Question

已知一线圈电感L=1H,电阻可以忽略,设流过线圈的电流i=√2sin(314t-60°)A.

Answer 1

（1）

∵i =√2sin(314t-60°)A

--> I = 1∠-60°; ω = 314 = 100π

--> X = ωL = 100π

--> Z = jX = j100π = 100π∠90°

--> U = IZ = 1∠-60° ×100π∠90° = 100π∠30° = 50√3π + j50π

（2）

此时i =√2sin(5000t-60°)A

--> I = 1∠-60°; ω = 5000

--> X = ωL = 5000

--> Z = jX = j5000 = 5000∠90°

--> U = IZ = 1∠-60° ×5000∠90° = 5000∠30° = 2500√3 + j2500