A.S. Graffeo Practice Exam - Impedance Relay question

[akyip]

Hi guys,

First time poster here, I'm practicing for the PE Power exam. I have a question about a distance (impedance) relay question based on the A.S. Graffeo PE Power practice exam (Question 69 on the A.S. Graffeo Afternoon Exam).

The question provides the following parameters regarding a distance relay for a transmission line:

PT Ratio = 2000

CT Ratio = 400

Vrelay = 50∠0 V

Irelay = 50∠-35 A

In the solution explanation for this question, A.S. Graffeo seems to use the primary voltage and primary current for calculating Zrelay, and I'm a bit confused on why that is. In his solution, he has:

Vprimary = Vrelay * PT ratio = 50∠0 V * 2000 = 100000∠0 V

Iprimary = Irelay * CT ratio = 50∠-35 A * 400 = 20000∠-35 A

Zrelay = Vprimary / Iprimary = 100000∠0 V / 20000∠-35 A = 5∠35 ohms

I'm just confused as to why in this question, the solution uses Vprimary and Iprimary values to calculate the Zrelay.

Can anyone provide an explanation for this? Or is it possible that this solution is wrong?

Thank you to anyone who spends his/her time helping me!

 

[Orchid PE]

I don't have the book with me, so what is the question asking exactly?

If it is asking what impedance value the relay sees on the line side, then he is correct. But if it is asking what impedance value the relay sees on the relay side, then he is incorrect.