Esercitazione1 ELT CANOVA

7/25/2019 Esercitazione1 ELT CANOVA

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Electrical Basics Course Prof. Aldo Canova

Electrical Basic Course

Exercise 1

Resistive network solution byConstitutive and KVL and KCLEquations


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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


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B

R1

Example 1

A

C

e

R2ig


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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


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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.


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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


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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


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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


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The compacted linear system in the matrix form is:

=

0

0110

0011

2

1

ii

i

g

0100 3

4

2

1

vR


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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


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Solution of circuit by KVL, KCL and constitutive equationsExample 2

C

R1

D

A B

e1

5g

3

R2 e2

R4

E

F


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C

R1

D

A B

e1

5 g3

R2 e2

R4

E

F


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C

R1

D

A B

e1

5 g3

R2 e2

R4

N=L=

N-1= KCLL-N+1= KVL


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C

R1

D

i3 i5

i6

A B

e1

5g

3

R2 e2

R4

i1

i4

i2

DB


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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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