MESH AND SUPERMESH ANALYSIS

1

Methods of Analysis

1. Mesh Analysis2. Supermesh Analysis

Process of Mesh Analysis1. Identify every mesh in the circuit.2. Label each mesh with a mesh current. It is recommended that all mesh

currents be labeled in the same direction (either clockwise (CW) or counter-clockwise (CCW)).

Mesh Analysis (Cont’d)4. Within a particular mesh of interest, use

Ohm’s law to express the voltage across any element within that mesh as the difference between the two mesh currents of continuous meshes shared by the element times the element impedance. The current within the mesh of interest is always considered to be larger than the rest of the mesh currents.

Mesh Analysis (Cont’d)5. Apply KVL to sum all voltages in that

mesh of interest. The resulting algebraic equation (called mesh equation) has all mesh currents as its unknowns.

6. Repeat steps 4 and 5 until all meshes are accounted for. The number of equations must be equal to the number of mesh currents.

Mesh Analysis (Cont’d)8. If a current source exists only in one

mesh, the mesh current is equal to the source current, and KVL is not applied to this mesh.

9. Solve the resulting simultaneous mesh equations to obtain the values of the unknown mesh currents.

10. Use the values of mesh currents above to find voltages and/or currents throughout the rest of the circuit.

7

Example

Number of nodes, n = 7

Number of branches, b = 10

Number of loops, l = 4

1l b n= − +

8

Example

Apply KVL to each mesh

2 1 7 5 0sV v v v− + + − =

2 6 7 0v v v− − =

15 3 0sv v v+ + =

Mesh 1:

Mesh 2:

Mesh 3:

14 8 6 0sv v V v+ − + =Mesh 4:

9

2 1 1 1 2 7 1 3 5( ) ( ) 0sV i R i i R i i R− + + − + − =

2 2 2 4 6 2 1 7( ) ( ) 0i R i i R i i R+ − + − =

13 1 5 3 3( ) 0si i R V i R− + + =

Mesh 1:

Mesh 2:

Mesh 3:

14 4 4 8 4 2 6( ) 0si R i R V i i R+ − + − =Mesh 4:

Mesh 1:

Mesh 2:

Mesh 3:

Mesh 4:

21 5 7 1 7 2 5 3( ) sR R R i R i R i V+ + − − =

7 1 2 6 7 2 6 4( ) 0R i R R R i R i− + + + − =

15 1 3 5 3( ) sR i R R i V− + + = −

16 2 4 6 8 4( ) sR i R R R i V− + + + =

10

Mesh 1:

Mesh 2:

Mesh 3:

Mesh 4:

21 5 7 1 7 2 5 3( ) sR R R i R i R i V+ + − − =

7 1 2 6 7 2 6 4( ) 0R i R R R i R i− + + + − =

15 1 3 5 3( ) sR i R R i V− + + = −

16 2 4 6 8 4( ) sR i R R R i V− + + + =

2

1

1

1 5 7 7 5 1

7 2 6 7 6 2

5 3 5 3

6 4 6 8 4

0

00

0 0

0 0

s

s

s

VR R R R R i

R R R R R i

VR R R i

R R R R i V

+ + − − ÷ ÷ ÷− + + − ÷ ÷ ÷= ÷ ÷ ÷ −− + ÷ ÷ ÷ ÷− + +

SupermeshIf a current source exists between two

contiguous meshes, the two meshes are collapsed into a single mesh called a supermesh, and the current source and any elements connected in series with it removed. However, KVL must still be satisfied within a supermesh using the old mesh current labels. Also, the removal of the current source provides another mesh equation.

Contd…

Properties of a Supermesh1. The current is not completely ignored

provides the constraint equation necessary to solve for the mesh current.

1. Several current sources in adjacency form a bigger supermesh.

6

20146

21

21

−=−=+

ii

ii

If a supermesh consists of two meshes, two equations are needed; one is obtained using KVL and Ohm’s law to the supermesh and the other is obtained by relation regulated due to the current source.

Example

Similarly, a supermesh formed from three meshes needs three equations: one is from the supermesh and the other two equations are obtained from the two current sources.

0102)(8

5

06)(842

443

432

21

24331

=++−−=−−=−

=+−++

iii

iii

ii

iiiii

THANK YOU