Hi All,

I have a 110V, 20A household electrical circuit that has current but no voltage. If I shut down the circuit break and no voltage or current is present at a plug (checked with Volt/OHM meter as well as a Voltage/Current tester) and then turn on an overhead light (three-way switch) in the same room, current appears at the plug but still no voltage. If I disconnect the wire carrying the current, the overhead light goes out; if I then turn the circuit breaker on I have a Hot/Neutral Reverse condition at the plug. I am in the process of taking the wall down so I can trace the circuit at that time. I was just curious; I thought you needed voltage to have current. Any ideas as to way this is happening would be appreciated and will help shorten my troubleshooting time.

Thanks,

zats

It is a principal of electricity that you cannot have current flow without voltage.

Voltage pushes current through the conductor..

In Ohm's law this is symbolized by the following formula:

E = R/I.

What amount of current are you seeing?

Don,

Thanks for the quick reply; I thought that was the case. There is enough current to power the six double fluorescent lights connected to the "other"circuit. I will measure the exact Amperage draw later today and let you know.

Much appreciated,

zats

Neutral carries current but does not have voltage in relation to ground.

The two circuits may be sharing a neutral. Which would explain the current and lack of voltage. It also explains why lights on a different circuit would turn off if a wire at the receptacle is disconnected.

Neutral carries current but does not have voltage in relation to ground.

The two circuits may be sharing a neutral. Which would explain the current and lack of voltage. It also explains why lights on a different circuit would turn off if a wire at the receptacle is disconnected.

Tev,

I think you are correct, it is the neutral wire that is carrying the current and that one neutral definitely affects both circuits. Probably why all the radios in the vicinity buzz when the lights are turned on... :( Any reason why this would be done? Could they have been trying to take a short-cut when installing the two-way switch in the room? There is plenty of capacity at the breaker box and it is fairly close.

Thanks for the help, I think I'll split the circuits off and give them both their own neutral.

-Dan-

If you are getting that much current, then there has to be a voltage on the line.

Set your meter to AC Voltage say around the 200 VAC range.

Now measure between the Black (red probe) and White (Black probe) I would expect that you see 120 Vac at a minimum.

Neutral is the return leg for current. Black carries the current to the load and white returns it to the electrical system.

Make sure you are measuring the correct black line to the correct white.

We really need much more information regarding what failure you are seeing and how you are testing for current and voltage.

If you are getting that much current, then there has to be a voltage on the line.

Set your meter to AC Voltage say around the 200 VAC range.

Now measure between the Black (red probe) and White (Black probe) I would expect that you see 120 Vac at a minimum.

Neutral is the return leg for current. Black carries the current to the load and white returns it to the electrical system.

Make sure you are measuring the correct black line to the correct white.

We really need much more information regarding what failure you are seeing and how you are testing for current and voltage.

Hi Don,

I know what you mean and you would assume voltage on the line, but that is not the case, and what prompted my question.

In fact there must be voltage…as you state…it is just coming from another source/circuit.

I am a little familiar with the use of a Volt/OHM meter in it's testing of both AC and DC circuits for voltage/current/resistance/continuity. My Volt/OHM meter is a SCOPE DVM-632 Digital Multitester, with probes; I supplement my testing with an EXTECT Instruments DVA30 CAT-III Voltage and Current tester, a non-contact type and a GB Electrical Inc. GRT-500 circuit tester, a plug-in type.

All three testers indicate the same thing; no voltage. With the receptacle circuit breaker shut off, the overhead lights shut off at the switch (they are on a different circuit remember), and my SCOPE tester set in the 200V range, I get; .78V (less than 1V) using any combination of white (neutral) and black (hot) wires, and .00V (none) on white to white or black to black combinations. There are only two black Hot and two white Neutral wires at the outlet so it is easy to test all combinations; I can also see the cable so I am sure of which two go together. There is no amperage draw at that time (yes, I broke the circuit and used my probes for testing), the EXTECT confirms this. When I turn on the overhead lights the voltage does not change at the receptacle; no voltage, but the amperage draw changes to 14A on my SCOPE and the EXTECT goes nuts. Again no volts, lots of current.

Some additional information; the two-way switch the activates the overhead lights seem normal in all conditions of on or off, with the exception of a voltage drop when the lights are powered on; from 124V in the off position to 94V in the on position.

I do think Tev is onto something with his “shared neutral” theory and would indeed be the reason I see no voltage but current at the receptacle.

I hope this helps, please let me know what you think, any assistance is appreciated.

-Dan-

The voltage is there, so current can flow. It's just a matter of which ungrounded wire you use to measure it. Use the ungrounded that has the breaker off and the voltage will read near zero. Use the ungrounded that has the breaker turned on and you'll read normal voltage.

Do yourself a favor when working on either of these circuits, turn both breakers off. With one on you could have the white wire running to the panel in one hand and the white wire running to the light in the other and the circuit becomes complete through you. If I were you I'd put these two breakers next to each other and use handle ties so they must be operated together. Or preferably since you are opening the wall anyway, get rid of the shared neutral so the circuits are completely separate.

Donf:

In post #2 you state that E=R/I; That's a big OOPS. V=IR or E=IR for the old folks who use Electromotiveforce rather than Voltage.

The voltage is there, so current can flow. It's just a matter of which ungrounded wire you use to measure it. Use the ungrounded that has the breaker off and the voltage will read near zero. Use the ungrounded that has the breaker turned on and you'll read normal voltage.

Do yourself a favor when working on either of these circuits, turn both breakers off. With one on you could have the white wire running to the panel in one hand and the white wire running to the light in the other and the circuit becomes complete through you. If I were you I'd put these two breakers next to each other and use handle ties so they must be operated together. Or preferably since you are opening the wall anyway, get rid of the shared neutral so the circuits are completely separate.

Thanks Tev,

I guess it doesn't matter why it was done but it's not good. I never handle electrical wiring with both hands, in fact I wear rubber insulated gloves just to prevent any problems, but I'll gladly take your advice. Connecting both breakers together is a good idea but won't be necessary, as I mentioned earlier and you eluded too in your last post; I will be totally separating the two circuits. I knew this was a problem and would not have proceeded until I corrected it, and I appreciate your help in identifying the problem.

Thanks everyone and have a Merry Christmas and a safe and Happy New Year!

-zats-

Donf:

In post #2 you state that E=R/I; That's a big OOPS. V=IR or E=IR for the old folks who use Electromotiveforce rather than Voltage.

As you already know; you are correct Sir, thanks.

-zats-

<snip>In post #2 you state that E=R/I; That's a big OOPS. V=IR or E=IR for the old folks who use Electromotiveforce rather than Voltage.

Hey! I resemble... err... resent that. I don't consider myself old, but I use E for voltage. Learned it that way in school back in the 70's. Well.. maybe to some I would qualify. :p

EPM

I would not necessarily say big oops,


Donf:

In post #2 you state that E=R/I; That's a big OOPS. V=IR or E=IR for the old folks who use Electromotiveforce rather than Voltage.

Since E=IR is the basis of the formal Ohm's Law expression, and Don may be showing his seniority.

However, I would agree not to use E when explaining theory to laypeople, and stick with Voltage.

I would also like to add that Resistance is seldom used by electricians in everyday Power, or should I say Watts, calculations.

And what about Intensity that means Amperes?


I don't want to bring up Volt-Amps here. May just confuse the issue.

tk:
You needed to put your glasses on. The operator used by don was divide, rather than multiply and the quantity solved for was E. e.g E = R divided by I. That is a BIG oops.

We always talk about and IR drop, so that should be easy to pick up. But, we all have bad days, including me.

Jeez Kiss, I just saw my mistake, but your too fast for me.

Your right, the " / " did not jump out at me.

I thought you were on Don for E and V substitution.

I came back to to retract my statement, as of course I did not see it at first, only after open mouth insert foot, and damn your quick.

Ok I deserve it.

Forgive the typo, for that was actually a typo. The point is that regardless of what is being said by the poster, you cannot have a current flow without Voltage (EMF) pushing the current.

Either this is a multiwire circuit or the circuit he is testing is being backfed.

Strictly speaking the "you cannot have current without voltage" is true 99.9% of the time and is true here.

The short circuit current parameter of a solar cell is indeed the current when the voltage is zero. It can be forced to operate there. The voltage is zero and the current is negative.

Somebody said "you cannot have a current flow without Voltage (EMF) pushing the current". What about the panel grounding conductor for the electrical service in my house? I've measured over 5 amps on it, but there is no voltage there. Now I know, it is the low side of a circuit, but it seems that the above quote should be qualified.

OK,

Set up an ohms law equasion with 0 volts and demonstrate for me a greater than 0 amperage, please.

I does not matter if there is even a trace of voltage, there still needs to be voltage to push a current through resistance.

Don, the calculations will be irrelevant in dealing with AC power that is using a grounded conductor (AKA Neutral).

This is why standard theory and troubleshooting will only be on the Hot conductor.

There will always be measurable current in a Neutral conductor of an operating circuit in use, same as measured in the Hot conductor.

This is why a Neutral is considered as a Current Carrying Conductor by the National Electric Code.

However, since this is a Grounded conductor,there will not be, or should not be in a properly grounded/wired system with no wiring faults, any measurable voltage.

This wire has been grounded to insure the voltage in this conductor has been brought to Zero or Earth Potential, but still will have current flowing, as it the return for the circuit.


Once an open neutral develops, all bets are off, as the circuit is now operating abnormally, and troubleshooting methods need to change to the condition.

Troubleshooting AC circuits with a grounded conductor requires the knowledge of how and why the system is wired.

I still am not sure if I am explaining this sufficiently to help understand the why's and wherefores of current but no voltage in a Neutral, and why it is so important to have this understanding in giving advice on troubleshooting and theory.

An electrician understands without going into some much detail, it is something we just not accept but get, but so hard for us explain to laypeople without just saying Because We Say So.

We all got enough of that comment from our parents growing up.

This may help a bit:

May-June 2006: Current in the Grounded Conductor (http://www.iaei.org/subscriber/magazine/06_c/johnston.html)

TK

Thanks for the article.

I guess the difference between electricity and electronics rears its ugly head once more.

For the present, I'm willing to accept that this is a property of "Electrical Systems". I've got several books handy regarding electrical theory and I promise I will do "Due Diligence" to better understand. Are you open to more questions regarding this topic or would you prefer me to let it alone because it is an accepted fact, period.

Don, of course I am open to questions to this or any topic.

The smarter I get you and any others interested will make my life easier in answering questions accurately.

Believe me, I much rather be just answering posters questions than correcting answers or clearing up concepts.

TKRussel, Yeah, I have problems explaining that one to a layman too.

Don, Current must be looked at in the context of the whole circuit, not just one conductor or point on a conductor. If you look at any given point on the hot conductor you can measure current without having any voltage change there. Voltage drop occurs wherever work is done in the circuit. Current is the flow of electrons. It is the same throughout the whole circuit. Voltage is how we measure the potential difference between two points.

Take any point on EITHER current carrying conductor and apply the P=V*A (using common abbreviations) equation. Voltage change at that point for all intents and purposes is 0. Amps can be whatever you want. Which gives 0 power at that POINT. There is no work being done there. The equation holds with V/R=A too. If you have 0 volts change divided by 0 ohms, the Amp side of the equation is completely indeterminant. It can be anything you want at that POINT. Of course the work done in the complete circuit will determine the magnitude of the current.

Example: Take a 10 foot section of current carrying conductor in a 120v circuit. Lets make it #10 copper and put 10A of current on it. The resistance of that section will be somewhere around .01 ohms, and so the voltage measured from end to end of that section will be lost in the noise of your meter, and yet you will easily be able to measure 10 amps of current. I think the confusion comes when we try do divide the circuit into discrete parts and only think of the voltage as the supply voltage rather than the voltage measured across work done in that section. For the sake of the discussion here we don't have work done (and therefore voltage drop) on the current carrying conductors, and yet there is current there.

Of course when you run 600 feet of 16 gauge extension cord, and wonder why your saw doesn't work right, then we can talk about voltage drop (work done/heat developed) on the current carrying conductors!

EPMiller

I have to disagree, but certainly does not mean a reddie is in order:


TKRussel, Yeah, I have problems explaining that one to a layman too.


Well done!

Would this happen if you used 12/3 wire for example to feed two circuits sharing a common neutral? Like many kitchens are...

Codyman,

The neutral conductor in a correctly implemented shared neutral circuit situation only carries the differential current between the two halves of the circuit. In other words, If the one breaker for the one half of the circuit is supplying 14 amps and the other breaker is supplying 9 amps, the neutral will be carrying 5 amps.

This is why you MUST run those circuits on a double pole breaker. (There are other reasons too, but we aren't discussing that here.) If you split it up on 2 single pole breakers and get them both on the same leg of the service (an even number of spaces separating them in the panel) the neutral conductor carries the SUM of the current in both halves of the circuit. So from the above example, the neutral would be carrying 23 amps. That's too much even for a 20 amp kitchen circuit. And it could go as high as 40 amps in that situation. I've corrected that on more than a few panels where the homeowner knew just enough to be dangerous.

Let's not go into 3-phase here. That's even more fun.

EPM

EPM,

Right because you are using two phases with the two poll breaker. I do understand the proper installation of such a circuit as I have read past posts on it. Although given I just bought 250' of 12/2 romex for a ridiculously cheap price I doubt I will use this setup when I (eventually) re-wire the kitchen. It will be more cost effective to buy additional breakers than the wire I would need.

But would this explain why you might read zero voltage but still pass current? Do the phases run unversed to each other meaning when one was at +120V the other would be at -120V on the neutral wire? That would mean the neutral wire would be passing amps but read at 0V if tested on the grounding wire?

I am sure that I am missing something or many things  thanks just want to discuss.

- JC

Read my post #20 above for an explanation on why a neutral carries current but is at earth potential of zero.

I guess it's my turn:

A couple of things:

The two 120 volts "phases" are derived from a transformer, whose center tap is at the 0 volt potential. The 120 V is really 155 * sin (wt) so it really goes through a peak and valley of +-155 volts. Meaning 120 VAC is equivalent to about 310 Vpeak to peak. (Vp-p)

So, you statement of negative and positive is correct except you have to be talking about p-p values. In a real world the voltages won't be exactly "identical".

-----------------> 120 (L1)
*
*
*---------------->0, (N), (Earth), L0 (Just a made up term)
*
*
------------------>120 (L2)

The "*"'s are transformer windings.

If you have ONLY 240 V loads, there is no current on L0

If you have 10 A on L1 with respect to L0 (120 v load)
and 10 A on L2 with respect to L0 (120 V load)
The currents cancel and there would be none on L0

If you have an unbalanced load, then (N) caries the difference. The sign may be different depending on your reference.


Now back to the age old problem of current with no voltage.

Depending on where a voltmeter is inserted and it's reference will determine if there is appreciable voltage.

A stupid example:
Say you had a 12 V battery and a non-contact DC amp probe.
If you measure from the - terminal to the +terminal you will have 12 V.
If you measure from the + to the -, you will have -12V
The current will change sign based on the orientation of the AMP probe.

In AC there is no sign change in terms of the measured value because RMS or Root-Mean_Square means the square root of the square of the average. RMS values of voltages have the same "heating" value as a DC voltage of the same potential and ohm's law works for resistive loads. Not motors.

Cheap meters are RMS reading, average responding. I'll stop here before it gets to complicated. This means that the sinusoid is flipped, so that both half cycles are positive, averaged and multiplied by a constant.

If you measure the current with a probe close to the reference, the voltage at the same spot would be nearly zero.

KISS,

Wow okay I was with you up until the battery/motor part. Ahh, I am just going to stop here before my head explodes. You guys sure as hell know your S***

<snip> But would this explain why you might read zero voltage but still pass current? Do the phases run unversed to each other meaning when one was at +120V the other would be at -120V on the neutral wire? That would mean the neutral wire would be passing amps but read at 0V if tested on the grounding wire? <snip>
- JC

For any detail freaks out there, let me state that I generally am ignoring the small resistance of the copper wire and its accompanying details that are not necessary for this type of explanation. I am also ignoring counter EMF and reactive loads.

Codyman,

The neutral conductor in a correctly implemented shared neutral circuit only carries the differential current between the two hot sides of the circuit (restating my first post for emphasis). This is a statement about current, it has NOTHING to do with voltage. Reread that earlier post after you are done with this one.

In a residential electrical service, the neutral conductor (grounded conductor, not to be confused with the bare grounding conductor) has 0 volts on it everywhere. Think about it. It is copper the whole way from the low side of the load to the ground stake.

Voltage is always measured across a power source or across the load (power sink) in a circuit. Since the wiring from the panel to the load does not consume any power, there is no voltage change on the wires. So you would be able to measure 120 volts anywhere along the hot wire right up to the load. And you would be able to measure 0 volts anywhere on the neutral wire too. The voltage is dropped across the load. If you could measure from the exact middle of the load to ground you would get 60 volts.

Now on the other hand current requires a complete circuit to "travel" through and therefore can be measured anywhere in that circuit regardless of the voltage potential at the point of measurement.

You can think of voltage as potential and current as movement. Wrap your head around Amps = Voltage / Resistance and Watts = Voltage * Amps. Using basic algebra, or that table that gets posted every so often around here, you can figure any parameter out if you have two known parameters.

It took me a while to see the whole picture too. I am self taught, there's hope for others too. Get a book on basic electricity and read it. It will use electronics to teach you, but the fundamentals are the same everywhere. Residential electricians are just working with the special case of a center grounded 60 Hz AC transformer power supply with enough wallop to kill you without any warning.