河师师范学院电路基础真题解答

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