Recent content by fetaker

Exam taken: Power October 2013 (1st time) Educational background: BSEE (Power) - 2004 How long did you study: 3 hrs. daily for 3 months. Approx. 5hrs on Saturdays Did you take a prep course? No State: CA List of materials you brought into the exam: Self-made binder, NEC, CI (1-4), NCEES...

I thought I would not make it, even bought some books after the exam for next round of preparation! But its a big relief indeed. Nasir: If I want to talk about certainty I would like to think I got 70% certain in the morning and about 50% in the afternoon. Now I need to get rid of the books...

Damn! I passed.

Agreed with knight1fox3. Another example I think of it is the voltage drop calculation. The impedance we consider is per phase.

It might be little too early to buy or sell. Anyways, where do you have it on sale? How's "Electrical Code Drill Book"?

Morning session was good, but afternoon was something else! Nothing that I could have expected. I am starting to think that the afternoon session is going to drag down my score below the threshold. It is frustrating to go through all this after all those months of dedication. But question...

Thank you for the link!

Single phasing occurs at a 3 phase motor. The output will still be a sinusoidal wave. Shouldn’t the answer be option "B" (which shows a sine wave)? I just don’t understand this question...

I got VD as 2.65V. And the voltage at load terminal117.35V

Yep, my version has the wrong answer. I have calculated as 0.94% drop.

Question: What is the voltage drop at a distance of 175 ft from the secondary of delta -wye connected transformer (34.5kV-480V). The line impedance per 1000 ft is (0.062+j 0.041)? Line current is 200<25 Amp. -I could never get it right. Seems CI has the wrong answer.....

Iahim, I appreciate your effort on explaining the concepts.I went through this problem with several methods to check my understanding. I had forgotten the concept of per unit along my preparation. Based on my understanding there are n+1 base voltages in a per unit calculations (n being number...

oops.... continued... Zg = 0.053pu (calculated) And Ztransformer = 0.13pu Then total Z = 0.183 pu And Ipu = 5.47 (calculated) Fault current = Ipu * IB = 3202 Amp.=Answer The thing is if i get rid of the voltage ratio (12.5 /34.5)2 I get the correct answer. And i'm not convinced that...

My calc is as below. SB=35MVA and VB=34.5kV .IB = 585.7 Amp (calculated) Zg, new = 0.23 x (35/20) x (12.5 /34.5)2 =

CI: 2#51. I agree. To get the answer what the question was asking for, the question should have been more clear; specifying on which side it seeks to have the calculation done. CI: 3#54 MVA method could be easier here but I need to know, if there is a flaw in my calculation or concepts. Or if...