James L
Senior Member
- Location
- Kansas Cty, Mo, USA
- Occupation
- Electrician
Small house that doesn't have rickety dressers, bookcases, cd towers, tv stands and the like in front of every receptacle
Bad Neutral?
- Thread starter Hv&Lv
- Start date
- Status
James L
Senior Member
- Location
- Kansas Cty, Mo, USA
- Occupation
- Electrician
I hope you don't go through all that trouble just to find you've overlooked a shoddy crimp sleeve connection in the vanity light.
oldsparky52
Senior Member
- Location
- Wilmington, NC USA
This doesn't make sense to me. 6 outlets before the hairdryer and only 25' of wire? That doesn't seem possible.
That wasnāt right.. itās about 25ā max to the outlet from the breaker. Probably closer to 75ā of wire.
Iāve now found out from the HO someone else āadded some thingsā under the house...
Guess Iām crawling under the house now.
I hate crawlspaces..
ptonsparky
Tom
- Occupation
- EC - retired
Now that you bring it up, maybe a box or two buried in the attic/crawlspace with a spider web out from them?
ptonsparky
Tom
- Occupation
- EC - retired
Oh, those little details.
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
210415-2029 EDT
Hv&Lv:
You list yourself as an engineer, and thus I would expect much better information on your problem.
I believe you have indicated the circuit is 15 A and on that basis I assume the wire is #14 copper or equivalent resistance per foot. You are indicating a voltage drop of 9 V at 12.4 A at the load. This alone does not tell us what kind of voltage drop occurs at the main panel with a load change of 12.4 A. But we can assume the total source impedance back to somewhere is 9/12.4 = 0.726 ohms. If the voltage back at the main panel has little voltage change for a 12.4 A load change, then most of the impedances is in the circuit from the main.
It appears you have read 123.5 V on the output of the AFCi breaker that feeds the circuit in question. And that the other phase is 124.2 V. I don't know whether your 12.4 A load was on or off at the breaker output when voltage was measured. I would like to know the main panel voltage at the panel input terminals on the phase of interest with and without the 12.4 A load. Also, I want to know the two voltages on the other phase, and their relation to the voltages of the first phase. The voltage on the opposite phase should increase when the voltage on the first phase drops. If the first phase, the loaded phase, drops by 0.5 V, then I might expect the other phase to rise by about 0.25 V.
If there is no great change in the main panel input voltage from the 12.4 A load change, then you are looking for a high resistance in the 12.4 A loading circuit. You first need to make sure there is no appreciable voltage drop across the output of the AFCI breaker associated with the 12.4 A load.
Next you connect a long test lead to the main panel neutral bus bar to what ever locations at which you what to test voltage drop. For checking only the neutral wire voltage drop you might getaway with using the EGC. But to check both the neutral and hot wires, then an extension cord is a good long test lead.
With a knowledge of what happens at the main panel with the 12.4 A load change, then you can go to the various outlets on the circuit and see what both the neutral and hot voltage drops are at the various outlets. The 12.4 A load change should be on the last outlet of the circuit. The various voltage drops should tell you where the problem(s) are.
A single line voltage drop for #14 copper, and 12.4 A is 2.5 ohms for 1000 ft or 9 V at 12.4 A is 1000*0.363/2.5 = 1000*0.145 = 145 ft.
The above should be about correct if I didn't make any mistakes.
If you really have a long cable before the first outlet, and then closely space outlets, and your high resistance is at the first outlet, then the difference between outlets might not be much, but there would be a large drop to the first outlet. You failed to provide any good specifics on your measurements.
.
Hv&Lv:
You list yourself as an engineer, and thus I would expect much better information on your problem.
I believe you have indicated the circuit is 15 A and on that basis I assume the wire is #14 copper or equivalent resistance per foot. You are indicating a voltage drop of 9 V at 12.4 A at the load. This alone does not tell us what kind of voltage drop occurs at the main panel with a load change of 12.4 A. But we can assume the total source impedance back to somewhere is 9/12.4 = 0.726 ohms. If the voltage back at the main panel has little voltage change for a 12.4 A load change, then most of the impedances is in the circuit from the main.
It appears you have read 123.5 V on the output of the AFCi breaker that feeds the circuit in question. And that the other phase is 124.2 V. I don't know whether your 12.4 A load was on or off at the breaker output when voltage was measured. I would like to know the main panel voltage at the panel input terminals on the phase of interest with and without the 12.4 A load. Also, I want to know the two voltages on the other phase, and their relation to the voltages of the first phase. The voltage on the opposite phase should increase when the voltage on the first phase drops. If the first phase, the loaded phase, drops by 0.5 V, then I might expect the other phase to rise by about 0.25 V.
If there is no great change in the main panel input voltage from the 12.4 A load change, then you are looking for a high resistance in the 12.4 A loading circuit. You first need to make sure there is no appreciable voltage drop across the output of the AFCI breaker associated with the 12.4 A load.
Next you connect a long test lead to the main panel neutral bus bar to what ever locations at which you what to test voltage drop. For checking only the neutral wire voltage drop you might getaway with using the EGC. But to check both the neutral and hot wires, then an extension cord is a good long test lead.
With a knowledge of what happens at the main panel with the 12.4 A load change, then you can go to the various outlets on the circuit and see what both the neutral and hot voltage drops are at the various outlets. The 12.4 A load change should be on the last outlet of the circuit. The various voltage drops should tell you where the problem(s) are.
A single line voltage drop for #14 copper, and 12.4 A is 2.5 ohms for 1000 ft or 9 V at 12.4 A is 1000*0.363/2.5 = 1000*0.145 = 145 ft.
The above should be about correct if I didn't make any mistakes.
If you really have a long cable before the first outlet, and then closely space outlets, and your high resistance is at the first outlet, then the difference between outlets might not be much, but there would be a large drop to the first outlet. You failed to provide any good specifics on your measurements.
.
It is 14 guage
.47V drop at the main panel with the dryer running, so yes, I agree most (all) impedances are from main to outlet. Suspect cheap backwired devices here..
The load was on the questioned circuit when measured.
voltage at the main panel is equal on both sides (180Ā° Split phase, not 120Ā° network) at ~124.5V.
The voltage on the opposite phase did not fluctuate ( it was within .25V) with this load as I expected when I was there, and you also expected by your comment above.
There is no great change at the panel and for this circuit my suspicion is backwired devices with the possibility of āsomething elseā as noted in a later post by the HO. Iām investigating that this weekend.
I did that with a #12 thwn wire.
This follows Larryās method with the drop cord also. In the postings I thought there was something else involved with his method which is why I asked for clarification.
Agreed. With above.
This should have been a simple ābad neutralā with the descriptions given by the HO.
After arriving and going through, I find out there may have been the possibility of lightning strike back in December(?), and recently found out there may have been some āother workā done by someone, which is why I guess Iām heading in the crawl space this weekend, as much as I hate it...
Tuesday we cleaned and resqueezed the neutral connections at the tap takeoff pole, re-terminated all connections at the transformer pole and the associated underground service drop all the way to the meterbase to eliminate the POCO as being the problem.
I now have a Guardian recorder in place at the meterbase, and a Dranetz HDPQ recorder set up on the circuit in the house.
A lot of my confusion came from the fact that things are burning up on different circuits all over the house, which as I stated before, led me to believe this was a simple bad neutral at the common points, ie, meterbase combo panel, or POCO transformer.
ptonsparky
Tom
- Occupation
- EC - retired
Ah Gee Whiz a wilikers. Possible lightning damage 4-5 months ago? That should have been, what, the second or third thing out of their mouths after Name, address and phone number?
Agreed...
Thatās where most of the confusion lies.
This was one of those āhey Iāve got this problemā on passing one day last weekend..
Like I said, the description sounded so easy.
Itās growing more and more each time I talk to them..
This is reminding me of Tom Bakers escape room thread...
What clues do I need to find to get out alive?
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
210418-1420 EDT
Following is some real data that provides some information on what you might expect in a residential application. This from my home.
Originally my pole transformer was on a pole about half way along the side of my yard. This is about 100 ft from a pole near the front of my yard near the street. The street pole is where my new pole transformer, 50 kVA, is now located. The original pole transformer was much smaller. There is now 100 ft of wimpy wire from the new transformer to the pole where the original transformer was located. The service drop from the original pole transformer location is about 20 ft down and into the ground about 2.5 ft. Then underground the service runs about 45 ft to the house, about 7 ft from underground to the power company meter. This meter is a single phase electronic bolt on type. From the meter to the main panel main lugs is about 6 to 7 ft of wire.
The main panel is a 200 A SqD unit with a main fuse block. When the fuse is pulled there is about a 6 " air gap between the input service and the load breaker bus bars. When that fuse block is pulled you know you are disconnected from the power company.
At the main panel input lugs and with whatever nominal house load exists the voltage measurements are:
---------- No added load ---- 12 A load on phase 1 ---- Change in voltage from the 12 A load on phase 1
Phase 1 ---- 122.2 V ---------------- 121.4 V -------------------- -0.8 V
Phase 2 ---- 122.4 V ---------------- 122.8 V -------------------- +0.4 V
Note the 2 to 1 ratio for the change in voltage, and the difference in change direction. The only difference in source impedance between neutral, and hots is transformer, and meter. Transformer is big, therefore low impedance, compared to the wires. I suspect meter impedance is small.
Next we move to my work bench. Now we add a breaker at the main panel, fuses are probably unimportant, but I did not measure their drop. The work bench is roughly 50 ft from the main panel, and two different wire sizes exist in the path. There is more than 50 ft of wire. At the bench there are two breakers in series. These breaker voltage drops are why in the following measurements there is more than a 2 to 1 ratio difference.
---------- No added load ---- 12 A load on phase 1 ---- Change in voltage from the 12 A load on phase 1
Phase 1 ---- 121.7 V ---------------- 118.4 V -------------------- -3.3 V
Phase 2 ---- 122.4 V ---------------- 123.7 V -------------------- +1.3 V
Neutral to EGC voltage difference. Note there are other loads on the bench that are on.
Phase 1 ---- 0.375 V ---------------- 1.36 V
-------------------- Does not quite correlate with the 1.3 V above, but keep in mind there are other loads on the bench.
.
Following is some real data that provides some information on what you might expect in a residential application. This from my home.
Originally my pole transformer was on a pole about half way along the side of my yard. This is about 100 ft from a pole near the front of my yard near the street. The street pole is where my new pole transformer, 50 kVA, is now located. The original pole transformer was much smaller. There is now 100 ft of wimpy wire from the new transformer to the pole where the original transformer was located. The service drop from the original pole transformer location is about 20 ft down and into the ground about 2.5 ft. Then underground the service runs about 45 ft to the house, about 7 ft from underground to the power company meter. This meter is a single phase electronic bolt on type. From the meter to the main panel main lugs is about 6 to 7 ft of wire.
The main panel is a 200 A SqD unit with a main fuse block. When the fuse is pulled there is about a 6 " air gap between the input service and the load breaker bus bars. When that fuse block is pulled you know you are disconnected from the power company.
At the main panel input lugs and with whatever nominal house load exists the voltage measurements are:
---------- No added load ---- 12 A load on phase 1 ---- Change in voltage from the 12 A load on phase 1
Phase 1 ---- 122.2 V ---------------- 121.4 V -------------------- -0.8 V
Phase 2 ---- 122.4 V ---------------- 122.8 V -------------------- +0.4 V
Note the 2 to 1 ratio for the change in voltage, and the difference in change direction. The only difference in source impedance between neutral, and hots is transformer, and meter. Transformer is big, therefore low impedance, compared to the wires. I suspect meter impedance is small.
Next we move to my work bench. Now we add a breaker at the main panel, fuses are probably unimportant, but I did not measure their drop. The work bench is roughly 50 ft from the main panel, and two different wire sizes exist in the path. There is more than 50 ft of wire. At the bench there are two breakers in series. These breaker voltage drops are why in the following measurements there is more than a 2 to 1 ratio difference.
---------- No added load ---- 12 A load on phase 1 ---- Change in voltage from the 12 A load on phase 1
Phase 1 ---- 121.7 V ---------------- 118.4 V -------------------- -3.3 V
Phase 2 ---- 122.4 V ---------------- 123.7 V -------------------- +1.3 V
Neutral to EGC voltage difference. Note there are other loads on the bench that are on.
Phase 1 ---- 0.375 V ---------------- 1.36 V
-------------------- Does not quite correlate with the 1.3 V above, but keep in mind there are other loads on the bench.
.
kwired
Electron manager
- Location
- NE Nebraska
- Occupation
- EC
If you have ~10 volts of drop and 10-15 amps of load that is 100 to 150 watts being given up wherever this has gone bad. That is a lot if just one termination and I expect that to not last long at all before the termination burns itself out. Poor (very poor) back stab connections? maybe still looking at 5-10 watts per connection - still enough that it would get pretty hot at the connection(s) I would think. Look how hot a small soldering iron gets that is only 5-10 watts. It is watt density in such a small area that makes it get so hot. 10 watts over 75 feet of conductor isn't near the watt density, but is still a lot for good continuous 14 AWG copper conductor, so something else is a factor here and in series with the current.
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
210418-1921 EDT
To continue what kwired said.
Consider #14 copper wire at 20 C the resistance is 2.5 ohms per 1000 linear feet of a single wire. Thus, for a foot of Romex it is 0.005 ohms for the loop resistance. At 15 A of load current this is 225*0.005 = 1.13 W of power dissipation per foot of Romex cable. The wire will heat some for this current, and thus the power dissipation will be somewhat greater because of the resistance rise from the increase of resistance from the current. A 100 ft cable run would be 113 W. A lot when you view it from this perspective, but not much cable temperature rise.
On the other hand if you take an Ohmite 10 W power resistor at full rated power and 15 A it would be 0.044 ohms and you would burn yourself if you touched it. Its voltage drop would be 0.66 V. The physical size of this resistor is about 1.9 x 0.4 inches. Reduce the physical size for this amount of power dissipation, and you produce much higher temperatures at the location of the power dissipation.
In the problem home of this thread one needs to measure the voltage drop along both the hot and neutral lines to try to identify where larger drops occur than expected. It is not likely that every receptacle has the same voltage drop on both the hot and neutral sides, and for each receptacle.
It might even be useful to use the first receptacle of the run as the reference point for voltage change measurements.
.
To continue what kwired said.
Consider #14 copper wire at 20 C the resistance is 2.5 ohms per 1000 linear feet of a single wire. Thus, for a foot of Romex it is 0.005 ohms for the loop resistance. At 15 A of load current this is 225*0.005 = 1.13 W of power dissipation per foot of Romex cable. The wire will heat some for this current, and thus the power dissipation will be somewhat greater because of the resistance rise from the increase of resistance from the current. A 100 ft cable run would be 113 W. A lot when you view it from this perspective, but not much cable temperature rise.
On the other hand if you take an Ohmite 10 W power resistor at full rated power and 15 A it would be 0.044 ohms and you would burn yourself if you touched it. Its voltage drop would be 0.66 V. The physical size of this resistor is about 1.9 x 0.4 inches. Reduce the physical size for this amount of power dissipation, and you produce much higher temperatures at the location of the power dissipation.
In the problem home of this thread one needs to measure the voltage drop along both the hot and neutral lines to try to identify where larger drops occur than expected. It is not likely that every receptacle has the same voltage drop on both the hot and neutral sides, and for each receptacle.
It might even be useful to use the first receptacle of the run as the reference point for voltage change measurements.
.
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
210418-2132 EDT
It may be worthwhile to change out the first receptacle in the circuit to a high quality unit with good screw clamps to the wires. Then you have a good reference point to make measurements relative to the other receptacle, and it becomes a good reference point to look for problems before this first receptacle.
.
It may be worthwhile to change out the first receptacle in the circuit to a high quality unit with good screw clamps to the wires. Then you have a good reference point to make measurements relative to the other receptacle, and it becomes a good reference point to look for problems before this first receptacle.
.
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
210419-1644 EDT
No one has commented on my 2 to 1 voltage change ratio when loading one phase. Do any of you understand where this comes from, and what assumptions need to be made for it to occur?
.
No one has commented on my 2 to 1 voltage change ratio when loading one phase. Do any of you understand where this comes from, and what assumptions need to be made for it to occur?
.
ptonsparky
Tom
- Occupation
- EC - retired
Hv&Lv, we are waiting for that report back.
Hope you made it out of that crawl space. Nasty place to be when looking for electrical problems.
Hope you made it out of that crawl space. Nasty place to be when looking for electrical problems.
oldsparky52
Senior Member
- Location
- Wilmington, NC USA
I'll play.
L1 drops 0.4 V in relation to L2 or N. N drops 0.4 V in relation to L1 which moves it closer to L1 and farther away from L2 by 0.4V. Since L1 and N are each losing 0.4V that means they are moving closer together so you add the two 0.4's and have a total drop of 0.8V. This looks like an add to L2 because when the N moves towards L1 by 0.4 V it's moving away from L2 by 0.4 V.
This only works this way on single phase.
---------- No added load ---- 12 A load on phase 1 ---- Change in voltage from the 12 A load on phase 1
Phase 1 ---- 122.2 V ---------------- 121.4 V -------------------- -0.8 V
Phase 2 ---- 122.4 V ---------------- 122.8 V -------------------- +0.4 V
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
210420-1221 EDT
oldsparky52:
Your response implies to me that you generally understand what is going on. But I do think that you need two different Ns in your description. That there have been no other responses somewhat implies most electricians do not understand certain aspects circuit analysis.
Fundamental to circuit analysis is that the sum of the instantaneous voltages around a closed loop is zero, and the sum of the instantaneous currents at a point is zero. These are known as Kirchhoff's laws. See
en.wikipedia.org
If you move from the end of a circuit, as mentioned above, to the transformer terminals, then the results are somewhat different.
.
oldsparky52:
Your response implies to me that you generally understand what is going on. But I do think that you need two different Ns in your description. That there have been no other responses somewhat implies most electricians do not understand certain aspects circuit analysis.
Fundamental to circuit analysis is that the sum of the instantaneous voltages around a closed loop is zero, and the sum of the instantaneous currents at a point is zero. These are known as Kirchhoff's laws. See
Kirchhoff's circuit laws - Wikipedia
en.wikipedia.org
If you move from the end of a circuit, as mentioned above, to the transformer terminals, then the results are somewhat different.
.
- Status