voltage drop question

[Designer69]

so if I have a run of cable from source to load that is of varying sizes, (see attached sketch) will the larger size portion of the cable make a difference on voltage drop or should voltage drop be calculated as if the entire run is of the smaller cable size?

Thank you

 

[Smart $]

Larger will make a difference. Calculate voltage drop for each section. Total voltage drop will be the sum of all sections.

 

[George Stolz]

To add to Smart's reply, the voltage drop is the resistance from beginning to end. The larger conductors contribute less resistance (regardless of where they are in the run) so you get some credit for the effort.

 

[Crohnos01]

Agree with the above answers. Inquiring minds want to know: why do you have a smaller gauge conductor in the middle? Is this a retrofit application?

 

[GoldDigger]

Agree with the above answers. Inquiring minds want to know: why do you have a smaller gauge conductor in the middle? Is this a retrofit application?

#2/0 (also written 00 for those who prefer the more graphic form, and called either "two-oh", "two aught" or "double-aught") is larger than #4.
It is the next size up from #0.

 

[Crohnos01]

my bad.... you're right.... need... more... COFFEE!

 

[topgone]

so if I have a run of cable from source to load that is of varying sizes, (see attached sketch) will the larger size portion of the cable make a difference on voltage drop or should voltage drop be calculated as if the entire run is of the smaller cable size?

Thank you View attachment 9397

You'll need the lengths of each run to calculate each of the section voltage drops, as usual.

 

[kwired]

my bad.... you're right.... need... more... COFFEE!

Even if the middle section were smaller, putting some distances on each segment may make more sense as to why each section is a different size, what if the middle section were really short for some reason? Since the middle section is larger, it does seem logical that it possibly is much longer than the other two sections.

 

[kingpb]

Total resistance = #4 R (1st section) + #2/0 R+ #4 R (3rd section)

Use total resistance in standard VD calc formula. This only applies because there are no sources or loads between beginning and end. If there were you would need to do each segment in order; where the voltage (VD) at the end of the segment becomes the starting voltage of the next segment. But I digress.

 

[kwired]

Total resistance = #4 R (1st section) + #2/0 R+ #4 R (3rd section)

Use total resistance in standard VD calc formula. This only applies because there are no sources or loads between beginning and end. If there were you would need to do each segment in order; where the voltage (VD) at the end of the segment becomes the starting voltage of the next segment. But I digress.

Resistance of a section of conductor in the circuit is a load.

 

[roger]

Agree with the above answers. Inquiring minds want to know: why do you have a larger gauge conductor in the middle? Is this a retrofit application?

It could be for a number of reasons but, if the connection points (lugs or terminals) are sized for the smaller conductors it makes sense for that reason alone.

Roger

 

[GoldDigger]

It could be for a number of reasons but, if the connection points (lugs or terminals) are sized for the smaller conductors it makes sense for that reason alone.

Roger

:thumbsup:
And there may be a factor of ease of running the wires to the load and supply in an interior space versus large PVC or direct burial wire for the long haul run outdoors.

 

[Crohnos01]

Resistance of a section of conductor in the circuit is a load.

I get so confused.... Star Trek taught me "resistance is futile"...and all this time I thought it was V/I.

 

[charlie b]

.... Star Trek taught me "resistance is futile"...

And the Hitchhiker's Guide taught me that "resistance is useless."

 

[charlie b]

My approach to the problem would be to perform the calculation by first assuming that the smaller wire was run for the entire distance. If the results were satisfactory, that would end the calculation process. But if that approach gave a total VD that was higher than I wanted, then I would refine the calculation, using the method described by the others. What I mean is that if I can show that the VD is no more than, say, 1.7%, then why should I spend the extra effort to show that it is really closer to 1.6%?

 

[Crohnos01]

All kidding aside, I would approach it exactly the same way as Charles just described.

 

[ggunn]

My approach to the problem would be to perform the calculation by first assuming that the smaller wire was run for the entire distance. If the results were satisfactory, that would end the calculation process. But if that approach gave a total VD that was higher than I wanted, then I would refine the calculation, using the method described by the others. What I mean is that if I can show that the VD is no more than, say, 1.7%, then why should I spend the extra effort to show that it is really closer to 1.6%?

Yes, but my take on it is that the smaller wire wouldn't give an acceptable Vd, else why have the bigger wire in the middle section at all? You can lump the resistance and calculate a single Vd, or you can calculate Vd for the three runs separately and add them, and you'll get the same number either way, and either way it's a very simple calculation. Spend the extra 5 minutes (or less) and get it right.

 

[iwire]

During an office building remodel they eliminated a standard range from the employee break room and added a micro and a 120 volt water heater.

I reused the 6/3 cable that had been the range circuit to supply the micro with a 20 amp circuit and the water heater with a 30 amp circuit. I had 10 and 12AWG on each end of the 6 AWG.

 

[GoldDigger]

During an office building remodel they eliminated a standard range from the employee break room and added a micro and a 120 volt water heater.

I reused the 6/3 cable that had been the range circuit to supply the micro with a 20 amp circuit and the water heater with a 30 amp circuit. I had 10 and 12AWG on each end of the 6 AWG.

That particular use does not seem to be a case of oversizing conductors to reduce voltage drop, but the #6 is definitely oversized for the circuit load.
My provocative question is that although the EGC in the #6 is proportionally oversized with the ungrounded conductors, does the provision requiring scaling up the EGC also apply to the EGC in that portion of the circuit which is served by the 20A and 30A breakers and uses smaller wire to connect to the #6??
Or, if you are lucky you have a raceway EGC and it does not matter.

Logically, I would have to say no, but what does the letter of the code say to you about it?

 

[kwired]

That particular use does not seem to be a case of oversizing conductors to reduce voltage drop, but the #6 is definitely oversized for the circuit load.
My provocative question is that although the EGC in the #6 is proportionally oversized with the ungrounded conductors, does the provision requiring scaling up the EGC also apply to the EGC in that portion of the circuit which is served by the 20A and 30A breakers and uses smaller wire to connect to the #6??
Or, if you are lucky you have a raceway EGC and it does not matter.

Logically, I would have to say no, but what does the letter of the code say to you about it?

Ever since they emphasised that change in ungrounded conductor size needs a proportional change in grounding conductor size, I have disagreed it was always necessary, but as is worded that 6 AWG ungrounded on a 20 amp circuit needs complimented with a 6 AWG EGC, no matter what other conditions may be. Exception would be a raceway or cable sheath that qualifies as EGC.