Post on 28-Mar-2016
description
ACTIVIDADES PROPIAS DE LA SELECCIÓN DE DISPOSITIVOS ELECTRONICOS
LILIANA ARNACHE
GINA ESPITIA
MANUEL VASQUEZ
SENA REGIONAL NORTE DE SNATANDER CIES
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
EJERCICIO 3
EJERCICIO DE SUPERNODOS
EJERCICIO 1
EJERCICIO 2