Thread: 110 volts between netral and ground & 110 volts between hot and ground

I joined this forum after seeing it on the web and, although I am not an inspector I held a Masters license in Maine and a Journeyman's license in Massachusetts from the 1960s thru the 1970s: I have an electrical problem that I have yet to solve and my Dad with 50 years of licensed electrical experience under his belt has yet to run into this problem which is as follows.
Background: This is a 35 year old modular home (that should have been my first warning before we purchased it). ANyway, my wife turned on the air conditioner in her sewing room, just off a spare bedroom on the second floor of the home. All the lights and the ac went off. I checked the breaker to find it had not tripped. the next day I went to that room to find the lights on. I turned the AC on and everything shut off, lights, AC and all. Went down to the panel and found nothing was tripped, so I started at that room and began tracing things back. In my searching I used a polarity tester that plugs into receptacles; it exhibited an open neutral. From there I started taking apart a JB in the ceiling and found the following. I had a reading on my voltage tester between the black feed and ground, and the white neutral and ground; but no reading between the hot wire and the neutral. Puzzling since this condition would dictate that there is a dead short and the breaker should have tripped. Upon a couple of days of searching, I found that this particular circuit had the following on it: 2 lites in the cellar, 2 lites on the first floor, 2 outside lites and 4 lites and receptacles on the second floor and a large whole house exhaust fan in the hallway ( in the meantime, I pulled a new circuit from the cellar to feed the sewing room so as to get the AC and two receptacles running and isolated from the rest of the circuit). When I switched the power on I got what looked like half voltage to the lites and receptacles. I then connected a test lite to the neutral and the ground on a receptacle on the second floor. This lite also lit with partial voltage. I began taking apart the conglomeration of do-it-yourself electrical work done on the home during the 1st 25 years that I did not reside there. I found stuff that was so screwy, like 3 wire romex with 1 wire cut off wires with black feeds to single pole switches with red returns (all pretty screwy stuff). handy boxes in the walls where either new or old work boxes should be, etc. After taking things apart, I found that the test lite went out, signifying to me that the neutral wire lost voltage. By this time I was so confused that I threw up my hands an took a break. I have located the initial feed from the basement and will check this back to a JB that feeds a fluorescent lite in the basement but don't have much faith (by the way, the breaker on this circuit would trip if I shorted out the neutral(white) to a grounded JB) If anyone has any idea as to what is going on I would appreciate a personal email to chuck at striper5162@hotmail.com Thanks for this site which has some great stuff on it...

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A few things that come to mind:
What type and brand of service equipment is being used?

Have you checked the power that is coming into the home from the street?

It could be a bad neutral connection at the transformer?

Just a wild guess, but I would suspect an open neutral back at the panel or the service entrance and your voltage on the neutral is from your loads. I bet if you 1) kill the power, 2) test jumper the neutral to a known ground/neutral, 3)turn the power on and test, that all your readings will return to normal.
If the neutral is established back to the power company, there should not be any voltage present between it and ground since the polarity is the same by definition.
(Disclaimer - do not try this if you are not a professional)

I am thinking open neutral too, but ...

You may have some multiwire circuits which have been modified and in which the shared neutral is no longer being used with the hot conductors originally in the multiwire circuit.

That scenario can present some really screwy things.

One key, though, is where you said you had half voltage, which would indicate that circuits on the other bus from the panel would have 1-1/2 times voltage 120 + 120 = 240 as does 60 + 180 = 240 ... which is not uncommon at all when the neutral is open, loose, or a bad connection at the panel or at the service equipment.

One thing which throws off the open neutral, though, is that the 240 a/c went off, and 240 circuits and their appliances are not affected by open neutrals ... unless their low voltage transformers for control (thermostat, etc.) are off 120 volt, in which case 60 volts would take a 24 volt transformer down to 12 volts, not enough to operate 24 volt items.

I don't think he mentioned 240 A/C unit... but it is late and the post is hard to read.

Originally Posted by chuck gricus

I turned the AC on and everything shut off, lights, AC and all.

Originally Posted by Jim Luttrall

I don't think he mentioned 240 A/C unit... but it is late and the post is hard to read.

And, yes, the post was hard to read with no paragraph separations. Many times I do not even bother to read those posts - if the poster does not want to make or take the effort to make it easy for us to read, I don't want or take the effort to read it.

Making it easy to read and follow what is being said seems ... well ... like common sense and considerate ... to me.

Had some problems logging in but, no, AC is not 240 but 110 volts. Panel is a Cutler Hammer and to address some of the comments, no one has addressed why, when you get potential between the hot (black) and ground wire, that the breaker does not trip since this appears to be a dead short, no? I apologize for the lack of breaks in my dialogue but I was on a roll. Presently I have disconnected several neutrals in two JBs and the test light (I call it a test lite for lack of a better term) that is connected to the neutral and ground of a receptacle exhibits no light, leading me to theorize that the circuit has reverted back to normal. However, such is not the case since all lites on the circuit still show half brilliance.

If this problem began at the panel or service, wouldn't every circuit be affected, rather than just this one circuit? Again, this situation came about when turning on the 110 volt AC which drew an initial load, seemingly initiating the current situation where the plugs and lites on the circuit began exhibiting the behavior they now exhibit, that is, low brilliance and the potential between the black and ground, neutral and ground but: No potential between the hot leg and the neutral... I'm baffled

Originally Posted by chuck gricus

Had some problems logging in but, no, AC is not 240 but 110 volts.

That would upset a lot with an open neutral.

no one has addressed why, when you get potential between the hot (black) and ground wire, that the breaker does not trip since this appears to be a dead short, no?

No potential between the hot leg and the neutral... I'm baffled

Now you really have me confused: first there is potential between hot and ground (as there should be) then there is not.

No voltage hot to ground but there is voltage hot to neutral would indicate an open ground.

I believe I have always maintained that there is no reading between hot to neutral but potential between hot and ground and neutral (white) and ground... which signifies some sort of backfeed going on here, no? I think you'd have to be in the home to really see what is happening which is really strange. I am looking for some significant (or not so significant) visual explanation, i.e., burned wires, fused wires, etc. but have yet to find that...

Originally Posted by chuck gricus

I believe I have always maintained that there is no reading between hot to neutral but potential between hot and ground and neutral (white) and ground... which signifies some sort of backfeed going on here, no?

Depends on the tester you are using.

If you are using one which lights up (as you have indicated) then I would not rely on that tester.

Instead, I would use a multimeter, analog preferably, or a Wiggy, which actually places a slight (VERY slight) load on the circuit being tested and provides a better reading.

I think you'd have to be in the home to really see what is happening which is really strange. I am looking for some significant (or not so significant) visual explanation, i.e., burned wires, fused wires, etc. but have yet to find that...

You stated your tester indicated an open neutral, so, back to my question: Do you have any multiwire circuits? If no, then that eliminates many weird things, and leads one to what your tester found "an open neutral" - WHERE is what you would need to trace down, but I would start and the service equipment and work toward each branch circuit end from there.

Also, if you have a clamp on ammeter I would check the current flowing on the grounding electrode conductor from the service equipment, high current there could also indicate an open, loose, or poor connection at the neutral.

Yes I am using both a Wiggy and test lite on pigtail and I thought that a clamp-on ammeter would be helpful... thanks

Chuck, a suggestion: diagram the circuits. It may help you sort out what is happening. You may also double-check the integrity of the neutral and the ground at the service entrance and back to the power line. Some of the things you describe could be happening as the result of a neutral problem that is intermittent due to a 240 volt load being used intermittenly. Are you the only home on the step-down transformer from the power company? Have you checked from current on ground at the main grounding electrode? It may be easier to re-wire that entire circuit.

[QUOTE=chuck gricus;94432] ANyway, my wife turned on the air conditioner in her sewing room, just off a spare bedroom on the second floor of the home. All the lights and the ac went off. I checked the breaker to find it had not tripped. the next day I went to that room to find the lights on. /QUOTE]

Out of the entire post, this is the one comment that has me scatching my head. How did the power come back on? I'll let you guys figure out the voltage potentials, I just can't figure out how the circuit would re-set itself if nothing was tripped

That's what had me puzzled as well, but it seems that the turning on of the AC triggered this situation and trying to find out where the problem is has been unsucessful to date...

Originally Posted by chuck gricus

ANyway, my wife turned on the air conditioner in her sewing room, just off a spare bedroom on the second floor of the home. All the lights and the ac went off. I checked the breaker to find it had not tripped. the next day I went to that room to find the lights on.

Out of the entire post, this is the one comment that has me scatching my head. How did the power come back on? I'll let you guys figure out the voltage potentials, I just can't figure out how the circuit would re-set itself if nothing was tripped[/quote]

Originally Posted by chuck gricus

That's what had me puzzled as well, but it seems that the turning on of the AC triggered this situation and trying to find out where the problem is has been unsucessful to date...

That is the easy part ... the a/c turned out to be a 120 volt a/c unit, which produces a pretty good load, especially under starting up.

With an open neutral at the service the 120 volt loads balance themselves across the 240 volt legs depending on their load. Have two fairly equal 120 volt loads and each will get a fairly equal 120 volts across it.

However, if you have a light 120 volt load (a few lights) and a heavy 120 volt load (the a/c) trying to balance themselves off across the 240 volt legs, most of the voltage will be at the heavy load with little left over for the light loads - remember, with a floating neutral the two loads will always have 240 volts across them, so if one 120 volt load is drawing enough more current to get 200 volts across it, that only leaves 40 volts for the light load, in this case the lights, and 40 volts is barely going to make the lights glow.

I am surprised that it has not burned the a/c out.

Let's say the a/c is drawing 15 amps and the lights are drawing 3 amps, the voltage across the a/c will be 5 times the voltage across the lights, or 40 volts x 5 = 200 volts and 200 volts + 40 volts = 240 volts, which is right at what my two numbers grabbed out of thin air in the above are in my example, how about that for a coincidence.

Thus, the lights "did not go out", the lights "just appeared to go out".

When the lights went out, the a/c was likely shut down too and his wife left the room. The next day, the lights came back on, the breaker never tripped.

I pulled a new circuit to the sewing room and isolated that room, including the AC from the problem circuit which is still a problem...

[/quote]That is the easy part ... the a/c turned out to be a 120 volt a/c unit, which produces a pretty good load, especially under starting up.

With an open neutral at the service the 120 volt loads balance themselves across the 240 volt legs depending on their load. Have two fairly equal 120 volt loads and each will get a fairly equal 120 volts across it.

However, if you have a light 120 volt load (a few lights) and a heavy 120 volt load (the a/c) trying to balance themselves off across the 240 volt legs, most of the voltage will be at the heavy load with little left over for the light loads - remember, with a floating neutral the two loads will always have 240 volts across them, so if one 120 volt load is drawing enough more current to get 200 volts across it, that only leaves 40 volts for the light load, in this case the lights, and 40 volts is barely going to make the lights glow.

I am surprised that it has not burned the a/c out.

Let's say the a/c is drawing 15 amps and the lights are drawing 3 amps, the voltage across the a/c will be 5 times the voltage across the lights, or 40 volts x 5 = 200 volts and 200 volts + 40 volts = 240 volts, which is right at what my two numbers grabbed out of thin air in the above are in my example, how about that for a coincidence.

Thus, the lights "did not go out", the lights "just appeared to go out".

When the lights went out, the a/c was likely shut down too and his wife left the room. The next day, the lights came back on, the breaker never tripped.[/QUOTE]




Just so I understand, you're saying the unbalanced load created the appearance of the lights going out, but once the AC was shut back down, wouldn't the lights immediately gone back on? Maybe Chuck can shed some light on this (sorry)

I vote for calling a good qualified electrical contractor and pay them to find and repair the problem.

but once the AC was shut back down, wouldn't the lights immediately gone back on?[/quote]

Yes ... unless the lights were turned off in the meantime while trying to figure out what happened.

Originally Posted by Jerry Peck

Out of the entire post, this is the one comment that has me scratching my head. How did the power come back on? I'll let you guys figure out the voltage potentials, I just can't figure out how the circuit would re-set itself if nothing was tripped

That is the easy part ... the a/c turned out to be a 120 volt a/c unit, which produces a pretty good load, especially under starting up.

With an open neutral at the service the 120 volt loads balance themselves across the 240 volt legs depending on their load. Have two fairly equal 120 volt loads and each will get a fairly equal 120 volts across it.

However, if you have a light 120 volt load (a few lights) and a heavy 120 volt load (the a/c) trying to balance themselves off across the 240 volt legs, most of the voltage will be at the heavy load with little left over for the light loads - remember, with a floating neutral the two loads will always have 240 volts across them, so if one 120 volt load is drawing enough more current to get 200 volts across it, that only leaves 40 volts for the light load, in this case the lights, and 40 volts is barely going to make the lights glow.

I am surprised that it has not burned the a/c out.

Let's say the a/c is drawing 15 amps and the lights are drawing 3 amps, the voltage across the a/c will be 5 times the voltage across the lights, or 40 volts x 5 = 200 volts and 200 volts + 40 volts = 240 volts, which is right at what my two numbers grabbed out of thin air in the above are in my example, how about that for a coincidence.

Thus, the lights "did not go out", the lights "just appeared to go out".

When the lights went out, the a/c was likely shut down too and his wife left the room. The next day, the lights came back on, the breaker never tripped.[/quote]

Jerry I'm confused . If the neutral opens the circuit becomes a series circuit with both the lights and the a/c now on 240v. The a/c would have the lower resistance than the lights and the current through the a/c would also go through the lights. The lights would now glow like small novas. This would happen through the whole house. For your scenario to even come close, both of the neutrals (a/c and bedroom lights) would have to be tied together, on oposite sides of the buss, and loose at the neutral. Still wouldn't fly!

Look to any 3 or 4 way switches on this circuit.

There are two 3-way switches wired as single pole switches, I changed them both to single pole since that is what they actually are. I'm diving into this today... A local electrician suggested I start at the panel and tighten up all the neutrals which I will do, he also thought I shoulf check the meter but if there were a problem wouldn't that manifesty itself in alol the circuits in the home?

Chuck, it's a little hard to visualize all that is happing on this circuit from just reading posts. I think the a/c and the lights are somewhere switched into being in series and still on 120v. Everything most likely worked when the mis-wire happened, but then some combination of switch configuration has given you this result. The most likely place for such a mis-wire would be at 3 or 4 way switches, but could have happened at a junction box.

Originally Posted by Vern Heiler

Jerry I'm confused . If the neutral opens the circuit becomes a series circuit with both the lights and the a/c now on 240v. The a/c would have the lower resistance than the lights and the current through the a/c would also go through the lights. The lights would now glow like small novas.

Nope.

voltage across a load equals the resistance of the load times the current through the load

Agreed?

While you do have the same current flowing through the entire circuit, the VOLTAGE across each series connected load will vary depending on the resistance of each load as compared to all other loads.

Agreed?

With the lamp you could have higher current and lower voltage and not the light as a mini-supernova. A filament is just a variable resistor in which the resistance varies with the current through it as produced by the voltage across it - or where the voltage across it is produced by the current through it.

That E=IR formula MUST always balance out.

E=I(R+R+R+R) where E is constant and I is constant and there are more than one R loads

IR=E+E+E+E where I is constant and R is variable depending on the variable E with each E being variable based on its own IR. Confusing? Yep. Especially in this case since I is also variable being interdependent on the variables R and E.

We have a variable resistance (impedance really) in the a/c, and a variable resistance in each lamp, with the same, but variable, current through each, with the voltage self-adjusting across each, based on R self-adjusting based on I self-adjusting based on E self-adjusting.

In other words ... the circuit will find its balance point, and quickly too.

Now turn something else on or turn something off ... the circuit will automatically readjust, and again, do so quickly. Just as those lamps do with their resistive filaments self-adjusting to the current through them and the voltage across them and EACH ONE must still maintain its own E=IR relationship.

We are presuming, of course, that the a/c is on one phase bus and the lamps on the other phase bus (which is what gives the 240 volts across the circuit with an ungrounded center point neutral). If the a/c and lamps were all on the same phase bus, then it would depend on loads on the other bus in relation to the loads on the bus with the lights and the a/c.

Just way too many unknowns here to try to provide the answer.

Originally Posted by chuck gricus

There are two 3-way switches wired as single pole switches, I changed them both to single pole since that is what they actually are. I'm diving into this today... A local electrician suggested I start at the panel and tighten up all the neutrals which I will do, he also thought I shoulf check the meter but if there were a problem wouldn't that manifesty itself in alol the circuits in the home?

Chuck,

Have you gone through the house with the meter or Wiggy and checked voltages with all appliances and lights 'off' (turn all 120 volt appliances and lights on, turn all single pole breakers off) to make sure you have 120 volts at all locations?

Then go back and turn all the breakers on one bus on and remeasure.

Then go back and turn all the breakers on the other bus on and remeasure.

Then turn off have the appliances and lights off which came on when you turned the second bus breakers on - that will change the load on the circuits for the loads on the other bus - checking voltages as you do so.

If you have a loose neutral you WILL see a fluctuation in voltage, usually pretty high fluctuation.

If not and everything is basically always 120 volts, then it almost has to be what Vern is saying - something was mis-wired somewhere along the way and it would either be 3-way or 4-way switches or (in my opinion) more likely in a junction box where things were accidentally wired in series instead of in parallel.

First, though, I would want to eliminate the probability of a loose neutral as a loose neutral can be harmful to 120 volt appliances.

One other thought came to mind the other night: what result is there when testing the outlet that the AC is plugged in to? Diagramming the problem circuit should help figure it out. Start at the main panel, and go from there. You will probably find your problem, even if it is as simple as a ground laying across a bare neutral. Definitely continue to pursue it, though. It is a fire hazard. Lastly: a SECOND for calling a LICENSED electrician to sort it out and correct the problem.

Originally Posted by Jerry Peck

Nope.

voltage across a load equals the resistance of the load times the current through the load

Agreed?

Just way too many unknowns here to try to provide the answer.

Uh, no! The constants that are known are voltage (240v) and resistance (well not actually known but a constant) so the curreint is a result of voltage (known and constant) devided by resistance (a/c, lights, etc.)

Sorry, did not read the rest of your explanation as I have to go.

Originally Posted by Vern Heiler

Uh, no! The constants that are known are voltage (240v)

Nope.

The voltages I am referring to has been consistently stated as the voltages across the 120 volt loads.

While the 240 volts is constant across ALL the 120 volt loads, with a floating neutral (open neutral) the voltages across the 120 volt loads ARE NOT constant, the voltages vary just like the resistance varies and the current varies.

and resistance (well not actually known but a constant) so the curreint is a result of voltage (known and constant) devided by resistance (a/c, lights, etc.)

Nope, again, you are trying to apply the constant factor of the 240 volts to the variable factor of the 120 volts which WILL vary depending on the resistance and inductive capacities of the loads, making all three factors (voltage, current, and resistance) variable depending on what is on and what is not on.

With an open neutral at the service the 120 volt loads balance themselves across the 240 volt legs depending on their load. Have two fairly equal 120 volt loads and each will get a fairly equal 120 volts across it.

However, if you have a light 120 volt load (a few lights) and a heavy 120 volt load (the a/c) trying to balance themselves off across the 240 volt legs, most of the voltage will be at the heavy load with little left over for the light loads - remember, with a floating neutral the two loads will always have 240 volts across them, so if one 120 volt load is drawing enough more current to get 200 volts across it, that only leaves 40 volts for the light load, in this case the lights, and 40 volts is barely going to make the lights glow.

(Still haven't figured out the multiple quotes )

Jerry that is so much dribble!

The light load (lamps) will have the highest voltage drop measured across them. Higher resistance = lighter load. Current through higher resistance = higher voltage drop.

If the neutral opens then two or more seperate 120v circuits become series with each other. The voltages you refer to are the result of current through a load and measured across each load. The summ of which must equal the supply or 240.

If there is a 120v a/c unit that draws say 12amps it has a 10 ohm impedance and is using 1440 watts. If there is a 60w lamp on the other 120v buss that lamp has a 240 ohm resistance with .5 amps draw. When the neutral opens the a/c and the lamp become series with 240v across the circuit. The total current through both the a/c and the lamp are now .96 amps. The 60w lamp now has almost double the current through it (nova).

Only way the OP can happen is if the two circuits are on the same 120v leg and are somehow switched into series. No current flows. Lights and a/c are off.

Originally Posted by Vern Heiler

Jerry that is so much dribble!

Vern,

Okay, YOU answer these questions: (see attached photo)

- How much voltage is across EACH load?

- How much current is through EACH load?

- How much resistance is each load?

Originally Posted by Jerry Peck

Vern,

Okay, YOU answer these questions: (see attached photo)

- How much voltage is across EACH load?

- How much current is through EACH load?

- How much resistance is each load?

Obviously I can not give you the answer without one more value, either resistances or voltages measured across each load.

I can tell you that if one of the resistances represents a 120v compressor and fans and the other represents a 60w 120v light bulb, the light bulb will drop the most voltage and the compressor/fans will drop the lessor. The total of the two will equal 240v.

If we assume the values I gave in my previous post, which I believe are reasonable, the light will have almost twice as much current through it as it would have if it were across its own 120v source. If there are more than one 60w bulbs on the circuit, they are in parallel and the current is even higher.

This is not something I am making up. I have seen this happen. Open neutral and all small radios, clocks, and light bulbs smoked.

Look at the diagram in Code Check Elec. pg. 14. fig. 45 has a 100w and a 500w bulb in series. The 100w bulb has 1.4a of current just like the 500w does. The power formula shows that the 100w bulb must dissipate 282w of power and the 500w must dissipate 57w of power.

Originally Posted by Vern Heiler

Obviously I can not give you the answer without one more value, either resistances or voltages measured across each load.

Precisely.

And because the neutral point is floating (open neutral) as those loads balance out the resistance, voltage, and current are all variables - they WILL NOT BE as stated on the name plate ... i.e., a 60 watt lamp will NOT be 60/120=0.5 amp ... WILL NOT BE ... not unless the other load was IDENTICAL, and we know it was not, we know it was an a/c, which contains a motor, actually three motors - a compressor, a fan for the condenser coil and a fan for the evaporator coil, and THEY WILL NOT be drawing as stated on the name plate.

When the neutral point is grounded to a ground reference and the 240 volts is grounded to that same reference point with 120 volts to both sides, then and only then will the name plate ratings apply as then and only then will the name plate ratings be met (one of the name plate ratings is 120 volts, and only when the neutral point is grounded will that be met) with two loads which are not identical.

Let's say the a/c has a name plate of 8.5 amps at 120 volts, and let's say that is the load on the right. And let's say the load on the left is a 60 watt lamp 120 volt rated.

What will the voltages be and what will the current be?

Originally Posted by Jerry Peck

Precisely.

And because the neutral point is floating (open neutral) as those loads balance out the resistance, voltage, and current are all variables - they WILL NOT BE as stated on the name plate ... i.e., a 60 watt lamp will NOT be 60/120=0.5 amp ... WILL NOT BE ... not unless the other load was IDENTICAL, and we know it was not, we know it was an a/c, which contains a motor, actually three motors - a compressor, a fan for the condenser coil and a fan for the evaporator coil, and THEY WILL NOT be drawing as stated on the name plate.

When the neutral point is grounded to a ground reference and the 240 volts is grounded to that same reference point with 120 volts to both sides, then and only then will the name plate ratings apply as then and only then will the name plate ratings be met (one of the name plate ratings is 120 volts, and only when the neutral point is grounded will that be met) with two loads which are not identical.

Let's say the a/c has a name plate of 8.5 amps at 120 volts, and let's say that is the load on the right. And let's say the load on the left is a 60 watt lamp 120 volt rated.

What will the voltages be and what will the current be?

Did you read what I posted? You stated the lights would not have enough power to fully light, right? I gave you everything you need including example drawings in the Code Check to show that the lights will glow VERY bright!

Originally Posted by Vern Heiler

Did you read what I posted? You stated the lights would not have enough power to fully light, right? I gave you everything you need including example drawings in the Code Check to show that the lights will glow VERY bright!

I did read what you posted.

Did you read what I posted?

You did not answer those questions, I see that.

Originally Posted by Jerry Peck

I did read what you posted.

Did you read what I posted?

You did not answer those questions, I see that.

ok. Here it is as you requested.

on the left is a 60w lamp. 60w and 120v = 240 ohms and .5 amps.

on the right is the a/c . 8.5a at 120v means impedance is about 28 ohms. Impedance is a product of frequency and inductance for the most part and will not change significantly due to voltage increase.

Now we series the two resistances and hit them with 240v.

240 volts divided by 268 ohms give approx. .9a.

Both resistances carry .9 amps.

The 60w lamp drops approx 216v and must dissapate 194 watts.

The a/c unit drops approx 25v and doesn't run with only 25w of power.

Got that slide rule smoken

Originally Posted by Vern Heiler

ok. Here it is as you requested.

on the left is a 60w lamp. 60w and 120v = 240 ohms and .5 amps.

on the right is the a/c . 8.5a at 120v means impedance is about 28 ohms. Impedance is a product of frequency and inductance for the most part and will not change significantly due to voltage increase.

Now we series the two resistances and hit them with 240v.

240 volts divided by 268 ohms give approx. .9a.

Both resistances carry .9 amps.

The 60w lamp drops approx 216v and must dissapate 194 watts.

The a/c unit drops approx 25v and doesn't run with only 25w of power.

"on the right is the a/c . 8.5a at 120v means impedance is about 28 ohms. Impedance is a product of frequency and inductance for the most part and will not change significantly due to voltage increase."

The impedance will vary depending on the inductive reactance which varies depending on the current through the inductance, it is not just frequency based.

Thus your calculations are based on an incorrect assumption that the a/c inductive reactance and impedance will remain constant and it does not - look at the start up current draw of a motor versus the running current draw of a motor ... the frequency remains the same, the voltage remains the same (because of the grounded neutral point), but the current and impedance changes due to the characteristics of the motors.

In my example, though, the neutral is not grounded so there is no constant center point reference to keep the voltage constant, thus the voltage will also change, and with the voltage changing, the current and impedance will also change.

Which means one cannot use the 120 volt current to calculate the impedance. There is no longer 120 volts there.

I'm glad this post generated some great theoretical discussion, however, at ground zero, my house, I am looking for a spot where the white neutral and black hot wire are touching each other since there is still potential between the white neutral and ground and black hot wire and ground which means that they are, indeed, touching unless there is another explanation... (question still remains as to why the breaker will not trip). Currently i am changing the bathroom GFI from this problem circuit to the bathroom circuit which is a separate circuit. This, at least narrows my search to several receptacles, switches and lights which are connected to this problematic circuit.

Chuck,

Now back on topic ... you may find a reverse polarity connection someplace which then was crossed over to the neutral, such as a white switch leg going to a switch (which is "hot) and then getting connected to other whites (which are not supposed to be "hot") but, as the Big Box store guys stress "white-to-white and black-to-black and you will be okay" - yeah, right!

Originally Posted by Jerry Peck

"on the right is the a/c . 8.5a at 120v means impedance is about 28 ohms. Impedance is a product of frequency and inductance for the most part and will not change significantly due to voltage increase."

The impedance will vary depending on the inductive reactance which varies depending on the current through the inductance, it is not just frequency based.

Thus your calculations are based on an incorrect assumption that the a/c inductive reactance and impedance will remain constant and it does not - look at the start up current draw of a motor versus the running current draw of a motor ... the frequency remains the same, the voltage remains the same (because of the grounded neutral point), but the current and impedance changes due to the characteristics of the motors.

In my example, though, the neutral is not grounded so there is no constant center point reference to keep the voltage constant, thus the voltage will also change, and with the voltage changing, the current and impedance will also change.

Which means one cannot use the 120 volt current to calculate the impedance. There is no longer 120 volts there.

More DRIBBLE!

Jerry I thought you were more knowledgable on electronics. The formula for inductive reactance is XL = 2πfL, no E in there! The reason there is a start up current higher than run is due to inrush, which has to do with EMF and Counter EMF generated in the motor windings as it comes up to speed. Too deep to get into on this board. The values I gave you are correct, go check it out.

Originally Posted by Vern Heiler

Jerry I thought you were more knowledgable on electronics. The formula for inductive reactance is XL = 2πfL, no E in there! The reason there is a start up current higher than run is due to inrush, which has to do with EMF and Counter EMF generated in the motor windings as it comes up to speed. Too deep to get into on this board. The values I gave you are correct, go check it out.

Vern,

Now you are trying to make me go back 40 years into those old brain cells stored on the hard drive which defrag calls "rarely used files" and the old 286 system has a very difficult time trying to search the larger storage capacity drives, so, while the old 286 inside was click-clacking away, I had to do some searching on Google, which finds things MUCH FASTER ...

Vern, you do understand that the dampening effect of impedance in a inductor is caused by the oscillating CURRENT of ac and that that oscillating CURRENT is what causes the MAGNETIC FIELDS to build and collapse and it is those building and collapsing MAGNETIC FIELDS which creates the dampening effect (resistance effect) to the flow of ac current, right?

You also know that CURRENT is driven by something called electromotive force (VOLTAGE), right?

Thus, it is the oscillating VOLTAGE of ac which drives the oscillating CURRENT which creates the MAGNETIC FIELDS which causes the resistance to current flow, or impedance.

And, yes, motors have coils and it is the current flowing through those coils are what creates the magnetic fields which causes the impedance, and the current is driven by voltage - ALWAYS. Thus, if you have current and resistance or impedance you will have voltage, otherwise you would not have current.

Impedance in a inductor is Z=jwL where j=square root of (-1) and w= 2*pi*(frequency of voltage source) in radians/sec and L=inductance in Henrys.

And, yes, I got that off the internet as my old 286 brain is still searching its hard drive.

Originally Posted by Vern Heiler

The formula for inductive reactance is XL = 2πfL, no E in there!

The formula for inductive reactance I found is: e=L(di/dt) ... which does contain e=instantaneous voltage:
AC inductor circuits : REACTANCE AND IMPEDANCE -- INDUCTIVE

I believe your formula came from a page similar to the link which follows, which shows XL=2pifL ... and in reading the paragraph above that formula it states "Reactance, then, increases with an increase of frequency and with an increase of inductance." What increases that inductance? Why current flow, which is driven by voltage.
AC inductor circuits : REACTANCE AND IMPEDANCE -- INDUCTIVE

No, I'm not doing that math, I did my share of that 40 years ago when I worked in a Standards Lab calibrating oscilloscopes, volt meter, signal generators, and all equipment used in a large defense plant which manufactured microwave tubes used for everything from battleships, futuristic fighter jets, to simple radar and TV stations, and experimental microwave heating devices. One of my supervisors was one of the people on the team at Bell labs that invented/developed the transistor.

I have forgotten more about "electronics" (as opposed to "electrical") than I want to try to relearn or remember - so the floor is yours, my head is feeling funny from all that searching in those old gray matter cells back there, besides, I just received an error message from my brain which said "Fatal Error - Insufficient Memory - Reboot Required".

Originally Posted by Jerry Peck

Vern,

Now you are trying to make me go back 40 years into those old brain cells stored on the hard drive which defrag calls "rarely used files" and the old 286 system has a very difficult time trying to search the larger storage capacity drives, so, while the old 286 inside was click-clacking away, I had to do some searching on Google, which finds things MUCH FASTER ...

Vern, you do understand that the dampening effect of impedance in a inductor is caused by the oscillating CURRENT of ac and that that oscillating CURRENT is what causes the MAGNETIC FIELDS to build and collapse and it is those building and collapsing MAGNETIC FIELDS which creates the dampening effect (resistance effect) to the flow of ac current, right?

You also know that CURRENT is driven by something called electromotive force (VOLTAGE), right?

Thus, it is the oscillating VOLTAGE of ac which drives the oscillating CURRENT which creates the MAGNETIC FIELDS which causes the resistance to current flow, or impedance.

And, yes, motors have coils and it is the current flowing through those coils are what creates the magnetic fields which causes the impedance, and the current is driven by voltage - ALWAYS. Thus, if you have current and resistance or impedance you will have voltage, otherwise you would not have current.

Impedance in a inductor is Z=jwL where j=square root of (-1) and w= 2*pi*(frequency of voltage source) in radians/sec and L=inductance in Henrys.

And, yes, I got that off the internet as my old 286 brain is still searching its hard drive.



The formula for inductive reactance I found is: e=L(di/dt) ... which does contain e=instantaneous voltage:
AC inductor circuits : REACTANCE AND IMPEDANCE -- INDUCTIVE

I believe your formula came from a page similar to the link which follows, which shows XL=2pifL ... and in reading the paragraph above that formula it states "Reactance, then, increases with an increase of frequency and with an increase of inductance." What increases that inductance? Why current flow, which is driven by voltage.
AC inductor circuits : REACTANCE AND IMPEDANCE -- INDUCTIVE

No, I'm not doing that math, I did my share of that 40 years ago when I worked in a Standards Lab calibrating oscilloscopes, volt meter, signal generators, and all equipment used in a large defense plant which manufactured microwave tubes used for everything from battleships, futuristic fighter jets, to simple radar and TV stations, and experimental microwave heating devices. One of my supervisors was one of the people on the team at Bell labs that invented/developed the transistor.

I have forgotten more about "electronics" (as opposed to "electrical") than I want to try to relearn or remember - so the floor is yours, my head is feeling funny from all that searching in those old gray matter cells back there, besides, I just received an error message from my brain which said "Fatal Error - Insufficient Memory - Reboot Required".

" What increases that inductance? Why current flow, which is driven by voltage.

Nope! To change the inductance you make more turns, bring the coils closer together or further appart. Voltage is used to determine the current through the coil using ohms law with XL as R.

And as your post stated:Impedance in a inductor is Z=jwL where j=square root of (-1) and w= 2*pi*(frequency of voltage source) in radians/sec and L=inductance in Henrys.
Coils have a know value, measured in Henerys.

Jerry, I did not look at the link you provided, as the formula is looking to solve "E". I could go to the site, but I too have forgotten more electronics theory than I care to admit. But I don't try to change ohms law, or the basic formulas that I learned 40 some years ago. Electromagnetic fields of force are the dominant current limiting factor in motors and are found by the formula XL = 2πfL, inductive reactance or the resistance to current flow. XL is used in the place of R in alternating circuits. If you know two of the three variables R, I, E, you can calculate the third. In our example we knew I & E, well I calculated I using the power formula for the light. All the formulas are there and they work, not to mention (as I did) that I saw this same scenario happen. And yes, the low power (high resistance) devices on the circuits smoked. None of the lights glowed dim, they all went brilliant and then died as did clocks, radios,etc.

Originally Posted by Vern Heiler

In our example we knew I & E

That's my point, I disagree, we do not know E.

However, you do your engineering calculations and I will go Huh? As I have forgotten all those things many years ago.

Yes, I know Ohm's Law, and, no, I am not trying to change it. As soon as you start applying voltage across the coils and creating magnetic fields in those coils, the calculations change.

How much they change I do not know and to figure it out I would need to create what I used to do all the time ... devise a test setup to test what was being tested, and attach the necessary equipment (calibrated equipment) to make the readings we wanted to know, all this done in either the Test Lab, the Standards Lab, or the R&D Lab where I also worked some of the time - or ... now ... in my garage.

I'm not up for that, but if you are, design the test setup, get the equipment together (does not need to be calibrated, I will accept the use of "everyday equipment" ), run the test, make the readings, and report the results here.

I look forward to your results.

Until then, my friend, have fun playing around with that, it will be interesting to see what your results are. If they confirm your position, I will accept that too.

Originally Posted by Jerry Peck

That's my point, I disagree, we do not know E.

E was 120v prior to the open neutral and 240v after it opened.

And the test was already done at a co-workers house 5 or 6 years ago when the electricians helper took the neutral loose at a sub-panel with power on. Over three thousand in damages.

Originally Posted by Vern Heiler

E was 120v prior to the open neutral and 240v after it opened.

Vern,

Incorrect again.

The voltage was never "120v prior to the open neutral and 240v after it opened".

Look, the test step-up is real simple to make: install a double pole breaker, run one wire to one receptacle, run the other wire to the another receptacle, run the neutral wire from both receptacles to a switch then to the neutral bar.

Turn the switch in the neutral wire 'ON', turn the breaker on there is now 120 volts / 120 volts at those receptacles.

Plug a light into one receptacle and turn the light on. Connect a voltmeter to the unused receptacle in that duplex receptacle to monitor the voltage at the light - the voltage at the drill or fan is 240 minus this reading.

Plug a drill or cheap fan into the other receptacle and turn it on.

Place a clamp on ammeter over each hot wire and not the reading.

You are now ready for the test.

All you have to do is switch the switch to 'OFF', read the voltage on the meter and the current on either meter (both will, should be, the same) and watch what happens.

This is not quite what he is describing as he still had "some" neutral connection which allowed the lights to turn on without the motor being on, either that or there was enough stuff "ON' the other bus which provided a current path when the light was 'ON'.

Originally Posted by Jerry Peck

Vern,


Turn the switch in the neutral wire 'ON', turn the breaker on there is now 120 volts / 120 volts at those receptacles.

Plug a light into one receptacle and turn the light on. Connect a voltmeter to the unused receptacle in that duplex receptacle to monitor the voltage at the light - the voltage at the drill or fan is 240 minus this reading.

Jerry, what are you smoken . You just discribed the normal circuit in every house and said the voltage across 120v outlets depends on what is plugged into the other buss. With the neutral connected there will be 120 across each receptacle regardless of what is plugged in.

Your set-up describes the diagram in Code Check Elec. pg. 14 fig. 45 to a tee. I know you have it, look at it. Then look at what happened as opposed to fig. 42, neutral connected. The current through the 100w bulb went from .833a to 1.4a when the neutral was broken. You could look at this as the 500w bulb representing the a/c and the 100w as the light.

For the lurkers...