# Current of balanced three phase

#### bwd111

Joined Jul 24, 2013
117
This is the question. What is the normal operating current of a balanced three phase 15kw heater (delta connected) when applied to balanced voltage 440/3/60

I got 19.7 A

Calc I used-- volts x amps x 3 sq/1000 = 19.7

Last edited:

#### WBahn

Joined Mar 31, 2012
29,164
This is the question. What is the normal operating current of a balanced three phase 15kw heater (delta connected) when applied to balanced voltage 440/3/60

I got 19.7 A

Calc I used-- volts x amps x 3 sq/1000 = 19.7
How did you use that calculation? No where are you using the power at all. What is "sq" mean? "square", "square root"? Square or square root of what?

On the right hand side you have 19.7, which you apparently mean 19.7A.

Does the left hand side yield units of amps?

Since you did get the correct answer, I'll show you how you might consider presenting your work:

Power in a balanced three-phase circuit:

P = V*I*sqrt(3)

where V is the phase-to-phase voltage and I is the line current. Thus

I = P/(V*sqrt(3)) // EDIT: Parens in denominator added to fix the mistake that LDC3 caught.

I = 15kW/(440Vrms*sqrt(3)) = 19.7Arms

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#### LDC3

Joined Apr 27, 2013
924
P = V*I*sqrt(3)

I = P/V*sqrt(3)
Now who is getting sloppy?
I believe it's:
I = P/(V*sqrt(3))

#### WBahn

Joined Mar 31, 2012
29,164
Now who is getting sloppy?
I believe it's:
I = P/(V*sqrt(3))
You are absolutely correct!

Interestingly, I originally wrote

I = V*sqrt(3)/P

and then went

I = 440Vrms*sqrt(3)/15kW = 19.7Arms

I had calculated the 19.7Arms on a calculator previously to verify the OP's answer and did the algebra mentally and so did not actually evaluate the above equation numerically.

But then, as is my engrained habit, I checked the units and clearly V/W does not yield A, so I caught the major mistake immediately. This happens of a very regular basis, both when I am writing a post and when I am working a live problem that I'm getting paid for. I patched it by simply copying the denominator from the right to the left and repositioning the division sign - and introduced the mistake you caught.

In my own defense, I have frequently pointed out that this particular mistake is a very natural and easy one to make and that, no matter how diligent you are, you are virtually guaranteed to still make it from time to time. The diligence and habit merely push the occurance rate down into the mud.

#### LDC3

Joined Apr 27, 2013
924
I blame the computer, it won't let me make a horizontal line for easily doing mathematics.

#### bwd111

Joined Jul 24, 2013
117
How did you use that calculation? No where are you using the power at all. What is "sq" mean? "square", "square root"? Square or square root of what?

On the right hand side you have 19.7, which you apparently mean 19.7A.

Does the left hand side yield units of amps?

Since you did get the correct answer, I'll show you how you might consider presenting your work:

Power in a balanced three-phase circuit:

P = V*I*sqrt(3)

where V is the phase-to-phase voltage and I is the line current. Thus

I = P/(V*sqrt(3)) // EDIT: Parens in denominator added to fix the mistake that LDC3 caught.

I = 15kW/(440Vrms*sqrt(3)) = 19.7Arms
This is the symbol I meant (3√)

#### WBahn

Joined Mar 31, 2012
29,164
So what would 16√9 be? would it be 4*9=36 or 16*3=48?

The convention I've always seen (could be different elsewhere, I realize) is that the thing being taken the square root of goes under the radical and the radical has that check-mark looking part on the left side. Thus, if you only have that character is a text only situation, the thing that you are taking the square root of goes to the right. This is also consistent with the usual convention used with infix notation in which unary operators, like operate on a single operand that is located to the right of the operator. So √9 would be 3 and 9√ would be undefinted, just like 9- would be undefined.

#### bwd111

Joined Jul 24, 2013
117
So what would 16√9 be? would it be 4*9=36 or 16*3=48?

The convention I've always seen (could be different elsewhere, I realize) is that the thing being taken the square root of goes under the radical and the radical has that check-mark looking part on the left side. Thus, if you only have that character is a text only situation, the thing that you are taking the square root of goes to the right. This is also consistent with the usual convention used with infix notation in which unary operators, like operate on a single operand that is located to the right of the operator. So √9 would be 3 and 9√ would be undefinted, just like 9- would be undefined.
It would be 48

#### WBahn

Joined Mar 31, 2012
29,164
Okay, so then if 16√9 means 16 multiplied by the square root of 9, what does 16√ mean?

#### bwd111

Joined Jul 24, 2013
117
Okay, so then if 16√9 means 16 multiplied by the square root of 9, what does 16√ mean?
I think its on the wrong side of the number

#### WBahn

Joined Mar 31, 2012
29,164
Agreed. So hopefully that means you will stop using that notation since you now appear to agree that it is confusing.

#### WBahn

Joined Mar 31, 2012
29,164
Shifting gears a bit, if you haven't already done so, it might be good for you to see if you can derive that equation that you used, namely

P = V*I*sqrt(3)

for a balanced three phase load and show that it applies for both wye and delta connected generators and loads.

#### bwd111

Joined Jul 24, 2013
117
Shifting gears a bit, if you haven't already done so, it might be good for you to see if you can derive that equation that you used, namely

P = V*I*sqrt(3)

for a balanced three phase load and show that it applies for both wye and delta connected generators and loads.
25kw/√3 *440=19.7A

#### bananabrain

Joined Dec 19, 2012
6
19.7 is what I've gotten too

#### bananabrain

Joined Dec 19, 2012
6
Anyone know why I cant seem to be able to post to this thing? I need help with 2 circuit analysis questions but cant upload questions via a new thread??

#### JoeJester

Joined Apr 26, 2005
4,390
Did you press the Go Advanced button?

#### bananabrain

Joined Dec 19, 2012
6
Go advanced button? I am new to the site I am studying electronics and require some assistance with a couple of questions I have on PDF and want to upload for help....

#### bertus

Joined Apr 5, 2008
22,218
Hello,

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Bertus

#### WBahn

Joined Mar 31, 2012
29,164
Anyone know why I cant seem to be able to post to this thing? I need help with 2 circuit analysis questions but cant upload questions via a new thread??
What is the error message (or behavior) that you are encountering.

There is a known problem with thread titles that are too long. There's a sticky at the top of one of the forums explaining that one.

#### WBahn

Joined Mar 31, 2012
29,164
25kw/√3 *440=19.7A
Units, units, UNITS!

The question I am asking here is for you to DERIVE the basic equation. In otherwords, can YOU establish why and where that √3 comes from?