Question

请问这道题怎么做呀？

Answer 1

既然无源网络 N 是由电阻（R'）与电容（C）串联组成的，那么，整个电路就是由这 4 个负载串联而成的：

R、R'、L 和 C
它们的总组抗 z :
z = (R + R') + zL + zC
= (R + R') + (jωL) + 1/(jωC)
= (3 + R')Ω + j[ωL - 1/(ωC)]
= (3 + R')Ω + j[2000rad/s * 2H - 1/(2000rad/s * C)]
因为 z = u/i = 30cos(ωt)/[5cos(ωt)] = 6Ω
说明整个电路为阻性负载，感性负载与容性负载刚好抵消。那么：
z = R + R' = 3Ω + R' = 6Ω
ωL - 1/(ωC) = 0
所以：
R' = 3Ω
C = 1/(ω²L) = 1/(2000² * 2) F = 1/(8000000) F = 0.125 μF