# 3-phase ans single phase apparent power equations



## Redskinsdb21 (Feb 9, 2013)

EX: I have a S= 19 KVA nonlinear load supplies by a 208/120, 4-wire, 3-phase, wye connected system.....what is the current delivered to the load?

I did like this I = S3phase/ (V_LL*sqrt(3)) = 19000/(208*1.732) = 52.7 Amps, is this correct?

Also, does this mean its the current delivered by all 3 phases? If so, how would I determine line current of each phase coming to the load?

Also, if this were a single phase load, would I just do this: I = S1phase/(V_LL) = 19000/208 ? And the result would simply be the line current of that one phase?

Thanks ahead of time.


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## Flyer_PE (Feb 9, 2013)

If the load is a balanced 3-phase load, the line current is equal to 19kVA/(sqrt3*208V) = 52.7 Amps. If it is a single-phase load, the current will be different depending on whether it is a 208V or 120V load. If it is a 208V load, the current will be on the two phases to which it is connected. If it's a 120V load, the current will be on the phase supplying the load and the neutral conductor.


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## Redskinsdb21 (Feb 9, 2013)

Im not sure what you mean depending on whether its 208 V or 120 V...to me..V_ll = 208....V_ln = 120 V....and if you have one phase, you model it phase to neutral using the 120 V....If it were on 208 V, that would be either V_ll for a 2 or 3 phase load..bout not a single phase....which is what I was asking for to make sure I could go from 3 phase to 1 phase and vice versa


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## Wael (Feb 9, 2013)

when dealing with load between 2 phases, your formula will be: S(between the 2 phases) / (V-ll). in your example it would be 208v.

if your load was line to nutral, you use the same formula with V-ln =120v.

draw the system before applying the formula if you get confused and use basic circuit analyses.


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## Flyer_PE (Feb 9, 2013)

Redskinsdb21 said:


> Im not sure what you mean depending on whether its 208 V or 120 V...to me..V_ll = 208....V_ln = 120 V....and if you have one phase, you model it phase to neutral using the 120 V....If it were on 208 V, that would be either V_ll for a 2 or 3 phase load..bout not a single phase....which is what I was asking for to make sure I could go from 3 phase to 1 phase and vice versa




It is possible to have a single phase load connected line-line. It's connected across two phases, but it is still a single-phase load.


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## power62 (Feb 10, 2013)

Flyer PE is right.

You may also want to review the video lessons provided in Complex Imaginary web site. It explains the use Sq. root 3 and single phase load calculations supplied by 3 phase source.


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## Redskinsdb21 (Feb 13, 2013)

Flyer_PE said:


> If the load is a balanced 3-phase load, the line current is equal to 19kVA/(sqrt3*208V) = 52.7 Amps. If it is a single-phase load, the current will be different depending on whether it is a 208V or 120V load. If it is a 208V load, the current will be on the two phases to which it is connected. If it's a 120V load, the current will be on the phase supplying the load and the neutral conductor.




Is this 52.7 amps of line current in all three phases of the balanced 3 phase system? Does it mean the total current being delivered to the load is 3*52.7 amps = 158.1 amps?


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## Flyer_PE (Feb 13, 2013)

Not really. You have to remember that the currents in question are not in phase with one another. It's vector addition and the angles matter.


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## Redskinsdb21 (Feb 13, 2013)

Ok, I see what your saying about being out of phase. But the 52.7 amps are on each phase correct? Also, Even if the currents are not in phase, is their a way to calculate how much current is being deliver to the load at any given time? Is there a min current the 3 phases must provide to the load at all times to keep the load operating?


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## Flyer_PE (Feb 13, 2013)

Yes, the magnitude of the current in each phase will be 52.7 amps. From Kirchoff's Law, the total current supplied to the load will always be 0. In order to keep a load operating, available current alone isn't the question. What is required is a source that can deliver sufficient current at an acceptable voltage.


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## Redskinsdb21 (Feb 14, 2013)

Flyer_PE said:


> Yes, the magnitude of the current in each phase will be 52.7 amps. From Kirchoff's Law, the total current supplied to the load will always be 0. In order to keep a load operating, available current alone isn't the question. What is required is a source that can deliver sufficient current at an acceptable voltage.




Is the line current the rms current or is it the peak current?


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## Flyer_PE (Feb 14, 2013)

Unless indicated otherwise, voltages and currents are typically rms values.


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## Redskinsdb21 (Feb 14, 2013)

Flyer_PE said:


> Yes, the magnitude of the current in each phase will be 52.7 amps. From Kirchoff's Law, the total current supplied to the load will always be 0. In order to keep a load operating, available current alone isn't the question. What is required is a source that can deliver sufficient current at an acceptable voltage.




oK, i SEE NOW THAT THE SUM OF THE RMS CURRENTS DELIVERED TO THE LOAD = 0 FOR A 3-PHASE SYSTEM...BASICALLY CANCELING EACH OTHER OUT WITH THEIR NEGATIVE AND POSITIVE HALY CYCLES OF EACH PHASE....BUT WHAT ABOUT SINGLE PHASE SYSTEMS? WHEN THE SINGLE PHASE CURRENT IS ALTERNATING HOW DOES THE LOAD RUN WHEN THE ONE PHASE IS AT 0 CURRENT SINCE THEIE ARE NO OTHER PHASES TO SUPPLYE THE POWER?


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## DK PE (Feb 15, 2013)

Hate to answer your question with another question but try this out.

Given a 115V RMS AC single phase system (i.e. std wall outlet) hooked to a 120W incandescent light bulb, what does the current waveform look like and describe the current flow? Then we can go from there.


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