Respuestas a las preguntas del examen de circuito básico de la Universidad Normal

z = 50 (100 j200)*(-j400)/(100 j200-j400)

= 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