 + Z4) I1 + (Z4) I2 – V = 0 (2.1)
It is worth noting that while applying KVL in mesh1, the difference of currents (I1 –
I2) flows through Z4, the direction of which is taken same as I1 presuming I1 has
greater magnitude as compared to I2. The same rule applies in mesh2 and mesh3.
Apply KVL in Mesh2
Z2 I2 + Z5 (I2 – I3) + Z4(I2 – I1) = 0
(Z4) I1 + (Z2 + Z5 + Z4) I2 + (Z5) I3 = 0 (2.2)
Apply KVL in Mesh3
Z3 I3 + Z5 (I3 – I2) = 0
(Z5) I2 + (Z3 + Z5) I3 = 0 (2.3)
Writing Eqs. (2.1), (2.2) and (2.3) in matrix form.
𝑍1 + 𝑍4 −𝑍4 0 𝐼1 𝑉
[ −𝑍4 𝑍2 + 𝑍4 + 𝑍5 −𝑍5 ] [𝐼2 ] = [ 0 ]
0 −𝑍5 𝑍3 + 𝑍5 𝐼3 0
𝑍1 + 𝑍4 −𝑍4 0
∆ = [ −𝑍4 𝑍2 + 𝑍4 + 𝑍5 −𝑍5 ]
0 −𝑍5 𝑍3 + 𝑍5
𝑉 −𝑍4 0 𝑍1 + 𝑍4 𝑉 0
∆1 = [ 0 𝑍2 + 𝑍4 + 𝑍5 −𝑍5 ] ∆2 = [ −𝑍4 0 −𝑍5 ]
0 −𝑍5 𝑍3 + 𝑍5 0 0 𝑍3 + 𝑍5
1
, 𝑍1 + 𝑍4 −𝑍4 𝑉
∆3 = [ −𝑍4 𝑍2 + 𝑍4 + 𝑍5 0 ]
0 −𝑍5 0
The mesh currents are
∆1 ∆2 ∆3
𝐼1 = ; 𝐼2 = ; 𝐼3 =
∆ ∆ ∆
The following points are worth noting in mesh current analysis.
(i) Mesh current method is developed by applying KVL around meshed in the circuit.
(ii) A mesh is a loop which does not contain any other loop/ loops within it.
(iii) Loop/ mesh analysis results in a system of linear equations which are solved for
unknown currents.
(iv) It reduces the number of equations to the number of meshes.
Example 2.1
Determine the power dissipation in 4 resistor of the circuit shown using mesh analysis.
5 2 6
Fig. 2.2 50V I1 I2 4 I3 10V
3
Solution
Loop 1: 50 = 5I1 + 3I1 – 3I2 or 50 = 8I1 – 3I2 (i)
Loop 2: 0 = 2I2 + 4I2 – 4I3 + 3I2 – 3I1
0 = 3I1 + 9I2 – 4I3 (ii)
Loop 3: 0 = 6I3 + 10 + 4I3 – 4I2
10 = 4I2 + 10 I3 (iii)
10 4I 2
From (iii) I3 (iv)
10
Putting (iv) in (ii), thus,
30I1 – 74 I2 = 40 (v)
Solving (i) and (v), we get.
I1 = 7.13 Amp; I2 = 2.35 Amp; I3 = 0.06 Amp.
Power Dissipation in 4 ohm = (2.35 – (0.06))2 x 4 = 23.23 watts.
Example 2.2
Using mesh analysis, determine the voltage Vs which gives a voltage, drop of 50 volts
across 10 ohm resistor as shown in Fig. 2.3
Fig. 2.3
2
, Solution:
From Fig. 2.3
50
𝐼4 = 10 = 5 A
Loop 1:
60 = 3 I1 + I1 – I2 or 4 I1 – I2 = 60 (i)
Loop 2:
5[I2 + I4] + 2[I2 – I3] + [I2 – I1] = 0
I1 + 8I2 – 2I3 = 25 (ii)
Loop 3:
50 = 4I3 + 2[I3 – I2]
2I2 + 6I3 = 50 (iii)
Loop 4:
Vs = 10 I4 + 5[I4 + I2] or Vs = 75 + 5I2 (iv)
From (iii) 6I3 = 50 + 2I2
50 2 I 2
I3
6
(v)
Putting (v) in (ii).
50 2 I 2
I1 8I 2 2 25
6
I1 7.333I 2 8.33
4 I1 29.332 I 2 33.32
(vi)
Adding (i) and (vi), we get
I2 = 0.94169 A; Vs = 79.7 V
Supermesh
When a current source is common to two meshes, we use the concept of supermesh. A
supermesh is created from two meshes that have a current source in common branch.
Example 2.3
Determine the current in 5 resistor in the network shown in Fig. 2.4
a b e
10
Mesh 2 2
Fig. 2.4 I2
2A
Mesh 1
c f
50V
3
I1
Mesh 3 1
5 I3
d
3
In mesh analysis, all the current sources are preferably transformed into their equivalent
voltage sources. The clockwise referenced mesh currents are assumed and finally KVL is
applied to each mesh. It involves systematic application of Ohm’s Law and KVL. A mesh
is defined as a loop which does not have any other loop in it. However, a loop may have
several meshes in it. Thus a mesh is necessarily a loop but a loop is not necessarily a
mesh. Figure 2.1 shown below gives the clear illustration of a mesh and a loop. As shown
in figure, ABCD, BCEF and EGHF are meshes whereas ABEGHFCD, ABEFCD and
BEGHFC are loops.
Fig. 2.1
While applying mesh analysis, the number of meshes are initially identified. The
directions of mesh currents and hence sign conventions at each circuit element are
assumed arbitrarily.
Apply KVL in Mesh1
V = I1 Z1 + (I1 – I2) Z4
V = (Z1 + Z4) I1 + (Z4) I2
or
(Z1 + Z4) I1 + (Z4) I2 – V = 0 (2.1)
It is worth noting that while applying KVL in mesh1, the difference of currents (I1 –
I2) flows through Z4, the direction of which is taken same as I1 presuming I1 has
greater magnitude as compared to I2. The same rule applies in mesh2 and mesh3.
Apply KVL in Mesh2
Z2 I2 + Z5 (I2 – I3) + Z4(I2 – I1) = 0
(Z4) I1 + (Z2 + Z5 + Z4) I2 + (Z5) I3 = 0 (2.2)
Apply KVL in Mesh3
Z3 I3 + Z5 (I3 – I2) = 0
(Z5) I2 + (Z3 + Z5) I3 = 0 (2.3)
Writing Eqs. (2.1), (2.2) and (2.3) in matrix form.
𝑍1 + 𝑍4 −𝑍4 0 𝐼1 𝑉
[ −𝑍4 𝑍2 + 𝑍4 + 𝑍5 −𝑍5 ] [𝐼2 ] = [ 0 ]
0 −𝑍5 𝑍3 + 𝑍5 𝐼3 0
𝑍1 + 𝑍4 −𝑍4 0
∆ = [ −𝑍4 𝑍2 + 𝑍4 + 𝑍5 −𝑍5 ]
0 −𝑍5 𝑍3 + 𝑍5
𝑉 −𝑍4 0 𝑍1 + 𝑍4 𝑉 0
∆1 = [ 0 𝑍2 + 𝑍4 + 𝑍5 −𝑍5 ] ∆2 = [ −𝑍4 0 −𝑍5 ]
0 −𝑍5 𝑍3 + 𝑍5 0 0 𝑍3 + 𝑍5
1
, 𝑍1 + 𝑍4 −𝑍4 𝑉
∆3 = [ −𝑍4 𝑍2 + 𝑍4 + 𝑍5 0 ]
0 −𝑍5 0
The mesh currents are
∆1 ∆2 ∆3
𝐼1 = ; 𝐼2 = ; 𝐼3 =
∆ ∆ ∆
The following points are worth noting in mesh current analysis.
(i) Mesh current method is developed by applying KVL around meshed in the circuit.
(ii) A mesh is a loop which does not contain any other loop/ loops within it.
(iii) Loop/ mesh analysis results in a system of linear equations which are solved for
unknown currents.
(iv) It reduces the number of equations to the number of meshes.
Example 2.1
Determine the power dissipation in 4 resistor of the circuit shown using mesh analysis.
5 2 6
Fig. 2.2 50V I1 I2 4 I3 10V
3
Solution
Loop 1: 50 = 5I1 + 3I1 – 3I2 or 50 = 8I1 – 3I2 (i)
Loop 2: 0 = 2I2 + 4I2 – 4I3 + 3I2 – 3I1
0 = 3I1 + 9I2 – 4I3 (ii)
Loop 3: 0 = 6I3 + 10 + 4I3 – 4I2
10 = 4I2 + 10 I3 (iii)
10 4I 2
From (iii) I3 (iv)
10
Putting (iv) in (ii), thus,
30I1 – 74 I2 = 40 (v)
Solving (i) and (v), we get.
I1 = 7.13 Amp; I2 = 2.35 Amp; I3 = 0.06 Amp.
Power Dissipation in 4 ohm = (2.35 – (0.06))2 x 4 = 23.23 watts.
Example 2.2
Using mesh analysis, determine the voltage Vs which gives a voltage, drop of 50 volts
across 10 ohm resistor as shown in Fig. 2.3
Fig. 2.3
2
, Solution:
From Fig. 2.3
50
𝐼4 = 10 = 5 A
Loop 1:
60 = 3 I1 + I1 – I2 or 4 I1 – I2 = 60 (i)
Loop 2:
5[I2 + I4] + 2[I2 – I3] + [I2 – I1] = 0
I1 + 8I2 – 2I3 = 25 (ii)
Loop 3:
50 = 4I3 + 2[I3 – I2]
2I2 + 6I3 = 50 (iii)
Loop 4:
Vs = 10 I4 + 5[I4 + I2] or Vs = 75 + 5I2 (iv)
From (iii) 6I3 = 50 + 2I2
50 2 I 2
I3
6
(v)
Putting (v) in (ii).
50 2 I 2
I1 8I 2 2 25
6
I1 7.333I 2 8.33
4 I1 29.332 I 2 33.32
(vi)
Adding (i) and (vi), we get
I2 = 0.94169 A; Vs = 79.7 V
Supermesh
When a current source is common to two meshes, we use the concept of supermesh. A
supermesh is created from two meshes that have a current source in common branch.
Example 2.3
Determine the current in 5 resistor in the network shown in Fig. 2.4
a b e
10
Mesh 2 2
Fig. 2.4 I2
2A
Mesh 1
c f
50V
3
I1
Mesh 3 1
5 I3
d
3
