Fundamentals of Electric Circuit - Chapter 3: Methods of analysis

    
 Mesh II: i i i i i
2 1 2 3 0
2( ) 8( ) 10 0    
 Set of equations:
2Ω
8Ω4Ω
10i0
i2
i i i
1 2 3
6 2 4 20  
i i
0 3

 Mesh III:
i i i
1 2 3
2 10 18 0

   

i i i i i
3 1 3 2 3
4( ) 8( ) 6 0     i i i
1 2 3
4 8 18 0   
i i i
i i i
i i i
1 2 3
1 2 3
1 2 3
6 2 4 20
2 10 18 0
4 8 18 0
  

   
    
i A
i A
i A
1
2
3
3,21
9,64
5
 

  
  
i A
0
5  
Fundamentals of Electric Circuits – Viet Son Nguyen - 2011
R 1 R 3
1 0 V
R 2
28
III.2. Mesh analysis with current sources
Chapter 3: Methods of analysis
6Ω
 In general, the presence of the current sources
reduces the number of equations in the mesh
analysis.
i1
 A current source exists only in one mesh mesh current = current source
3Ω4Ω
5A
i2
i i i
i A
1 1 2
2
4 6( ) 10
5
  

 
 Consider two cases:
i A
i A
1
2
2
5
 
 
 
 A current source exists between two meshes  create a super-mesh by
excluding the current source and any elements connected in series with it
A super-mesh results when two meshes have a (dependent or independent) 
current source in common.
Fundamentals of Electric Circuits – Viet Son Nguyen - 2011
R 1
R 3
R 3
2 0 V
R 2
29
III.2. Mesh analysis with current sources
Chapter 3: Methods of analysis
4Ω
Ex 1: Find the branch currents using mesh analysis.
0
10Ω
2Ω
6A
i2
i i i
1 2 2
20 6 10 4 0    
 Create a super-mesh.
6Ω
i1
i1
i2
 There is a current source 6-A between two mesh.
4Ω
10Ω6Ω
R 1
R 3
R 3
2 0 V
i2i1
 Applying KVL to the super-mesh gives:
i i
1 2
6 14 20  
 Applying KCL to the node in the current source branch :
i i
2 1
6 
 Note that:
 The current source in the super-mesh provides
the constraint equation to solve for the mesh
currents
 A super-mesh has no current of its own
 A super-mesh requires the using of both KVL
and KCL
Fundamentals of Electric Circuits – Viet Son Nguyen - 2011
R 1
R 3
R 4
R 2
1 0 V
R 5
30
III.2. Mesh analysis with current sources
Chapter 3: Methods of analysis
Ex 2: Find i1, i4 using mesh analysis.
Q
3i0
6Ω
5A
i i i i i
1 3 3 4 2
2 4 8( ) 6 0    
2Ω
i2
i2
 There are two super-meshes
8Ω
2Ω4Ω
i i
2 1
5 
i3
P
i1
i0
i4i3i2
i1
 Two super-meshes intersect form a
larger super-mesh
 Applying KVL to the larger super-
mesh:
i i i i
1 2 3 4
3 6 4 0    
 Applying KCL to the node P:
 Applying KCL to the node Q: i i i
2 3 0
3  i i i i i0 4
2 3 4
3
  
 Applying KVL in mesh 4: i i i4 4 32 8( ) 10 0    i i4 35 4 5   
 These equations give:
i A
1
7.5  i A
2
2.5 
i A
3
3,93 i A
4
2.14
Fundamentals of Electric Circuits – Viet Son Nguyen - 2011
R 1
R 3
R 4R 2
6 V
R 5
31
III.2. Mesh analysis with current sources
Chapter 3: Methods of analysis
Ex 3.3: Find i1, i2, i3 using mesh analysis.
4Ω
i i i i i
1 3 2 3 2
6 2( ) 4( ) 8 0      
2Ω
 Applying KVL to super-mesh:
8Ω
2Ω
1Ω
A
i2
i3
i1
 Applying KCL to node A gives:
 We have a set of equations:
3A
i i i
1 2 3
2 12 6 6   
i i
1 2
3 
 Applying KCL to the mesh III:
i i i i i
3 1 3 2 3
2( ) 4( ) 2 0     i i i
1 2 3
2 4 8 0   
i i i
i i
i i i
1 2 3
1 2
1 2 3
2 12 6 6
3
2 4 8 0
  

 
   
i A
i A
i A
1
2
3
3,47
0,47
1,11


 
 
Fundamentals of Electric Circuits – Viet Son Nguyen - 2011
32
III.3. Mesh analysis by inspection
Chapter 3: Methods of analysis
 In general, if a circuit (with only independent voltage sources) has N meshes,
the mesh current equations can be expressed in terms of the resistances as:
N
N
N N NN N N
R R R i v
R R R i v
R R R i v
11 12 1 1 1
21 22 2 2 2
1 2
...
...
...
     
     
     
     
     
     
where:
 R i v
 Rkk : Sum of the resistances in mesh k.
 Rkj = Rjk : Negative of the sum of the resistances in common with meshes k
and j, k ≠ j.
 ik : Unknown mesh current for mesh k in the clockwise direction.
 vk : Sum taken clockwise of all independent voltage sources in mesh k, with
voltage rise treated as positive.
Fundamentals of Electric Circuits – Viet Son Nguyen - 2011
R 1 R 3
R 4R 2
6 V
R 5
R 7
R 6
1 0 V
4 V
R 8
R 9
1 2 V
R10
33
III.3. Mesh analysis by inspection
Chapter 3: Methods of analysis
Ex 3.4: Write the mesh current equations.
3Ω
5Ω
2Ω
i2
i1
i5
4Ω 3Ω
1Ω1Ω
4Ω
2Ω 2Ω
i4
i3
 There are 5 meshes
R R R R R i v
R R R R R i v
R R R R R i v
R R R R R i v
R R R R R i v
11 12 13 14 15 1 1
21 22 23 24 25 2 2
31 32 33 34 35 3 3
41 42 43 44 45 4 4
51 52 53 54 55 5 5
     
     
     
     
     
     
          
R R R R
11 6 8 10
9    
R R R R R R
22 2 4 5 6 7
10      
R R R R
33 7 8 9
9     R R R R
44 1 2 3
8     R R R
55 3 4
4   
Fundamentals of Electric Circuits – Viet Son Nguyen - 2011
R 1 R 3
R 4R 2
6 V
R 5
R 7
R 6
1 0 V
4 V
R 8
R 9
1 2 V
R10
34
III.3. Mesh analysis by inspection
Chapter 3: Methods of analysis
Ex 3.4: Write the mesh current equations.
3Ω
5Ω
2Ω
i2
i1
i5
4Ω 3Ω
1Ω1Ω
4Ω
2Ω 2Ω
i4
i3
 There are 5 meshes
R R R R R i v
R R R R R i v
R R R R R i v
R R R R R i v
R R R R R i v
11 12 13 14 15 1 1
21 22 23 24 25 2 2
31 32 33 34 35 3 3
41 42 43 44 45 4 4
51 52 53 54 55 5 5
     
     
     
     
     
     
          
R R R
12 21 6
2     
R R R
13 31 8
2     
R R
14 41
0 
R R
15 51
0 
R R R
23 32 7
4     
R R R
24 42 2
1     
R R R
25 52 4
1     
R R
34 43
0 
R R
35 53
0 
R R R
45 54 3
3     
Fundamentals of Electric Circuits – Viet Son Nguyen - 2011
R 1 R 3
R 4R 2
6 V
R 5
R 7
R 6
1 0 V
4 V
R 8
R 9
1 2 V
R10
35
III.3. Mesh analysis by inspection
Chapter 3: Methods of analysis
Ex 3.4: Write the mesh current equations.
3Ω
5Ω
2Ω
i2
i1
i5
4Ω 3Ω
1Ω1Ω
4Ω
2Ω 2Ω
i4
i3
 There are 5 meshes
R R R R R i v
R R R R R i v
R R R R R i v
R R R R R i v
R R R R R i v
11 12 13 14 15 1 1
21 22 23 24 25 2 2
31 32 33 34 35 3 3
41 42 43 44 45 4 4
51 52 53 54 55 5 5
     
     
     
     
     
     
          
v V
1
4
v V
2
10 4 6  
v V
3
6 12 6   
v
4
0
v V
5
6 
i
i
i
i
i
1
2
3
4
5
9 2 2 0 0 4
2 10 4 1 1 6
2 4 9 0 0 6
0 1 0 8 3 0
0 1 0 3 4 6
      
    
        
       
    
      
          
Fundamentals of Electric Circuits – Viet Son Nguyen - 2011
R 1
R 3
R 2
R 4
R 6R 5
1 2 V
2 4 V
R 7
1 0 V
36
III.3. Mesh analysis by inspection
Chapter 3: Methods of analysis
Ex 3.5: Write the mesh current equations.
60Ω
50Ω 30Ω
i2
i1
i5
40Ω
80Ω
20Ω10Ω
i4
i3
 There are 5 meshes
R R R R R i v
R R R R R i v
R R R R R i v
R R R R R i v
R R R R R i v
11 12 13 14 15 1 1
21 22 23 24 25 2 2
31 32 33 34 35 3 3
41 42 43 44 45 4 4
51 52 53 54 55 5 5
     
     
     
     
     
     
          
R R R R
11 1 5 7
170    
R R R R
22 1 2 3
80    
R R R
33 2 4
50    R R R
44 3 5
90    R R R
55 4 6
80   
Fundamentals of Electric Circuits – Viet Son Nguyen - 2011
R 1
R 3
R 2
R 4
R 6R 5
1 2 V
2 4 V
R 7
1 0 V
37
III.3. Mesh analysis by inspection
Chapter 3: Methods of analysis
Ex 3.5: Write the mesh current equations.
60Ω
50Ω 30Ω
i2
i1
i5
40Ω
80Ω
20Ω10Ω
i4
i3
 There are 5 meshes
R R R R R i v
R R R R R i v
R R R R R i v
R R R R R i v
R R R R R i v
11 12 13 14 15 1 1
21 22 23 24 25 2 2
31 32 33 34 35 3 3
41 42 43 44 45 4 4
51 52 53 54 55 5 5
     
     
     
     
     
     
          
R R R
12 21 1
40     
R R
13 31
0 
R R R
14 41 5
80     
R R
15 51
0 
R R R
23 32 2
30     
R R R
24 42 3
10     
R R
25 52
0 
R R
34 43
0 
R R R
35 53 4
20     
R R
45 54
0 
Fundamentals of Electric Circuits – Viet Son Nguyen - 2011
R 1
R 3
R 2
R 4
R 6R 5
1 2 V
2 4 V
R 7
1 0 V
38
III.3. Mesh analysis by inspection
Chapter 3: Methods of analysis
Ex 3.5: Write the mesh current equations.
60Ω
50Ω 30Ω
i2
i1
i5
40Ω
80Ω
20Ω10Ω
i4
i3
 There are 5 meshes
R R R R R i v
R R R R R i v
R R R R R i v
R R R R R i v
R R R R R i v
11 12 13 14 15 1 1
21 22 23 24 25 2 2
31 32 33 34 35 3 3
41 42 43 44 45 4 4
51 52 53 54 55 5 5
     
     
     
     
     
     
          
v V
1
24
v
2
0
v V
3
12 
v V
4
10
v V
5
10 
i
i
i
i
i
1
2
3
4
5
170 40 0 80 0 24
40 80 30 10 0 0
0 30 50 0 20 12
80 10 0 90 0 10
0 0 20 0 80 10
      
    
       
       
    
      
         
Fundamentals of Electric Circuits – Viet Son Nguyen - 2011
39
IV. Nodal versus Mesh analysis
Chapter 3: Methods of analysis
 Nodal and mesh analyses provide a systematic way of analyzing a complex
network.
 Question: Given a network, which method is better or more efficient ?
 Mesh analysis many series-connected elements, voltage sources, or
super-meshes; nodal analysis  parallel-connected elements, current
sources, or super-nodes.
 Circuits have n < l nodal analysis; but l < n mesh analysis.
 Select method that results in the smaller number of equations
 Information required:
 Node voltages are required nodal analysis.
 Branch or mesh currents are required mesh analysis.
 Particular problems: Analyzing transistor , op amp circuits, or non-planar
circuit.