Methods of Analysis

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Transcript Methods of Analysis

Methods of Analysis

Basic Nodal and Mesh Analysis

Al-Qaralleh

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Methods of Analysis

• Introduction • Nodal analysis • Nodal analysis with voltage source • Mesh analysis • Mesh analysis with current source • Nodal and mesh analyses by inspection • Nodal versus mesh analysis Methods of Analysis

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Steps of Nodal Analysis

1. Choose a reference (ground) node.

2. Assign node voltages to the other nodes.

3. Apply KCL to each node other than the reference node; express currents in terms of node voltages.

4. Solve the resulting system of linear equations for the nodal voltages.

EEE 202

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Common symbols for indicating a reference node, (a) common ground, (b) ground, (c) chassis.

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1. Reference Node

I

1 500 W

+

500 W

V

–

500 1k W W 500 W

I

2

The reference node is called the ground node where V = 0

EEE 202

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Steps of Nodal Analysis

1. Choose a reference (ground) node.

2. Assign node voltages to the other nodes.

3. Apply KCL to each node other than the reference node; express currents in terms of node voltages.

4. Solve the resulting system of linear equations for the nodal voltages.

EEE 202

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2. Node Voltages

I

1 1

V

1 500 W

V

2 500 W 2 500 W 1k W

V

3 3 500 W

I

2 EEE 202

V

1 , V 2 , and V 3 are unknowns for which we solve using KCL Lect4

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Steps of Nodal Analysis

1. Choose a reference (ground) node.

2. Assign node voltages to the other nodes.

3. Apply KCL to each node other than the reference node; express currents in terms of node voltages.

4. Solve the resulting system of linear equations for the nodal voltages.

EEE 202

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EEE 202

Currents and Node Voltages

V

1 500 W

V

2

V

1 

V

2 500 W

V

1 500 W

V

1 500 W

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I

1

3. KCL at Node 1

V

1 500 W

V

2 500 W

I

1 

V

1 

V

2 500 W 

V

1 500 W EEE 202

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EEE 202

3. KCL at Node 2

V

1 500 W

V

2 500 W

V

3 1k W

V

2 

V

1 500 W 

V

2 1 k W 

V

2 

V

3 500 W  0

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3. KCL at Node 3

V

2 500 W

V

3 500 W

I

2

V

3 

V

2 500 W 

V

3 500 W 

I

2 EEE 202

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Steps of Nodal Analysis

1. Choose a reference (ground) node.

2. Assign node voltages to the other nodes.

3. Apply KCL to each node other than the reference node; express currents in terms of node voltages.

4. Solve the resulting system of linear equations for the nodal voltages.

EEE 202

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I

1

4. Summing Circuit Solution

500 W

+

500 W

V

–

500 1k W W 500 W

Solution: V = 167I 1 + 167I 2

I

2 EEE 202

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Typical circuit for nodal analysis Methods of Analysis

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i

1

i

2

i

3

I

1 

I

2 

i

1 

i

2   

i

I v

2 

i

2 higher  

v i

3 lower

R v

1 

R

1 0 or

i

1

v

1 

R

2

v

2 or

i

2

v

2 

R

3 0 or

i

3   

G

1

v

1

G G

3 2

v

2 (

v

1 

v

2 ) Methods of Analysis

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I

1

I

2  

I

2

v

1 

v R

1 

v

2 1

R

2 

v

1 

R

2

v

2 

v R

2 3 

I I

1 2  

I

2  

G

2 (

G v

1 1

v

1  

v

2

G

2 )  (

v

1

G

3 

v

2

v

2 )   

G

1  

G G

2 2

G

2 

G

 2

G

3    

v v

2 1    

I

1

I

2

I

2   Methods of Analysis

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Calculus the node voltage in the circuit shown in Fig. 3.3(a)

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At node 1

 5 

i

1

v

1 

i

2 

v

2 

i

3  4

v

1  2 0 Methods of Analysis

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At node 2

 5

i

2  

v

2

i

4  

v

1

i

1 4  

i

5

v

2 6  0 Methods of Analysis

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In matrix form:

    1 2   1 1 4 4 1 6  1  4 1 4      

v v

1 2      5 5   Methods of Analysis

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Practice

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Determine the voltage at the nodes in Fig. below

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At node 1,

 3  3

v

1 

i

1 

v

3 

i x

 4

v

1 

v

2 2 Methods of Analysis

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At node 2

v

1 

v

2

i x

 2 

i v

2 2 8 

i

3 

v

3 

v

2 4  0 Methods of Analysis

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At node 3

v

1 

v

3 4

i

 1 

i

2

v

2 

v

3 8  2

i x

 2 (

v

1 

v

2 ) 2 Methods of Analysis

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In matrix form:

        3  4 1 2 3 4  1 7 2  8 9 8  1  4 1 3 8 8             

v v v

1 2 3            3 0 0      Methods of Analysis

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3.3 Nodal Analysis with Voltage Sources

Case 1: The voltage source is connected between a nonreference node and the reference node : The nonreference node voltage is equal to the magnitude of voltage source and the number of unknown nonreference nodes is reduced by one.

Case 2: The voltage source is connected between two nonreferenced nodes : a generalized node ( supernode ) is formed.

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3.3 Nodal Analysis with Voltage Sources

A circuit with a supernode.

v

1 

v

2 2 

i

1

v

i

4 1  

v

3  4

v

2

i

2  

v

3 

i

3

v

2  8   5 0 

v

3 6  0 Methods of Analysis

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A supernode is formed by enclosing a (dependent or independent) voltage source connected between two nonreference nodes and any elements connected in parallel with it .

The required two equations for regulating the two nonreference node voltages are obtained by the KCL of the supernode and the relationship of node voltages due to the voltage source.

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Example 3.3

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

2  7 

i

1 

i

2  0 2

v

1  7 

v

2 

v

1   2  2

v

2 4  0 i1 Methods of Analysis i2

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Find the node voltages in the circuit below.

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At suopernode 1-2,

v

3 

v

2 6

v

 10  1 

v

2

v

1  3  20

v

4 

v

1 2 Methods of Analysis

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At supernode 3-4,

v

1  3

v v

4 3  

v

4

v

3 

v

2  6 3 (

v

1  

v

4 1

v

4 ) 

v

3 4 Methods of Analysis

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3.4 Mesh Analysis

Mesh analysis: another procedure for analyzing circuits, applicable to planar circuit.

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

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(a) A Planar circuit with crossing branches, (b) The same circuit redrawn with no crossing branches.

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A nonplanar circuit.

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1.

2.

3.

Steps to Determine Mesh Currents: Assign mesh currents

i

1 , i 2 , .., i

n

to the n meshes.

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

Solve the resulting n simultaneous equations to get the mesh currents.

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Fig. 3.17

A circuit with two meshes.

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Apply KVL to each mesh. For mesh 1,

V

1 

For mesh 2,

(

R

1  

R

1

i

1

R

3  )

i

1

R

3 (

i

1 

R

3 

i

2

i

2 )  

V

1 0

R

2

i

2 

R

3

i

1 

V

2   (

R

2

R

3 (

i

2 

R

3  )

i

2

i

1 )   0 

V

2 Methods of Analysis

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Solve for the mesh currents.

 

R

1  

R

3

R

3

R

2 

R

 3

R

3    

i i

2 1      

V

1

V

2   

Use i for a mesh current and I for a branch current. It’s evident from Fig. 3.17 that

I

1 

i

1 ,

I

2 

i

2 ,

I

3 

i

1 

i

2 Methods of Analysis 41

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Find the branch current I 1 , I 2 , and I 3 analysis.

using mesh

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For mesh 1,

 15  

For mesh 2,

5

i

1 3  10 (

i

1

i

1  2

i

2 

i

2  1 )  10  0 6

i

2  4

i

2 

i

1 10 ( 

i

2 2

i

2 

i

1 )  1  10  0 

We can find i 1 and i 2 Cramer’s rule. Then, by substitution method or

I

1 

i

1 ,

I

2 

i

2 ,

I

3 

i

1 

i

2 Methods of Analysis 43

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Use mesh analysis to find the current I 0 circuit.

in the

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Apply KVL to each mesh. For mesh 1,

 24 

For mesh 2,

 10 (

i

1 11

i

1   5

i i

2 ) 2   12 (

i

1 6

i

3   12

i

3 )  0 24

i

2   4 (

i

2 5

i

1 

i

3 )  19

i

2  10 (

i

2  2

i

3  

i

1 ) 0  0 Methods of Analysis

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For mesh 3,

4

I

0 At node 4 (

i

1 

i

2 ) A,

I

 0 12 (

i

 3

I

1  

i

1 )

i

2 ,  4 (

i

3 

i

1  12 (

i

3 

i

2  2

i

 3

i

1  ) 0  4 (

i

3 

i

2 )  0 

i

2 )  0 

In matrix from become

   11   5 1  5 19  1   2 6 2   

we can calculus i 1 , i 2 and i

  

i i i

1 2 3

3

     12   0 0   

by Cramer’s rule, and find I 0 .

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3.5 Mesh Analysis with Current Sources

A circuit with a current source.

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Case 1

●

Current source exist only in one mesh

i

1   2 A ●

One mesh variable is reduced

Case 2

●

Current source exists between two meshes, a super mesh is obtained.

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

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Properties of a Supermesh

1.

●

The current is not completely ignored provides the constraint equation necessary to solve for the mesh current.

2.

3.

A supermesh has no current of its own bigger supermesh.

.

Several current sources in adjacency form a

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For the circuit below, find i 1 analysis.

to i 4 using mesh

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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

6

i

1

current source.

i

1  

i

2 14

i

2    6 20 Methods of Analysis

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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.

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Methods of Analysis 2

i

1  4

i

3  8 (

i

3

i i

1 2 8 (

i

i

2 

i

3 3   5  

i

4 ) 

i

4  2

i

4 

i

4 )   10  6

i

2 0  0

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3.6 Nodal and Mesh Analysis by Inspection

The analysis equations can be obtained by direct inspection (a)For circuits with only resistors and independent current sources (b)For planar circuits with only resistors and independent voltage sources Methods of Analysis

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the circuit has two nonreference nodes and the node equations

I

1 

I

2  

I G

2 2  (

v G

1

v

1 1 

MATRIX

v

2 )

G

2 (

v

1 

G

3

v

2 

v

2 )  

G

1  

G G

2 2

G

2 

G

 2

G

3    

v v

1 2    ( ( 3 3 .

.

8 7 ) ) 

I

1

I

2

I

2   Methods of Analysis

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In general, the node voltage equations in terms of the conductances is

or simply Gv = i 

G

11    

G G

 21

N G

12

G

1

G

22

N

2    

G G

 

G

1 2

N N NN

     

v

1    

v v

 2

N

      

i

1    

i i

2 

N

     where

G

: the conductance matrix,

v

: the output vector,

i

: the input vector Methods of Analysis

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The circuit has two nonreference nodes and the node equations were derived as

 

R

1  

R

3

R

3

R

2  

R

3

R

3    

i i

2 1     

v

1 

v

2   Methods of Analysis

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In general, if the circuit has N meshes, the mesh current equations as the resistances term is

or simply Rv = i      

R R R

11 21

N

    1

R

12

R

22

R N

2 

R

R

1 2 

R NN N N

           

i i i

 1 2

N

            

v v v

 1 2

N

      where

R

: the resistance matrix,

i

: the output vector,

v

: the input vector Methods of Analysis

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Write the node voltage matrix equations

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The circuit has 4 nonreference nodes, so

G

11  1 5  1 10  0 .

3 ,

G

22  1 5  1 8  1 1  1 .

325

G

33  1  1  1  0 .

5 ,

G

44  1  8 8 4 8

The off-diagonal terms are

 1 2  1 1  1 .

625

G

12   1 5   0 .

2 ,

G

13 

G

14  0

G

21

G

31

G

41     0 .

2 ,

G

23 0 ,

G

32    1   0 .

125 ,

G

24 8  0 .

125 ,

G

34   0 .

125 0 ,

G

42   1 ,

G

43   0 .

125   1 1   1 Methods of Analysis

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The input current vector i in amperes

i

 3 ,

i

  1  2   3 ,

i

 0 , 1 2 3 

The node-voltage equations are

i

4  2  4  6        0 0.2

0 0 .3

   0.2

1 .325

 0.125

1  0 0.125

0.125

 0 .5

 0 1 0.125

1 .625

           

v v v v

1 2 3 4               3 3 0 6       Methods of Analysis

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Write the mesh current equations

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The input voltage vector v in volts

v

1  4 ,

v

2  10  4  6 ,

v

3   12  6   6 ,

v

4  0 , 

The mesh-current equations are

v

5   6        9    2 2 0 0 1    2 0 4   2 4 9  1 0  0 1 0 3  1 0 8  0 0 1 4 3        

i

1  

i

2    

i i i

3 4 5               4 6   0 6 6       Methods of Analysis

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3.7 Nodal Versus Mesh Analysis

Both nodal and mesh analyses provide a systematic way of analyzing a complex network.

The choice of the better method dictated by two factors.

●

First factor : nature of the particular network. The key is to select the method that results in the smaller number of equations .

●

Second factor : information required .

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3.10 Summery

1.

●

Nodal analysis: the application of KCL at the nonreference nodes A circuit has fewer node equations 2.

A supernode: two nonreference nodes 3.

●

Mesh analysis: the application of KVL A circuit has fewer mesh equations 4.

A supermesh: two meshes

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