Recent content by Jonjo

johnlittle864 My case , I was out the school for at least 20 years , and when I decided go for my license the only thing I did was study from the F.E book the NCEES offer, that book has also a CD for practice , go and do it all, actually time yourself , go and check your school books. If I...

I took the last year on October in Florida and the results come up on time (one of the first ones , after Xmass), don't worry they will be up soon at least for P.E, good luck everyone !!!! Just get prepare for jokes like this, The results are up!! , I got my results !! ....and .......!!!

vdubEE , I agree with your friend at work , NCEES examples are good enough + some books as backup for theory , at the exam do not forget your NEC 2008 and if you have the handbook will be best . Good luck everyone !!

Fed EE , I graduated on 1990 , EE back in Peru ,I am licensed Engineer back there ( Peru) , I was out my career for a while working in different areas, then I took back again and I start study 2009 for my EIT in USA , taking once and pass , then take the P.E in USA taking once and passed ...

Here it is the info Reles codes.pdf

I copy and paste and did not work !! , still need help for upload

Hi all , I have info for ElecPwrPEOct11 but with this new form, I don't know how upload , need help !

The wave equation has this shape Cos(wt +Φ) , or Sin(wt +Φ) ; then you just have Φ = 45° , which is the phase angle You looking for w (rad/sec), from the data you just gave to us ,there is no way to find w

EEVA : r = (Vrms/Vavg) X 100 the Vrms is the value for AC components and the r = (√((Vrms/Vavg)²-1)) X 100 , missing the power of 2 and do not requiere the Vrms for the components . Both are equavalent just wacth what number you put on it

For any config this formula is valid , after you apply this formula you have to see if your trafo side you solve for is in Delta or Wye config and then you can figure the current you need. Because the formula above always show the Line current ( I line ) If is for the Delta side, just divided...

Another way to tackle this problem is , Power of the primary = Power of the secondary 1.7321 x V primary(Line to Line) x I primary (Line) = 1.7321 x V secondary ( Line to Line ) x I Primary (Line) 1.7321 x 15(kV) x 62.7(Amp) = 1.7321 x 6.6 (kV) x I (Amp). And that is your I you looking for...

Check out NJ board ( Go online)

Wildsoldeir : Just apply "de Moivre theorem" , look at your Math book (Trig) , here is the link in wikipedia , http://en.wikipedia.org/wiki/De_Moivre%27s_formula Jonjo

Yeah ,I had that too, long , long time ago , my teacher for Circuits II use the same analogy for explain the Reactive Power . Jonjo