Ingenieria

Universidad Autónoma de Nuevo León












Practica #2
Comprobación de Leyes de Kirchoff
Medir;
R1= 330Ω
R2= 470 Ω
R3= 1k ΩR4= 2.2k Ω
R5= 680 Ω
Vr1= 17.5v
Vr2= 13.4v
Vr3= -4.1
Vr4= 12.4v
Vr5= 16.5v
Ir1= 56.7mA
Ir2= 28.1mA
Ir3= -4mA
Ir4= 56.5mA
Ir5= 24.2mA
It= 79.9mA
Ley de Lazos o Voltajes |-30v = Vr1 + Vr4 -Vr2 +Vr5 = 30v Vr1 + Vr3 + Vr5 = 30v … -30v = 17.5v + 12.4v -13.4v -16.5v = -30v 17.5v - 4.1v – 16.5v = 30v |
Ley de Nodos o Corrientes |
“A” It = Ir1 + Ir2 “B” It= Ir2 + Ir3+ Ir4 “C” It= Ir4 + Ir5 … “A” It = 56.7mA +28.1mA “B” It = 28.1mA – 4mA + 56.5mA “C” It = 56.5mA + 24.2mA |
Practica # 3
Comprovacion de metodo de mallas
Procedimiento…
“i1 = It”
“i2 =Ir2”
“i3 = Ir5”
*Establecer ecuaciones de Mallas*
# De Mallas = # De Ecuaciones
“i1” = (R1 + R4)i1 – R1i2 – R4i3 = 30v
“i2” = (R1 + R2 + R3)i2 – R1i1 – R3i3 = 0v
“i3” = (R3 + R4+ R5) i3 – R4i1 –R3i2 = 0v
…
30v = (330Ω + 2.2kΩ)79.9mA – (330Ω)(28.1mA) – (2.2kΩ)(24.2mA)
0v = (330Ω + 470Ω + 1kΩ)28.1mA – (220Ω)(79.9mA) – (1kΩ)(24.2mA)
0v = (1kΩ + 2.2kΩ + 680Ω)24.2mA – (2.2kΩ)(79.9mA) –(1kΩ)(28.1mA)


Ir1 = i1 – i2, Ir2 = i2, Ir3 = i2 – i3, Ir4 = i1 – i3, Ir5 = i3









Practica # 4
Comprobaciones del método de voltajes de Nodo
Procedimiento…
# deNodos = # de Ecuaciones
℮A = 30 , ℮B = 12.4, ℮C = 16.5
*Ec. Auxiliar* ℮A - ℮R =30v
* ℮B * (1/R1 + 1/R3 + 1/R4) ℮B – 1/R1(℮A) – 1/R3(℮C) – 1/R4(℮R) = 0
* ℮C * (1/R2 + 1/R3 + 1/R5) ℮C – 1/R2(℮A)– 1/R3(℮B) – 1/R5(℮R) = 0
…
℮A = 30v = 30v
℮B = (1/330 + 1/1000 + 1/2200)12.4 – 1/330(30) – 1/1000(16.5) = 0
℮C = (1/470 + 1/1000 + 1/680)16.5 – 1/470(30) – 1/1000(12.4) = 0

℮B = Vr4
℮C =Vr5
Vr1= ℮A - ℮B = 30 – 12.4 = 17.6
Vr2 = ℮A - ℮C = 30 – 16.5 = 13.5
Vr3 = ℮C - ℮B = 16.5 – 12.4 = 4.1
Vr4 = ℮R = 0
Vr5 = ℮C = 16.5




Practica #5
Comprobación de...