Nodal and Mesh Analysis

Nodal and Mesh Analysis

Nodal Analysis– KCL– Consider

• Node current• Node voltage

Need to assign one node as a reference node.Normally it will be a ground node

R1, R2, R3, Vs, Is => electric elements

R1 R2

Vs Is

R1, R2, R3, Vs, Is => electric elements

R1 R2

Vs Is

Vi Vj

i j

vi & vj are node voltage

Branch voltage vij=vi-vj

Vi Vj

i j

vi & vj are node voltage

Branch voltage vij=vi-vj

Example 1

Given => node current

Determine => node voltage =>Find voltage at node 1,2 and 3

-8A. 1Ω 5Ω

7Ω

3Ω

4Ω

-25A.

-3A.

12

3

4

Steps of Nodal Analysis

5Ω

4Ω

-3A.

7Ω

-25A.

V1V2 V3

1Ω

-8A.

3Ω

1. Assign a reference node => a node having the most electric elements connecting to it or a ground node if it is given

2. Apply KCL to each node except the reference node.

3. Solving the simultaneous equations.

constant,,,,

constant,,

33131211

321

3333232131

2323222121

1313212111

aaaa

bbb

bxaxaxa

bxaxaxa

bxaxaxa

BAX

KCL@Node 1

V1 V23Ω

-3A.

4Ω

V3

-8A.

i1

i2 )2(1134

)1(08334

3

44

083

2131

2131

212

31131

21

vvvv

vvvv

vvi

vvvi

ii

KCL@Node 2

-3A.

3Ω 7Ω

1Ω

V1V2 V3

i2

i3

i4

)3(373

3173

03

23212

23221

432

vvvvv

vvvvv

iii

KCL@Node 3

From (2), (3) and (4) we get

v1 = 5.412 V , v2 = 7.736 V , v3 = 46.32 V

4Ω

7Ω

5Ω

-25A.

V1

V2 V3 )4(2554731323

vvvvv

Example 2

4Ω

2Ω 8Ω

4Ω2ix3A

.

(1)(2)

(3)

ix

Determine v1, v2 and v3

KCL @ node 1

)1(324

2131

vvvv

KCL @ node 2

)2(048223212

vvvvv

KCL @ node 3

)3(04

284

22313

vvvvv

From (1), (2) and (3) we getv1 = 4.8 V , v2 = 2.4 V , v3 = -2.4 V

Example 3

4Ω

3Ω

1Ω-8A.

-25A.

-3A.

22V.V3

V1V2

5Ω

Supernode

Determine v1, v2 and v3

Supernode internal relationship)1(2223 vv

KCL @ node 1

)2(1134

2131

vvvv

KCL @ supernode

)3(02554

313

313212

vvvvvv

From (1), (2) and (3) we getv1 , v2 , and v3

Example 4 (1)

0.5Ω 2Ω

0.2vy2.5Ω

12V.

14A.

0.5vx

1Ωvy

vx

V2

V3

V4

Ref.

+

-

-

+

Supernode

(2) 2.0

)1(V12

43

1

yvvv

v

Determine the voltage of the unknown node to reference voltages.

We know that

KCL @ node 1 Because it connects to reference node, the node equation has not to determine.

KCL @ node 2

)3(02

145.0

3212

vvvv

Example 4 (2)

0.5Ω 2Ω

0.2vy2.5Ω

12V.

14A.

0.5vx

1Ωvy

vx

V2

V3

V4

Ref.

+

-

-

+

Supernode

Determine the voltage of the unknown node to reference voltages.

KCL @ supernode (3,4)

)6(

)5(

)4(05.21

5.02

14

12

14423

vvv

vvv

vvvv

vv

y

x

x

From (1) - (6) we getv1 = -12 V, v2 = -4 V, v3 = 0 V and v4 = -2 V

Summarize of Nodal Analysis

1. Select the node to which the highest number of branches is connected as the referenced node.

2. Set up KCL equations for other nodes by expressing the unknown currents as a function of the node voltages measured with respect to the referenced node.

3. If the given circuit contains voltage sources, KCL equations of those two nodes connected by a voltage source are combined to eliminate the redundancy of KCL equations since the additional information is available through the node voltage.

4. Solve KCL equations to determine the node voltages.

Mesh Analysis

Mesh = a smallest loop

(a loop does not contain other loops inside)

Mesh analysis can be used only with planar-network.

Figure 3.5 Examples of planar and nonplanar networks. (a) and (c) are planar networks whereas (b) is a nonplanar network.

Figure 3.5 Examples of planar and nonplanar networks. (a) and (c) are planar networks whereas (b) is a nonplanar network.

Mesh Analysis Procedure

1. Assign all mesh currents in the same direction.

2. Set up KVL equation for each mesh. Use Ohm’s law to express the voltages in term of the mesh currents

3. If the given circuit contains current sources on the perimeter of any meshes. That is, two meshes share current sources in common. Such meshes form a supermesh to eliminate the redundancy of KVL equations.

4. Solve KVL equations to determine the mesh currents

Example 5

3Ωi1 i242V. 10V.

6Ω 4Ω)1(42)(36 211 iii

)2(10)(34 122 iii

Determine the current flowing through 3 Ω resistor using mesh analysis.

KVL @ mesh 1

A4

A6

2

1

i

i

KVL @ mesh 2

From (1) + (2),

current flowing through 3 Ω = 6 – 4 = 2 A. from top to bottom.

Example 6

2Ω

1Ω

6V.

2Ω

7V.

1Ω

3Ωi1

i2

i3

)1(123

122

7)(26)(

321

3121

3121

iii

iiii

iiii

)2(036

0)()(32

321

12322

iii

iiiii

Find i1, i2 and i3

KVL @ mesh 1

KVL @ mesh 2

KVL @ mesh 3

)3(6632

6)(2)(3

321

13323

iii

iiiii

A3 and,A2,A3 321 iii

Example 6

3Ωi1 i210V. 5A.

6Ω 4Ω

ix

9

40

)5(9

59

59

)5(310

9

310, Substitute

10)(36

5

21

212

211

2

iii

iii

Viii

Ai

xFind ix using mesh analysis

Because 5A source is at perimeter,The value i2 is already know

Example 7

A8.2andA2.3

(2), and (1) From

)2(20146

eshKVL@superm

)1(6

iprelationsh internalsupermesh

21

21

12

ii

ii

ii

20V.

6A.

4Ω

10Ω6Ω

2Ωi1 i2

20V.

10Ω6Ω

i1 i2

Find i1 , i2

Supermesh

Example 8 (1)

i1

i2i3

9V.

2.2kΩ

10kΩ

3.3kΩ4.7kΩ

2.2kΩ

1mA.

3mA.

Find current flowing through 4.7 k Ω and 2.2 k Ω.

i1

i2i3

10kΩ

3.3kΩ4.7kΩ

2.2kΩ

3mA.

mA3

mA1

32

1

ii

i

02.210)(3.37.4 32123 kikiiikki

Mesh analysis

Example 8 (2)

i1

i2i3

9V.

2.2kΩ

10kΩ

3.3kΩ4.7kΩ

2.2kΩ

1mA.

3mA.

Find current flowing through 4.7 k Ω and 2.2 k Ω.

9V.

2.2kΩ

10kΩ

3.3kΩ4.7kΩ

2.2kΩ

1mA.

3mA.

V2 V1

Node analysis