#### Clint B

##### New member

- Location
- Tomball,tx

- Thread starter Clint B
- Start date

- Location
- Tomball,tx

- Location
- Chapel Hill, NC

- Occupation
- Electrical Contractor

Also look at 250.122(B) for sizing the egc

- Location
- NE Nebraska

Your short piece of #12 likely has less resistance than the long run of larger conductor and will have very little voltage drop across it in comparison to the long larger piece.

- Location
- Chapel Hill, NC

- Occupation
- Electrical Contractor

There are many online VD calculators online. Here is one https://www.calculator.net/voltage-drop-calculator.html

- Occupation
- Electrical Contractor

- Location
- The Motor City, Michigan USA

"Transferable" is the wrong word and might be causing confusion.

What you have here is a series circuit with six elements: Source, load, two long & fat conductors, and two short & thin conductors.

The total resistance can be calculated by calculating the resistance of each element, then adding them.

The current can be calculated by dividing the supply voltage by the total resistance.

(which is the same everywhere because there's only one current path)

The voltage drop across each element can be calculated by multiplying the current by each individual resistance.

The total voltage drop can be calculated by adding the voltage drop across the four individual conductors.

If the load has a large reactive (inductive or capacitive) component, these calculations will be somewhat inaccurate, but probably close enough.

The "thin" conductors have sufficient ampacity to carry the load current. Regardless of how long or short they are, or what else is connected to them, the breaker will open before they will overheat.

What you have here is a series circuit with six elements: Source, load, two long & fat conductors, and two short & thin conductors.

The total resistance can be calculated by calculating the resistance of each element, then adding them.

The current can be calculated by dividing the supply voltage by the total resistance.

(which is the same everywhere because there's only one current path)

The voltage drop across each element can be calculated by multiplying the current by each individual resistance.

The total voltage drop can be calculated by adding the voltage drop across the four individual conductors.

If the load has a large reactive (inductive or capacitive) component, these calculations will be somewhat inaccurate, but probably close enough.

The "thin" conductors have sufficient ampacity to carry the load current. Regardless of how long or short they are, or what else is connected to them, the breaker will open before they will overheat.

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- Location
- Seattle, WA

With all due respect to you fellow master electrician, this is utter nonsense. It is not resistance that creates a risk of a wire burning up. It is the heat generated by current flowing through the wire. That heat is generated at a rate that depends on resistance and current. You have 15 amps of current flowing through a wire (#12 AWG) that is rated to handle 20 amps (more, actually, but we are required to protect it at 20 amps). There is no way your current will result in overheating the wire.He says you cannot do this because the resistance would burn up the smaller conductor.

Welcome to the forum.

- Location
- Richmond, Virginia

I explained that, if testing line-to-line, which causes the greatest tester current, would not damage the tester, then testing in series with the load would not either. It dawned on him when I explained the difference between a voltage tester and a current tester.

In this discussion, having a wire with lower resistance does not cause the line current to somehow rise above that of the load. It merely reduces the loss of current which would otherwise be caused by the increased total circuit resistance of using a smaller wire.

- Location
- Ann Arbor, Michigan

Others have answered the question for you.

Basically one wants to prevent overheating the insulation. For each incremental length of wire, for example 1" or 1 ft, there is an amount of power dissipated in the wire. Temperature rise in the wire and thus the insulation adjacent to the wire is proportional to the power dissipated in that increment of wire. So the resistance of said length times I^2 thru that wire segment determines the power dissipated in that unit length. For a given fixed current thru the wire the temperature rise will be less for a larger wire because the resistance of the wire decreases as the diameter is increased.

As an approximation wire resistance changes by a factor of about 2 for a change of 3 wire sizes. #10 copper is about 1 ohm per 1000 ft. #12 about 1.5 ohms, and #14 about 2.5 ohms. #13 about 2 ohms. See Cirris https://www.cirris.com/learning-center/calculators/133-wire-resistance-calculator-table

At 20 A #12 copper dissipates about 400*1.59*1/1000 = 0.64 W/foot. Look what happens when you go to 30 A in that same wire, 1.43 W/foot. Note 1.5*1/5 = 2.25 and that is why power goes up so much.

.

- Location
- Placerville, CA, USA

- Location
- Vancouver, WA

Upsize EGC

I agree that upsizing to a larger conductor for the 1000ft run to lower the voltage drop is not a problem and solves the issue. However, NEC requires that the EGC be upsized if the conductors are upsized to reduce voltage drop. So, if you run #12 AWG for hot, neutral and ground then switch to 1/0 for the 1000ft, then the hot, neutral and ground all need to be upsized to 1/0.

For a 1000ft run, you are better off stepping up to 600V then back down again. But, that is not the question here.

- Location
- chicago, il, USA

You also are allowed to do that splice in the panel itself, you don't need to add a gutter.

- Location
- NE Nebraska

Unless there isn't sufficient space to do so. Small 100 amp panel and 350 kcmil conductors because of voltage drop - may not be even be deep enough panel for the required raceway to enter the cabinet let alone any room to do anything with those conductors if you do get them in there.You also are allowed to do that splice in the panel itself, you don't need to add a gutter.

- Location
- Austin, TX, USA

- Occupation
- Electrical Engineer - Photovoltaic Systems

Not necessarily. The ground needs to be upsized proportionately from its minimum size as dictated by code, not necessarily from what it is in the small wire run.I agree that upsizing to a larger conductor for the 1000ft run to lower the voltage drop is not a problem and solves the issue. However, NEC requires that the EGC be upsized if the conductors are upsized to reduce voltage drop. So, if you run #12 AWG for hot, neutral and ground then switch to 1/0 for the 1000ft, then the hot, neutral and ground all need to be upsized to 1/0.

Additive or cumulative might be better words for what amounts to a series circuit?"Transferable" is the wrong word and might be causing confusion..

- Location
- NE Nebraska

Thing is normally a 20 amp circuit uses 12 AWG circuit conductors and requires 12 AWG EGC. Since they are both already same size, an increase because of voltage drop means both are still same size afterwards if they were increased the same proportion. Only when the EGC was smaller to begin with will result in still having smaller EGC, but still proportionally larger.Not necessarily. The ground needs to be upsized proportionately from its minimum size as dictated by code, not necessarily from what it is in the small wire run.

- Location
- Ann Arbor, Michigan

Look at the problem from a circuit perspective.

Assume a breaker or fuse opens within 1/60 second at 10 times its rating. Just to make things simple. Also assume our criteria is circuit interruption within 1/60 second with a dead short circuit at the circuit end.

Consider a #12 copper circuit 1000 ft long with #12 copper for the EGC protected at 20 A. Loop resistance is about 3 ohms. Short circuit current at 120 V input is 40 A. Does not meet our criteria of 200 A for instantaneous trip. So independent of voltage drop we can not use #12.

Next assume hot and neutral are increased to #6. A six number size change or about 1/4 the resistance, 1/2*1/2. So 1.5/4 makes the hot about 3/8 ohm. Loop resistance for a dead short at end of hot to EGC is 0.375 + 1.5 = 1.875, or a 120 V current of 64 A. Does not meet our criteria.

Next make the EGC equal to the hot, and loop resistance becomes 0.375 + 0.375 = 0.75 ohms. Now 120 V short circuit current is 120 / (3/4) = 160 A. This still does not meet our criteria, but it is a lot closer.

To meet our assumed criteria relative to short circuit current our wire size for all of hot, neutral (common or grounded is a better term), and EGC needs to be larger. Note, this is independent of needs for voltage drop.

I have not rechecked my calculations, but the theory illustrates the point.

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- Location
- Austin, TX, USA

- Occupation
- Electrical Engineer - Photovoltaic Systems

Of course. I was speaking to the general case, i.e., if the CCC's and EGC are all the same size, increasing the size of the CCC's for Vd does not necessarily mean you have to increase the size of the EGC. The EGC may already be oversized.Thing is normally a 20 amp circuit uses 12 AWG circuit conductors and requires 12 AWG EGC. Since they are both already same size, an increase because of voltage drop means both are still same size afterwards if they were increased the same proportion. Only when the EGC was smaller to begin with will result in still having smaller EGC, but still proportionally larger.

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As the current for a short in #12 copper 1000 ft long circuit is only 40A, it may be considered a overload current instead of short circuit current for the thermal-magnetic CB and the tripping time may be found from the overload characteristic curve of the CB accordingly.

Look at the problem from a circuit perspective.

Assume a breaker or fuse opens within 1/60 second at 10 times its rating. Just to make things simple. Also assume our criteria is circuit interruption within 1/60 second with a dead short circuit at the circuit end.

Consider a #12 copper circuit 1000 ft long with #12 copper for the EGC protected at 20 A. Loop resistance is about 3 ohms. Short circuit current at 120 V input is 40 A.Does not meet our criteria of 200 A for instantaneous trip. So independent of voltage drop we can not use #12.

Next assume hot and neutral are increased to #6. A six number size change or about 1/4 the resistance, 1/2*1/2. So 1.5/4 makes the hot about 3/8 ohm. Loop resistance for a dead short at end of hot to EGC is 0.375 + 1.5 = 1.875, or a 120 V current of 64 A. Does not meet our criteria.

Next make the EGC equal to the hot, and loop resistance becomes 0.375 + 0.375 = 0.75 ohms. Now 120 V short circuit current is 120 / (3/4) = 160 A. This still does not meet our criteria, but it is a lot closer.

To meet our assumed criteria relative to short circuit current our wire size for all of hot, neutral (common or grounded is a better term), and EGC needs to be larger. Note, this is independent of needs for voltage drop.

I have not rechecked my calculations, but the theory illustrates the point.

.

- Location
- Ann Arbor, Michigan

Sahib:

Your statement is correct, but did you bother to read my assumptions which were there to define or frame the background for the scope of my discussion?

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