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You cant change disciplines, th eone you registered in is the one you will get in the afternoon. so check your account in the NCEES examinee acount to make sure you are prepared for the right one. all the best guys.

Guys the sqrt(3) is there because it is 3 phase formula. similar to P= sqrt(3) VIPF. from voltage level you can tell it is 3 phase problem if not said clearly.

For me, when dealing with impedance, this tells me to deal with (phase values instead of line values) . Then, when I am given the power factor, regardless of the configuration (delta or wye), it takes care of the angle difference of phase angle values (θv- θi) which is the same angle of...

it has been included in the PF angle, since theta = phase Voltage angle - I angle.

EItoPE, I didn't get the solution. lets start by a) how did you conclude that G1 is 1.5MW at 60HZ. what formula did you use?

My understanding to phase shift is that it results when you go from different configuration of transformer Y to delta or vice versa. phase shift has nothing to do with dot convention. Dots will play role in Mutual inductance and the flux between the windings.

for this problem: single phase system. At load side: 3.3KV, 75KW @ 0.9 PF lagging. Z feeder to load = 0.2 + j0.5. Find V at source? In the solution: 1- I = P/(V*PF) = 25.3 @ theta= -25.8 2- V source = IZ + V load and the answer 3010 since it is single phase system: shouldn't V source =...

Ill echo what Flyer mentioned: for such high Voltage and Load, you can tell that it is 3 phase system.

It is the Apparent Power (KVA) which is divided by the voltage to get the current. S= 120+j65 kva is S not Z S=VI* All the best

Sorry, Seems i forgot to add the PF in P formula. So: P in three phase= Vline x I line x Sqrt(3) x PF

check your math again: 2.4K * 6.6 * cos -22 = 17.1KW

1- it applies to both. for most problem you calculate this to find the Power consumption at Load side 2- PF= cos (Theta), E=V: so [P= VI PF] is same as [EI cos (theta)] ( given that you are dealing with phase values) 3 P in three phase= Vline x I line x Sqrt(3) 2 and 3 applies for Delta and...

thats right !

I think you have to learn the power triangle. If you review that, you will figure out that: these formulas for 1 phase for simplicity. Multiply by sqrt 3 for 3 phase: S(KVA)=V x I* P(KW)= S x PF = V x I* x PF it is crucial to have this concept cleared out prior the exam. All the best.

check this: http://engineerboards.com/index.php?showtopic=21007

you need to be familiar a little bit with the ampacity tables. the tables with 75C are based on 30C Ambient Temperature. Therefore, only answer D is applicable for this question.

you use these tables based on the foot (Base temperature defined in each Ampacity tables' titles. For easiness, just follow the foot notes on the bottom of each each of Ampaciity tables. For example: T.310.15 (B) (16) is based on 30 C Ambient T. so you go to 310.15(B)(2)(a) While T.310.15...

when dealing with load between 2 phases, your formula will be: S(between the 2 phases) / (V-ll). in your example it would be 208v. if your load was line to nutral, you use the same formula with V-ln =120v. draw the system before applying the formula if you get confused and use basic circuit...

Yes they are the same.

For Auto transformers formula, open page 342 in the CI exam book. i am referring to problem 30 in Volume 3. In general: V1(high side) is across both coils (N1 +N2), V2 (low side) is across N2. Apply the formula V1/V2 = (N1+N2)/N2 = [(N1/N2)+1] this formula is based on (Vp/Vs = Np/Ns) All...