chap3

1. Methods of Analysis Eastern Mediterranean University 1
Methods of
Analysis
Mustafa Kemal Uyguroğlu

2. Methods of Analysis
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Methods of Analysis 2
• 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

3. 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.
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Methods of Analysis 3

4. Figure 3.1
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Methods of Analysis 4
Common symbols for indicating a reference node,
(a) common ground, (b) ground, (c) chassis.

5. Figure 3.2
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Methods of Analysis 5
Typical circuit for nodal analysis

6. Eastern Mediterranean University
Methods of Analysis 6
3
2
2
2
1
2
1
i
i
I
i
i
I
I





R
v
v
i lower
higher 

2
3
3
3
2
3
2
1
2
2
2
2
1
2
1
1
1
1
1
1
or
0
)
(
or
or
0
v
G
i
R
v
i
v
v
G
i
R
v
v
i
v
G
i
R
v
i











7. Eastern Mediterranean University
Methods of Analysis 7
3
2
2
2
1
2
2
2
1
1
1
2
1
R
v
R
v
v
I
R
v
v
R
v
I
I








2
3
2
1
2
2
2
1
2
1
1
2
1
)
(
)
(
v
G
v
v
G
I
v
v
G
v
G
I
I














 


















2
2
1
2
1
3
2
2
2
2
1
I
I
I
v
v
G
G
G
G
G
G

8. Example 3.1
Calculus the node voltage in the circuit shown in
Fig. 3.3(a)
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Methods of Analysis 8

9. Example 3.1
At node 1
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Methods of Analysis 9
2
0
4
5 1
2
1
3
2
1







v
v
v
i
i
i

10. Example 3.1
At node 2
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Methods of Analysis 10
6
0
4
5 2
1
2
5
1
4
2








v
v
v
i
i
i
i

11. Example 3.1
In matrix form:
Eastern Mediterranean University
Methods of Analysis 11



























5
5
4
1
6
1
4
1
4
1
4
1
2
1
2
1
v
v

12. Practice Problem 3.1
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Methods of Analysis 12
Fig 3.4

13. Example 3.2
Determine the voltage at the nodes in Fig. 3.5(a)
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Methods of Analysis 13

14. Example 3.2
At node 1,
Eastern Mediterranean University
Methods of Analysis 14
2
4
3
3
2
1
3
1
1
v
v
v
v
i
i x








15. Example 3.2
At node 2
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Methods of Analysis 15
4
0
8
2
2
3
2
2
1
3
2








v
v
v
v
v
i
i
ix

16. Example 3.2
At node 3
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Methods of Analysis 16
2
)
(
2
8
4
2
2
1
3
2
3
1
2
1
v
v
v
v
v
v
i
i
i x









17. Example 3.2
In matrix form:
Eastern Mediterranean University
Methods of Analysis 17










































0
0
3
8
3
8
9
4
3
8
1
8
7
2
1
4
1
2
1
4
3
3
2
1
v
v
v

18. 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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Methods of Analysis 18

19. 3.3 Nodal Analysis with Voltage Sources
5
6
0
8
0
4
2
3
2
3
2
3
1
2
1
3
2
4
1














v
v
v
v
v
v
v
v
i
i
i
i
Eastern Mediterranean University
Methods of Analysis 19
Fig. 3.7 A circuit with a supernode.

20. 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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Methods of Analysis 20

21. Example 3.3
For the circuit shown in Fig. 3.9, find the node
voltages.
2
0
4
2
7
2
0
2
1
7
2
2
1
2
1











v
v
v
v
i
i
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Methods of Analysis 21
i1 i2

22. Example 3.4
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Methods of Analysis 22
Find the node voltages in the circuit of Fig. 3.12.

23. Example 3.4
At suopernode 1-2,
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Methods of Analysis 23
20
2
3
10
6
2
1
1
4
1
2
3







v
v
v
v
v
v
v

24. Example 3.4
At supernode 3-4,
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Methods of Analysis 24
)
(
3
4
1
6
3
4
1
4
3
3
4
2
3
4
1
v
v
v
v
v
v
v
v
v
v









25. 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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Methods of Analysis 25

26. Fig. 3.15
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Methods of Analysis 26
(a) A Planar circuit with crossing branches,
(b) The same circuit redrawn with no crossing branches.

27. Fig. 3.16
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Methods of Analysis 27
A nonplanar circuit.

28.  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.
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Methods of Analysis 28

29. Fig. 3.17
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Methods of Analysis 29
A circuit with two meshes.

30. Apply KVL to each mesh. For mesh 1,
For mesh 2,
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Methods of Analysis 30
1
2
3
1
3
1
2
1
3
1
1
1
)
(
0
)
(
V
i
R
i
R
R
i
i
R
i
R
V








2
2
3
2
1
3
1
2
3
2
2
2
)
(
0
)
(
V
i
R
R
i
R
i
i
R
V
i
R










31. Solve for the mesh currents.
Use i for a mesh current and I for a branch current.
It’s evident from Fig. 3.17 that
Eastern Mediterranean University
Methods of Analysis 31
























2
1
2
1
3
2
3
3
3
1
V
V
i
i
R
R
R
R
R
R
2
1
3
2
2
1
1 ,
, i
i
I
i
I
i
I 




32. Example 3.5
Find the branch current I1, I2, and I3 using mesh
analysis.
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Methods of Analysis 32

33. Example 3.5
For mesh 1,
For mesh 2,
We can find i1 and i2 by substitution method or
Cramer’s rule. Then,
Eastern Mediterranean University
Methods of Analysis 33
1
2
3
0
10
)
(
10
5
15
2
1
2
1
1








i
i
i
i
i
1
2
0
10
)
(
10
4
6
2
1
1
2
2
2







i
i
i
i
i
i
2
1
3
2
2
1
1 ,
, i
i
I
i
I
i
I 




34. Example 3.6
Use mesh analysis to find the current I0 in the
circuit of Fig. 3.20.
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Methods of Analysis 34

35. Example 3.6
Apply KVL to each mesh. For mesh 1,
For mesh 2,
Eastern Mediterranean University
Methods of Analysis 35
12
6
5
11
0
)
(
12
)
(
10
24
3
2
1
3
1
2
1









i
i
i
i
i
i
i
0
2
19
5
0
)
(
10
)
(
4
24
3
2
1
1
2
3
2
2









i
i
i
i
i
i
i
i

36. Example 3.6
For mesh 3,
In matrix from Eqs. (3.6.1) to (3.6.3) become
we can calculus i1, i2 and i3 by Cramer’s rule, and
find I0.
Eastern Mediterranean University
Methods of Analysis 36
0
2
0
)
(
4
)
(
12
)
(
4
,
A,
node
At
0
)
(
4
)
(
12
4
3
2
1
2
3
1
3
2
1
2
1
0
2
3
1
3
0

















i
i
i
i
i
i
i
i
i
i
I
I
i
i
i
i
I































0
0
12
2
1
1
2
19
5
6
5
11
3
2
1
i
i
i

37. 3.5 Mesh Analysis with Current Sources
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Methods of Analysis 37
Fig. 3.22 A circuit with a current source.

38. Case 1
● Current source exist only in one mesh
● One mesh variable is reduced
Case 2
● Current source exists between two meshes, a super-
mesh is obtained.
Eastern Mediterranean University
Methods of Analysis 38
A
2
1 

i

39. Fig. 3.23
a supermesh results when two meshes have a
(dependent , independent) current source in
common.
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Methods of Analysis 39

40. 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.
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Methods of Analysis 40

41. Example 3.7
For the circuit in Fig. 3.24, find i1 to i4 using mesh
analysis.
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Methods of Analysis 41

42.  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. 6
20
14
6
2
1
2
1





i
i
i
i
Eastern Mediterranean University
Methods of Analysis 42

43. 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 43

44. 0
10
2
)
(
8
5
0
6
)
(
8
4
2
4
4
3
4
3
2
2
1
2
4
3
3
1















i
i
i
i
i
i
i
i
i
i
i
i
i
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Methods of Analysis 44

45. 3.6 Nodal and Mesh Analysis by
Inspection
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Methods of Analysis 45
(a)For circuits with only resistors and
independent current sources
(b)For planar circuits with only resistors and
independent voltage sources
The analysis equations can be
obtained by direct inspection

46. In the Fig. 3.26 (a), the circuit has two
nonreference nodes and the node equations
Eastern Mediterranean University
Methods of Analysis 46





 

























2
2
1
2
1
3
2
2
2
2
1
2
3
2
1
2
2
2
1
2
1
1
2
1
)
8
.
3
(
)
(
)
7
.
3
(
)
(
I
I
I
v
v
G
G
G
G
G
G
MATRIX
v
G
v
v
G
I
v
v
G
v
G
I
I

47. In general, the node voltage equations in terms of
the conductances is
Eastern Mediterranean University
Methods of Analysis 47





































N
N
NN
N
N
N
N
i
i
i
v
v
v
G
G
G
G
G
G
G
G
G









2
1
2
1
2
1
2
22
21
1
12
11
or simply
Gv = i
where G : the conductance matrix,
v : the output vector, i : the input vector

48. The circuit has two nonreference nodes and the
node equations were derived as
Eastern Mediterranean University
Methods of Analysis 48
























2
1
2
1
3
2
3
3
3
1
v
v
i
i
R
R
R
R
R
R

49. In general, if the circuit has N meshes, the mesh-
current equations as the resistances term is
Eastern Mediterranean University
Methods of Analysis 49





































N
N
NN
N
N
N
N
v
v
v
i
i
i
R
R
R
R
R
R
R
R
R









2
1
2
1
2
1
2
22
21
1
12
11
or simply
Rv = i
where R : the resistance matrix,
i : the output vector, v : the input vector

50. Example 3.8
Write the node voltage matrix equations in Fig.3.27.
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Methods of Analysis 50

51. Example 3.8
The circuit has 4 nonreference nodes, so
The off-diagonal terms are
Eastern Mediterranean University
Methods of Analysis 51
625
.
1
1
1
2
1
8
1
,
5
.
0
4
1
8
1
8
1
325
.
1
1
1
8
1
5
1
,
3
.
0
10
1
5
1
44
33
22
11















G
G
G
G
125
.
0
,
1
,
0
125
.
0
,
125
.
0
,
0
1
1
1
,
125
.
0
8
1
,
2
.
0
0
,
2
.
0
5
1
43
42
41
34
32
31
24
23
21
14
13
12


























G
G
G
G
G
G
G
G
G
G
G
G

52. Example 3.8
The input current vector i in amperes
The node-voltage equations are
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Methods of Analysis 52
6
4
2
,
0
,
3
2
1
,
3 4
3
2
1 








 i
i
i
i














































6
0
3
3
.625
1
0.125
1
0
0.125
.5
0
0.125
0
1
0.125
.325
1
0.2
0
0
0.2
.3
0
4
3
2
1
v
v
v
v

53. Example 3.9
Write the mesh current equations in Fig.3.27.
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54. Example 3.9
The input voltage vector v in volts
The mesh-current equations are
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Methods of Analysis 54
6
,
0
,
6
6
12
,
6
4
10
,
4
5
4
3
2
1












v
v
v
v
v



























































6
0
6
6
4
4
3
0
1
0
3
8
0
1
0
0
0
9
4
2
1
1
4
0
1
2
0
0
2
2
9
5
4
3
2
1
i
i
i
i
i

55. 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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56. BJT Circuit Models
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Methods of Analysis 56
(a)An npn transistor,
(b) dc equivalent model.

57. Example 3.13
For the BJT circuit in Fig.3.43, =150 and VBE =
0.7 V. Find v0.
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58. Example 3.13
Use mesh analysis or nodal analysis
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59. Example 3.13
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Methods of Analysis 59

60. 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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61. Homework
Problems 7, 12, 20, 31(write down required
equations only), 39, 49, 53(write down required
equations only)
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