# current phasors

Discussion in 'General Electronics Chat' started by Sparky, Feb 4, 2011.

1. ### Sparky Thread Starter AAC Fanatic!

Aug 1, 2005
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Greetings,

Can you explain regarding a 3-phase delta system: delta source connected to a delta load, why are the current phasors 30 degrees off from the voltage phasors?

Meaning, I was told that for unity power factor, balanced load the current phasors are 30 degrees off from the voltage. (no problem with 120 degrees apart for the voltages)

I have been told the statement but in looking back over old textbooks I cannot find a good answer.

I do see in the books that the angle of the load affects the angle of the current.

Can you provide a good derivation  mathematical explanation as to why the current has to be 30 degrees?

My first attempt that I was told I was wrong on was that if there was a unity power factor then the current would be in phase with the voltage  wrong L

Thanks
Sparky

2. ### shespuzzling Active Member

Aug 13, 2009
88
0
When you're dealing with delta connected systems there are two currents that are identified: the line currents and the delta (phase) currents that circulate within the delta (but in a balanced system will sum to 0).

In a 3 Phase Wye-Delta system, the phase currents in the delta load will be shifted by 30° from the line currents. So the line current Ia will be shifted 30° from the delta current Iab. It's easy to see this if you draw out your 3 phase system, label all of the nodes and perform KVL on each loop. This will give you the currents in the delta. Then use KCL to determine the line currents and you'll see the 30° phase shift between line currents and delta currents (this website is a good resource to see what I'm talking about: http://eece.ksu.edu/~starret/581/3phase.html).

If this is a balanced system, then it won't make a difference what the power factor is; the phase currents will always be shifted by 30° from the line currents. However, if you're dealing with unity PF, then the delta currents will be shifted 30° from the source phase-N voltages, but in phase with the line to line voltages. The line currents will be shifted 30 from the line voltages, but in phase with the phase voltages. Drawing it out really helps.

In a 3 Phase Delta-Delta system the line currents will also be shifted by 30° from the phase currents, the same way they were in the Wye-Delta example. This result is independent of what kind of elements are in the load (as long as they're balanced).

If you have a unity PF (purely resistive load), then the line voltages will be in phase with the phase currents, and thus the line currents are also shifted 30° from the line voltages.

This can be confusing when you just read it and really the best way to understand it is by drawing out the systems and working through them with KVL and KCL.

Last edited: Feb 4, 2011
3. ### Sparky Thread Starter AAC Fanatic!

Aug 1, 2005
75
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I have read through the link you provided and think I have equations describing the 3 currents and their phases (30 degrees) - I get 30 or -30 depending on how I start the KCL, 150, and -90.

Regarding the pdf I have attached - notice there is a line brought out splitting the A and B phase - how can I write a KVL to use to derive an equation for V(c-to-neutral) showing the phase angle?

Meaning I want Vcn = Xcos(wt +- angle) when the math is finished.

I see from the link you provided how it should work out but I need the KVl to start.

Thanks
Sparky

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4. ### Sparky Thread Starter AAC Fanatic!

Aug 1, 2005
75
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I have attached an attempt at some KVL's around the Delta

Still need some help

Thanks
Sparky

File size:
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5. ### shespuzzling Active Member

Aug 13, 2009
88
0
A delta configuration does not have a neutral. I'm unclear on what you're after becuase I don't think Vcn will have any significance in this delta system.

If you wanted to determine the phase-to-neutral source voltage from your delta-delta connection, the only way I could think of doing so would be to figure out what the equivalent phase voltage would be if your source were connected in Wye. You could do this by making the line to line voltages of the delta equal to the line to line of the Wye and then, using the relation for V(line to line)=sqrt(3)*V(phase), and the fact that line voltages will lag phase voltages by 30° in a Wye-Delta connection. Knowing your phase voltage are 120 out of hpase, you could determine the actual angles. In your case Vcn would be at an angle of -90° (-120°+30°) and the other 2 phases would follow by 120°.

Last edited: Feb 7, 2011
6. ### Sparky Thread Starter AAC Fanatic!

Aug 1, 2005
75
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The configuration Im attempting to show is a 4-wire Delta its how you can get 120v by splitting 240.

see:
http://en.wikipedia.org/wiki/File:High_leg_delta_transformer.png

The pseudo neutral divides a transformer coil  240vAC across A and B and then between A and this neutral gives a 120 that can be used. I agree its not a neutral  like a common point in a wye arrangement. But it acts like a neutral when measuring the 120v between n and A or B.

Im wanting an expression for the voltage between the C and this new splitting line:

Vcn = Xcos(wt+ phase)

I am thinking I could get there by voltage expressions and then trig identities  (I hope)

Thanks
Sparky

7. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
The voltage between L3 and N on your diagram would be √3/2 Vline.

If V(L1,N) has zero phase angle (by some arbitrary assignment), then V(L3,N) would lead by 90°.

Your statement about the pseudo-neutral "N" in this case is correct - it's not a neutral reference in the traditional 3-phase ac sense. In fact the phase difference between V(L1,N) and V(L2,N) is 180° - rather than the usual 120° for a 3-phase system.

8. ### shespuzzling Active Member

Aug 13, 2009
88
0
I hope tnk answered your question. I'm unfamiliar with a 4-wire delta but seeing it acutally brought up a question I have that maybe you know the answer to. Why doesn't grounding the center tap of a delta cause a short circuit? I don't see why it wouldn't. The same question applies to corner groudning too (which I just learned about when thinking about your question). Don't mean to hijack your thread, but since you're working on a 4 wire delta I thought you might know.