When calculating voltage drop for paralleled conductors using the basic formula do you divide by the total circular mill area of the parallel conductors or do you take have the total circuit current and divide by the circular mills of one of the conductors, there seems to be some debate and could not find the correct answer hoping someone could help.
Voltage Drop Parallel Conductors
- Thread starter Mike01
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hardworkingstiff
Senior Member
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- Wilmington, NC
I don't know what the correct answer is, but I take the total amperage, divide by the number of conductors per phase and then figure the voltage drop for one of those conductors. Curious, do you get different answers with the 2 methods?
- Location
- Lockport, IL
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- Retired Electrical Engineer
Without knowing what forumula(s) you are working with, I cannot give you an answer. But I will say that if you use a formula that has CM area as a variable in the calculation of VD, then you use the total area, including the area of all parallel conductors.
follow up
follow up
Thanks everyone Charlie b here is the formula with example.
VD=(√3*K*I*D)/Cmills (k=12.9 for copper, I=load, D=distance)
For example 400A feeder if you were to utilize 500's or 600's it's pretty straight forward, if you were to use a parallel 4/0 conductors let's say 350A of load would you use the 350A for the ampacity and 423,200 for the circular mills (4/0*2) or would you use 125A of load (half the calculated load) and 211,600 circular mills for the area (1-conductor). if you were to assume a 200' run (utilizing the formula above) if you use the combine number you get approx. 3.6 individually with half the load you get 2.6 per conductor)? just confused ?
follow up
Thanks everyone Charlie b here is the formula with example.
VD=(√3*K*I*D)/Cmills (k=12.9 for copper, I=load, D=distance)
For example 400A feeder if you were to utilize 500's or 600's it's pretty straight forward, if you were to use a parallel 4/0 conductors let's say 350A of load would you use the 350A for the ampacity and 423,200 for the circular mills (4/0*2) or would you use 125A of load (half the calculated load) and 211,600 circular mills for the area (1-conductor). if you were to assume a 200' run (utilizing the formula above) if you use the combine number you get approx. 3.6 individually with half the load you get 2.6 per conductor)? just confused ?
- Location
- Lockport, IL
- Occupation
- Retired Electrical Engineer
You are not going to like this, Mike. But there is nothing else for it. Try the math again, but this time let us use 175 as being half of 350. 
3.6 it is
3.6 it is
charlie my fault (I should have checked my math) and why would I not like this? so 3.6 it is so by default for a 400a OCPD if you use 500, better voltage drop (lower), 600's better , 4/0 ok for shorter installations where 500's & 600's would be tough to install. thaks again charlie b you are the man.
3.6 it is
charlie my fault (I should have checked my math) and why would I not like this? so 3.6 it is so by default for a 400a OCPD if you use 500, better voltage drop (lower), 600's better , 4/0 ok for shorter installations where 500's & 600's would be tough to install. thaks again charlie b you are the man.
Volta
Senior Member
- Location
- Columbus, Ohio
I think your answer will be closer to the actual drop if you add the Cmils and calculate with the total rather than using half twice.
hardworkingstiff
Senior Member
- Location
- Wilmington, NC
Why do you think that?
I would think if everything else in the formula stayed the same and you halved the CM and halved the amps the answer would be the same.
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