Recent content by VectrenEng

As an employee of a power company, I also struggled with this topic. Of all the topics, this was the most scary for me. The other thread linked in a previous post is helpful as is the sample NCEES test and PPI material. I used the 2008 NEC Handbook which has commentary on the code throughout...

On a budget, I recommend sources on the Internet. You can make quite a reference using a 3-ring binder. From your list, I used #3, 7, 8, and 9. I intentionally avoided Kaplan and always recommend against using it. I also used the accompanying PPI books. If you can find it, a great reference...

I agree - Using the methods outlined in the link, my calculations match yours. (I let W1=WC and W2=WB)

I posted a message for a thread similar to this, see this link. That quote alone answers this problem. Additionally, Setting the two equal to one another, and the use of a trig identity gives: WA = WC VIcos(30º - Φ) = VIcos(30º + Φ) cos(30º - Φ) = cos(30º + Φ) cos(30º)cos(Φ) +...

Sorry for the double-post, but I did some more research. I hate when I have unclear, or conflicting sources of information, especially this close to the test. This is a good source for the two-wattmeter method. Two-Wattmeter Method For this problem, I changed the notation for PB to WC, since...

I had that epiphany (i.e. always get positive angle) last night. Too much studying....

Again, I don't trust the Kaplan answer. I think it is 0.80 lagging. In the Camara reference book - can't remember the section (42 Instrumentation?) or page number because I don't have my book in front of me - is the following equation: tan Φ = SQRT(3)(PH - PL) / (PH + PL) where PH is the...

Is everyone in agreement that the "N" next to "B" represents neutral or ground? If so, I am trying to think of an application on a wye configuration to ground one phase; it is somewhat common on delta networks.

Yes - I concur. I bought a Kaplan book a few years ago while I was studying for the FE...and used it once (it wasn't worth my time). Ever since then I have swore off Kaplan.

I wish the NEC questions were free points! Those are what scare me the most. I am a power engineer for a utility, i.e. no real-world experience with the NEC. Thanks for the info though.

Is the methodology incorrect? On the utility side, I rarely use the NEC, so I am trying to learn/understand these problems.

The input power of the motor is the load "seen" by the source. The motor's output is 15hp, but requires more given the efficiency of the motor.

This is my first post, so here we go. OK - I was able to get the answers. Your work for Part A is correct (at least it matches mine). Break this problem into 3 loads: motor, cap bank, and heater (each with its own power triangle) and then bring them back together at the end. Motor: P = 75hp...