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  1. Dude99

    Solve This Problem and Get Paid

    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...
  2. Dude99

    Solve This Problem and Get Paid

    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 😁
  3. Dude99

    Solve This Problem and Get Paid

    0
  4. Dude99

    Just got an NCEES email saying October PE exams are canceled.

    Are results available quicker with the CBT format?
  5. Dude99

    Cram for Exam Vol 4 Question 10 - 4-Phase Systems and Polyphase Systems with more than 3 phases

    V ph LL = 1/0 - 1/ 360/ph V3 = 1/0 - 1/120 V4 = 1/0 - 1/90
  6. Dude99

    Synchronous Machine - Zero-Sequence Impedance, Parallel Circuit Method

    This may help you know V and I https://www.delorenzoglobal.com/documenti/prodotti/140417_Prova_sincrona.pdf
  7. Dude99

    NCEES Sample exam question 108

    I should read the problem! Load is S not Z
  8. Dude99

    Watts Per Square Foot by Building Type

    Or ASHRAE https://elearning.ashrae.org/design-for-building-electrical-systems.html
  9. Dude99

    Voltage Drop Calculations

    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...
  10. Dude99

    COVID-19

    I don't understand why they are automatically refunding rather than asking if you want one or rescheduled for Oct.
  11. Dude99

    Oh oh

    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...
  12. Dude99

    Cram for the PE Sample Test 1 - Problem #13

    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...
  13. Dude99

    Voltage Drop Calculations

    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.
  14. Dude99

    Voltage Drop Calculations

    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.
  15. Dude99

    Voltage Drop Calculations

    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
  16. Dude99

    Voltage Drop Calculations

    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)?
  17. Dude99

    Voltage Drop Calculations

    What were the answer choices?
  18. Dude99

    Voltage Drop Calculations

    The real part of 1.13/-24.4 = 1.03 exactly 1.0291, same as the NEC 1.0291
  19. Dude99

    Voltage Drop Calculations

    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...
  20. Dude99

    Voltage Drop Calculations

    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
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