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Put in the hours of study. Repetition is a big part of it. It helped me to remember such that I didn't need to crack a book very often during the exam, and finished up early so I could go back through it (at which time you can open your books). This forum was a God send. If you are using...

Passed. score - 75 Electrical (Power)

Another question on this one. The power factor is 0.9, and the NEC says that any power factor other than 0.85 requires correct by Rxpf + XL(sin(arccos(pf) why didn't they use the correction factor?

Okay, don't understand the solution to this one. It wants to know the pahse to neutral value of the complex impedance of a 400 kcmil, 75 deg cable at 500 ft. I used table 9 of the NEC to get the resistance and reactance. Then I did the following: Z=(0.035+0.049j)(500/1000)(2) , where the...

It appears they used the current divider formula.

It's the Power Sample Exam booklet

Thanks for the response. In regards to the sqrt of 3, Camara has it outside of the V2/Q, and not with V. That's what is confusing me. I got 9 uF as well.

By the way, after finding Xc, he converts to capacitance and gets 5.31 x 10-6 uF.

For those who don't have the book: A 60 Hz, three phase, 208 V system carries a total load of 200+j150 VA. What is most nearly the total capacitance required to correct the power factor? A) 0.001 uF B) 5 uF C) 9 uF D) 500 uF I used the equation Xc = V2 / Q, but the book had Xc = (sqrt 3)...

I'm using the Casio fx-115ES. I really like how it works.

I see I misquoted in my first statement. My question should have read "why WOULDN'T it be at angle of -30 degrees. The answer in the back of the book shows Van to be 7.62 at angle 0. After converting from line to phase, I thought it should have been 7/62 at angle -30.

Another way to look at it: P=V x I x PF 0.5 = 0.6 x 1.2 x PF PF = (0.5)/(0.6 x 1.2) = 0.69444 I got the same as you on Z, but in a little different manner.

I'm planning to take the Power PE Exam in October 2012. My reference material is: Electromechanical Energy Devices and Power Systems by Yamayee (this has been a great help) Power Reference Manual for Electrical and Computer PE by Camara (seems okay) Ugly's NEC NCEES Sample Questions What...

I understand the use or non use of a negative, but I'm wondering why the angle didn't shift 30 degrees going from VAB to VAN

We are given VAB as 13.2 kV for a 3 phase, 4 wire, wye connected utility line. Two loads are placed between line A to N and B to N. So, we are going to need to convert the line voltage VAB to phase voltages. I understand the sqrt 3, but why would VAN be at an angle of -30 degrees, and VBN be...

To better explain, PF=cosX, or in terms of right triangle, cosx=adjacent/hypotenuse, which is 0.5/0.72 so, PF=0.5/0.72=0.69444

Here is what I did to find the power factor: True power was 0.5 pu Apparent power is S=VI, or S=(1.2 puA)(0.6 puV)=0.72 0.5/0.72=0.694444444 Not a 100% on the way I did it, but it got the right answer.

Okay... think I answered my own question. For three phase S=VI * sqrt 3, which in turn gives I=S/(V*sqrt 3). I guess I was thinking in terms of single phase.

The answer requires the following equation: I= S/(V*sqrt3) Why and when is the sqrt 3 used? In the problem, you have the line voltage, so if anything, I would assume you would divide V by the sqrt 3 to get phasze voltage. Any help is appreciated.

THANKS!!!