Mesh Analysis

Examples

Example 1

2 6.364A

40V6.364A

20V1.818A

4

6I1 I2

Using Mathcad

mesh 1: 20 4I1 6 I1 I2 0

mesh 2: 2I2 6 I2 I1 40 0

Find I1 I2 1.818

6.364

Using PSpice

2 6.364A

40V6.364A

20V

1.818A

41.818A

0

6

4.546A

I1 = -1818 A I2 = -6.364 A

Example 2: By Inspection

8.863A

12V6.549A

46.235A 1

1

24V

18V

30 1

2

6V 2

I2I1

I3

Using Mathcad: By InspectionR I V

R

4 3( )

0

3

0

1 1 2( )

1

3

1

2 3 1 1( )

V

18 6

24 18

12 24

I lsolve R V( ) I

6.235

8.863

6.549

Using PSpice

0

12V6.549A

4

6.235A1

8.863A

1

24V

18V

3 1

2

6V 2

Example 3

4k2mA

3V

4k

6k

2k

4mA

-2mA

4mA

I1

Using Mathcad

2 k 4 m 2k 4k 6k( ) I1 4k( ) 2 m( ) 3 0

Find I1 250

Mesh 1:

Using PSpice

4k

6.000mA

2mA

2.000mA

3V

250.0uA

0

4k

2.250mA

6k

250.0uA2k

3.750mA4mA

4.000mA

I1 = 250 A

Example 4: Supermesh

3

20V 3

6A

46

I2I1

Mesh 1 and mesh 2 form supermesh

I1 = -1.6 A I2 = 4.4 AAns:

Using Mathcad

Supermesh 1& 2: 20 4I1 6I2 0

Auxilliary : I2 I1 6

Find I1 I2 1.6

4.4

Using PSpice

34.400A

20V

1.600A

3

4.400A

6A

6.000A

41.600A

0

6

6.000A

I1 = -1.6 A I2 = 4.4 AAns:

Example 5: Supermesh

2.5Ib20V

2Ia+-

8

80

40

5A

Ia

Ib

Example 5: Supermesh

2.5Ib20V

2Ia+-

8

80

40

5A

Ia

Ib

I1

5A I2

I3

I1 = 5.431A I2 = -3.621A I3 = -1.81AAns:

Using Mathcad: equations

Controlling variable 1: Ib I2Controlling variable 2: I1 I3 Ia

Supermesh 1&2: 80 I1 5 8 I1 I3 40 I2 I3 20 0

Auxilliary: I2 I1 2.5 Ib

mesh 3: 8 I3 I1 40 I3 I2 2Ia 0

Using Mathcad: Results

Find I1 I2 I3 Ia Ib

5.431

3.621

1.81

7.241

3.621

Using PSpice

G1

2.5*I(Vb)

9.052A

20V

3.621A

Vb0Vdc

3.621A

2*I(Va)1.810A

+-

0

8

7.241A

80

431.0mA

40

1.810A

5A

5.000A

Va

0Vdc

I1 = 5.431A I2 = -3.621A I3 = -1.81AAns:

Find Va and Ix

Problem 1

i1

3A

6V

-

2 2

4

81

i2

i3

+

Va

Ix

+

-

Using Mathcad

Supermesh 1&2: 6 2 I1 I3 4 I2 I3 8I2 0

Auxilliary: I1 I2 3

mesh 3: 2 I3 I1 2I3 4 I3 I2 0

Find I1 I2 I3 3.474

0.474

1.105

Ix = -I1 = -3.474 A

To Find Ix

i1

3A

6V

-

2 2

4

81

i2

i3

+

Va

Ix

+

-

Mesh 1: -6+(I1-I3)2+(1 3)+Va =0

To Find Va

i1

3A

6V

-

2 2

4

81

i2

i3

+

Va

Ix

+

-

+

-

+ -

+

-

V8

V4

V1

V2

-

+

Mesh 2: -(1 3)+ (I2-I3)4+8I2-Va =0 orVa = -1.737V

Problem 2

Find Vxy, Vc, and the power absorbed by all elements.

x

2.5Ib20V

2Ia +-

8

80

40

5A

Ia

Ib

I1

5A I2

I3

10

y

+

-

Vc

Using Mathcad: Equations

Controlling variable 1: Ib I2Controlling variable 2: I1 I3 Ia

Supermesh 1&2: 80 I1 5 8 I1 I3 40 I2 I3 20 0

Auxilliary: I2 I1 2.5 Ib

mesh 3: 8 I3 I1 40 I3 I2 2Ia 10I3 0

Using Mathcad: Results

Find I1 I2 I3 Ia Ib

5.625

3.75

1.563

7.188

3.75

To Find Vxy

Vxy, Vc

x

2.5Ib20V

2Ia +-

805A

Ib

I1

5A I2

I3

10

y

+

-

Vc

+ -

-Vxy-2Ia+10I3= 0

Vxy

Vxy = -2Ia+10I3

Vxy 2 Ia 10I3 Vxy 30

To Find Vc

x

2.5Ib20V

2Ia +-

805A

Ib

I1

5A I2

I3

10

y

+

-

Vc

+ -

+Vc+(I2-I3)40-20= 0

V40

Vc = -(I2-I3)40+20

40

Vc I2 I3 40 20 Vc 107.5

Power absorbed by active elements

x

2.5Ib20V

2Ia +-

8

80

40

5A

Ia

Ib

I1

5A I2

I3

10

y

+

-

Vc+

-

V5A

555 AA VPbcIb IVP 5.25.2

)(2020 bV IP

)(2 32 IIP aIa

For the active elements, we have to follow PSC to calculate the power absorbed.

Power absorbed by passive elements

x

2.5Ib20V

2Ia +-

8

80

40

5A

Ia

Ib

I1

5A I2

I3 10

+

-

Vc+

-

V5A

102310 IP

40)( 23240 IIP

80)5( 2180 IP

828 aIP

The power absorbed by the resistor is always positive; we can always use P = I2R

Power absorbed by all elements

P5A P2.5Ib P20V P2Ia P80 P8 P10 P40 0.000

P40 191.406P40 I2 I3 2 40

P10 24.414P10 I32 10

P8 413.281P8 Ia2 8

P80 31.25P80 5 I1 2 80

P2Ia 22.461P2Ia 2 Ia I3

P20V 75P20V 20 I2

P2.5Ib 1.008 103P2.5Ib Vc 2.5 Ib

P5A 250P5A V5A 5

V5A 50V5A I1 5 80