Current of balanced three phase
- Thread starter bwd111
- Start date
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?
Look at your units.
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
Last edited:
You are absolutely correct!Now who is getting sloppy?
I believe it's:
I = P/(V*sqrt(3))
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.
This is the symbol I meant (3√)
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.
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
I think its on the wrong side of the number
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.
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??
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....
I am studying electronics and require some assistance with a couple of questions I have on PDF and want to upload for help....
Hello,
Please start a new thread if you have questions of your own.
Here is how to handle the attachments:
From the FAQ:
Please start a new thread if you have questions of your own.
Here is how to handle the attachments:
From the FAQ:
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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.
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?
