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Question states : "A 3-phase, 22 kV, 500 MVA star-connected synchronous generator has a reactance Xs of 1.5 pu and supplies 0.8 pu real power (P) at the rated voltage, with a lagging power factor of 0.6 to the load.Calculate the magnitude of the actual complex power consumed by the load in MVA."...

Ball Pen is OK. Make sure you don't have pencil markings on your notes. They don`t check each page one by one, but we came so far for this exam so why take risk. I use frixion ball pen for marking on my books if necessary so that i can erase easily. Thanks to frixion technology.

They will provide you a mechanical pencil with eraser in it and year and month of exam with its logo. No sharpener. Enough leads to complete your AM and PM exam. This is the same one i got one. Leads were pretty good at least i didn't have any problem.

Font size pretty much similar to NCEES Practice Exam. You will have enough space for your calculation beneath the question set like in NCEES. Ans yes you can bring your one extra backup calculator with you. Are allowed to have on your exam table. 

Cut sheets are taken form old thread (link at the end) Question) Ans: (only initial description, no calculation) Any idea how PU impedance of generator is 0.1 pu? I couldn't understand how identical X/R ratios leads to ratio of powers that provides PU impedance of GEN. Please go through the...

eatsleep: my point is that if 2&4 breaker opens, load C will be disconnected. While if 1&3 breaker opens, load A,B &C will be disconnected. Isn't it confusing?

If there's a fault at point D and breaker 5 fails to trip, which breakers should trip next? (Ans: says 1 & 3 would trip if 5 failed to trip.) Can somebody explain this to me? Source: 

Thank you ellen. Now it make sense.

12. Can someone please explain the answer to me. Ans:

Looking for Power PE Materials cheap here in Japan. like: - Complex Imaginary - Camera PPI Practice Exam / problems Is there way to buy softcopies of these?

If possible lets create a group, drop our notes and share our thoughts, as we know sharing is learning.

Is there any one who can help me with this one?

If i change the P (hp) to P (watt) and use the formula P=sq.rt(3) x VI, the value of current will be 93.6 A. This is different from the answer in the example as, I=600Ph / E. Am i missing something here. Cuz in exam for these kind of question, the first thing i will remember is P=sq.rt(3) x...

Can someone please help how Pmax derived here?

Can someone please explain how Pmax is as shown in the attachment (Source: Graffeo Book)

Where is the link?

I agree with you. VAN shall be 277<-30. These kind of things we have to be careful during calculation. 

Vload = vsource - (Iload)(Zload/line) = ( (480∠-30° / √3) - (75∠-cos-10.85)(0.1 + j0.02)(1750 / 1000) )) ---> (L-N) = 263.8 <-30.48 ---> (L-N) = √3 x 263.8 <(-30.48+30)---> (L-L) =457<-0.48V---> (L-L)  √3 there in the solution is just conversion from L-N to L-L.

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