Originally posted by gar
View Post
Announcement
Collapse
No announcement yet.
Is resistance transferable ?
Collapse
X


1905202114 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]
Leave a comment:

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

Originally posted by Sahib View PostFor 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.
Leave a comment:

Originally posted by gar View Post1905200948 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.
.
Leave a comment:

Originally posted by kwired View PostNow 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.
Leave a comment:

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.
Leave a comment:

1905200948 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.
.
Leave a comment:

Originally posted by gar View Post1905200739 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?
.
Leave a comment:

1905200739 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?
.
Leave a comment:

Originally posted by gar View Post1905181626 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.
.
Leave a comment:

Originally posted by kwired View PostThing 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.Last edited by ggunn; 051919, 05:24 PM.
Leave a comment:

1905181626 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.
.
Leave a comment:

Originally posted by ggunn View PostNot 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.
Leave a comment:
Leave a comment: