Fixed my porch light, not sure how

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One thing that you can also try with the analog is to set it to the highest voltage scale. If you show anything that is less than the next scale down, switch to that scale, the meter will stay almost the same physical place if a 'phantom voltage', you can then go to the next lower scale and the meter will not move much from the same physical point. The actual voltage the meter reads will be differant, but the physical point will not move much.
This digital vers analog comes up almost as much as the WD-40 being oil or not.
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wrote:

Test the bad switch with an ohm meter

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wrote:

That works - but you need to turn the power off for that test. The voltmeter test requires removing the switch plate - and that is all -unless you can't get your test lead to the terminals without loosening the switch from the box - but the troubleshooting can all be done live.

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Well, I tried again.
I actually do know how wiring normally works, but that doesn't prevent me from having a brain fart in the middle of a project.
So, bulb (CFL) in, take some measurements off the two terminals of the switch. (note, neither switch nor box appears to be grounded; with switch off I can't pull any voltage from either terminal to box or ground screw on switch, and only two wires enter the box, both look to be #12 solid, one black and one white)
Switch on. Both terminals should be at same potential, meters should read zero. Simpson 270 reads 0, R/S digital reads roughly 10 - 14, floats a bit.
Switch off. One terminal should be hot, one terminal should be grounded through the lamp. Simpson reads 124, digital reads 121.
Okay, now unscrew the bulb and repeat.
Switch on, meters should read 0. Yup, both do.
Switch off. What should they read? One is hot with respect to ground, but there is no ground in the box. The other is nothing, it is theoretically not grounded through the lamp but just a piece of metal in thin air. No current could flow, so intuitively the voltage difference should be zero, but I guess since we really don't know what potential that wire is sitting on it could be anything.
R/S digital reads 84 V. Simpson reads 25 V.
Which, if either, is real? The R/S will sometimes give me 10 V on just one probe, if the other is capacitatively coupled, but I've never seen phantom voltages off the analog.
I guess the test would be to put a load on and see if I could draw a current, but I've spent enough time on this and there are other honeydoes to get to.
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wrote:

zero, replaced the bulb and checked the voltage across the switch, which read 80 with the switch on and off - then you replaced the switch and got 80 and zero, and the light worked.
If, as you state now, the 80 volts was with the bulb removed, capacitive coupling was giving you the 80 volts. (which happens not only with digital meters, but with "sensitive" analogs as well.
You DO understand that you were not using the meter correctly? The CORRECT measurement is line to neutral - or in absense of an available neutral connection, line to ground. With a high impedence meter (like most digitals) you can substitute your damp finger for ground and usually end up within a couple of volts of accurate due to the high capacitance of your body coupling to ground (and NO danger of a shock).
A good switch will read line votage on the line side, both on and off, and zero volts (or extremely close) on the load side with switch off and load connected, and line voltage with switch on - load or no load. I specify load connected on the OFF position because you can get those "phantom" voltages on a "disconnected" lead - which will dissapear if the wire is grounded through the load.
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On May 16, 11:40pm, snipped-for-privacy@snyder.on.ca wrote:

I'm following you, and agree, but with a caveat or two, because you're making assumptions that are not explicit.
One is that there is no real voltage in the circuit. That is not a given. There can be sources connected accidentally that give real voltages between the 0 and 120 references. E.G., transformers for doorbells (it is at the back door),
Another is, coupling to what? Hot is 120 referenced to neutral (there is no such thing as voltage, only voltage difference) but while ground SHOULD be the same as neutral, unless forced it may not be. The coupling can be to ground which may be at various potentials, or at some other object in the area - my aluminum screen door right next to it, etc.
I didn't know analog meters could have that problem, I've always assumed when I was getting weird readings off a digital it was better to dig out the Simpson. And cart around ten pounds of meter instead of five ounces. Hee, hee.

Well, yeah, I'll concede that one. When I opened the box I figured I'd check both sides of the switch to ground, and I'd find one leg hot and the other not, but of course there was no ground. And since you're supposed to switch only a hot, no neutral either. (I knew there was no power at the fixture, I wanted to check power to the switch, hoping I wouldn't have to trace back any further) I guess in hindsight I should have run a jumper to the nearest solid ground and checked that way instead.
But it's not really correct to say I was using it as an ammeter. An ammeter measures current in a circuit. With the meter in series, that 10 million ohm resistance is effectively an open circuit, no current can flow. And with the bulb removed, the circuit should have been broken at an additional place.
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You could have just used a jumper across the switch as a test and skipped all the meter drama.
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wrote:

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.
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wrote:

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.
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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.
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wrote:

with the load. A voltmeter is connected ACROSS the load.
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On May 17, 8:45pm, snipped-for-privacy@snyder.on.ca wrote:

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?
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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.
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On 5/18/2012 11:36 AM, Ralph Mowery wrote:

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": http://www.nema.org/stds/eng-bulletins/upload/Bull_88_reaffirmed_12_15_11.pdf

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.
--
bud--

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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 = 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.
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On 5/19/2012 4:14 PM, TimR wrote:

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).
--
bud--



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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.
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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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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.
Not 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.
Show 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.
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wrote:

mention the possibility that he measured 80 *milli*volts rather than 80 volts. If the meter is auto-ranging, that is a distinct possibility and often missed by those dambling in electrical work. The display says 80.0 and a tiny little mv is displayed on the side instead of V.
Pat
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