Download Chapter 3 Methods of Analysis

Chapter 3
Methods of Analysis
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So far, we have analyzed relatively simple circuits by
applying Kirchhoff’s laws in combination with Ohm’s law.
We can use this approach for all circuits, but as they
become structurally more complicated and involve more
and more elements, this direct method soon becomes
cumbersome. In this chapter we introduce two powerful
techniques of circuit analysis: Nodal Analysis and Mesh
Analysis.
These techniques give us two systematic methods of
describing circuits with the minimum number of
simultaneous equations. With them we can analyze almost
any circuit by to obtain the required values of current or
voltage.
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Nodal Analysis
• Steps to Determine Node Voltages:
1. Select a node as the reference node(ground), define the node voltages
1, 2,… n-1 to the remaining n-1nodes . The voltages are
referenced with respect to the reference node.
2. Apply KCL to each of the n-1 independent 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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Fig. 3.2 Typical circuit for nodal
analysis
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i
vhigher  vlower
R
So at node 1 and node 2, we can get the following
equations.
v1 v1  v2
I1  I 2  
R1
R2
I1  I 2  i1  i2
I 2  i2  i3
v1  v2 v2
I2 

R2
R3
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In terms of the conductance, equations become
I1  I 2  G1v1  G2 (v1  v2 )
I 2  G2 (v1  v2 )  G3v2
Can also be cast in matrix form as
 G2   v1   I1  I 2 
G1  G2

 G




G2  G3  v2   I 2 
2

Some examples
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Fig. 3.5 For Example 3.2: (a) original circuit, (b)
circuit for analysis
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Nodal Analysis with Voltage Sources(1)
•
Case 1
If a voltage source is connected
between the reference node and a
nonreference node, we simply set
the voltage at the nonreference
node equal to the voltage of the
voltage source. As in the figure
right:
v1  10V
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Nodal Analysis with Voltage Sources(2)
•
Case 2
If the voltage source (dependent or
independent) is connected between
two nonreference nodes, the two
nonreference nodes form a
supernode; we apply both KVL and
KCL to determine the node
voltages. As in the figure right:
i1  i4  i2  i3
v1  v2 v1  v3 v2  0 v3  0
or



2
4
8
6
and v2  v3  5
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Nodal Analysis with Voltage Sources(3)
• Case 3
If a voltage source
(dependent or independent) is
connected with a resistor in
series, we treat them as one
branch. As in the figure right:
R1
i V
11
1
V1
V22
V11  V22  V1
i
R1
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Nodal Analysis with Voltage Sources(3)
Example
P113
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Mesh Analysis
• 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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Fig. 3.17 A circuit with two meshes
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 V1  R1i1  R3 (i1  i2 )  0
R2i2  V2  R3 (i2  i1 )  0
or
( R1  R3 )i1  R3i2  V1
 R3i1  ( R2  R3 )i2  V2
In matrix form:
 R1  R3

  R3
 R3  i1   V1 
   

R2  R3  i2    V2 
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Fig. 3.18 For Example 3.5
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Mesh Analysis with Current Sources(1)
•
Case 1
When a current source exists only
in one mesh: Consider the figure
right.
i2  2 A
 10  4i1  6(i1  i2 )  0
 i1  2 A
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Mesh Analysis with Current Sources(2)
•
Case 2
When a current source exists
between two meshes:Consider the
figure right.
2 solutions:
1. Set v as the voltage across the
current source, then add a
constraint equation.
2. Use supermesh to solve the
problem.
R1
ib
R2
+
V1
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v
R4
-
R3
V2
Is
ic
R5
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R1
Solution 1:
ib
R2
+
V1
ia
v
R4
-
R2 (ia  ib )  v  R4ia  V 1
R3
V2
Is
ic
R5
R1ib  R3 (ib  ic )  R2 (ib  ia )  0
R5ic  v  R3 (ic  ib )  V 2
and
supermesh
ia  ic  Is
Solution 2:
R 2(ia  ib )  R3(ic  ib )  V 2  R5ic  R 4ia  V 1  0
Note:
1, The current source in the supermesh is not completely ignored; it provides the constraint
equation necessary to solve for the mesh current.
2, A supermesh has no current of its own.
3, A supermesh requires the application of both KVL and KCL.
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Fig. 3.24 For Example 3.7
2i1  4i3  8(i3  i4 )  6i2  0
and
i2  i1  5 i2  i3  3i0
i0  i4
2i4  8(i4  i3 )  10  0
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Fig. 3.31 For Example 3.10
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Fig. 3.32 For Example 3.10; the schematic of the circuit in Fig. 3.31.
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Nodal Versus Mesh Analysis
• Both provide a systematic way of analyzing a complex network.
• When is the nodal method preferred to the mesh method?
1. A circuit with fewer nodes than meshes is better analyzed using nodal
analysis, while a circuit with fewer meshes than nodes is better
analyzed using mesh analysis.
2. Based on the information required.
Node voltages required--------nodal analysis
Branch or mesh currents required-------mesh analysis
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