Divisor de voltaje en AC

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Simulacion

Potencia Promedio

V*I+cos(α)= 2,848*10^-3

Divisor de voltaje en AC

Encontrar VR, VL y VC

VR=(1*10^3+0.889*10^3)(3*10^3)

VL=(1*10^3+0.889*10^-3i)(2*10^3)

VC= (1*10^3+0.889*10^-3i)(-1*10^3I)

VC= 0.889i = 1.338<48.36°

VL= -1.778+2i = 2.67<48.36°

VR= 3+2.66i = 4.014<41.56°

E= 6/√2=4.24<60°

I= E/2T = 1*10^-3+0.889*10^-3

i = 1.34*10^-3<41.63°

e = 6sen(314t + 60°)

f= w/2π = 314/2π = 49.87°

Xc= -i/ 2πcf = 1*10^-3i = -i/2πc*49.97

C= -i/ 2π(-1*10^3i)(49.87)= 3.185*10^-6F

XL= L(2πfi)= L = XL/2πfi = 6.37Hz

α= tan^-1(4*10^3/3*10^3) = 18.43°

Encuentre ZT

ZT= 3*10^3+(2*10^3i-110^3)-3*10^3+1*10^3i