Methods of Analysis

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Transcript Methods of Analysis

Basic Nodal and
Mesh Analysis
Al-Qaralleh
Methods of Analysis
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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
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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.
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Lect4
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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
500W
500W
+
I1
500W
V
1kW
500W
I2
–
The reference node is called the ground node where
V=0
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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.
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Lect4
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2. Node Voltages
V1
500W V 500W
2
1
I1
2
500W
V3
3
1kW
500W
I2
V1, V2, and V3 are unknowns for which we solve
using KCL
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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.
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Lect4
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Currents and Node Voltages
V1
500W
V2
V1
500W
V1  V2
500 W
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V1
500 W
Lect4
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3. KCL at Node 1
V1
I1
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500W
V2
V1  V2
V1
I1 

500 W 500 W
500W
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3. KCL at Node 2
V1
500W
V2 500W
V3
1kW
V2  V1 V2 V2  V3


0
500 W 1kW 500 W
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3. KCL at Node 3
V2 500W
V3
500W
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I2
Lect4
V3  V2
V3

 I2
500 W 500 W
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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.
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4. Summing Circuit Solution
500W
500W
+
I1
500W
V
1kW
500W
I2
–
Solution: V = 167I1 + 167I2
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Typical circuit for nodal analysis
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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
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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
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 G2   v1   I1  I 2 

G2  G3  v2   I 2 
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 Calculus the node voltage in the circuit shown in
Fig. 3.3(a)
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 At node 1
i1  i2  i3
v1  v2 v1  0
5

4
2
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 At node 2
i2  i4  i1  i5
v2  v1 v2  0
5

4
6
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 In matrix form:
1 1

2 4
 1
 
 4
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1 
   v  5
1
4
 



1 1 v2  5
 
6 4
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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  i1  ix
v1  v3 v1  v2
3

4
2
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 At node 2
ix  i2  i3
v1  v2 v2  v3 v2  0



2
8
4
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 At node 3
i1  i2  2ix
v1  v3 v2  v3 2(v1  v2 )



4
8
2
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 In matrix form:
 3
 4
 1

 2
 3
 4
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1

2
7
8
9

8
1
 
4  v1  3

1    
  v2   0 
8
3  v3  0
8 
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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.
i1  i4  i2  i3 
v1  v2 v1  v3 v2  0 v3  0



2
4
8
6
 v2  v3  5
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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  i1  i 2  0
v v
27 1  2  0
2 4
v1  v2  2
i1
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Find the node voltages in the circuit below.
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 At suopernode 1-2,
v3  v2
v1  v4 v1
 10 

6
3
2
v1  v2  20
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 At supernode 3-4,
v1  v4 v3  v2 v4 v3

 
3
6
1 4
v3  v4  3(v1  v4 )
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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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 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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 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
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 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 ,
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I 3  i1  i2
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 Find the branch current I1, I2, and I3 using mesh
analysis.
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 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
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 Use mesh analysis to find the current I0 in the
circuit.
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 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
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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 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.
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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
i1  2A
● One mesh variable is reduced
 Case 2
● Current source exists between two meshes, a supermesh 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. A supermesh has no current of its own.
3. Several current sources in adjacency form a
bigger supermesh.
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 For the circuit below, find i1 to i4 using mesh
analysis.
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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
current source.
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6i1  14i2  20
i1  i2  6
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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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2i1  4i3  8(i3  i4 )  6i2  0
i1  i2  5
i2  i3  i4
8(i3  i4 )  2i4  10  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
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 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
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 G2   v1   I1  I 2 

G2  G3  v2   I 2 
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 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
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 The circuit has two nonreference nodes and the
node equations were derived as
 R1  R3
  R3
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 R3   i1   v1 

R2  R3  i2   v2 
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 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
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 Write the node voltage matrix equations
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 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
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 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   
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 Write the mesh current equations
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 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
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 2  2 0 0  i1
10  4  1  1  i2

 4 9 0 0  i3
1
0 8  3  i4

1
0  3 4  i5
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
 4
 6
    6
 0
   6
  
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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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