Mesh and Nodal analysis introduction

1. Any closed path formed by branches in a network
is called as mesh. The network shown in Fig(1) ABDA and
BCDB are the two mesh and we need to find the value of
each mesh current.
The number of equations is equal to the number of meshes in the circuit. The
procedure used in applying the mesh current method to linear circuit is given below
1. Define each mesh currents in a clockwise direction.
2. Apply KVL around each mesh expressing voltage across each element in terms of one
or more mesh currents flowing through it.
3. Solve the resulting linear equations with mesh currents as dependent variables.
MESH:
1
D
 
Fig 1
2
V
1
R

1
V 
1
I 2
I
A C
B



2
R
3
R
The number of independent mesh equations, ‘m’ is related to branches ‘b’ and nodes ‘n’ in
an electrical network by the equation
m=b-n+1

2. Steps to solve the mesh current method
By applying KVL to the network shown in fig(1),
we get the equations,
For mesh ABDA
Mesh Current Method:
 
1 1 1 2 2 1
1 1 2 2 2 1
-I R -(I -I )R +V =0
I (R +R ) I R =V 1
 
2
2
V
1
R

1
V 
1
I 2
I
A C
B
D
 
Fig 1



2
R
3
R
1 2
(I -I )

1
I

 2
I

For mesh BCDB
 
2 3 2 2 1 2
1 2 2 2 3 2
-I R -V -(I -I )R =0
I R I (R +R )=V 2
 
The mesh currents are find out by solving the equations (1) and (2)

3. Example problem:
3
30

1
V 120V
 
1
I 2
I
A C
B
D
 
Fig 2



50
10
1 2
(I -I )

1
I

 2
I

2
V 60V

 
1 2 1
1 2
-30I -10(I -I )+120=0
40I 10I 120 1
  
1.Find the mesh current for the network shown in fig (2)
Solution:
First assign mesh currents to the meshes ABDA and
BCDB
For Mesh ABDA
For Mesh BCDB
 
2 2 1
1 2
-50I -60-10(I -I )=0
-10I 60I 60 2
   
By solving equation (1) and (2) we get the two mesh currents

4. Multiplying equation (2) by 4 and adding it to equation (1), we get
1
2
I 0.521
I = 2.86A
 
4
 
 
1 2
1 2
2
40I 10I 120 1
4*(2) ( ) -40I 240I 240 3
I 0.521A (4)
  
     
  
Substitute equation (4) in the equation (1) and we can find the value of 2
I
 
 
1
2 1
1
1
1 40I 10( 0.521) 120
I =-0.521A (1) 40I 120 5.21
120 5.21
I
40
I 2.86A 5
   
  


 
The mesh currents in the network given is
Similarly we can solve more number of meshes with same set of procedures.

5. Super Mesh:
5
In an electrical circuit if there is any common
independent current source for two meshes, then it is
difficult to write and analysis the mesh current method.
In such a case, it is difficult to assign a voltage to
the independent current sources since a current source
will supply rated current at all voltages. In order to solve
such a electrical network we introduce the concept of
‘super mesh’.
If there is two current sources in a circuit then by
source conversion techniques we can convert one current
source into voltage source(adding resistance in series)
and apply the super mesh technique.
 
Fig 3
2
d
b

10
3
I

5
3
V
50
1
1
I
2
I
2A
f
c
e
a
In fig (3) the current
source is common to mesh
‘befcb’ and mesh ‘cfdc’, so
now we can form the super
mesh for this network. For the
circuit shown in fig (3) fid the
current in the 5Ω resistor by
using super mesh.

6. 6
 
1 2 2 1 3
1 2 3
10(I -I )+2I +5(I -I ) =50
15I 10I 5I 50 1
   
Solution:
Apply KVL for the mesh ‘abcda’ we get 2
d
b

10
3
I

5
3
V
50
1
1
I
2
I
2A
f
c
e
a
From the second(befcb) and third(cfdc) meshes, we can
form a super mesh
 
2 1 2 3 3 1
1 2 3
10(I -I )+2I +I +5(I -I ) =0
-15I 12I +6I 0 2
  
The current source is equal to the difference between Second and third mesh currents, that
is given by
 
2 3
I -I =2A 3

Solving equation (1), (2) and (3) we have
 
 
 
1
2
3
I =19.99A 4
I =17.33A 5
I =15.33A 6



The current in the 5Ω resistor is
 
1 3
I -I =4.66A 7


7. Nodal analysis:
7
A node is defined as a junction of three or more branches. The nodal method is
used to analyze multisource(current source) circuits.
Kirchhoff’s current law is used to develop the method referred to as nodal analysis.
Steps in nodal analysis
• Determine the number of nodes within the
network.
• Pick a reference node, and label each remaining
node with a subscript value of voltage: V1, V2,
and so on.
• Apply Kirchhoff’s current law at each node
except the reference node.
• Solve the resulting equation to get the voltages
at each node.
1
I 2
I
1
R 2
R
3
R
4
R 5
R
1
V 2
V 3
V
(1) (2) (3)

8. 8
Steps in nodal analysis (super node)
In some occasion there will be independent voltage sources in the
network to which nodal analysis is to be applied. If so, convert the voltage
source to a current source if a series resistor is present, If not use the supernode
approach.
• Assign a nodal voltage to each independent node of the network.
• Replace independent voltage sources with short-circuits.
• Apply KCL to the defined nodes of the network.
• Relate the defined nodes to the independent voltage source of the network,
and solve for the nodal voltages.

9. 9
1
I 2
I
1
R
2
R
3
R
1
V 2
V
(1) (2)
(3)
node
Ref.
1
I 2
I
R3
I
R1
I R2
I
Consider the above circuit shown, in this circuits there are three nodes so it is
possible to write 2 (n-1) equations.
Nodal analysis: Formation

10. 10
Applying KCL at node 1 gives
R3
R1
1 I
I
I 

0
I
I
I R3
R1
1 



Where
3
2
1
R3
1
1
R1
R
V
V
I
R
V
I



Hence
0
R
V
V
R
V
I
3
2
1
1
1
1 



Applying KCL at node 2 gives
R3
R2
2 I
I
I 

0
I
I
I R2
R3
2 



Where
3
2
1
R3
2
2
R2
R
V
V
I
R
V
I



Hence
0
R
V
R
V
V
I
2
2
3
2
1
2 



(1) (2)
By rearranging and solving equation 1 and 2, nodal voltages V1 and V2 are
obtained.

11. 11
Using nodal analysis, determine the current flowing through 20Ω resistor
1A 2A

20
1
V 2
V
(1) (2)
(3)
node
Ref.
1
I 2
I
R3
I
R1
I R2
I

5

10
Nodal analysis: Example

12. 12
Applying KCL at node 1 gives
R3
R1
1 I
I
I 

Where
10
V
V
I
20
V
I
2
1
R3
1
R1



Hence
10
V
V
20
V
1 2
1
1 


Applying KCL at node 2 gives
R3
R2 I
I
2 

Where
10
V
V
I
5
V
I
2
1
R3
2
R2



Hence
5
V
10
V
V
2 2
2
1



(1) (2)
By rearranging and solving equation 1 and 2, nodal voltages V1 and V2 are
obtained.
Solution

13. 13
2
1
2
1
2
1
1
2
1
1
V
2
3V
20
20
V
2
3V
1
20
V
2
20
2V
20
V
1
10
V
10
V
20
V
1
1
equation














(3)
Solution
2
1
2
1
2
2
1
2
2
1
V
3
V
20
10
V
3
V
2
10
2V
10
V
10
V
2
5
V
10
V
V
2
2
equation














(4)
Solving (3) and (4) you get
5.714V
V
2.858V
V
2
1



Current flowing through 20Ω resistor 0.1429A
20
2.858
20
V
I 1
R1 

