I am having a difficulty time wrapping my head around a formula that I found on a voltage drop calculator. In the calculator, in both 3 phase and single phase, there is a "Current voltage drop to NEUTRAL" which uses the standard VD calculations (Vd=2KxLxI/Cm or Vd=1.73KxLxI/Cm). Now my issue is with the next calculation, which states "Current Voltage Drop PHASE to PHASE" and it take the VD from the standards formula and multiplies it by .866. The VD percentage uses that VD (.866xVd) figure divided by voltage. Now I don't know where they get the .866 figure and how it works. Can anyone help with this? Thanks.
Voltage drop questions
- Thread starter bracemc
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drbond24
Senior Member
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- Barboursville, West Virginia
I can answer half of that question. 0.866 is the square root of 3 divided by 2.
drbond24
Senior Member
- Location
- Barboursville, West Virginia
Oh ok. I thought you were saying that you didn't recognize the number at all.
I just jotted the formulas down and noticed something silly. If I understand you correctly, the phase-to-phase voltage drop was just the regular (single phase) formula multiplied by 0.866. Well, since 0.866=sqrt(3)/2, if you multiply that by the single phase voltage drop formula the 2s cancel and you get the three phase voltage drop formula.
Could you post a link to the calculator you found?
I just jotted the formulas down and noticed something silly. If I understand you correctly, the phase-to-phase voltage drop was just the regular (single phase) formula multiplied by 0.866. Well, since 0.866=sqrt(3)/2, if you multiply that by the single phase voltage drop formula the 2s cancel and you get the three phase voltage drop formula.
Could you post a link to the calculator you found?
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