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imho my solution is correct using V.  I don't care about the $.   As noted the I's in each xfmr prim and sec are equal in magnitude and opposite in sign summing to 0 at the node between each xfmr.  This makes the V's also opposite and equal summing to 0. If you did have a resultant V you would...

VpT1 = 4/7 • V                              VsT1 = 1/4 •4/7 • V = V/7 VpT2 = 3/7 x V                              VsT2 = 1/3 • 3/7 • V = V/7 subtractive VsT1 - VsT2 = V/7- V/7 = 0 Is = 0/Z = 0 therefore Ip = 0 😁

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Are results available quicker with the CBT format?

V ph LL = 1/0 - 1/ 360/ph V3 = 1/0 - 1/120 V4 = 1/0 - 1/90

This may help you know V and I https://www.delorenzoglobal.com/documenti/prodotti/140417_Prova_sincrona.pdf

I should read the problem! Load is S not Z

Or ASHRAE https://elearning.ashrae.org/design-for-building-electrical-systems.html

Wrong...again that is not a trick, that is seeing if you can separate the wheat from the chaff.  Most real world problems require gather all information and eliminating the 'noise'.  Obviously it is important to know what the 'question is asking'. lol The 'best' approach varies by individual...

I don't understand why they are automatically refunding rather than asking if you want one or rescheduled for Oct.

Email Due to increased concerns related to the coronavirus disease 2019 (COVID-19), NCEES is canceling the April 2020 exam administration. This decision is based on the recommendations of public health officials to help contain and slow the spread of the virus and to help ensure the health and...

This is a good example: THINK before grabbing the pencil and calculator. can't be 0 unless it is the faulted line or inf bus no and no Z. can't be 13800 unless NO Z except a hi NGR. same for 12090, nrg must be >>>> X1 that leaves 6990, which makes sense since L-N is ~7970   imo...

You also thought 0.046 was 4 awg cmil. It's easy if they give you LOAD pf. But you don't need to go that far.

More than 1 way to skin a cat.  In my opinion the test is looking for minimum competency, comprehension/understanding of concepts, not memorization or math skills.  They are not trying to 'trick' anyone.

Who's to say?I've passed PE EE (general, decades ago)  EE power and PE controls both within the last 10 yrs, and am taking the PE ME HVAC. any of the methods will get you there adjusted for pf

imo if they give conduit type and reference the NEC use their approximation method if they give Z in complex form use it isn't system pf = cos(line ang - load ang)?

What were the answer choices?

The real part of 1.13/-24.4 = 1.03 exactly 1.0291, same as the NEC 1.0291

The way I do it X = 0.024, R = 0.155, Z Line phase = 8.8 deg, old pf = 0.85 31.79 deg, new 0.9 25.84 deg Z at 0.85 = 0.145 from table at 0.85 pf Z at 0.9 = Z x cos(Z line - Z load old/(cosZ line - Z load new) plugging Z = 0.1506  any method will get you 'close enough) if a motor load ph...

imho the NEC note is confusing they give you R and X hence the line ph ang = atan x/r assume 500' #4 R = 0.155 and X = 0.024 Z = R x cos(atan x/r) + j X sin(atan  /r) = 0.153/8.8 deg using the NEC method of 0.9 x R + X x sin(acos 0.9) = 0.150 moot on 100 A load: 15.3 vs 15.0 drop