Electrical Suppy and Installation questionsss (15marks)

2008-05-13 4:37 pm

Three identical 0.5H inductors with negligible resistance (0 Ohms)are connected to a 380V, 50Hz, 3-phase, 4-wire power suply. Calculate:

i) the impedance of the inductor;
ii)the phase coltage while the inductors are connected in star;
iii)the phase current while the inductors are connected in star;
iv)the line current while the inductors are connected in star;
v)the neutral current while the inductors are connected in star;
vi)the phase voltage while the inductors are connected in delta;
vii)the phase current while the inductors are connected in delta; and
viii)the line current while the inductors are connected in delta.

回答 (1)

用戶8430

2008-05-15 7:12 am

✔ 最佳答案

L = 3 x 0.5H = 1.5H
R = 0Ω
UL = 380V
F = 50Hz
Ph = 3-phase
XL = 2лFL = 2*3.14*50*1.5=471Ω (2*3.14*50*0.5=157Ω each )
i. Z = √X + R = √471*471+0 = 471Ω
ii. VP = VL /1.732 = 380/1.732=220V
iii. IP = IL = 1.4A
iv. I = √3V/Z = (1.732*380)/471 = 658/471=1.4A
v. In = If 3 L is balance In = 0A
vi. VP = VL 380V
vii. IP = VL / Z = 380 / 157 = 2.42A
viii. IL = IP x 1.732 = 2.42*1.732 = 4.2A

圖片參考：http://sub.allaboutcircuits.com/images/12115.png

In balanced Δ circuits, line voltage is equal to phase voltage, while line current is equal to phase current times the square root of 3.

圖片參考：http://sub.allaboutcircuits.com/images/12118.png

Δ-connected three-phase voltage sources give greater reliability in the event of winding failure than Y-connected sources. However, Y-connected sources can deliver the same amount of power with less line current than Δ-connected sources.
希望可以幫到你

2008-05-14 23:16:49 補充：
改正，顯示問題 ² 顯示唔到

i. Z = √X² + R² = √471*471+0 = 471Ω

參考: http://www.allaboutcircuits.com/vol_2/chpt_10/5.html

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