Electricidad metodo de Nudo y Malla
Enviado por EMIGDIO VELASCO PALACIOS • 15 de Abril de 2017 • Ensayo • 431 Palabras (2 Páginas) • 159 Visitas
METODO DE MALLA[pic 1][pic 2][pic 3]
-V1 + R2i1 + V2 + R3(i1+i2) + R4(i1+i2) + R1i1 = 0
-V1 + R2i1 + V2 + R3i1 + R3i2 + R4i1 + R4i2 + R1i1 = 0
-14 + 3.5i1 + 9.5 + 0.5i1 + 0.5i2 + 5.5i1 + 5.5i2 + 0.5i1 = 0
-14 + 10i1 + 6i2 + 9.5=0
10i1 + 6i2 = 14 - 9.5
10i1 + 6i2 = 4.5 (ecuacion 1)
V2 + R5i2 + R4(i2+i1) + R3(i2+i1) = 0
V2 + R5i2 + R4i2+R4i1 + R3i2 + R3i1 = 0
V2 + 2i2 + 5.5i2 + 5.5i1 + 0.5i2 + 0.5i1 = 0
V2 + 8i2 + 6i1 = 0
9.5 + 8i2 + 6i1 = 0
8i2 + 6i1 = -9.5 (ecuacion 2)
6I1 = -9.5 - 8i2
(Ecuacion 3)[pic 4]
Sutituyendo ecuacion 3 en ecuacion 1
10i1+6i2=4.5
[pic 5]
-15.83-13.33i2+6i2=4.5
-15.83-7.33i2=4.5
-7.33i2=4.5+15.83
[pic 6]
I2 = -2.77 Amp.
Sustituyo I2 en ecuacion 2
8i2 + 6i1 = -9.5
8(-2.77) + 6i1 = -9.5
-22.16 + 6i1 = -9.5
6i1 = -9.5 + 22.16[pic 7]
[pic 8]
I1=2.11 Amp
METODO DE NODO
[pic 9]
[pic 10]
Malla Superior
VA - VB= -R1i1 + V1 -R2i1 (ecuacion 1)
VA= -R1i1 -R2i1 + V1
VA= -0.5i1 - 3.5i1 + 14
VA= -4i1 + 14
[pic 11]
Malla Central
VA=R4i2+R3i2+V2 (Ecuacion 2)
VA=5.5i2+0.5i2+9.5
VA=6i2+9.5
[pic 12]
Malla Inferior
VA – VB = R5i3
VA = R5i3
VA = 2i3
[pic 13]
Aplicando I1 + I 2 + I3 = 0
[pic 14]
[pic 15]
[pic 16]
[pic 17]
[pic 18]
[pic 19]
VA = 5.54
[pic 20]
[pic 21]
[pic 22]
METODO DE MALLA
[pic 23][pic 24]
Malla 1
V1 + R1i1 + R2(i1+i2) = 0
40 + 200i1 + 80i1 + 80i2 = 0
40 + 280i1 + 80i2 = 0 (ecuación 1)
[pic 25]
Malla 2
V1 + R2(i2+I1) + R3(i2+i3) – V2 = 0
40 + 80i2 + 80i1 + 20i2 + 20i3 - 360=0
40 + 100i2 + 80i1 + 20i3 - 360=0
80i1 + 100i2 + 20i3 = 320 (ecuacion 3)
Sustituyendo ecuacion 2 en ecuacion 3
80(-0.142-0.285i2) + 100i2 + 20i3 = 320
-11.42 - 22.8i2 + 100i2 + 20i3 = 320
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