# 3 Phase Load Question



## knight1fox3 (Oct 14, 2010)

In the attached PDF, if I have 3 resistors connected in a delta configuration to a 480V 3-phase source as shown, what will be the total current draw in each phase?

Circuit_Load.pdf


----------



## Flyer_PE (Oct 14, 2010)

I assume you're looking for the line current being supplied to the network:

ILine = sqrt(3)*IPhase = sqrt3*30 = 51.96 Amps


----------



## knight1fox3 (Oct 14, 2010)

Flyer_PE said:


> I assume you're looking for the line current being supplied to the network:
> ILine = sqrt(3)*IPhase = sqrt3*30 = 51.96 Amps


You are correct. So 51.96A is being supplied in each phase.

I had a discussion with a colleague of mine who said that 60A was being drawn in each phase. I said I disagreed and wanted to confirm my understanding of this circuit.


----------



## LMAO (Oct 14, 2010)

knight1fox3 said:


> In the attached PDF, if I have 3 resistors connected in a delta configuration to a 480V 3-phase source as shown, what will be the total current draw in each phase?


your schematic is correct, just make sure you are not confusing the line values with phase values. When you are converting a three phase circuit to a single phase like the way you did, the line and phase currents are equal in per phase schematic but when you convert it back to three phase you have to subtract the phase currents sharing the common line to find the target line current, and due to the 120 degree phase shift, line current magnitude turns out to be (3^.5) times phase current.


----------



## knight1fox3 (Oct 14, 2010)

So is what I said in my 2nd post correct then?


----------



## Flyer_PE (Oct 14, 2010)

knight1fox3 said:


> So is what I said in my 2nd post correct then?


Yes.

Another way to look at it is this: The currents are not scalar quantities, they are vectors. If you apply KCL to any of the three corners of the delta, the incoming line current will be equal to the sum of the two phase currents. In a balanced system, these phase currents will be separated by 120o.

Using your example of a 30 Amp phase current, you will get a line current = (30/0o)-(30/120o) = 51.96/-30o.

Edit: I would also suggest switching the direction of the arrow you show for current between phases A and C.


----------



## knight1fox3 (Oct 14, 2010)

Excellent! Thank you all.


----------

