Is resistance transferable ?

[Sahib]

190520-2114 EDT

[COLOR=#000000]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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[/COLOR]

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 [COLOR=#000000] post #17 with regard to [/COLOR][COLOR=#000000]the problem of not using an EGC of adequate size ?[/COLOR][COLOR=#000000] [/COLOR]

[gar]

190520-2114 EDT

[COLOR=#000000]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.

.

[/COLOR]

[LarryFine]

The previous posts illustrate why the suggestion to use a higher voltage pops up so often.

[kwired]

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.

[Sahib]

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.

[Sahib]

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.

[kwired]

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.

[gar]

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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[Sahib]

190520-0739 EDT

[COLOR=#000000]Sahib:

[/COLOR]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]

190520-0739 EDT

[COLOR=#000000]Sahib:

[/COLOR]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?

.

[Sahib]

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.

[ggunn]

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.

[gar]

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.

.

[kwired]

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.

[Besoeker3]

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

Additive or cumulative might be better words for what amounts to a series circuit?