# Is resistance transferable ?

#### Clint B

##### New member
I am having a debate with another master on a tricky subject . We have 1000 foot of conductor with a 15 amp load at the end. The D rating word meaning that the conductor need it would be too large to fit under a standard breaker . So my thoughts are that you could use a standard number 12 conductor out of the breaker and Tie in to the larger conductor through a gutter. He says you cannot do this because the resistance would burn up the smaller conductor . My thoughts are that resistance is only calculated through the length of wire and is only impartrd as current travels through the conductor . For example if you cross a stream of water From side to side that is only 2 foot wide there would be very little resistance. If you tried to cross the same stream perpendicular for a mile the resistance would be great . I know it is alternating current but . Would not the increased conductor size relieve the resistance as the current is traveling through that conductor and then the smaller conductor only carries their resistance for 2 feet ? To look at it another way resistance is only being applied to the section of wire that the current is passing through and the resistance does not pass to the smaller wire . In addition the larger conductor would relieve the resistance end it would not build beyond the smaller conductors capabilities ? Help please.

#### Dennis Alwon

##### Moderator
Staff member
You are correct. Resistance is going to be low thru the run if you upsize the conductor so adding a short piece of #12 at the end will not cause any issues if the load is 20 amps or less.

Also look at 250.122(B) for sizing the egc

#### kwired

##### Electron manager
Your two conductors are connected in series.

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.

#### Dennis Alwon

##### Moderator
Staff member
Think of it this way-- do a VD calculation of the 15 amp load at 999 feet and then do the same calculation for the 1' of #12 wire. Add them together and the #12 will add virtually no VD to the answer.

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

#### ptonsparky

##### Senior Member
Change in conductor size is done all the time. The only time there is a problem is when the current exceeds that of the smaller. (PP workmanship not included.)

#### drcampbell

##### Senior Member
"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.

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#### charlie b

##### Moderator
Staff member
He says you cannot do this because the resistance would burn up the smaller conductor.
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.

Welcome to the forum.

#### LarryFine

##### Master Electrician Electric Contractor Richmond VA
I had a slightly similar discussion here a few years ago, when I mentioned using my solenoid tester to find an open fuse by testing from line to load. Someone was aghast because he thought the load would force its full current through the tester and destroy it.

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.

#### gar

##### Senior Member
Your "another master" totally lacks an understand of electrical circuit theory and thermal theory. I would suggest he should not be a "master".

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.

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#### GoldDigger

##### Moderator
Staff member
The only thing of even potential concern is that you now have two terminations (at the breaker and at the joint) very close to each other on the same short length of wire. Localized heating which is in excess of what is expected and allowed (workmanship, design, etc.) at both of these points can result in a higher wire temperature even though the individual substandard terminations might not cause a problem by themselves.

#### beanland

##### Senior Member
Upsize EGC

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.

#### rlundsrud

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

#### kwired

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

#### ggunn

##### PE (Electrical), NABCEP certified
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.
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.

#### Besoeker3

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

#### kwired

##### Electron manager
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.
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.

#### gar

##### Senior Member
190518-1626 EDT

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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#### ggunn

##### PE (Electrical), NABCEP certified
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.
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.

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#### Sahib

##### Senior Member
190518-1626 EDT

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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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.

#### gar

##### Senior Member
190520-0739 EDT

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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#### Sahib

##### Senior Member
190520-0739 EDT

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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Sorry, gar. Your assumption is flawed: a short circuit assumes negligible resistance and 3 ohms is not negligible.

#### gar

##### Senior Member
190520-0948 EDT

Sahib:

The 3 ohms was the resistance of the 1000 ft of #12 copper branch circuit wire. A dead short was placed at the end of the bramch circuit producing a 40 A load thru the circuit.

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#### kwired

##### Electron manager
Now getting back to OP's application and questioning of a short length of 12 AWG at each end for termination, it is presumed the increased conductor size over the bulk of the run is low enough impedance it isn't much impact on magnetic trip functioning, and the short length of 12 AWG just to help facilitate connections, is pretty much a negligible resistance in this circuit.

#### Sahib

##### Senior Member
Now getting back to OP's application and questioning of a short length of 12 AWG at each end for termination, it is presumed the increased conductor size over the bulk of the run is low enough impedance it isn't much impact on magnetic trip functioning, and the short length of 12 AWG just to help facilitate connections, is pretty much a negligible resistance in this circuit.
For long circuits such as 1000ft as stated by gar only thermal trip ie overload element of CB would operate and not the magnetic element of the CB for a short at the end of the circuit due to considerable resitance/impedance in the circuit.

#### Sahib

##### Senior Member
190520-0948 EDT

Sahib:

The 3 ohms was the resistance of the 1000 ft of #12 copper branch circuit wire. A dead short was placed at the end of the bramch circuit producing a 40 A load thru the circuit.

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That 40A current is not short circuit current (even though it results from a short) but overload current due to considerable resistance in the circuit and so your conclusion that #12 copper wire is not suitable may not be correct since the thermal element of the CB would still operate and disconnect the circuit without any damage.

#### kwired

##### Electron manager
For long circuits such as 1000ft as stated by gar only thermal trip ie overload element of CB would operate and not the magnetic element of the CB for a short at the end of the circuit due to considerable resitance/impedance in the circuit.
Yes even with larger conductors to reduce voltage drop, there still can be enough resistance in entire run to limit fault current to a level below magnetic trip level of the breaker. This sort of can lead us into AIC rating discussions and incident energy discussions for arc flash issues. Conductors do have resistance which is current limiting.

#### LarryFine

##### Master Electrician Electric Contractor Richmond VA
The previous posts illustrate why the suggestion to use a higher voltage pops up so often. :happyyes:

#### gar

##### Senior Member
190520-2114 EDT

Sahib:

In my post #17 I made specific assumptions, and I stated them. This was to try to avoid all sorts of if, and, and but stuff.

I will further clarify that a real world voltage source consists of an ideal voltage source with a series internal impedance. Further assume linear components.

I consider a source to be where you connect a load. If I talk about a dead short somewhere, then that is some impedance that is very small compared to its source, and it is the load. For the problem I was illustrating the wiring to the load was part of the source, and not part of the load. Roughly speaking a dead short will have near zero voltage drop across it with current flow.

You can define a load at any two terminal point you want in a circuit. By virtue of my use of the term "dead short circuit" I defined where those two terminals were.

The purpose of my post was to try to define a simple circuit that would illustrate the problem of not using an EGC of adequate size. I used the trip time of 1/60 as a way of defining a specific trip characteristic.

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#### Sahib

##### Senior Member
190520-2114 EDT

Sahib:

In my post #17 I made specific assumptions, and I stated them. This was to try to avoid all sorts of if, and, and but stuff.

I will further clarify that a real world voltage source consists of an ideal voltage source with a series internal impedance. Further assume linear components.

I consider a source to be where you connect a load. If I talk about a dead short somewhere, then that is some impedance that is very small compared to its source, and it is the load. For the problem I was illustrating the wiring to the load was part of the source, and not part of the load. Roughly speaking a dead short will have near zero voltage drop across it with current flow.

You can define a load at any two terminal point you want in a circuit. By virtue of my use of the term "dead short circuit" I defined where those two terminals were.

The purpose of my post was to try to define a simple circuit that would illustrate the problem of not using an EGC of adequate size. I used the trip time of 1/60 as a way of defining a specific trip characteristic.

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Thanks for your response, gar. I still have the impression that you considered only magnetic trip only 20 A CB, leaving out thermal-magnetic 20A CB. Supposing a 20 A thermal-magnetic CB is present, how will it change your reasoning in post #17 with regard to the problem of not using an EGC of adequate size ?