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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

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    \$\begingroup\$ Draw a schematic. Are you discussing changing the secondary from star to delta? If so you will need to change the number of secondary turns otherwise the voltage between phases or between phase and neutral on delta will be \$ \frac {1}{\sqrt 3} \$ times that on star. \$\endgroup\$ – Transistor Nov 21 '18 at 15:17
  • \$\begingroup\$ I don't understand what are you trying to prove. First of all, what is a "star transformer"? you mean a Y-Y (primary/secondary) transformer? and in what concept you are trying to prove that Y and Δ transformers provide the same amount of power? \$\endgroup\$ – thece Nov 21 '18 at 16:15
  • \$\begingroup\$ A picture will be added in the description in few minutes... \$\endgroup\$ – Cherry Nov 21 '18 at 17:55
  • \$\begingroup\$ @Cherry: Please read carefully through the comments. They indicate that your question is badly worded or phrased. Try to (in your question) supply all the information requested. \$\endgroup\$ – Transistor Nov 21 '18 at 19:35
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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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