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.
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.
Doug Miller (alphageek at milmac dot com)
I've had only HORRIBLE experience with digital voltmeters (plural).
Please do say more about the problems with them.
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
(At least that's my experience.)
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
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
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
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:
[clang] Bring out your dead!
I'm not dead!
Nothing. Here's your ninepence.
I'm not dead!
'Ere. He says he's not dead!
Yes, he is.
Well, he will be soon. He's very ill.
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.
On Tue, 24 Oct 2006 17:34:36 GMT, email@example.com (Doug Miller)
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.
---------------- question about *this* post:
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?
(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!).)
I was in Lowe's recently and heard this over the loud speaker:
"mr smith to aisle 6, wirecutting emergency"
"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
Enterprise Manager do something like this when adding a server
hitting the "..." button to select the server.)"--John Fisher
On Wed, 22 Nov 2006 02:00:50 +0000 (UTC), firstname.lastname@example.org (David
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.
An effect almost unnoticable when the voltages source has low
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
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).
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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