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Loose neutral

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I run across a lot of loose neutrals some on the utility side some and just loose in the panel. I know from experience what the problem is but I wish I had a tool that would show it. I know Georgia power uses a St 800 for the service coming in.
Is there any tool we can use inside the house to show there's a loose neutral? Most of the time there's no problem while we are there ,it reads great while we're there it only comes and goes while we're not there.
 

Joethemechanic

Senior Member
Location
Hazleton Pa
Occupation
Electro-Mechanical Technician. Industrial machinery
You need a little load bank, Maybe something like a hair dryer with alligator clips instead of a plug, If you need to pull a heavier load than that you could make up something with burners from an electric cooktop for load.
 

don_resqcapt19

Moderator
Staff member
Location
Illinois
Occupation
retired electrician
A hair dryer or small heater will work for this. You first measure the line to line and line to neutral voltages with no load. Then put you load on one leg and read the voltages again. Next you put the load on the other leg and repeat the voltage readings. A neutral problem on the line side of where you have connected your loads will result in differing line to neutral voltages.
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
230705-0841 EDT

shawelectric:

A single phase system feeding a transformer with a center tapped secondary generates two voltage that are phased 180 degrees apart. Both voltages will be approximately equal in voltage, and each 120 V loop will have about equal impedances for their source as seen at the main panel. This usually means that a typical system will have a neutral impedance equal to its hot line impedance. In my home this is about 1/20 ohm in the neutral, and 1/20 ohm in each of my hot supply lines.

If I place a 10 amp load on one phase to neutral.I get a total voltage change in that loop of about 1 V. There is no current change in the unloaded loop. Because most of my voltage drop is in power wiring rather than in the transformer this means I see about a 1 V drop in the loaded voltage side. This makes my unloaded side increase in voltage by about 1/2 V.

I can also do a 240 V measurement that excludes any neutral change. Here I should see abot a 1/2 V change.in the 240 V.readings.

With this expected information if the neutral voltage changes by more than about the expected 1/2 change, then I expect a high resistance in the neutral

.
 

Strathead

Senior Member
Location
Ocala, Florida, USA
Occupation
Electrician/Estimator/Project Manager/Superintendent
A hair dryer or small heater will work for this. You first measure the line to line and line to neutral voltages with no load. Then put you load on one leg and read the voltages again. Next you put the load on the other leg and repeat the voltage readings. A neutral problem on the line side of where you have connected your loads will result in differing line to neutral voltages.
I am thinking two loads, especially two similar loads. Put them on A to N and B to N. A quick read (change) of voltages before and after will quickly tell you if you have a loose neutral. This just makes it more obvious that the other methods discussed.
 

don_resqcapt19

Moderator
Staff member
Location
Illinois
Occupation
retired electrician
I am thinking two loads, especially two similar loads. Put them on A to N and B to N. A quick read (change) of voltages before and after will quickly tell you if you have a loose neutral. This just makes it more obvious that the other methods discussed.
You don't want the line to neutral loads on at the same time for this type of testing. You test with the load A to N, record the voltages and then test again with the load B to N. Equal loads on both A and B will not put load on the neutral and will not tell you anything about a neutral issue.
 

Strathead

Senior Member
Location
Ocala, Florida, USA
Occupation
Electrician/Estimator/Project Manager/Superintendent
You don't want the line to neutral loads on at the same time for this type of testing. You test with the load A to N, record the voltages and then test again with the load B to N. Equal loads on both A and B will not put load on the neutral and will not tell you anything about a neutral issue.
You are thinking incorrectly Don. If I put a load on A to N and B to N, and I have a loose neutral, I will read high voltage on one phase and low voltage on the other phase as I now have a load (resistance on the neutral). As so as I remove one load, the voltage will come back to close to nominal. Much more definitively because if one voltage goes up even a little, you know for a fact you have resistance on your neutral.
 

LarryFine

Master Electrician Electric Contractor Richmond VA
Location
Henrico County, VA
Occupation
Electrical Contractor
You are thinking incorrectly Don. If I put a load on A to N and B to N, and I have a loose neutral, I will read high voltage on one phase and low voltage on the other phase as I now have a load (resistance on the neutral). As so as I remove one load, the voltage will come back to close to nominal. Much more definitively because if one voltage goes up even a little, you know for a fact you have resistance on your neutral.
I have to disagree with you. Matching line-to-neutral loads is seen by the system as a single line-to-line load, and will have no effect on neutral current.

One line-to-neutral load will increase neutral current and expose neutral resistance.
 

don_resqcapt19

Moderator
Staff member
Location
Illinois
Occupation
retired electrician
You are thinking incorrectly Don. If I put a load on A to N and B to N, and I have a loose neutral, I will read high voltage on one phase and low voltage on the other phase as I now have a load (resistance on the neutral). As so as I remove one load, the voltage will come back to close to nominal. Much more definitively because if one voltage goes up even a little, you know for a fact you have resistance on your neutral.
You will only read different voltages on the two legs to neutral if the loads are not equal. Equal loads equals equal resistance and there will be no change on the line to neutral voltages. This is why you can disconnect the neutral where there are equal loads and the two equal loads work perfectly. If the loads are not equal, then you see the problem.

See sections 3.6 and 3.7 of Beast of Burden usage guide. This is a device that some utilities plug into the meter socket to check for problems, including neutral problems.
 

Strathead

Senior Member
Location
Ocala, Florida, USA
Occupation
Electrician/Estimator/Project Manager/Superintendent
You will only read different voltages on the two legs to neutral if the loads are not equal. Equal loads equals equal resistance and there will be no change on the line to neutral voltages. This is why you can disconnect the neutral where there are equal loads and the two equal loads work perfectly. If the loads are not equal, then you see the problem.

See sections 3.6 and 3.7 of Beast of Burden usage guide. This is a device that some utilities plug into the meter socket to check for problems, including neutral problems.
That is right, my thinking was bad in that way as well. But the concept of putting loads on both A and B to see the imbalance is valid. They definitely need to be different resistances though.
 

Strathead

Senior Member
Location
Ocala, Florida, USA
Occupation
Electrician/Estimator/Project Manager/Superintendent
I have to disagree with you. Matching line-to-neutral loads is seen by the system as a single line-to-line load, and will have no effect on neutral current.

One line-to-neutral load will increase neutral current and expose neutral resistance.
Unless I miss what you are saying, that would be when there are other loads on the system. My read from the OP is to test for this before putting the system in to operation, so there wouldn't be any other loads. If there are already loads on the system, simply read the voltage with loads, then turn off all loads. If the voltage goes down on either phase then you have a problem.
 

OldBroadcastTech

Senior Member
Location
Western IL
Occupation
Retired Broadcast Technician
230705-0841 EDT

shawelectric:

A single phase system feeding a transformer with a center tapped secondary generates two voltage that are phased 180 degrees apart. Both voltages will be approximately equal in voltage, and each 120 V loop will have about equal impedances for their source as seen at the main panel. This usually means that a typical system will have a neutral impedance equal to its hot line impedance. In my home this is about 1/20 ohm in the neutral, and 1/20 ohm in each of my hot supply lines.

If I place a 10 amp load on one phase to neutral.I get a total voltage change in that loop of about 1 V. There is no current change in the unloaded loop. Because most of my voltage drop is in power wiring rather than in the transformer this means I see about a 1 V drop in the loaded voltage side. This makes my unloaded side increase in voltage by about 1/2 V.

I can also do a 240 V measurement that excludes any neutral change. Here I should see abot a 1/2 V change.in the 240 V.readings.

With this expected information if the neutral voltage changes by more than about the expected 1/2 change, then I expect a high resistance in the neutral

.
"A single phase system feeding a transformer with a center tapped secondary generates two voltage that are phased 180 degrees apart."

Sorry, have to disagree. The two voltages (ends of the secondary, lets call them L-1 and L-2) are of opposite polarity. Which in a perfectly balanced secondary winding, is indistinguishable from a 180 degree phase shift.

(Best Twilight Zone voice) 'Imagine if you will........... a wave-form to the input of the power-company transformer consisting of a positive-going (half) sine wave followed by a negative-going square wave. A 180-degree phase shift would have the L-1 terminal of the secondary showing a positive-going (half) sine followed by a negative-going square wave; the L-2 terminal would have a negative-going square followed by a positive-going (half) sine wave.'

Again, with 60-Hz power waveforms, '180 degrees out of phase' and 'opposite polarities' are indistinguishable.
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
230707-1925 EDT

OldBroadcastTech:

If you have a center tapped secondary on a transformer with a single primary, then the instantaneous output of an ideal transformer on one side of the secondary center tap is of reverse polarity of the other half of the secondary winding. Move the center tap to some non-center point and you have two waveforms that are both replicas of the primary except that magnitudes may be different, and one is the inverse of the other,or a 180 degree phase shift. Note: phase shift does not mean there is or isn't a time delay between waveforms. Phase shift is simplly a measure of a difference in two or more waveforms.

.
 

don_resqcapt19

Moderator
Staff member
Location
Illinois
Occupation
retired electrician
There only appears to be a 180 degree phase shift because of the reference points when we take a voltage measurement. It is absolutely impossible for the output of a single winding to be out of phase with itself. The addition of a center tap does not change this.
Look at two batteries in series. If you measure from left to center and then from center to right keeping the center lead in place, you will show the opposite polarity. This is exactly the same as what we see with a center tapped transformer secondary.
 

mtnelect

HVAC & Electrical Contractor
Location
Southern California
Occupation
Contractor, C10 & C20 - Semi Retired
The United States of America is planning to go "All Electric". The weakness of our grid system will eventually become fully apparent.
 
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