Resistive neutral on a 240/120 single phase system

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NSTech

Member
Greetings. Often I end up at a customer site where people and/or animals are experiencing shocks. I will sometimes find an "open" neutral that is the culprit. But even though the neutral might be effectively disconnected at the meter enclosure or transfer switch there is still almost always a connection between the center tap of the utility transformer and the neutral bar of the electrical service panel through ground rods. So here is my question: Lets say there are 120 V loads on Line 1 with a total resistance of 10 ohms and Line 2 is at 30 ohms. With an open neutral the voltage and amperage on this voltage divider would be easy to calculate. BUT here is where I am confused: The center tap of the transformer is connected to the utility ground and their neutral is connected to multiple grounds effectively bringing their neutral on the supply side to near zero ohms. Then lets say the building that is fed with the "open" neutral has a single ground rod of 25 ohms. Now the simple voltage divider becomes more complex because there is a 25 ohm resistance from the neutral bar on the service panel to the center tap of the transformer. Can someone explain to me the formula that will take into account all 3 resistances? Thanks!
 

Smart $

Esteemed Member
Location
Ohio
... Can someone explain to me the formula that will take into account all 3 resistances?
Well, to get a discussion underway... I don't know of any, and it seems nobody else does either.

Curious... Is there some practical purpose you want to know?
 

NSTech

Member
Thanks for the reply. I spend much of my time checking/repairing stray voltage for several utilities. I would like to be able to calculate the possible voltage at the panels and grounding electrodes with an open neutral (open but with earth as a resistor). I think many electricians assume when they find an open neutral at a 3 wire service equipment panel there is absolutely no connection back to the center tap of the voltage source. But usually there is through the existing electrode grounding system which I have found holds the "open" neutral close enough to ground to prevent a fire/shock hazard (not every time). But this resistive connection is also preventing me from easily calculating what the voltage to earth is at the premises grounded equipment.
It is easy enough to calculate the total load resistance on Line 1 and Line 2. I can also measure the ground resistance of the premises using my Fluke 1630 ground tester once I restore the neutral connection to the utility ground. Seems Kirchoff and/or nodal analysis might be needed but it's been a long time since I've done either.
 

mivey

Senior Member
A loop analysis would do it:

For:
E = V_LG where both legs have the same voltage magnitude
Z_1 = Leg #1 impedance
Z_2 = Leg #2 impedance
Z_L1 = Load #1 impedance
Z_L2 = Load #2 impedance
Z_N = Neutral impedance

We get these formulas:

I_1 = (2*E)*(Z_2+Z_L2+2*Z_N)/((Z_1+Z_L1)*(Z_2+Z_L2+2*Z_N)+(Z_2+Z_L2)*(Z_1+Z_L1+2*Z_N))

I_2 = (2*E)*(Z_1+Z_L1+2*Z_N)/((Z_1+Z_L1)*(Z_2+Z_L2+2*Z_N)+(Z_2+Z_L2)*(Z_1+Z_L1+2*Z_N))

V_L1 = I_1 * ZL1

V_L2 = I_2 * ZL2
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
120726-1547 EDT

You have two loops. Write the equation for each loop. Then solve the two equations.

Loop 1.
10*I10 + 25*(I10 - I30) = 120

Loop 2.
30*I30 - 25*(I10 - I30) = 120

Add the two equations and the result is
10*I10 + 30*I30 = 240

From this
I10 = 240/10 - 30*I30/10
I10 = 24 - 3*I30

Next substitute I10 into the loop 2 equation
30*I30 - 25*(24 - 3*I30 - I30) = 120
30*I30 + 25*I30 - 25*24 - 25*3*I30 = 120
30*I30 + 25*I30 + 25*3*I30 = 120 + 600
130*I30 = 720
I30 = 720/130 = 5.538462 A
V30 = 30*5.538462 = 166.154 V

I10 = 24 - 3*5.538462 = 7.384615 A
V10 = 10*7.384615 = 73.846 V

166.154 + 73.846 = 240


I25 = 7.384615 - 5.538462 = 1.84615 A
V25 = 25*1.84615 = 46.15383 V

46.15383 + 73.846 = 120
166.154 - 46.15383 = 120

Check my math.

.
 

Smart $

Esteemed Member
Location
Ohio
... Seems Kirchoff and/or nodal analysis might be needed but it's been a long time since I've done either.
If you have access to the service conductors, putting an amp clamp around all three conductors should indicate current "leaking" through other paths. However, the indicated current may not originate from that service. "Leaking" current could be from one or more [faulting] services returning on the measured service's grounding system.
 
Last edited:

Phil Corso

Senior Member
NSTech...

The assumption that currents are resistive, may not yield a correct solution.

I call this type of problem "Neutral Interruptus!" It would help if you could measure the voltage between the "two" alledged neutrals locations?


Regards, Phil Corso
 

kingpb

Senior Member
Location
SE USA as far as you can go
Occupation
Engineer, Registered
My only comment (to be consistent with my ideology) :

240/120V is not single phase, it is a three phase delta system. In that regard; I think the OP may have intended to say 120/240V.

Carry on...........................:angel:
 

Smart $

Esteemed Member
Location
Ohio
NSTech...

The assumption that currents are resistive, may not yield a correct solution.

I call this type of problem "Neutral Interruptus!" It would help if you could measure the voltage between the "two" alledged neutrals locations?
I agree that a non-unity power factor caused by reactive loads will make the calculation posed inaccurate. Nonetheless, measuring voltage between neutral locations may not provide a direct indication of a resistive connection. The OP knowingly posed a scenario with an imbalance. However, the L-N loads could be well balanced. The only real way to check is to make certain there is a major L-N imbalance and measure to see where the neutral current goes.

Ohh! If one has to create the imbalance, its best to turn off the main, then all L-L and L1 or L2 to Neutral loads, then turn the main back on. This switching method will help to prevent frying any equipment due to overvoltage. Unfortunately, it will not prevent smoking any equipment due to undervoltage, so consider taking further precautions.
 

Phil Corso

Senior Member
NSTech...
'
Using the 3 resistance values you presented, and referring to the "two" groundED conductors as the Suppy Neutral-Conductor and the Load Neutral-Conductor, then the latter will be "elevated" in potential to about 74 Volts, above the former!

Essentially the "ground" path between the two grounding-electrodes is ineffective... presenting a hazard to humans and animals.


Regards, Phil Corso
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
120728-1350 EDT

Phil:

The voltage drop from the transformer center tap to the node point between the three resistors is 46.154 V. That is the voltage drop along the high resistance neutral path.

.
 

Phil Corso

Senior Member
Gar...

I don't want to start a PEing contest, but, here is my solution: Source line-current L1 and source line-current L2 are 7.4A and 5.5A, respectively, and neutral-current is 1.8A!

Phil :D
 

steve066

Senior Member
Greetings. Often I end up at a customer site where people and/or animals are experiencing shocks. I will sometimes find an "open" neutral that is the culprit. But even though the neutral might be effectively disconnected at the meter enclosure or transfer switch there is still almost always a connection between the center tap of the utility transformer and the neutral bar of the electrical service panel through ground rods. So here is my question: Lets say there are 120 V loads on Line 1 with a total resistance of 10 ohms and Line 2 is at 30 ohms. With an open neutral the voltage and amperage on this voltage divider would be easy to calculate. BUT here is where I am confused: The center tap of the transformer is connected to the utility ground and their neutral is connected to multiple grounds effectively bringing their neutral on the supply side to near zero ohms. Then lets say the building that is fed with the "open" neutral has a single ground rod of 25 ohms. Now the simple voltage divider becomes more complex because there is a 25 ohm resistance from the neutral bar on the service panel to the center tap of the transformer. Can someone explain to me the formula that will take into account all 3 resistances? Thanks!

Looking at the original question, althogh the utility transformer may still be grounded, this has no effect on the voltage divider that would happen if the neutral to a load or receptacle is open.

Think of Kirchoffs current law - current must flow in a complete loop. There is a connection to ground at the transformer, but if any current flows to ground, it must return back to the transformer through a load.

In a properly connected wiring system, there is no other way for this current to get back to the transformer. Therefore, no current flows to ground, and therefore, the ground connection has no effect on the voltages measured across the loads.

So to answer your original question, you calculate the voltages the same way you showed. No need to consider the neutral.

Steve
 

Smart $

Esteemed Member
Location
Ohio
... Therefore, no current flows to ground, and therefore, the ground connection has no effect on the voltages measured across the loads.

...
From a real world perspective, current can and does flow through the grounding system... even on properly connected wiring systems. Current takes all paths, not just the one least impeding. Current on those paths is inversely proportional to each path's impedance.
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
120729-1411 EDT

NSTech:

Your original question provides an interesting exercise for one in circuit analysis, but not much use in practical troubleshooting other than providing an understanding of what happens in the circuit.

The resistance of a ground rod or grounding electrode of some form can range all over the map from a fractional part of an ohm to hundreds of ohms. This will be an unknown to you.

You can estimate the neutral path resistance (includes both the neutral wire and ground resistance) by a simple measurement.

First, monitor simultaneously the voltages from both legs to neutral. Two 25 or 15 W incandescent light bulbs will work. Use the same wattage with one bulb on one phase, and the other on the opposite phase. Both at the main panel.

If both are about the same brightness and there is a load change and both brighten or dim together, then this is not a neutral problem.

If one brightens slightly and the other dims slightly, then this results from a 120 V load and the change in current thru the neutral from the load change. Neutral is probably OK. But the expected change is of course a function of the magnitude of the load change.

The eye can see a volt or two change in a 120 V supply to an incandescent.

Two voltmeters might be better in some ways, but not in other ways. Two recording voltmeters would be ideal.

If a small load change produces a big difference in the voltages, then you have a neutral problem.


Second, if it is not obvious that a neutral problem exists, then use a voltmeter and a test load.

Find the power company transformer. Run a long test lead from the ground rod at the transformer. You might measure the current in the transformer grounding wire. It should not be very much. This wire will be a moderately heavy wire relative to expected normal current, and whatever current is in the wire won't produce much voltage drop from the transformer center tap to the ground rod. If there is high current, then you know there is some kind of problem.

This test lead is taken to the main panel. Note: there could be a large voltage between the test lead wire and the main panel neutral. Be careful and use suitable equipment.

Measure the voltage difference between the main panel neutral bus and the transformer ground rod. This is the voltage drop across the transformer neutral to main panel neutral, assuming not much voltage drop on the transformer grounding conductor. If the first experiments with the light bulbs or meters did not indicate a large voltage change with normal load changes, then while monitoring the neutral path voltage connect a 100 W load to one phase. Note: the change in neutral voltage difference. At 8 A at my home I get about 0.2 V change. At 200 A this would extrapolate to about 5 V. If with the 100 W load you get a very small change, then try a 1500 heater, about 12 A.

Note: my calculated neutral path resistance is about 0.2/8 = 0.025 ohms.

.
 
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