Parcial fasores y diagrama fasorial

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I1 I2 
2Ω 
C = 
1 
𝐹 
2 
Universidad Politécnica de Valencia 
Valencia edo. Carabobo 
Nombre: Norbelia Heras 
Unidad Curricular: Análisis de circuitos eléctricos 
Tema: Fasores y Diagrama fasorial 
 
Parcial #2 
 
1.- Para la red eléctrica de la figura, encuentre I2, siendo V1=Cos2t. Además, 
dibuje el diagrama fasorial para I2,V1 ,VL ,V2 ,I1 , y el diagrama temporal. 
 
 
L=1 Hr 
 
 
 
 
 
V1 = cos2t = 1 𝑠𝑒𝑛(2𝑡 + 90°) → V1= 0,707 ∟ 90° 
√2 
 
 
 
ZL= jXL = jwL=2j(1Hr) → ZL= 2Ωj = 2 ∟ 90°Ω 
 
Zc = 
−j 
= 
−j 
= 
j → Zc= -1Ωj= 1∟ - 90° Ω 
Xc wc 
 
 
 
Z 
2(0,5F) 
 
Zr = 2Ω 
−j1Ω . 2Ω −2jΩ 2Ω+1Ωj −4Ωj−2Ωj2 
 P= Zc//Zr = 
2Ω−1Ωj 
→ 
2Ω−1Ωj 
x 
2Ω+1Ωj 
= 
(2Ω)2−(1Ωj)2 
Zi 
 Zc Zr 
Forma polar 
 
C=√(0.4)2 − (1.2)2 = √1.6=1.265 
α = arctag( ) 
1.2 
0.4 
α = arctag(3)= 71.565° 
1.265∟71.565° 
Zc 
I1 
−2(−1)Ω−4Ωj 
= 
2Ω+4Ωj 
4Ω−1(−1)Ω 5Ω 
 
ZP= (0.4-0.8j)Ω 
 
ZT= ZL+ ZP = 2Ωj+(0.4-0.8j)Ω 
ZT=(0.4+1.2j)Ω 
 
 
I1= 
V1
= 
0.707L90°𝑣 
ZT (0.4+1.2j)Ω 
 
I1= 0.707L90° → = (0.707) (90° − 71.565°) 
1.265L71.56° 1.265 
I1= 0.56∟ 18.435° A 
 
 
 
 
 
2 2 
▪ 
R=0.25cos(81.87°) 
J=0.25sen(81.87°) 
(0.035+0.247j) A 
▪▪ 
0.56∟18.435° A 
R=0.56cos(18.435°) 
J= 0.56sen(18.435°) 
I1=0.531+0.177j 
Ir=( Zc )I1=( 1Ωj )0.56∟18.435° A▪▪ 
Zc+Zr 1Ωj+2Ω 
 
1jΩ 
2Ω+1Ωj 
x 
2Ω−1Ωj
= 
2Ωj−1Ωj2 
= 
(1+2j)Ω 
2Ω−1Ωj (2Ω) −(1Ωj) 5Ω 
=(0.2+0.4j)Ω 
 
= 0.447∟63.435° 
 
Tenemos que 
 
I2=Ir=0.447∟63.435° x 0.56∟18.435° 
I2= 0.25∟81.87° A ▪ 
I1=Ic+Ir 
Ic=I1-Ir 
Ic=(0.531+0.177j) – (0.035+0.247j) 
Ic= 0.5-0.07j= 0.5∟-7.96° A 
V2=Ir.ZP = (0.447)(0.8)=0.357v VL=Ir.ZL=(0.447)(2)= 0.894 v 
V1=VL+V2= (0.894)(0.357)=0.319 v 
V2= 0.357 cos2t ∟90° v VL=0.894 cos2t V 
ZL 
 Zc Zr 
Forma polar 
 
C=√(0.2)2 − (0.4)2 = 
√0.2=0.447 
α = arctag( ) 
0.4 
0.2 
α = arctag(2)= 63.435° 
0.447∟63.435°