= 50 (800-j400)*(1 J2)/((1-J2)*(1 J2) )
= 370 j240
≈ 441∠32.97 Ω
cosθ = 370 / 441 = 0.839
I = U / Z
= 220∠0 / 441∠32.97
= 0.5∠- 32.97
s = U * I = 220 * 0.5 = 110va
p = S * cosθ= 110 * 0.839 = 92.3 W
q = S * sinθ= 110 * 0.545 = 59.9 Var
Respuestas a las preguntas del examen de circuito básico de la Universidad Normal
z = 50 (100 j200)*(-j400)/(100 j200-j400)