Recent content by GabeM

Does anyone have a good feel for when the license search website will be updated?

Are the questions in the FE online practice exams (http://www.ncees.org/Exams/Study_materials...actice_exam.php) different from the question in the FE sample questions and solutions books (http://www.ncees.org//Exams/Study_materials.php?exam=FE)?

Alaska, 12/29/10, Email Arkansas, 12/29/10, State Board Website Arizona, 12/29/10, NCEES Website Colorado 12/29/10 NCEES Website Connecticut 12/29/10 NCEES Website Florida -- 12.29.10 --NCEES Website Idaho, 12/29/10, IPELS website roster search Illinois, website Indiana, 12/29/10, NCEES...

Do you mean 68% or 68/80? 68/80 seems rather high for a failing score.

I read this article a little bit and I think I get it now. There is a tap setting for the current transformer itself and there is also a tap setting for the relay coil. NCEES is asking for the tap setting on the relay coil.

Did they use that term correctly? Now that I think about it, it makes more sense if it means what "tap" you connected the cables to, like cableguy's example of multiple taps on the high side of a current transformer. I think the problem should ask for the "activation point setting" since we...

I talked to a couple people about this and they both did not recommend bringing any IEEE books with them to the Power exam. I just printed out that page 97 from the red book in case I get confused about the voltage drop with low power factor equation again.

Is the per unit voltage base always phase voltage? That's the only way your equation (1 pu volt)/(j.7975+j.075) = 1.146 per unit amps could work.

I don't have the problem in front of me right this minute, but I recall the problem stating "the tap setting is 5 A", then it goes on to ask you what it is. I think what it means the first time by "tap setting" is where you connect the ammeter wires to the secondary winding. So by saying that...

I believe the solution to 513 is dividing volts per unit by ohms per unit to get amps per unit. It then multiplies by 4811 amps to convert per unit amps to actual amps. The solution is confusing because it omits volts per unit. It can do this because volts per unit is 1. However, this means that...

This might sound crazy this close to the exam, but I would forget about per unit and learn how to use the MVA method linked in the following thread: MVA method I can't answer your question because I don't even know how to use per unit. I don't know if this is going to bite me in the exam, but...

500 kV is the line voltage; you need to use the voltage drop across the reactance, which is phasor Va minus phasor Vb. A couple people here mentioned that above 90 degrees between Va and Vb, the system goes unstable. How does the system go unstable?

I see, you just have to make sure you set "0" at the beginning of a pulse for your equation to work.

Given Kaplan's reputation at this point, it would be easy to write Kaplan's equations off as errors. Also, your equation only works if the waveform is a sine function. For example, it wouldn't work if the function was cosine.

knight1fox3's solution is correct. You mentioned you have the IEEE books in another post. Check the IEEE red book page 97. That phasor diagram was the only way I could understand that equation, even after reading many posts on this forum trying to explain it.