Is resistance transferable ?

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Sahib

Senior Member
Location
India
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
Location
Ann Arbor, Michigan
Occupation
EE
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
Location
NE Nebraska
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
Location
India
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
Location
India
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
Location
NE Nebraska
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
Location
Henrico County, VA
Occupation
Electrical Contractor
The previous posts illustrate why the suggestion to use a higher voltage pops up so often. :happyyes:
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
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.

.

 

Sahib

Senior Member
Location
India
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.

.

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 ?
 
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