chapter 3 mesh and nodal analysis

27
ELECTRICAL TECHNOLOGY ENT 188/3 Methods of Analysis

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Page 1: Chapter 3 mesh and nodal analysis

ELECTRICAL TECHNOLOGY ENT 188/3

Methods of Analysis

Page 2: Chapter 3 mesh and nodal analysis

• Introduction

• Nodal Analysis

• Nodal Analysis with Voltage Sources

• Mesh Analysis

• Mesh Analysis with Current Sources

Page 3: Chapter 3 mesh and nodal analysis

INTRODUCTION

Two powerful techniques for circuit analysis:

1. Nodal analysis ( application of KCL)

2. Mesh analysis ( application of KVL)

Page 4: Chapter 3 mesh and nodal analysis

NODAL ANALYSIS

1. In nodal analysis, we are interested in finding the node voltages.

2. Steps to determine node voltage:

1. Select reference node.

2. Assign voltage to the remaining n-1 nodes (with respect to reference node)

3. Apply KCL to each nonreference nodes. Use Ohm’s Law to express the branch currents.

4. Solve the resulting simultaneously equations to solve for node voltage

,1,2,1 ...... −nvvv

Page 5: Chapter 3 mesh and nodal analysis

The first step is selecting a node as the reference node. The reference node is commonly called the ground since it is assumed to have zero potential.

NODAL ANALYSIS

Page 6: Chapter 3 mesh and nodal analysis

NODAL ANALYSIS

Consider the above figure as an example:

1. Ground has been chosen as the reference node

2. Assign v1 and v2 as node 1 and 2 respectively. (node voltage = voltage of node with respect to the reference node).

3. Apply KCL:

322

2121

:2:1

iiINodeiiIINode

=+++=

Page 7: Chapter 3 mesh and nodal analysis

NODAL ANALYSISApply Ohm’s Law, where current flows from a higher potential to a lower potential in a resistor:

3

23

2

212

1

11

0,,0R

viR

vviR

vi −=

−=

−=

Rvv

i lowerhigher −=

3

2

2

212 R

vR

vvI =−

+

2

21

1

121 R

vvRvII −

++=

We obtain:

Substitute into equation:

Solve for v1 and v2 using elimination technique/ Cramer’s rule.

Page 8: Chapter 3 mesh and nodal analysis

NODAL ANALYSIS

Example 3.1:

Calculate the node voltages in the circuit shown in Figure 1.

1. References node

2. Nodes Voltage,

v1 & v2

3.Applying KCL

(refer next slide)

Page 9: Chapter 3 mesh and nodal analysis

NODAL ANALYSISSolution:

Using the elimination technique

Eq. (3.1.1)

Page 10: Chapter 3 mesh and nodal analysis

NODAL ANALYSIS

Page 11: Chapter 3 mesh and nodal analysis

NODAL ANALYSISExample 3.2:

Determine the voltages at the nodes in Figure shown.

Solution:

Page 12: Chapter 3 mesh and nodal analysis

NODAL ANALYSIS

We have three simultaneous equation to solve to get the node voltages v1,v2, v3.

We can get the answer by Using elimination technique

Answer :

Page 13: Chapter 3 mesh and nodal analysis

Vvv

vvinSubstitute

vvvv

Fromvv

vv

vvvvvv

vvv

4.2125

12914)4()6(

)6......(..........242

)5()5..(..........042

)3()2()4...(..........1297

)2()1()3....(..........032)2(..........074)1..(..........1223

2

2

22

21

21

21

21

321

321

321

==

=−

==

=+−+

=−−

=+−=−+−

=−−

Continue with v1 & v3..!!!!!!!

Answer : v1 = 4.8V

v2 = 2.4V

v3 = -2.4V

Page 14: Chapter 3 mesh and nodal analysis

NODAL ANALYSIS WITH VOLTAGE SOURCES

Possibilities:

1. If a voltage source is connected between the reference node and a nonrefencenode, simply set the voltage at the nonreference node equal to the voltage of the voltage source

2. If the voltage is connected between two nonreference nodes, the two nonreference nodes from a generalized node or supernode; apply both KCL and KVL to determine the node voltages.

Supernode: formed by enclosing voltage source connectedbetween two nonreference nodes and any elements

connected in parallel with it.

Page 15: Chapter 3 mesh and nodal analysis

From the figure:

• v1 = 10V

• Nodes 2 and 3 form a supernode

• KCL at supernode:

60

80

42323121 −

+−

=−

+− vvvvvv

505 3232 =−⇒=++− vvvvKVL at supernode:

3241 iiii +=+

NODAL ANALYSIS WITH VOLTAGE SOURCES

Page 16: Chapter 3 mesh and nodal analysis

NODAL ANALYSIS WITH VOLTAGE SOURCES

Practice Problem:

Find v and i in the circuit in Fig 3.11:

Answer: -0.2V, 1.4A

At node 1

)4......(....................6

02

03

04

33221 −+

−+

−=

− vvvvv

i1

i2

i3

i41 2 3 )1.......(....................71 Vv =

At node 2

)2.....(..........4321 iiii ++=

Node 2 & node 3 form super node

)3.....(....................303

23

32

+==+−−

vvvv

From (2)

To be continued……….

Page 17: Chapter 3 mesh and nodal analysis

NODAL ANALYSIS WITH VOLTAGE SOURCES

Practice problem:

For the circuit shown in Figure 3.9, find the node voltages.

Answer: v1= -7.333V, v2= -5.333V

Page 18: Chapter 3 mesh and nodal analysis

MESH ANALYSIS

1. Mesh analysis is also known as loop analysis or the mesh-current method.

2. Mesh is a loop which does not contain any other loops within it.

3. Application: to find unknown currents

4. Only capable to a planar circuit

5. Planar circuit: can be drawn in a plane with no branches crossing one another.

Page 19: Chapter 3 mesh and nodal analysis

MESH ANALYSIS

Fig. 3.15 a) a Planar circuit with crossing branches.

b)The same circuit redrawn with no crossing branches b)

Page 20: Chapter 3 mesh and nodal analysis

MESH ANALYSIS

Steps in determining node voltage:

1. Assign mesh currents i1,i2,…in to the n meshes

2. Apply KVL to each of the n meshes. Use Ohm’s Law to express the voltages in terms of the mesh currents.

3. Solve the resulting n simultaneously equations to get the mesh currents.

Page 21: Chapter 3 mesh and nodal analysis

MESH ANALYSIS

Consider the figure below:1. Assign i1 and i2 as meshes 1 and 2.

2. Apply KVL to each mesh:

3. Solve for mesh currents i1 and i2

0)(:20)(:1

123222

213111

=−++=−++−

iiRViRMeshiiRiRVMesh

i3

i1i2

I3= i1 - i2

Page 22: Chapter 3 mesh and nodal analysis

MESH ANALYSIS

Example: For the circuit in Figure, find the branch currents I1, I2 and I3 using mesh analysis.

i3

i1i2

I3=i1-i2 or I3 = -(i2-i1)

Page 23: Chapter 3 mesh and nodal analysis

MESH ANALYSIS

Practice Problem:

Calculate the mesh currents i1 and i2 in the circuit of Figure shown.

Answer: i1= 2/3 A, i2=0A

Page 24: Chapter 3 mesh and nodal analysis

MESH ANALYSIS WITH CURRENT SOURCESPossibilities:

1. When a current source exists only in one mesh – set the current as equal to the source.

• set i2 = -5A

• Mesh equation:

Aiiii 2,0)(6410 1211 −==−++−

Page 25: Chapter 3 mesh and nodal analysis

MESH ANALYSIS CURRENT SOURCES

2. When the current source exists between two meshes – create a supermesh (by excluding the current source and any elements connected in series with it).

A supermesh results when two meshes have a (dependent or independent) current source in common.

Fig: a) Two meshes having a current source in common,

b) a supermesh, created by excluding the current source

Page 26: Chapter 3 mesh and nodal analysis

• From the above figure:

1. Apply KVL to supermesh:

2. Applying KCL to node 0:

3. Solving:

AiAi 8.2,2.3 21 =−=

0410620 221 =+++− iii

612 += ii

MESH ANALYSIS CURRENT SOURCES

Page 27: Chapter 3 mesh and nodal analysis

FURTHER READING

1. Fundamentals of Electric Circuits, 2nd Edition,McGrawhill Alexander, C. K. and Sadiku, M. N. O.

2. Electric Circuit, 8th Edition, Pearson, Nillson and Riedel