1 Mesh Analysis Discussion D2.4 Chapter 2 Section 2-8.

1

Mesh Analysis

Discussion D2.4

Chapter 2

Section 2-8

2

Mesh Analysis• Mesh analysis applies KVL to find unknown

currents. • It is only applicable to planar circuits (a circuit that

can be drawn on a plane with no branches crossing each other).

• A mesh is a loop that does not contain any other loops.

• The current through a mesh is known as the mesh current.

• Assume for simplicity that the circuit contains only voltage sources.

3

Mesh Analysis Steps

1. Assign mesh currents i1, i2, i3, … il, to the l meshes,

2. Apply KVL to each of the l meshes and use Ohm’s law to express the voltages in terms of the mesh currents,

3. Solve the l resulting simultaneous equations to find the mesh currents.

4

Example

DC

DC

1R

3R5R

7R

2R

6R

8R

4R

1v 2v

3v 4v

5v6v

7v

8v

+ +

+ +

++

+

+

-

-- -

-

-

-

-

1sV

2sV 1i 2i

3i 4i

Number of nodes, n =

Number of branches, b =

Number of loops, l =

1l b n

7

10

4

5

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:

DC

DC

1R

3R5R

7R

2R

6R

8R

4R

1v 2v

3v 4v

5v6v

7v

8v

+ +

+ +

++

+

+

-

-- -

-

-

-

-

1sV

2sV 1i 2i

3i 4i

6

DC

DC

1R

3R5R

7R

2R

6R

8R

4R

1v 2v

3v 4v

5v6v

7v

8v

+ +

+ +

++

+

+

-

-- -

-

-

-

-

1sV

2sV 1i 2i

3i 4i

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:

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:

Express the voltage in terms of the mesh currents:

7

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

8

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

9

Ri = v

R

v

iis an l x l symmetric resistance matrix

is a 1 x l vector of mesh currents

is a vector of voltages representing “known” voltages

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

10

•The matrix R is symmetric, rkj = rjk and all of the off-diagonal terms are negative or zero.

Writing the Mesh Equations by Inspection

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

The vk (the kth component of the vector v) = the algebraic sum of the independent voltages in mesh k, with voltage rises taken as positive.

The rkj terms are the negative sum of the resistances common to BOTH mesh k and mesh j.

The rkk terms are the sum of all resistances in mesh k.

DC

DC

1R

3R5R

7R

2R

6R

8R

4R

1v 2v

3v 4v

5v6v

7v

8v

+ +

+ +

++

+

+

-

-- -

-

-

-

-

1sV

2sV 1i 2i

3i 4i

11

MATLAB Solution of Mesh Equations

1i R 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

Ri = v

12

Test with numbers

DC

DC

1R

3R5R

7R

2R

6R

8R

4R

1

4

2

4

1

3

32

2V

4V 1i2i

3i4i

1

2

3

4

2 4 1 4 1 0 4

4 3 2 4 0 2 0

1 0 3 1 0 2

0 2 0 2 4 1 2

i

i

i

i

13

Test with numbers

1

2

3

4

2 4 1 4 1 0 4

4 3 2 4 0 2 0

1 0 3 1 0 2

0 2 0 2 4 1 2

i

i

i

i

1

2

3

4

7 4 1 0 4

4 9 0 2 0

1 0 4 0 2

0 2 0 7 2

i

i

i

i

Ri = v

14

MATLAB Run

DC

DC

1R

3R5R

7R

2R

6R

8R

4R

1

4

2

4

1

3

32

2V

4V 1i2i

3i4i

1

2

3

4

2 4 1 4 1 0 4

4 3 2 4 0 2 0

1 0 3 1 0 2

0 2 0 2 4 1 2

i

i

i

i

15

PSpice Simulation

MATLAB:

16

What happens if we have independent or dependent current sources in the circuit?

1. Write the mesh equations in the same way we did for circuits with only independent or dependent voltage sources.

2. Express the current of each independent or dependent current source in terms of the mesh currents.

3. Rewrite the equations with all unknown mesh currents on the left hand side of the equality and all known voltages on the r.h.s of the equality.

17

Example

Write mesh equations by inspection.

1

2

3

1 3 3 1 10

3 3 2 4 2 0

1 2 2 1 a

i

i

i v

DC 10V

1

3A

+ v -a

1i 2i

3i3 3i

18

1

2

4 3 0 7

3 9 0 6

1 2 1 9a

i

i

v

1

2

4 3 1 10

3 9 2 0

1 2 3 3 a

i

i

v

19

MATLAB Run

AAV

i1i2va

1

2

4 3 0 7

3 9 0 6

1 2 1 9a

i

i

v

DC 10V

1

3A

+ v -a

1i 2i

3i

20

PSpice Simulation

MATLAB:i1

va

i2

i1

va

i2

+ -

21

Nodal Analysis vs Mesh Analysis

• If the circuit is nonplanar we must use nodal analysis.

• If the circuit is planar, there are two principle considerations:

– The number of equations we need to write.– The information we need in the circuit. For

example, do we want to find a current or a voltage?

22

The Number of Mesh Equations

If the circuit has only voltage sources or current sources in parallel with resistances (that is, current sources that can be exchanged for voltage sources) then after we have exchanged the current sources, we have to write l mesh equations and the fundamental theorem of network topology tells us that:

( 1)l b n

23

The Number of Nodal Equations

We also know that if the circuit has only current sources or voltage sources in series with resistances (that is, sources that can be transformed to current sources) then after we have transformed the voltage sources, the number of nodal equations is (n - 1).

If we are not interested in the currents through voltage sources or voltages across current sources we may be able to reduce the number of equations we have to write.

24

1/8

si

3 yv

+ -

+-

yv

xv+

+ -

-

2 xv

1V2V

3V

4V

Example

Transform voltage sources into current sources

8S

S

S

2S

si

12 yv

1V

xv+

-

4 xv

2V

b =

n =

7

3

1 4 1 2( 2 )y xv V V V V v

25

Example

8S

S

S

2S

si

12 yv

1V

xv+

-

4 xv

2V

1

2

411 3

12 43 7s x

y x

i vV

v vV

1 2( 2 )y xv V V v

1xv V

Write nodal equations

26

Example

1

2

411 3

12 43 7s x

y x

i vV

v vV

1 2( 2 )y xv V V v

1xv V

1 1

2 1 2 1 1

411 3

12 12 24 43 7sV i V

V V V V V

1

2

7 3

13 19 0sV i

V

27

1

2

7 3

13 19 0sV i

V

Example

1 27 3 sV V i

1 213 19 0V V

2 1

13

19V V

1 1

3 137

19 sV V i

2 0.0756 sV i

1 0.1105 sV i

8S

S

S

2S

si

12 yv

1V

xv+

-

4 xv

2V

28

1/8

si

3 yv

+ -

+-

yv

xv+

+ -

-

2 xv

1V2V

3V

4V

Example

Transform current source into a current source and write mesh equations

1/8

8si

3 yv

+ -

+-

yv

xv+

+ -

-

2 xv

1V2V

3V

4V

+-

5V

2i

1i

1

8 8s

x

i iv 2

2y

iv

29

1/8

8si

3 yv

+ -

+-

yv

xv+

+ -

-

2 xv

1V2V

3V

4V

+-

5V

2i

1i

1

2

8 31 8 1 1 4 1

21 1 2 1s y

x

i vi

vi

Example

30

1

2

8 31.375 1

21 1.5s y

x

i vi

vi

1

8 8s

x

i iv

2

2y

iv

21

12

8 1.51.375 1

0.25 0.251 1.5s

s

i ii

i ii

1

2

81.375 0.5

0.251.25 1.5s

s

ii

ii

Example

31

1

2

81.375 0.5

0.251.25 1.5s

s

ii

ii

1/8

8si

3 yv

+ -

+-

yv

xv+

+ -

-

2 xv

1V2V

3V

4V

+-

5V

2i

1i2 0.0698 si i

1 0.1163 si i

Example