EJERCICIOS DE MALLAS

Para resolver los ejercicios debemos tener en cuenta lo siguiente.

1. Identificar el concepto.2. Plantear el problema3. 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 i 2 = - 2

-6 + 1i2 + 4 ( i2 – i1 ) = 0

- 6 + 1i2 + 4i2 – 4i1 = 0

– 4i 1 + 5 i 2 = 6

6 – 4Δ= -4 5 = 30 – [(-4) (-4)] Δ = 14 = 30 – 16 =14

-4 - 2Δi1 = 5 6 = -24 – (-10) = - 24 + 10 = -14 Δi1 = -14

6 -2Δi2 = -4 6 = 36 – 8 = 28 Δi2 = 28

i1 = Δi1 / Δ = -14/14 i 1 = 1 i2 = Δ i2 / Δ = 28 / 14 = 2 i 2 = 2

EJERCICIO 3.

-6 + 14i1 + 10 ( i1 – i2 ) = 0-6 + 14i1 + 10i1 – 10i2 = 024i 1 – 10i 2 = 6

5 + 10 ( i2 – i1 ) + 10i2 = 05 + 10i2 – 10i1 + 10i2 = 0-10i 1 + 20i 2 = -5

24 -10Δ = -10 20 = 480 – 100 = 380

6 -10Δi1 = -5 20 = 120 – 50 = 70

24 6Δi2 = -10 -5 = -120 + 60 = -60

i1 = Δi1 / Δ = 70/380 = i 1 = 0,18 A

i2 = Δi2 / Δ = -60/380 = i 2 = 0,15 A

EJERCICIOS DE NODOS

EJERCICIO 1.

-5 + VA / 2 + VA–VB / 2 = 0 -10 + V A + V A - V B = 0 2 V A – V B = 10

2

VB–VA / 2 + VB / 2 + VB-VC / 2 = 0 V B –V A + V B + V B –V C = 0

2

–V A + 3V B = - 4

2 -1Δ = -1 3 = 6 – 1 = 5

-1 10ΔVA = 3 – 4 = 4 – 30 = - 26

2 10ΔVB = -1 – 4 = - 8 + 10 = 2

VA = ΔVA / Δ = -26 / 5 = -5,2

VB = ΔVB / Δ = 2 / 5 = 0,4

VC = - 4

EJERCICIO 2.

V A = 12

V C = 12

VB – VA / 4 + VB / 6 + VB – VC / 3 = VB – 12 / 4 + VB / 6 + VB – 12 / 3 =

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

9VB = 84

VB = 84 / 9

V B = 9,3

EJERCICIO 3.

-3,1+ VA/ 2 + VA-VB/5 = 0 -31 + 5V A + 2V A - 2V B = 7V A - 2V B = 31 10

VB- VA/5 + VB/1 + (-1,4)/ 1= V B - V A + 5V B – 7 = -V A + 6V B = 7 5 7 -2Δ = -1 6 = 42- 2 = 40

31 -2ΔVA = 7 6 = 186 + 14 = 172

7 31ΔVB = -1 7 = 49 +31 = 80

ΔVA = ΔVA / Δ = 172/40= 4,3

ΔVB = ΔVB/ Δ = 80/40 =2

EJERCICIOS DE SUPERMALLAS

EJERCICIO 1

i 2 – i 1 = 1

12 + 4i2 + 3i1 = 0 3i 1 + 4i 2 = -12

-1 1 Δ = 3 4 = -4 -3= -7

1 1Δi1 = 12 4 = 4 -12 =8

- 1 1Δi2 = 3 12 = -12 – 3 = -15

i1= Δi1/ Δ = 8/-7 = 1,1

i2 = Δi2/ Δ = -15/-7 = 2,1

EJERCICIO 2

i 1 – i 2 = 2

8i1+ 2i2 + 3(i2-i3) + 2(i1-i3)=0 8i1+2i2+3i2-3i3+2i1-2i3=0 10i 1 +5i 2 -5i 3 =0

2i3-2i1+3i3-3i2-1 = 0 -2i 1 – 3i 2 + 5i 3 = 1

1 -1 0 10 5 -5Δ= -2 -3 5 = 1(25-15) – (-1) (50-10) + 0(-30+10) 1(10) + 1(40) + 0 = 50

2 -1 0 0 5 -5

Δi1= 1 -3 5 = 2 (25-15) – (-1) (0+5) + 0 20 + 5 + 0 = 25

1 2 0 10 0 -5Δi2= - 2 1 5 =1 ( 0 + 5) -2 (50-10) + 0 5 - 80= -75

1 -1 2 10 5 0Δi3= -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

i 2 – i 1 = 5

-100 + 3(i1-i3) + 2 (i2-i3) + 50 + 4i2 + 6i1 = 0- 100 + 3i1 - 3i3 + 2i2 – 2i3 + 50 + 4i2 + 6i1 = 09i + 6i – 5i = 100 – 509i 1 + 6i 2 – 5i 3 = 50

10i3 + 2(i3-i2) + 3(i3-i1) = 010i3 + 2i3 – 2i2 + 3i3 – 3i1 = 0-3i 1 – 2i 2 + 15i 3 = 0

-1 1 0 9 6 -5Δ= -3 -2 15 = -1(90-10) – 1 (135-15) + 0 -80 – 120 = - 200

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

-1 5 0 9 50 -5Δi2= -3 0 15 = -1 (750 + 0) -5(135 -5) + 0 -750 - 600 =1350

-1 1 59 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

V B – V A = 5

-4 + V A + V A – V B +V B – V A + V B – 9 = 0 2V A + 3V A – 3V B +3V B – 3V A + 6V B -13 1/2 1/3 1/3 1/6 1 1 1 1 1 1

=2V A + 6V B = 13

1 -1Δ= 2 6 =6 + 2 = 8

5 -1Δi1= 13 6 = 30 + 13 = 43

1 5Δi2= 2 13 = 13 – 10 = 3

V A = 43/ 8 = 5,3

V B = 3 / 8 = 0,3

EJERCICIO 2

V C – V B = 10

10 + V A – V B = 0 V A – V B = -40 4

V B – V A + 0 – V B + 0 – V C = 0 15V B – 15V A + 0 – 20V B + 0 - 12V C = 4 3 5 60

-15V A – 5V B – 12V C

EJERCICIO 3

V B – V C = 22

8 + V A – V B + V A – V C = 0 96+ 4V A - 4V B + 3V A – 3V C = 3 4 12

7V A – 4V B – 3V C = -96

V B – V A + V B + V C + V C – V A - 25 = 3 1 5 4

20V B – 20V A + 60V B + 12V C + 15V C – 15V A = -30V A + 80V B + 27V C = 1500 60