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7/25/2019 Esercitazione1 ELT CANOVA
1/15
Electrical Basics Course Prof. Aldo Canova
Electrical Basic Course
Exercise 1
Resistive network solution byConstitutive and KVL and KCLEquations
7/25/2019 Esercitazione1 ELT CANOVA
2/15
Electrical Basics Course Prof. Aldo Canova
Solution of circuit by KVL, KCL and constitutive equations
Given a circuit with L branches and N nodesThe solution of a circuit means the calculation of the unknowns: current and
voltage of each branch.The number of unknowns is equal to 2*L and so the number of equations hasto be write is equal to 2*LEach branch implies a constitutive equation: L equations of the system comesfrom constitutive e uations
The other L equations depends from the topology of the circuit which impliesconstraints on currents and voltagesThe constraint equations are KCL and KVL
The maximum number of independent linear equations can be write usingthe KCL is equal to N-1
The maximum number of independent linear equations can be write usingthe KVL is equal to L-N+1The sum is equal to: N-1 + L-N+1 = L
L equation are obtained from constitutive equationsL equation are obtained from KVL and KCL
7/25/2019 Esercitazione1 ELT CANOVA
3/15
Electrical Basics Course Prof. Aldo Canova
Solution of circuit by KVL, KCL and constitutive equations
B
R1
Example 1
A
C
e
R2ig
7/25/2019 Esercitazione1 ELT CANOVA
4/15
Electrical Basics Course Prof. Aldo Canova
Solution of circuit by KVL, KCL and constitutive equations
B
The circuit has 4 branches and three nodes (assuming for each component abranch).For each branch we assume as unknown the current and the voltage (the
orientation is arbitrary)
2v2
L=4 2*L=8
v1
A
C
1
3 4
v1
i1
i2
v3
i3 i4
v4
v2
v3v4i1i2
i3i4
L=4 eq. costN-1=3-1=2 KCLL-N+1=4-3+1=2 KVL
7/25/2019 Esercitazione1 ELT CANOVA
5/15
Electrical Basics Course Prof. Aldo Canova
Solution of circuit by KVL, KCL and constitutive equations
Constitutive equations
1
2
v1
i1 v2
i
e R1v =e v =R i
Source Conv. Load Conv.
34
v3
i3i4
v4ig R2
v4= - R2 i4i3= - ig
Load Conv. Source Conv.
7/25/2019 Esercitazione1 ELT CANOVA
6/15
Electrical Basics Course Prof. Aldo Canova
Solution of circuit by KVL, KCL and constitutive equations
B
KCL and KVL
2v2
i1-i2=0
i2-i3-i4=0KCL
A
C
1
3 4
v1
i1
i2
v3
i3 i4
v4
v3+v4=0
v1-v2-v3=0 KVL
7/25/2019 Esercitazione1 ELT CANOVA
7/15
Electrical Basics Course Prof. Aldo Canova
Solution of circuit by KVL, KCL and constitutive equations
Linear system in matrix form
0
00000100
0010000
00010000
2
1
1
i
e
i
i
i
R
=
0
0
0
0
0
11000000
01110000
00001110
00000011
1000000
4
3
2
1
42
v
v
v
v
iR
7/25/2019 Esercitazione1 ELT CANOVA
8/15
Electrical Basics Course Prof. Aldo Canova
Solution of circuit by KVL, KCL and constitutive equations
The reduction of the matrix dimension can be obtained by substituting theconstitutive equations inside the KCL and KVL equationsIn the case of resistors one have to chose if the voltage or the current will be the
unknown
( )
( )
=+
=
=
=
00
0
243
321
42
21
iRvviRe
iiiii
g
e un nowns are:1. The currents in the
resistor and voltagesources
2. The voltages of thecurrent sources
421 iii ,,
3v
7/25/2019 Esercitazione1 ELT CANOVA
9/15
Electrical Basics Course Prof. Aldo Canova
Solution of circuit by KVL, KCL and constitutive equations
The compacted linear system in the matrix form is:
=
0
0110
0011
2
1
ii
i
g
0100 3
4
2
1
vR
7/25/2019 Esercitazione1 ELT CANOVA
10/15
Electrical Basics Course Prof. Aldo Canova
Solution of circuit by KVL, KCL and constitutive equations
The system can be further reduced considering that the KCL equation on the nodeA is quite obvious and a unique current can be assumed as unknown:
iii == 21Substituting this equation:
( )
=+
=
=
0
0
423
31
4
iRv
viRe
g
7/25/2019 Esercitazione1 ELT CANOVA
11/15
Electrical Basics Course Prof. Aldo Canova
Solution of circuit by KVL, KCL and constitutive equationsExample 2
C
R1
D
A B
e1
5g
3
R2 e2
R4
E
F
7/25/2019 Esercitazione1 ELT CANOVA
12/15
Electrical Basics Course Prof. Aldo Canova
Solution of circuit by KVL, KCL and constitutive equations
C
R1
D
A B
e1
5 g3
R2 e2
R4
E
F
7/25/2019 Esercitazione1 ELT CANOVA
13/15
Electrical Basics Course Prof. Aldo Canova
Solution of circuit by KVL, KCL and constitutive equations
C
R1
D
A B
e1
5 g3
R2 e2
R4
N=L=
N-1= KCLL-N+1= KVL
7/25/2019 Esercitazione1 ELT CANOVA
14/15
Electrical Basics Course Prof. Aldo Canova
Solution of circuit by KVL, KCL and constitutive equations
C
R1
D
i3 i5
i6
A B
e1
5g
3
R2 e2
R4
i1
i4
i2
DB
7/25/2019 Esercitazione1 ELT CANOVA
15/15
Electrical Basics Course Prof. Aldo Canova
Solution of circuit by KVL, KCL and constitutive equationsAdvantage of the proposed method:It is systematicIt allow the calculation of all the electrical quantitiesDrawbacks
It requires the solution of a system even if only an electrical quantity is requiredIf the circuit is complex the system solution requires the use of a computer
?We need a method which allowsthe synthesis of parts of the circuit
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