Taller Estrella Delta Circuitos

Informe de teoría de Circuitos TALLER #3
Cristian Torres Limongi
02/06/2011

Objetivo.- Determinar el % de Error de las 5 resistencias seleccionadas de acurdo a su código de colores de cadaresistencias.
Procedimiento.-

Conversión de Estrella a Delta

R12A=R4*R3+R4*R12+(R12*R3)R12
R12A=5.6*40+5.6*20+(20*40)20=56.8 Ω

R3C=R4*R3+R4*R12+(R12*R3)R3
R3C=5.6*40+5.6*20+(20*40)40=28.4ΩR4B=R4*R3+R4*R12+(R12*R3)R4
R4B=5.6*40+5.6*20+(20*40)5.6=202.85Ω

Conversión de Delta a Estrella

R10A=R10*R4B(R10+R4B+R7)
R10A=7.5*202.85(7.5+202.85+690)=1.68Ω

R4BB=R10*R7(R10+R4B+R7)R4BB=7.5*690(7.5+202.85+690)=5.74Ω

R7C=R4B*R7(R10+R4B+R7)
R7C=202.85*690(7.5+202.85+690)=155.45Ω
R6,4BB=R6+R4BB=3300+5.74=3305.74Ω
Conversión de Delta a Estrella

R12AA=R1*R2(R12A+R2+R1)R12AA=10*690(56.8+690+10)=9.11KΩ

R2B=R2*R12A(R12A+R2+R1)
R2B=690*56.8(56.8+690+10)=51.78KΩ

R1C=R1*R12A(R12A+R2+R1)
R1C=10*56.8(56.8+690+10)=0.75KΩ

R9,12AA=R9+R12AA=36+9.11=45.11kΩR10,2AB=R2B+R10A=51.78+1.68=53.46kΩ

Conversión de Delta a Estrella
R9.12.10=R9.12AA*R10.2AB(R9.12AA+R6.4BB+R10.2AB)
R9.12.10=45.11*53.46(45.11+3305.74+53.46)=0.71KΩR10.6.4=R10.2AB*R6.4AB(R9.12AA+R6.4BB+R10.2AB)
R10.6.4=53.46*3305.74(45.11+3305.74+53.46)=51.91KΩ

R12.6.4=R9.12AA*R6.4AB(R9.12AA+R6.4BB+R10.2AB)
R12.6.4=45.11*3305.74(45.11+3305.74+53.46)=43.80KΩR5.12=R5+R12.6.4=1500+43.80=1543.80kΩ
R9,12=R1C+R9.12.10=0.75+0.71=1.46kΩ
R7.10=R10.6.4+R7C=51.91+155.45=207.36kΩ

Conversión de Delta a Estrella

R7.8=R8*R7.10(R8+R5.12+R7.10)
R7.8=5.6*207.36(5.6+1543.80+207.36)=0.66KΩR5.8=R5.12*R7.10(R8+R5.12+R7.10)
R5.8=1543.80*207.36(5.6+1543.80+207.36)=182.22KΩ

R5.7=R5.12*R8(R8+R5.12+R7.10)
R5.7=1543.80*5.6(5.6+1543.80+207.36)=4.92KΩ

CIRCUITO EN SERIE

XRC3.7.8=RC3+R7.8=28.4+0.66=29.06kΩ
Y R9,12.7.8=R9.12+R5.7=1.46+4.9=6.36kΩ

CIRCUITO EN PARALELO

Z Rxy=RC3.7.8*R9.12.7.8RC3.7.8+R9.12.7.8=
Z Rxy=29.06*6.3629.06+6.36=5.21kΩ

CIRCUITO EN SERIE...