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EJERCICIOS DE MALLAS

Para resolver los ejercicios debemos tener en cuenta lo siguiente.

1. Identificar el concepto. 2. Plantear el problema 3. Resolver el problema por el sistema de ecuaciones.

EJERCICIO 1.

10i – 6 +8i +12 = 0 18i + 6 = 0 i = -6/18 i = -1/3


EJERCICIO 2.

2 + 4 ( i1 – i2 ) + 2 i1 = 0 2 + 4 i1 – 4 i2 + 2 i1 = 0 6 i 1 – 4 i2 = - 2 -6 + 1i2 + 4 ( i2 – i1 ) = 0 - 6 + 1i2 + 4i2 – 4i1 = 0 – 4i1 + 5 i2 = 6

Δ=

6–4 -4 5

Δi1 =

-4 - 2 5 6

Δi2 =

6 -2 -4 6

= 30 – [(-4) (-4)] = 30 – 16 =14

Δ = 14

= -24 – (-10) = - 24 + 10 = -14

Δi1 = -14

= 36 – 8 = 28

Δi2 = 28

i1 = Δi1 / Δ = -14/14

i1 = 1


i2 = Δ i2 / Δ = 28 / 14 = 2

i2 = 2

EJERCICIO 3.

-6 + 14i1 + 10 ( i1 – i2 ) = 0 -6 + 14i1 + 10i1 – 10i2 = 0 24i1 – 10i2 = 6 5 + 10 ( i2 – i1 ) + 10i2 = 0 5 + 10i2 – 10i1 + 10i2 = 0 -10i1 + 20i2 = -5

24 -10 Δ = -10 20 = 480 – 100 = 380

6 -10 Δi1 = -5 20

Δi2 =

= 120 – 50 = 70

24 6 -10 -5 = -120 + 60 = -60

i1 = Δi1 / Δ = 70/380 =

i1 = 0,18 A


i2 = Δi2 / Δ = -60/380 =

i2 = 0,15 A EJERCICIOS DE NODOS

EJERCICIO 1.

-5 + VA / 2 + VA–VB / 2 = 0

-10 + VA + VA - VB = 0

2 VA – VB = 10

2 VB–VA / 2 + VB / 2 + VB-VC / 2 = 0

VB–VA + VB + VB –VC = 0 2

–VA + 3VB = - 4

Δ=

2 -1 -1 3 = 6 – 1 = 5

-1 10 ΔVA = 3 – 4

= 4 – 30 = - 26

2 10 ΔVB = -1 – 4 = - 8 + 10 = 2 VA = ΔVA / Δ = -26 / 5 = -5,2 VB = ΔVB / Δ = 2 / 5 = 0,4


VC = - 4 EJERCICIO 2.

VA = 12 VC = 12 VB – V A / 4 + V B / 6 + V B – V C / 3 =

VB – 12 / 4 + VB / 6 + VB – 12 / 3 =

3VB- 36 + 2VB + 4VB - 48 = - 84 + 9 VB /12 = 12 9VB = 84 VB = 84 / 9 VB = 9,3


EJERCICIO 3.

-3,1+ VA/ 2 + VA-VB/5 = 0

-31 + 5VA+ 2VA- 2VB 10

=

7VA- 2VB = 31

VB- VA/5 + VB/1 + (-1,4)/ 1= VB - VA + 5VB – 7 = -VA + 6VB = 7 5 7 -2 Δ = -1 6 = 42- 2 = 40 31 -2 ΔVA = 7 6 = 186 + 14 = 172

ΔVB =

7 31 -1 7

= 49 +31 = 80

ΔVA = ΔVA / Δ = 172/40= 4,3 ΔVB = ΔVB/ Δ = 80/40 =2


EJERCICIOS DE SUPERMALLAS

EJERCICIO 1

i2 – i1 = 1

12 + 4i2 + 3i1 = 0

Δ=

Δi1 =

Δi2 =

-1 1 3 4

3i1 + 4i2 = -12

= -4 -3= -7

1 1 12 4 = 4 -12 =8

-11 3 12

= -12 – 3 = -15

i1= Δi1/ Δ = 8/-7 = 1,1 i2 = Δi2/ Δ = -15/-7 = 2,1


EJERCICIO 2

i1 – i2 = 2 8i1+ 2i2 + 3(i2-i3) + 2(i1-i3)=0 2i3-2i1+3i3-3i2-1 = 0

Δ=

8i1+2i2+3i2-3i3+2i1-2i3=0

10i1+5i2-5i3=0

-2i1 – 3i2 + 5i3 = 1

1 -1 0 10 5 -5 -2 -3 5 = 1(25-15) – (-1) (50-10) + 0(-30+10)

1(10) + 1(40) + 0 = 50

Δi1=

2 -1 0 0 5 -5 1 -3 5 = 2 (25-15) – (-1) (0+5) + 0

20 + 5 + 0 = 25

Δi2=

1 2 0 10 0 -5 - 2 1 5 =1 ( 0 + 5) -2 (50-10) + 0

5 - 80= -75

1 -1 2 10 5 0


Δi3= -2 -3 1 =1 (5 +0) – (-1) (10 +0) + 2 (-30 + 10)

5 +10 – 40 = - 25

i1 = Δi1/ Δ = 25 / 50 = 2 i2 = Δi2 / Δ = -75 / 50 = - 1,5 i3 = Δi3 / Δ = -25 / 50 = 0,5

EJERCICIO 3

i2 – i1 = 5 -100 + 3(i1-i3) + 2 (i2-i3) + 50 + 4i2 + 6i1 = 0 - 100 + 3i1 - 3i3 + 2i2 – 2i3 + 50 + 4i2 + 6i1 = 0 9i + 6i – 5i = 100 – 50 9i1 + 6i2 – 5i3 = 50 10i3 + 2(i3-i2) + 3(i3-i1) = 0 10i3 + 2i3 – 2i2 + 3i3 – 3i1 = 0 -3i1 – 2i2 + 15i3 = 0

-1 1 0 9 6 -5 Δ= -3 -2 15 = -1(90-10) – 1 (135-15) + 0

-80 – 120 = - 200


5 1 0 50 6 -5 Δi1= 0 -2 15 = 5(90 -10) – 1 (750 + 0) + 0 =400 -750 = -350

-1 5 0 9 50 -5 Δi2= -3 0 15

= -1 (750 + 0) -5(135 -5) + 0

-750 - 600 =1350

-1 1 5 9 6 50 Δi3= -3 -2 0 = -1 (0 - 100) -1 (0+150) + 5(-18+18)= = -100 -150 +5 = -245

i1 = Δi1/ Δ = -350/-200 = 1,75 i2 = Δi2/ Δ = -1350/-200 = 6,75 i3 = Δi3/ Δ= -245/ -200 = 1,25


EJERCICIOS DE SUPERNODOS

EJERCICIO 1

VB – V A = 5 -4 + VA + VA – VB +VB – VA + VB – 9 = 0 1/2 1/3 1/3 1/6 =2VA + 6VB = 13 1 -1 Δ= 2 6

=6 + 2 = 8

Δi1=

5 -1 13 6 = 30 + 13 = 43

Δi2=

1 5 2 13

= 13 – 10 = 3

VA = 43/ 8 = 5,3

2VA + 3VA – 3VB +3VB – 3VA + 6VB -13 1 1 1 1 1 1


VB = 3 / 8 = 0,3

EJERCICIO 2

VC – VB = 10

10 + VA – VB = 0 4

VA – VB = -40

VB – V A + 0 – V B + 0 – V C = 0 4 3 5 -15VA – 5VB – 12VC

15VB – 15VA + 0 – 20VB + 0 - 12VC = 60


EJERCICIO 3

VB – VC = 22 8 + VA – V B + VA – V C = 0 3 4

96+ 4VA - 4VB + 3VA – 3VC = 12

7VA – 4VB – 3VC = -96

VB – VA + VB + VC + VC – VA - 25 = 3 1 5 4 20VB – 20VA + 60VB + 12VC + 15VC – 15VA = -30VA+ 80VB + 27VC = 1500 60


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