# AC measures 27volts

Can anyone tell me why I would get a 27 volt reading on a 120v circuit. I have a old home but some of the home has been re-wired. I am remodeling my bath room and removed old florescent lights. I tested the power at the connections and got a reading of 27v. I climbed in the attic and it looks like the wire is coming from a junction box with several other wires. It looks like the work was done by a pro. The upstairs has 2-20amp (connected together) circuits.
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snipped-for-privacy@yahoo.com wrote:

Probably because you're using a digital voltmeter, which is reading a very very low amperage induced current in the conductor you're testing. Try using an analog meter instead, and you'll probably see -zero- volts.
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Regards,
Doug Miller (alphageek at milmac dot com)
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Could be 27 milivolts, I've done that.
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Christopher A. Young
You can\'t shout down a troll.
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I've had only HORRIBLE experience with digital voltmeters (plural).
And if some are OK, or even EXCELLENT -- what is it that makes them that way -- as compared with the el-cheapo ones that seem reliable only on DC (batteries, etc). AC -- forget it.
(At least that's my experience.)
Thanks,
David
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snipped-for-privacy@yahoo.com wrote in

You are using a low impedance (resistance) meter (cheapo). It is not invisible to the circuit when you connect it. All meters have an internal resistance.
The lower the resistance of anything, the more current (amps) will flow through it. The more current that flows through it the higher the voltage drop across it.
The sum of the voltages in a loop with 120v source will be 120. How the 120 is divided among the components in the loop depends on the resistance of each device but it will always add up to 120.
I have a cheapo digital and an old high impediance analog meter. When I get a reading such as you did, I get out the analog. It always proves the cheapo wrong.
You have a few choices here:
a) Trust me (ill advised according to many). b) Get a high impedance digital or analog (needle type) meter. c) Get a cheap test lamp for 120. Compare brightness of lamp to an outlet working normally. d) Grab the alleged 27v wires. Do you get a tingle or jolt? (ill advised according to many)
Option d has the advantage of a free ekg if you survive. If you don't survive, as the Monty Python "Bring out your dead" skit went: ---------------------
Customer Here's one.
Cart Master Ninepence.
Cart Master What?
Cart Master 'Ere. He says he's not dead!
Customer Yes, he is.
Cart Master He isn't?
Customer Well, he will be soon. He's very ill. ---------------------

I'm no electrician but this sounds fishy.
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Al Bundy wrote:

Precisely backwards...a _high_ impedance meter can load a circuit and read "phantom" voltages. For such tests of household wiring circuits an inexpensive analog meter is probably more reliable than the digital.
To OP, need more. Were the lights functioning before and where/what are you measuring? Breaker on/off, at the switch or the feed.
What you're seeing at the upstairs junction box is probably simply a feed junction. Two circuits connected together is one circuit.
If you're actually measuring something that should be a true voltage, it implies a loose neutral or hot. If you're simply measuring the leads from the switch you just disconnected w/ the switch off, it is almost certain your _HIGH_ (not low) impedance digital meter is loading the circuit. Put a light bulb across the wires and measure again and it will undoubtedly be zero.
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dpb spake thus:

Nope, *you've* got it backwards: high impedance = high sensitivity. For a meter with an impedance in the megohm range, it's very easy to pick up stray, "phantom" voltage readings.
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Ummmm.... that's exactly what he said. What's "backwards"?
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Regards,
Doug Miller (alphageek at milmac dot com)
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Doug Miller wrote:

I usually place the terminals of a battery across my tongue. A sour taste implies the battery is good. Haven't tried that technique as a test for impedence. Maybe impudence.
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Doug Miller spake thus:

The part where he said "a high impedance meter can load a circuit". I shoulda been more clear; a high-impedance meter reads phantom voltages because it loads the circuit *less*, not more.
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Since when does a high impedance place less load on a circuit than a low impedance??
--
Regards,
Doug Miller (alphageek at milmac dot com)
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On Tue, 24 Oct 2006 13:35:09 GMT, snipped-for-privacy@milmac.com (Doug Miller) wrote:

Since "impedance" took on its current meaning.
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62 days until the winter solstice celebration

Mark Lloyd
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--
Regards,
Doug Miller (alphageek at milmac dot com)
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On Tue, 24 Oct 2006 17:34:36 GMT, snipped-for-privacy@milmac.com (Doug Miller) wrote:

Would you be saying you don't know what "impedance" means?
A high impedance source (such as that 27VAC probably is), is essentially a voltage source in series with a resistor. The effect of this is that any attempt to draw current from this source will lower the voltage, potentially to near zero. Phone lines are like this.
A high impedance load (such as a digital meter or VTVOM/FETVOM) will draw very little current from a source (not lowering the voltage very much). Note that such a meter is necessary for some sensitive electronic circuits (that would be disturbed by a low impedance load).
An analog meter on a high impedance source will give inconsistent readings on different ranges, since it has a different impedance on each range, loading the circuit differently. Adding a (120V 60W) light bulb in parallel will make the impedance MUCH lower.
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62 days until the winter solstice celebration

Mark Lloyd
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Not at all -- and if by "a high impedance ... loads the circuit less" you mean that a high impedance draws less current than a low impedance, I have no disagreement with you.
--
Regards,
Doug Miller (alphageek at milmac dot com)
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On Tue, 24 Oct 2006 20:18:25 GMT, snipped-for-privacy@milmac.com (Doug Miller) wrote:

That's what I meant.

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62 days until the winter solstice celebration

Mark Lloyd
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... ...
#1:

#2:
#3:
About #1, please explain what effect the impedance of the "source" has on this problem. (I'm sure it has one, just not sure what it is.)
About #2: You're trying to measure volts (as opposed to current); thus you're putting the two leads "across" the load.
eg, you have an extension-cord onto which is plugged three floor-lamps, a stereo, etc. So mentally you'd like to rub off the outside insulation on the two wires in the extension cord and touch your two leads there.
Obviously you want to NOT disturb the situation (by merely measuring it), you want your meter to have HIGH impedance -- so that only a TINY bit of current runs through it.
Whereas one with low impedance (resistance here) would get more (much more?) current running through it, "disturbing" the situation.
So that sort of "proves" (well, restates) #2.
About #3: Here you mix in the source-concept with the meter.
Could you perhaps explain this part a bit more?
Thanks!
David
(Way, WAY back when, I recall the name "Thevenin's Theorem", something to do with the source, maybe, but I surely have no concept now (not sure I ever did back then!).)
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David Combs wrote:

I was in Lowe's recently and heard this over the loud speaker:
"mr smith to aisle 6, wirecutting emergency"
mk5000
"The question is this: What is the best (or at least a good) way to search all the machines on a LAN for running instances of this COM object? (I've seen SQL Server's Enterprise Manager do something like this when adding a server registration and hitting the "..." button to select the server.)"--John Fisher
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On Wed, 22 Nov 2006 02:00:50 +0000 (UTC), snipped-for-privacy@panix.com (David Combs) wrote:

I forgot what they CALL that, but any real voltage source can be considered as an ideal voltage source (one that never changes voltage under any load) in series with some impedance. That impedance determines how much the voltage will drop when a load is applied.
BTW, one high impedance source people have is a phone line. 48VDC or so, dropping to 6-10V when a phone is in use.

Yes, that's how you measure voltage.

Right.
An effect almost unnoticable when the voltages source has low impedance.

I found the idea of source impedance difficult to understand once. Think of a 12V battery with an internal series resistor, where you have NO access to the point between the battery and resistor. You just measure the voltage at the terminals.
1. Measure the voltage with no load.
2. Measure the voltage with a load. This voltage will be lower than the no-load voltage. The amount of this drop depends on the resistance of that hidden resistor.
The load has resistance, which is in series with than internal resistance. Voltage is divided between the series resistors, according to resistance.
Real batteries aren't normally made with resistors, but they do have some internal resistance and will show a voltage drop with load.
Impedance is like resistance, but with AC. It considers that some components react differently to AC.
Where you read 120V with a DIGITAL meter, but 27V with an ANALOG meter, you've found a high impedance source. When a wire that's not connected is physically close to one with current in it, these wires act as a transformer (not a very good one).

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33 days until the winter solstice celebration

Mark Lloyd
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Doug Miller spake thus:

Instead of "impedance", substitute "resistance" (which is essentially the same in this context) and then re-think the question: I'm assuming you know enough about basic electricity to work it out for yourself.
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