Voltage drop to dedicated receptacles

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minesh21 said:
Yup that was my next guess. With single phase loads you multiply times 2, for 3 phase you multiply times 1.73. Although most calculators ask you for the ONE WAY distance of the run, because they automatically multiply times 2 for you...

To clarify I wasn't the one that got the low voltage drop. It was another user.
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I knew it wasn't you. Sorry if I led you to think that.
 
K8MHZ said:
I used an on line calculator. All the ones that I found used circuit length that had the two conductors already factored in. If hbiss was using a calculator or calculation that did not factor in both the hot and the neutral, the effect would be the same as halving the length of a two wire circuit.

Since his figures are very close to a 150' run that considers both conductors, I was thinking he didn't take the fact into consideration that there were two conductors that needed to be figured in. I remember this happening to many students in my apprenticeship on tests.
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I guess I'm confuses as to why 300' wouldn't be used.
The OP indicated the load was 300' away from the source.


JAP>
 
jap said:
I guess I'm confuses as to why 300' wouldn't be used.
The OP indicated the load was 300' away from the source.
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It depends on the formula being used.

300' home run is 600' of wire. Some use a formula that does that part for you. Some don't.

I agree to try to use a multi-wire circuit.
 
MAC702 said:
It depends on the formula being used.

300' home run is 600' of wire. Some use a formula that does that part for you. Some don't.

I agree to try to use a multi-wire circuit.
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I'm old fashioned

2x300x12x16= 115200/41740 (Cm for #4) = 2.76/100 = .03 or 3% Voltage Drop.

I may be all wet on this though.

JAP>
 
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