Electronic – Proof that Delta and Y transformer have the same power

transformery-to-delta

I'm trying to proove that a Delta and Y transformer both have the same power. Somewhere I must get wrong in my process…

I have a reading in voltage and current in a Star transformer: 600 V line-line (347 line-ground) and 5 A going through each resistances.

I obtain my power for 1 phase like this:

P1=U*I/square(3)

U=600V
I=5A

So P1 (for one phase) is 1734 W, approximatively 5 kw for all 3 phases.

Now I want to switch this transformer to a delta configuration. I calculate resistance: V/I = R -> 347V/5A = 69.4 ohms

Ok now I switch to the Delta configuration and find the current in one phase:
600V/69.4ohm=8.64A in each resistance (line-line current)

So I need to multiply by square(3) to obtain current line-ground. Let's call this current I

I= 8.646*sqrt(3) = 14.95A

So now the power in one phase is

P2 = I*U/square(3) = 14.95*600/square(3)

So P2 is 3*P1

Which means I have 3 times the power in the Delta configuration vs what I had with star configuration…

I've been banging my head on this for hours, checking with friends, I do not see where I go wrong…

So here I'm lost: Is my calculation incorrect OR am I wrong and Delta configuration power is 3 times the power from Star?

Here is a picture of my calculation, it shows that the Delta has triple power…

enter image description here

Best Answer

Which means I have 3 times the power in the Delta configuration vs what I had with star configuration

Yup, that's what happens. In Y formation, one primary receives \$\frac{1}{\sqrt3}\$ of the line voltage and, for a given resistive load output, there is \$\frac{1}{\sqrt3}\$ of load current. Because V x I = power, the power is only \$\frac{1}{3}\$ of what it is when connected in delta.

Application: a star/delta configuration is used to start induction motors. You begin in start (one third the power) and when it's close to full speed you switch the windings to delta.

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