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Little Experiment

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ptonsparky

ptonsparky

Tom
Location
NE (9.06 miles @5.9 Degrees from Winged Horses)
Occupation
EC - retired
and I use the term loosely.
Sheetrock was hung & taped in my OF garage last week. I'm all electric so I had three 1500-watt 120v heaters plus 4000 and 4800 watt 240-volt heaters to dry the mud.

I purposely turned all on to confirm VD calculations. Almost exact.
I loaded up one circuit with all three 1500 watt heaters. (34.5 amps trips a 20 amp Homeline in about 30 seconds).

I have a Ufer plus two rods. Current on the GEC was 1.7 amps for the above load.

What is the resistance my GES?
 
LarryFine

LarryFine

Master Electrician Electric Contractor Richmond VA
Location
Henrico County, VA
Occupation
Electrical Contractor
ptonsparky said:
I loaded up one circuit with all three 1500 watt heaters. (34.5 amps trips a 20 amp Homeline in about 30 seconds).
Click to expand...
I think many of us would like to know how long it would hold with two heaters at 23 amps.

ptonsparky said:
Current on the GEC was 1.7 amps for the above load.

What is the resistance my GES?
Click to expand...
I think we'd need more info. Why is there any current on it?
 
A

AC\DC

Senior Member
Location
Florence,Oregon,Lane
Occupation
EC
LarryFine said:
I think many of us would like to know how long it would hold with two heaters at 23 amps.


I think we'd need more info. Why is there any current on it?
Click to expand...
would not the fact its in parallel with the neutral at the service disconnect, allow some current even if its in milliamps normally.
 
Dsg319

Dsg319

Senior Member
Location
West Virginia
Occupation
Wv Master “lectrician”
ptonsparky said:
and I use the term loosely.
Sheetrock was hung & taped in my OF garage last week. I'm all electric so I had three 1500-watt 120v heaters plus 4000 and 4800 watt 240-volt heaters to dry the mud.

I purposely turned all on to confirm VD calculations. Almost exact.
I loaded up one circuit with all three 1500 watt heaters. (34.5 amps trips a 20 amp Homeline in about 30 seconds).

I have a Ufer plus two rods. Current on the GEC was 1.7 amps for the above load.

What is the resistance my GES?
Click to expand...
Just take off your GEC and connect 120volts and measure with an amp clamp than E/I=R.

:p. I say this because I don’t know how to go about figuring it the way you are asking.
 
H

Hv&Lv

Senior Member
Location
-
Occupation
Engineer/Technician
ptonsparky said:
This is a B question. IDK the answer. Voltage applied is 118. Current is 1.7. is it that simple?
Click to expand...
No.
current is 1.7
All that means is there is 1.7A on there, but you don’t know what was on other wires.
It could be that simple if ALL the current was going through it, and not sharing another path.
 
Another C10

Another C10

Electrical Contractor 1987 - present
Location
Southern Cal
Occupation
Electrician NEC 2020
LarryFine said:
I think we'd need more info. Why is there any current on it?
Click to expand...


Why is there any current on it? ..

That would be my concern, current on the grounding electrode conductor. nothing should be on that reference unless there's a problem.
 
W

wwhitney

Senior Member
Location
Berkeley, CA
Occupation
Retired
ptonsparky said:
Total current on the hot was 34.5. 1.7 is on the GEC. 32.8 on the neutral by subtraction.
FWIW, VD was 123.8 to 118.3 or 5.5 volts.
. . .
1/0 1/0 2 Al 300'. Buried in PVC.
Click to expand...
OK, if I understand your description properly, the transformer to load circuit looks like this: an ungrounded 1/0 Al conductor, the load, and a #2 Al neutral in parallel with the path utility GES - earth - service GES. So you won't be able to determine your service GES resistance, just the sum of the utility GES and service GES resistances.

Call the ungrounded conductor resistance R1, the neutral resistance R2, and the sum of the GES resistances R3. #1/0 has (almost) twice the area of #2, so we should have R2 = 2 * R1. As R2 and R3 are parallel paths, we should have 32.8 * R2 = 1.7 * R3, or R3 = 19.3 R2. Then the parallel return paths have a resistance of RN = R2 * (1/(1+1/19.3)) = 0.95 R2 = 1.9 R1.

Thus the total series resistance causing the 5.5V drop for 34.5A is 2.9 R1. That means R1 = 55 mOhms, R2 = 110 mOhms, and R3 = 2.1 Ohms. That resistance R3 seems implausibly small, so maybe there's a mistake in my calculations, or an unjustified assumption in the model? The model does ignore any impedance of the transformer.

Cheers, Wayne
 
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