Methods of
Analysis
Mustafa Kemal Uyguroğlu
Methods of Analysis
Eastern Mediterranean University
1
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
Eastern Mediterranean University
2
3.2 Nodal Analysis
Steps to Determine Node Voltages:
1. Select a node as the reference node. Assign voltage v1,
v2, …vn-1 to the remaining n-1 nodes. The voltages are
referenced with respect to the reference node.
2. Apply KCL to each of the n-1 nonreference nodes. Use
Ohm’s law to express the branch currents in terms of
node voltages.
3. Solve the resulting simultaneous equations to obtain
the unknown node voltages.
Methods of Analysis
Eastern Mediterranean University
3
Figure 3.1
Common symbols for indicating a reference node,
(a) common ground, (b) ground, (c) chassis.
Methods of Analysis
Eastern Mediterranean University
4
Figure 3.2
Typical circuit for nodal analysis
Methods of Analysis
Eastern Mediterranean University
5
I1  I 2  i1  i2
I 2  i2  i3
vhigher  vlower
i
R
v1  0
i1 
or i1  G1v1
R1
v1  v2
i2 
or i2  G2 (v1  v2 )
R2
v2  0
i3 
or i3  G3v2
R3
Methods of Analysis
Eastern Mediterranean University
6
v1 v1  v2
 I1  I 2  
R1
R2
v1  v2 v2
I2 

R2
R3
 I1  I 2  G1v1  G2 (v1  v2 )
I 2  G2 (v1  v2 )  G3v2
G1  G2


  G2
Methods of Analysis
 G2   v1   I1  I 2 

G2  G3  v2   I 2 
Eastern Mediterranean University
7
Example 3.1
 Calculus the node voltage in the circuit shown in
Fig. 3.3(a)
Methods of Analysis
Eastern Mediterranean University
8
Example 3.1
 At node 1
i1  i2  i3
v1  v2 v1  0
5

4
2
Methods of Analysis
Eastern Mediterranean University
9
Example 3.1
 At node 2
i2  i4  i1  i5
v2  v1 v2  0
5

4
6
Methods of Analysis
Eastern Mediterranean University
10
Example 3.1
 In matrix form:
1 1

2 4
 1
 
 4
Methods of Analysis
1 
   v  5
1
4
 



1 1 v2  5
 
6 4
Eastern Mediterranean University
11
Practice Problem 3.1
Fig 3.4
Methods of Analysis
Eastern Mediterranean University
12
Example 3.2
 Determine the voltage at the nodes in Fig. 3.5(a)
Methods of Analysis
Eastern Mediterranean University
13
Example 3.2
 At node 1,
3  i1  ix
v1  v3 v1  v2
3

4
2
Methods of Analysis
Eastern Mediterranean University
14
Example 3.2
 At node 2
ix  i2  i3
v1  v2 v2  v3 v2  0



2
8
4
Methods of Analysis
Eastern Mediterranean University
15
Example 3.2
 At node 3
i1  i2  2ix
v1  v3 v2  v3 2(v1  v2 )



4
8
2
Methods of Analysis
Eastern Mediterranean University
16
Example 3.2
 In matrix form:
 3
 4
 1

 2
 3
 4
Methods of Analysis
1

2
7
8
9

8
1
 
4  v1  3
1    
  v2   0 
8
3  v3  0
8 
Eastern Mediterranean University
17
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.
Methods of Analysis
Eastern Mediterranean University
18
3.3 Nodal Analysis with Voltage Sources
Fig. 3.7 A circuit with a supernode.
i1  i4  i2  i3 
v1  v2 v1  v3 v2  0 v3  0



2
4
8
6
 v2  v3  5
Methods of Analysis
Eastern Mediterranean University
19
 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.
Methods of Analysis
Eastern Mediterranean University
20
Example 3.3
For the circuit shown in Fig. 3.9, find the node
voltages.
2  7  i1  i 2  0
v1 v2
 0
2 4
v1  v2  2
27
i1
Methods of Analysis
i2
Eastern Mediterranean University
21
Example 3.4
Find the node voltages in the circuit of Fig. 3.12.
Methods of Analysis
Eastern Mediterranean University
22
Example 3.4
 At suopernode 1-2,
v3  v2
v1  v4 v1
 10 

6
3
2
v1  v2  20
Methods of Analysis
Eastern Mediterranean University
23
Example 3.4
 At supernode 3-4,
v1  v4 v3  v2 v4 v3

 
3
6
1 4
v3  v4  3(v1  v4 )
Methods of Analysis
Eastern Mediterranean University
24
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
Methods of Analysis
Eastern Mediterranean University
25
Fig. 3.15
(a) A Planar circuit with crossing branches,
(b) The same circuit redrawn with no crossing branches.
Methods of Analysis
Eastern Mediterranean University
26
Fig. 3.16
A nonplanar circuit.
Methods of Analysis
Eastern Mediterranean University
27
 Steps to Determine Mesh Currents:
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 simultaneous equations to get the
mesh currents.
Methods of Analysis
Eastern Mediterranean University
28
Fig. 3.17
A circuit with two meshes.
Methods of Analysis
Eastern Mediterranean University
29
 Apply KVL to each mesh. For mesh 1,
 V1  R1i1  R3 (i1  i2 )  0
( R1  R3 )i1  R3i2  V1
For mesh 2,
R2i2  V2  R3 (i2  i1 )  0
 R3i1  ( R2  R3 )i2  V2
Methods of Analysis
Eastern Mediterranean University
30
 Solve for the mesh currents.
 R1  R3
  R3
 R3   i1   V1 

R2  R3  i2   V2 
 Use i for a mesh current and I for a branch current.
It’s evident from Fig. 3.17 that
I1  i1 , I 2  i2 ,
Methods of Analysis
I 3  i1  i2
Eastern Mediterranean University
31
Example 3.5
 Find the branch current I1, I2, and I3 using mesh
analysis.
Methods of Analysis
Eastern Mediterranean University
32
Example 3.5
 For mesh 1,
 15  5i1  10(i1  i2 )  10  0
For mesh 2,
3i1  2i2  1
6i2  4i2  10(i2  i1 )  10  0
i1  2i2  1
 We can find i1 and i2 by substitution method or
Cramer’s rule. Then, I1  i1 , I 2  i2 , I 3  i1  i2
Methods of Analysis
Eastern Mediterranean University
33
Example 3.6
 Use mesh analysis to find the current I0 in the
circuit of Fig. 3.20.
Methods of Analysis
Eastern Mediterranean University
34
Example 3.6
 Apply KVL to each mesh. For mesh 1,
 24  10(i1  i2 )  12(i1  i3 )  0
For mesh 2,
11i1  5i2  6i3  12
24i2  4(i2  i3 )  10(i2  i1 )  0
 5i1  19i2  2i3  0
Methods of Analysis
Eastern Mediterranean University
35
Example 3.6
For mesh 3,
4 I 0  12(i3  i1 )  4(i3  i2 )  0
At node A, I 0  I1  i2 ,
4(i1  i2 )  12(i3  i1 )  4(i3  i2 )  0
 i1  i2  2i3  0
 In matrix from Eqs. (3.6.1) to (3.6.3) become
 11  5  6  i1  12
 5 19  2 i2    0 
  1  1 2  i   0 

 3   
we can calculus i1, i2 and i3 by Cramer’s rule, and
find I0.
Methods of Analysis
Eastern Mediterranean University
36
3.5 Mesh Analysis with Current Sources
Fig. 3.22 A circuit with a current source.
Methods of Analysis
Eastern Mediterranean University
37
 Case 1
● Current source exist only in one mesh
i1  2A
● One mesh variable is reduced
 Case 2
● Current source exists between two meshes, a supermesh is obtained.
Methods of Analysis
Eastern Mediterranean University
38
Fig. 3.23
 a supermesh results when two meshes have a
(dependent , independent) current source in
common.
Methods of Analysis
Eastern Mediterranean University
39
Properties of a Supermesh
1. The current is not completely ignored
●
provides the constraint equation necessary to solve for
the mesh current.
2. A supermesh has no current of its own.
3. Several current sources in adjacency form a
bigger supermesh.
Methods of Analysis
Eastern Mediterranean University
40
Example 3.7
 For the circuit in Fig. 3.24, find i1 to i4 using mesh
analysis.
Methods of Analysis
Eastern Mediterranean University
41
 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
current source.
Methods of Analysis
Eastern Mediterranean University
6i1  14i2  20
i1  i2  6
42
 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.
Methods of Analysis
Eastern Mediterranean University
43
2i1  4i3  8(i3  i4 )  6i2  0
i1  i2  5
i2  i3  i4
8(i3  i4 )  2i4  10  0
Methods of Analysis
Eastern Mediterranean University
44
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
Eastern Mediterranean University
45
 In the Fig. 3.26 (a), the circuit has two
nonreference nodes and the node equations
I1  I 2  G1v1  G2 (v1  v2 ) (3.7)
I 2  G2 (v1  v2 )  G3v2
(3.8)
 MATRIX
G1  G2
  G2
Methods of Analysis
 G2   v1   I1  I 2 

G2  G3  v2   I 2 
Eastern Mediterranean University
46
 In general, the node voltage equations in terms of
the conductances is
or simply
Gv = i
G11 G12  G1N  v1  i1 
G G  G  v  i 
2N
 21 22
 2    2 
        
 
G G  G  v  i 
 N
 N1 N 2
NN   N 
where G : the conductance matrix,
v : the output vector, i : the input vector
Methods of Analysis
Eastern Mediterranean University
47
 The circuit has two nonreference nodes and the
node equations were derived as
 R1  R3
  R3
Methods of Analysis
 R3   i1   v1 

R2  R3  i2   v2 
Eastern Mediterranean University
48
 In general, if the circuit has N meshes, the meshcurrent equations as the resistances term is
or simply
Rv = i
 R11 R12  R1N  i1  v1 
 R R  R  i  v 
2N
 21 22
 2    2 
        
 
 R R  R  i  v 
 N
 N1 N 2
NN   N 
where R : the resistance matrix,
i : the output vector, v : the input vector
Methods of Analysis
Eastern Mediterranean University
49
Example 3.8
 Write the node voltage matrix equations in Fig.3.27.
Methods of Analysis
Eastern Mediterranean University
50
Example 3.8
 The circuit has 4 nonreference nodes, so
1 1
1 1 1
G11    0.3, G22     1.325
5 10
5 8 1
1 1 1
1 1 1
G33     0.5, G44     1.625
8 8 4
8 2 1
The off-diagonal terms are
1
G12    0.2, G13  G14  0
5
1
1
G21  0.2, G23    0.125, G24    1
8
1
G31  0, G32  0.125, G34  0.125
G41  0, G42  1, G43  0.125
Methods of Analysis
Eastern Mediterranean University
51
Example 3.8
 The input current vector i in amperes
i1  3, i2  1  2  3, i3  0, i4  2  4  6
 The node-voltage equations are
0
0
 0.3  0.2
 v1   3 
 0.2 1.325  0.125  1
 v    3 

 2    
 0.125  v3   0
 0  0.125 0.5


 0 1

 6

0.125
1
.625
v

 4   
Methods of Analysis
Eastern Mediterranean University
52
Example 3.9
 Write the mesh current equations in Fig.3.27.
Methods of Analysis
Eastern Mediterranean University
53
Example 3.9
 The input voltage vector v in volts
v1  4, v2  10  4  6,
v3  12  6  6, v4  0, v5  6
The mesh-current equations are
 9
 2

 2
 0

 0
Methods of Analysis
 2  2 0 0  i1 
4





10  4  1  1 i2
   6 
 4 9 0 0  i3    6
 



0
1
0 8  3 i4

   6
1
0  3 4  i5   
Eastern Mediterranean University
54
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.
Methods of Analysis
Eastern Mediterranean University
55
BJT Circuit Models
(a)An npn transistor,
(b) dc equivalent model.
Methods of Analysis
Eastern Mediterranean University
56
Example 3.13
 For the BJT circuit in Fig.3.43, =150 and VBE =
0.7 V. Find v0.
Methods of Analysis
Eastern Mediterranean University
57
Example 3.13
 Use mesh analysis or nodal analysis
Methods of Analysis
Eastern Mediterranean University
58
Example 3.13
Methods of Analysis
Eastern Mediterranean University
59
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
Methods of Analysis
Eastern Mediterranean University
60
Homework
 Problems 7, 12, 20, 31(write down required
equations only), 39, 49, 53(write down required
equations only)
Methods of Analysis
Eastern Mediterranean University
61