TAREA: Lunes 03 de Noviembre/2010
Ley de Ohm
PARTE #1
R1=1000Ω R2= 1 = 3,589.74Ω
1/5600 + 1/10 000
RT= 1000+3,589.74= 4,589.74Ω
IT= 9V = 0.001960894 A
4,589.74Ω
V1= (0.001960894)(1000) = 1.96 V
V2= (0.001960894)(3,589.74) = 7.04V
PARTE #2
R1= 10 000
R2= 1 = 848.48Ω
1/1000+1/5600
RT= 10,848.48Ω IT= 9V = 0.000829609A
10,848.48Ω
V1= (0.000829609)(10,000) = 8.29V
V2= (0.000829609)(848.48) = 0.70V
DIVISOR DE CORRIENTE DE AMBAS PARTES
DIVISOR DE TENSIÓN DE AMBAS PARTES
EN EL CASO DE R= [5600Ω] ES AFECTADA POR DOS FUENTES, POR LO TANTO SE REALIZA LO SIGUIENTE:
VR[5600]= (7.04V) + (0.70V) = 7.74V
Mallas
Ec. 1= I1(1000Ω) + I1(5600Ω) - I2(5600Ω) = 9V
=I1(6600Ω) - I2(5600Ω) = 9V
Ec. 2= I2(5600Ω) +I2(10 000Ω) - I1(5600Ω) = -9V
= I2(15600Ω)- I1(5600Ω) = -9V
I1 I2
I1 I2
I1 I2
Vx=I1 - I2 (5600) = [(-0.001256983)-(0.000125698)] (5600) =
=(-0.001382)(5600)= 7.74V
Nodos:
V1 - 9V + V1 + V1-9V = 0
1000 5600 10 000
V1( 1/1000 + 1/5600 + 1/10 000) = 9V/10 000 + 9V/1000
V1(5600/5600000 + 1000/5600000 + 560/5600000) = V(9/10 000 + 90/10 000)
7160/5600000V1= 99/10 000 V
V1= 99/10 000 = 7.74V
7160/5600000
TAREA: Martes 12 de Octubre/2010
Solución por Método de Nodos
Ec.1 V1-12V + V1 + V1-V2 = 0
10Ω 5Ω 20Ω
V1(1/10+1/5+1/20)-V2(1/20)=(12/10)V
Ec1 0.35V1-0.05V2=1.2V
Ec. 2 V2-V1 + V2 =0
20Ω 10Ω
-V1(1/20)+V2(1/20+1/10)=0
Ec1 -0.05V1+0.15V2=0
= [(0.35)(0.15)]-[(-0.05)(-0.05)]=
=(0.0525)-(0.0025)=0.05
VOLTAJES
=[(1.2)(0.15)]-[(-0.05)(0)]
=(0.18)-(0)=0.18
V1= 0.18 = 3.6V
0.05
=[(0.35)(0)]-[(1.2)(-0.05)]=0-(-0.06)=.06
V2= 0.06 = 1.2V
0.05
