# Transformer exercise problem

The transformer of the picture is connected to a transmission line of 230kV.

As you can see the transformer is delta-star connected. Its turns ratio is a=1:10. I had a go at the solution but when checking the solution manual I found something wrong.

In my solution for the right side of the transformer and for single phases the voltage is $$E_{ph}=\frac{E_l}{\sqrt 3}=230/\sqrt 3=132.8kV$$

The solution manual does the same but when it comes to drawing the equivalent circuit of the single phase the following is drawn Note that Va'n' is the E_{ph} of my calculations. As you can see in the transformer 'box' appart from the 30 degree shift and the ratio 1/10 the voltage is also divided by the square root of 3.

Why is that? The left side is delta connected and we've already converted the voltage.

Idea: Could the voltage ratio be for the whole system? Meaning that this is not true $$a=\frac{Van}{Va'n'}$$ but this is instead $$a=\frac{Van}{\sqrt 3 Va'n'}$$

• What are the variables Ep, El, t, a, etc.? – Transistor Aug 5 '17 at 19:01
• Ep is the phase voltage, El the line voltage , a as I wrote is the turns ratio of the transformer coils and t is the value by which you multiply or divide to move a value from one side of the transformer to the other – John Katsantas Aug 5 '17 at 19:08
• I missed 'a' but what's 'h'? 't' looks inverted to me if the left side of the box is the low-voltage side. How have we got an 'n' on the left side when it's delta connected with no neutral reference. I'm glad I don't have to sort it out! The exercise seems a bit of a mess unless you've left out some context. – Transistor Aug 5 '17 at 20:09
• Sorry about the h , my bad ! I corrected it. As for the neutral reference you mentioned I chose to ignore the n's everywhere lol. Leaving this aside , I think that the ratio I'm given doesn't have to do with the turns. I was given something like 13.2/132 kV . I guess it just states the voltage conversion including the conversions for star delta. That's the last bit I included in my question. What do you think ? – John Katsantas Aug 5 '17 at 20:17
• Moving a voltage from the left side to the right you divide by t and that will give us a multiplication with the square root of 3. So it could be that t is given this way so you don't have to multiply by the square root of 3 later because of the star connection. – John Katsantas Aug 5 '17 at 20:25

I think that since the transfer function includes the term $e^{-j30°}$ that all voltages are either phase to phase or phase to neutral. (That would be the large triangle of (i) and the black of (iii).) This would also account for the $\sqrt{3}$ in the transfer function.