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Anonymous Poster #1

# Confused Student?

05/17/2017 12:17 PM

If my teacher says a 10 amp 3 phase load does that mean there i 3.333A going through each wire?

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#1

### Re: confused student?

05/17/2017 12:58 PM

No... it means that each leg of the 3 phase supply is flowing 10 amps of current.

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#2

### Re: confused student?

05/17/2017 1:01 PM

Well, yes.... but there is a little more to it.....

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#4

### Re: confused student?

05/17/2017 1:08 PM

Really? Like what?

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Anonymous Poster #1
#3

### Re: confused student?

05/17/2017 1:02 PM

So that means the supply would have to be rated for 30 amps?

That does not sound correct

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#5

### Re: confused student?

05/17/2017 1:09 PM

No it does not. It means the supply must be rated for a minimum of 10 amps, 3 phase.

You are confusing a single phase source and a three phase one.

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#12

### Re: confused student?

05/17/2017 4:06 PM

One cannot say, because the cabling design is not to hand.

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#18

### Re: confused student?

05/17/2017 4:30 PM

Let me ask you this, why would you think the source would have to be rated for 30 amps?

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#6

### Re: Confused Student?

05/17/2017 1:44 PM

I think the word "load" it is not so clear. The current is 10 A on each line but the power [=load] is sqrt(3)*VLL*10A. VLL=line-to-line[phase-to-phase] voltage. If you use VLN=line-to-neutral voltage the power will be 3*VLN*10. However, the result it is not W [or kW or HP] but VA [volt ampere].If-let's say-it is an induction motor supply, for instance, you need to know the power factor and the efficiency in order to get kW and to divide by 0.746 [or 0.736 in IEC World] in order to get HP.

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#7

### Re: Confused Student?

05/17/2017 1:47 PM

LOL... I am willing to bet our OP is now even more confused.

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#9

### Re: Confused Student?

05/17/2017 3:15 PM

I am on your side of the table when that money gets laid down.

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#8

### Re: Confused Student?

05/17/2017 3:14 PM

No, it does not mean that, go back and actually read the book, then do your own homework this time.

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#14

### Re: Confused Student?

05/17/2017 4:08 PM

Vulgar/Rude/Improper Behavior: This post was deleted because it did not adhere to the behavioral policies of the site. Please review Section 14 of the CR4 Site FAQ and the Rules of Conduct.

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#10

### Re: Confused Student?

05/17/2017 4:00 PM

Maybe easier to just go with watts....

Calculation with line to line voltage

The power P in watts (W) is equal to square root of 3 times the power factor PF times the phase current I in amps (A), times the line to line RMS voltage VL-L in volts (V):

P(W) = 3 × PF × I(A) × VL-L(V)

Calculation with line to neutral voltage

The power P in watts (W) is equal to 3 times the power factor PF times the phase current I in amps (A), times the line to neutral RMS voltage VL-N in volts (V):

P(W) = 3 × PF × I(A) × VL-N(V)

http://www.rapidtables.com/calc/electric/Amp_to_Watt_Calculator.htm

Haha,, yes 10 amps per leg....

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#11

### Re: Confused Student?

05/17/2017 4:05 PM

One could ask the teacher, of course...

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Anonymous Poster #1
#13

### Re: Confused Student?

05/17/2017 4:06 PM

actually you cant because he refuses to answer questions

Thanks for assuming you know everything about my school though

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#15

### Re: Confused Student?

05/17/2017 4:09 PM

Nothing is known about the school unless it is posted here. How is the forum supposed to know that the teacher won't answer questions (rhetorical question - NNTR)? After all, it is impossible to convey that information within a one-sentence posting.

Some teacher!

<unsubscribes>

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Anonymous Poster #1
#16

### Re: Confused Student?

05/17/2017 4:16 PM

i wonder if the engineering student thought to ask the teacher before the forum. I wonder.......

He made it all the way to engineering school but never thought once in his life to ask a teacher a question.

There is a rhetorical question

Are you familiar with sarcasm?

Let me introduce you

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Anonymous Poster #2
#17

### Re: Confused Student?

05/17/2017 4:23 PM

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#19

### Re: Confused Student?

05/17/2017 5:02 PM

Here is something for the OP to chew on.

In a three phase system, you have 3 voltage sources. Those 3 sources are "out of phase" to each other by 120 degrees. So that means the instantaneous voltage value of any phase at a "given" point in time is not the same as the instantaneous voltage level in the other two phases at that same "given" point in time.

When phase A reaches its highest voltage level, the peak voltage on phase B is 120 degrees behind phase A and the peak voltage on phase C is 120 degrees behind phase B.

All of this means that... Power = Volts x Amps.

So, in a single phase system...

120V x 10A means 1200 watts of power.

While in a three phase system...

(120v x 1.73) x 10A =2076 watts of power.

The "1.73" is the value of the square root of 3, which gives us the value that is used as the multiplier to determine the "boost" we get from a 3 phase system to deliver power over a single phase system.

There is another value that enters into these equations called "power factor". Let's ignore that for now.

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