Circuits Help ( 3 Phase System)?
In the balanced, 3-phase circuit below, |Van|=202VRMS and Zp=(2+0.8∗j)Ω. Find:
(a) |I_Line|
(b) P_AV supplied by the 3 Phase source
(c) Q supplied by the 3 Phase source
http://tinypic.com/r/25smln7/5
I found I_Line to be: 261.207-j104.48, which is correct; however I still don't understand what it means.
Comments
Wye-connected (Y) source to delta-connected (Δ) load. You have the math correct, so I'm not really sure I can explain it.
Let's say Van is reference. Vector Van = 202V ∠0°.
For a Y-connected source, the line voltage Vab = √(3) Van and will lead phase voltage by 30°. Vector Vab = 350V ∠30°.
We get to the Δ-connected load. The load phase voltage VZan = Vab and be in phase with Vab. Vector VZan = 350V ∠30°.
The load phase voltage is applied to a lagging load Zan = 2Ω + j0.8Ω = 2.15Ω ∠21.8°.
Load phase current will be Vector IZan = VZan / Zan = 350V ∠30° / 2.15Ω ∠21.8° = 162A ∠8.2°.
For a Δ-connected load, line current Ia = √(3) IZan and will lag phase currents by 30°. From the attached source "The phasor diagram shows that the load currents in a delta load lead the line currents by 30° and are 1/√(3) times the magnitude".
Vector Ia = 281A ∠-21.8° = 261.2A - 104.5A
At the Y-connected source, Vector Ian = Ia = 281A ∠-21.8°. The magnitude of the angle at the source is the same at the load, just a different polarity. That's as a result of the Y-Δ.
Not really sure if this helps you, but it does explain what you did, even if you did not know why you did it!
P = 158.3kW
Q = 63.3kVAR