Pretty incomplete test, though.
If I did that and the lamp worked, it would confirm the switch was bad. But it wouldn't tell me what breaker to use to turn off the power. True, you CAN replace the switch hot. But there's no point in doing so when not under production pressure.
If I did that and the lamp did not work, I would not know if I had a bad lamp, bad fixture, bad switch, or no power to the switch, and I wouldn't know what breaker to turn off.
Actually in this case you are sort of using it as an ammeter when you put it in series with the circuit. That is why I mentioned the 'trick' with the Simpson of going down scale to see if the meter movement stays about the same physical place. The impedance of the circuit is high enough that you are getting an almost constant current of a few micro amps. There is really no such thing as a pure ammeter or voltmeter. It is just in how you are using it and how it is calibrated.
If you look at most of the 'good quality' analog meters, they are using a meter movement of around 50 micro amps and resistors are switched in to make it show a full scale of whatever is desired. The Simpson has a meter movement of about 48 microamps (I think that is the number, or close) and one of the first calibration steps is to adjsut a pot inside the meter so it shows a full scale at 50 micro amps.
I don't care to do the math right now, but you can calculate how much impedance it takes to have 10 or 20 micro amps and at what voltage. If you knew how long the wires are and a few things about the insulation, you could calculate the capacitance of the wiring and what voltage you should see. Just way too much to get into here.
Each wire in turn, to neutral or ground. VoltMeter across a switch is a pretty lame attempt to troubleshoot the problem, because no matter what reading he gets, it doesn't mean SQUAT.
You are obviously unable to grasp the concept.
The meter WAS connected AS an ammeter. It was connected IN SERIES with the load. A voltmeter is connected ACROSS the load.
OK. EXACTLY what ic onnecting the meter across the switch telling you? If you get a non 120 volt reading either a)- there is no power to the switch B) the switch is shorted or closed, C) the bulb is open, D) there is a bad connection between the switch and the bulb, E) there is a bad connection between the bulb and the neutral buss on the panel, F) the meter is defective.
Take your pick - you really have not learned anything of value.
If on the other hand you get a 120 volt reading, you know the switch is open and there is a measure of continuity between the switch and the neatral buss, as well as between the switch and the fuse/breaker, which is supplying power.
If the test is done with no bulb in the socket - (no load) there is no valid reading that tells you ANYTHING.
And that's no bullcrap.
No, but in your case, it would have proved you had a bad switch - and if the jumper was left on when the light went out you would have known which breaker it was.
I hope you labelled the breaker now that you know what it controls???
You SHOULD have a detailed list showing what is on what breaker.
In hindsight, I did it wrong.
With no lamp in the fixture, reading across an open switch does not give meaningful results. Although I guess if it read 120 V then the line is shorted to ground somewhere between switch and lamp. If it read 0 V the line could be shorted to hot and the neutral broken - too many possibilities. (when troubleshooting, I ask 3 questions: what does it mean, what else could it mean, how can I tell the difference)
That the meter read anything at all is only due to the line (not the meter) being coupled to some sink somewhere between ground and hot. I say not the meter, because the meter didn't read 80 V until touched to the terminal. Getting that 80 V reading made it easy to find the breaker, but that was dumb luck; and getting a zero reading would not have guaranteed the circuit wasn't live.
But an ammeter? Nah. An ammeter is in series with the circuit. There was no circuit with no lamp. And I would argue there is also no circuit with the lamp in place and a 10 million ohm resistance (digital meter) breaking the circuit - that's effectively an open. Even the 1 million ohm of the analog is pretty close to an open circuit. Of course that assumes both are set to the voltmeter range; if either are set to amps then the meter closes the circuit and I'll agree it's an ammeter. Put a voltmeter into a normal outlet, does it become an ammeter? Why not, you just closed the circuit, and the meter is the load?
You may not have learned anything.
A competent technician can evaluate what is happening. The information is combined with the test below - if there is 120V with the switch open it leaves B.
If there is 120V in both positions the switch is bad. That is what the OP would have found if he had an incandescent bulb in the socket. A simple test.
You might not figure out anything. But a competent technician would evaluate no bulb and 80V in both positions as an open switch. I believe that is what the OP found.
I said 2 times have an incandescent bulb in the socket.
Only if your grasp of electrical troubleshooting is defective.
You have still not answered Hey's question. The OP said there is no neutral or ground at the switch.
You are obviously unable to grasp the concept.
One of the smartest electricians I know wanted to measure the current for a motor in a food processing plant. It was a possibly explosive atmosphere so there was a motor control center in a purged room. He defeated the interlock on the module door and amp-clamped a motor wire - an absolutely standard procedure. No one knows what happened - maybe a loose screw faulted the busbars, but there was an arc flash. He was badly injured (including condensed copper on him) and was in the hospital quite a while. The facility was buying distribution voltage power from the utility and one of the fuse holders was destroyed in the event - I have no idea how that could happen. (Arc flash was not an issue back then.)
At a trade show a manufacturer field engineer had a seminar. As an aside, he was working for a major company like Westinghouse and the client wanted a module for a motor control center or switchgear. The space was open so he was measuring the size with a steel tape. The next thing he knew he was on his back on the other side of the room. If he hadn't been thrown he might be dead. At least in that case he did something dumb.
Some of the required inservice classes have covered arc flash. They had an arcflash suit - pants, top, hood and gloves. (And the protection includes the natural fiber clothes inside.) Don't know how you do useful work in something like that. Someone brought in copies of the labels at his facility that are required to evaluate what protection is required. One of them said something like "No safe approach is possible" (with protective panels removed).
The default standard for worker electrical safety is NFPA 70E. Your plant must have a copy and I wouldn't be surprised if you have read it. Interesting information on what protection is required for different voltages and available fault currents.
I just retired as of yesterday. We have all the protective gear. Gloves, coat, face shield. While we seldom use it, we have a couple of special outfits that are rated for 100 cal. The coats we normally have are rated for 40 cal. I was almost tempted to bring my 40 caliber Glock on the last day and shoot at my arc flash coat and tell them it would not protect against a 40 caliber... Almost impossiable to do useful work with that junk on. Most of the time we can cut the power off and do a test and then take all the stuff off.
We have lables like that all over the plant. It is a large plant and covers many acers. A small building ( maybe 50 x 100 feet) has a motor control center and it had a sign on it that mainly stated we could not get within about 500 feet of it. That ment if any breaker on it tripped, it would have to stay off. Someone finally wised up and replaced that sign with one that was workable.
Most of the arc flash junk started when a company came up with some high dollar main breakers that were suspose to trip under certain conditions. As many companies did not want to replace them, the arc flash hazard was 'invented'.
Outside of them being marked volts or amps, how do you tell which is which ? any volt meter takes a certain ammout of amps (usually microamps) and any amp meter will have a certain volt burden. In your example, the high impedance of the voltmeter will protect the amp meter. The volt meter will probably read close to 120 volts and the amp meter will show a few microamps. Whatever the voltage shown on the voltmeter is devided by the resistance, minus a small ammount for the amp meter burden.
Case in point is the analog meters. They often have a meter of 50 micro amps for full scale. If that meter is placed across a source that drives it to full scale , it will always take the same voltage to drive it to full scale. It could also be marked for that voltage and not for 50 micro amps.
Simple enough to explain. There are is a constant ammount of parallel wires in the circuit. That makes a capacitor. It will have a constant impedance at 60 Hz.
If I hook up the Fluke meter I have, I could measuer the capacitance, calculate the impedance and then with the meter impedance of probably 10 meg ohms calculate the voltage. It could also be worked backwards. YOu meter is dropping 80 volts. Assuming 120 volts at the breaker and the copper wire having zero resistance, the capacitance reactance is dropping 40 volts. You can then calculate the capacitor value. This is the same as if you take a physical capacitor of the same value. Put one meter lead in the 120 volt socket, the other to the capacitor and the capacitor back to the neutral..
This is simply word games. To measuer the voltage potential differnace with a common digital or analog meter, there has to be some current flow, in this case a few micro amps. If there was no current flow, the meter would not work.
Except of course when you're using an instrument that has a high enough impedance in a way that you will see those effects. That is exactly what you are doing.
I have no idea that point you're trying to make here besides using instruments in very strange ways.
Because it's likely capacitive coupling on the cable run. I think that has been pointed out many times now, now? There is enough capacitance there in that open run of wire that together with the high impedance meter it forms an RC circuit.
=A0Not why I might get some phantom voltage reading,
What exactly do you think a "phantom voltage" is? Something that can't be explained by electrical circuits and instead by withcraft? It's only a phantom if you don't understand circuit basics.
=A0Show me
It's most simply modeled as a circuit with one resistor, that being the multimeter, one capacitor, that being the distributed capacitance of the cable, and a voltage source, that being
120V AC.
But unless you accept that the one conductor is acting as a small capacitor in an AC circuit, then you have no potential on that unconnected wire going to the empty light bulb socket. If it's just a perfect conductor with no capacitve effect, no part of a complete circuit, etc. then there is no potential difference and you have no explanation for the 80V you see. When you start to model it by what it really looks like, which is a distributed model with R, C, and L, then the voltage you see is no longer some mysterious "phantom" voltage, but is instead explainable by the circuit dynamics and the meter.
At his day job, one of the instructors of our nightly apprentice program at the International Brotherhood of Electrical Workers put a wiggy on the line terminals of a 4160V switch gear once. As a reminder, a wiggy is good for 600V.
This did not end well. I don't remember the details, but I have heard the story a thousand different times. /////// When I was an apprentice, my roommate was also an apprentice. He worked for a very small company. The company was dad, son and my roommate. The son was changing a buss fuse under load. When the two maintenance workers that were in the area because of the outage became aware of what the son was doing, they ran for the door. I am told that both maintenance workers were burned on their back half of each's bodies as they had made it to the door and were each half shielded by the door jam as they left the room. The son was killed. He was an experienced electrician. The rush to restore power to the building cause him to make the error.
Forget it bud. I do this diagnostic stuff a lot, and have for almost
50 years.
Forget it clare. I have been a licensed master electrician for about 40 years and do this diagnostic stuff a lot.
Still missing - an answer to Hey's question. Two wires, no ground, no neutral.
Minor comment - a 50 microamp meter is "20,000 ohms per volt". On AC ranges the same meter (including my Simpson 260) is "5,000 ohms per volt" which is 200 microamps full scale. That means the load on the circuit is even higher and phantom voltage will be less visible. I believe the current is higher on AC ranges to get into a more linear curve on the diodes.
An additional complication in calculating what is happening is that the voltages across the meter and capacitance are at 90 degrees.
I agree with both you and trader. It is so simple - once you understand it.
Also explained with a lot less detail but from a "reliable source":
Another way to look at it is the resistance of the meter and the capacitive reactance are in series and split the voltage (a voltage divider). The meter measures the voltage drop across the meter resistance. Might make Tim happier (but is no better than your explanation).
Congratulations (in some cases condolences) on retiring. Sounds like you had an interesting job.
On May 18, 1:36=A0pm, "Ralph Mowery" wrote: YOu meter is dropping 80 volts.
Some of what various posters have pointed out is correct. I did not think all the way through this problem.
But the above is dead wrong.
You cannot calculate the capacitor value from the above information.
You can assume the hot terminal is at 120 V referenced to neutral.
You CANNOT assume the other terminal is capacitatively coupled to a 0 reference voltage. It is coupled to something, but you don't know what. Therefore you can't assume 120 - 80 =3D 40.
The other wire may be coupled to neutral, or it may be coupled to ground (which may or may not be somewhere near neutral) or it may be coupled to something else entirely.
For this problem there is no difference between ground and neutral - they are at the same potential.
One has to make simplifying assumptions. I think what Ralph wrote is reasonable. I would assume one end of the capacitive reactance is at ground/neutral potential.
However, as I wrote, the voltage across the capacitance and the voltage across the meter resistance are at 90 degrees. If the capacitive reactance and meter resistance are equal, the voltage across each is about 84V. Adding 84V to 84V at 90 degrees phase displacement gives
120V. With voltage across the meter at 80V the voltage across the capacitance is around 89V.The meter resistance has to be known to calculate further (and I don't have the ambition to calculate).
There is no valid reason to make that assumption, and I have seen many real world examples where that assumption was wrong.
Actually, unless forced to be true by very, very good bonding and grounding practice, it is unlikely to be true.
Now, the ground WIRE in a box very well may be pretty close to the neutral on any given branch circuit. But this was clearly a case with no ground wire. And ground itself is a whole different story.
There is a valid reason to make that assumption and it's because Bud is trying to be realistic and explain what you are seeing using basic circuit theory. Sure, in many cases there is a small difference between ground and neutral, a volt or two. But typically not 80 volts which is what you are seeing. And seeing that kind of voltage when using a meter incorrectly isn't limited to your circuit, it happens frequently.
So, say it's a 10 volt difference. So, what? It doesn't change the analysis that 3 of us have given you that explains the AC coupling effect. It just means that on the other end of the modeling capacitor instead of being connected to 0 volts, it's now connected to
10 volts. So you have a small capacitor connected from 120V to 10 volts. Same analysis using a distributed R,C, L model for the wire still applies. In this case, we focused on just the C, because unless something very unusual is going on it's going to be the dominating factor.We've given you a logical explanation using basic circuit theory for what you are seeing. What is your explanation for "phantom voltage" that one sees when using a meter like you did, ie connecting it between hot and one end of a long length of AC cable that is open on the other end?
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