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1. Where do you shop for professional liability insurance (E&O)? Do you just google that? I was looking at CNA. Any other suggested one's based upon your experience? 2. For registering your company, can you have PO box number as your address in that state if you plan to work from home and...

I am electrical PE working for MEP consulting firm with more than 7 years experience. I would like to know the lessons learned, suggestions, pitfalls from people who quit their job to start their own consulting firm based upon their past experience. Sometimes I think I want to jump into it but...

I am electrical PE working for MEP consulting firm with more than 7 years experience. I would like to know the lessons learned, suggestions, pitfalls from people who quit their job to start their own consulting firm based upon their past experience. Sometimes I think I want to jump into it but...

Kentucky does not release the score, but it was enough to put PE after my name.

Passed PE Electrical Power in KY

I passed PE Electrical Power. I wanted to thank all the EB members and especially Flyer PE for always making an extra effort with the explaination to any problems and technical insight. Just wanted to let you all guys know that all the support and help provided is not forgotten.

I wanted to have this thread since in the past KY results were not posted on NCEES website. By the way, I passed.

I am still confused about this problem based upon the explanation and formula mentioned in Stevenson (page 108) which is; Q=Vt/Xd (Ei*cos delta - Vt) For this problem since delta =90, Q=Vt^2/Xd. With line voltage Q=square of 500kV/50=5000. What am i doing wrong here?

Can someone please explain, the relationship KVAR directly proportional to voltage square? The charge of the capacitor is Q = 1/2(CV^2). How that relates to KVAR?

It seems for this problem, the delta load is insignificant. Basically all the calculations are done for the transmission line. Even if the load had been Y, still you would end up with the same answer. Is that correct?

I would approach this by converting Z (Y) = 1/3 Z (Delta). Pph=Vph*Iph=208*(100/3)=6933 Watts P(Total)=3*Pph=20.8KW

Can someone please help me with this? Why is the stator current 1.05 angle (90 -18.19) and not 1.05 angle (-18.19)?

Can someone solve the following problem? 'A balanced delta connected resistive load of 8000KW is connected to the low-voltage, delta connected side of a Y-Delta transformer rated 10,000KVA, 138/13.8 KV. Find the load resistance in ohms in each phase as measured from line to neutral on the high...

Why can't I solve this with 50,000/[sqrt(3)*13.2]=2186A Is it because, the MVA and KV are just rated values of the Generator or is it because of the impedance of T.L. between the Gen & Transformer?

Thanks Flyer_PE. That's what I though too, but just didn't wanted to jump to the conclusion.