# Help: low voltage in circuit, switch off

• posted on June 10, 2006, 5:58 pm
I have a circuit in my house with a funny characteristic, and I'd like to track it down. In that circuit, I measure a slight voltage difference between power and neutral at a light fixture when the switch to that fixture is off. Yet, my meter shows no continuity when I try to measure resistance. Other circuits do not show any voltage when they are switched off.
I assume this means that the neutral on the peculiar circuit is not at a common ground with the 'real' ground. It does not strike me as inherently dangerous, yet it seems like I might as well sort out the problem. Anybody seen this and have a recommendation on troubleshooting?
thanks,
Steve

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• posted on June 10, 2006, 6:17 pm

It is probably nothing to worry about. If you are using a digital meter it is most like picking up some induced voltage from other conductors near it or the switch could be leaking a very small ammount. How many volts are you calling 'slight' ?

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• posted on June 10, 2006, 6:54 pm
I'm using a ~20 year old analog multimeter. I measure ~5V on the 30V scale, but about 0.5V on the 3V scale; so it's just enough to boost the meter off zero. I was wondering about induction, the circuit in question is only several meters from the service box, but next to a GFI outlet.
Steve
Ralph Mowery wrote:

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• posted on June 10, 2006, 8:00 pm
Steve wrote:

The other folks keep calling it "induced" voltage, but you only really couple inductively if there's a current in one wire running next to another one, not just voltage on the wire.
From your description it sounds like you have capacitive coupling across the switch. The capacitive reactance behaves like a very high impedance at 60 HZ. For the purpose of understanding what your meter is showing, you could consider it like a very high value resistor across the switch, conducting current from the 120 volt source through your meter's internal load resistance to ground.
Now, your conventional analog meter will have a resistance rating, something like "20,000 ohms per volt". That means, when set to it's 30 volt scale it will present a load of 600,000 ohms, but when set to its 3 volt scale it will present a load of 60,000 ohms.
That's why you see two different voltage readings, the current flow is pretty much constant, as the voltage across the capacitive reactance is nearly 120 volts and varies less than 5 volts between the two measurements, but the meter's load resistance is changing by 10:1, so that relatively constant current creates two significantly different voltages across the meter's input terminals, and that's why your two measurements differ.
Either way, it's nothing to worry about, if your meter is a 20,000 ohm per volt unit the current flow is approximately 5/600,000 amperes, about 8 microamps. You prolly couldn't feel that low a current even if you stuck your tongue on the wire while grabbing metal water pipes with both hands.
Jeff
--
Jeffry Wisnia
(W1BSV + Brass Rat \'57 EE)

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• posted on June 11, 2006, 3:20 pm
On Sat, 10 Jun 2006 16:00:43 -0400, Jeff Wisnia

Induction occurs when the current changes, something happening all the time with AC current. A constant DC current will not induce anything. That seems to be one reason for using AC so much, it allows you to use transformers.
[snip]
--
Mark Lloyd
http://notstupid.laughingsquid.com

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• posted on June 11, 2006, 4:22 pm
Mark Lloyd wrote:

Correct of course, I should have said "AC current". I was cought up in the OP's subject and I don't normally think of DC in regard to household power wiring. <G>
Jeff
--
Jeffry Wisnia
(W1BSV + Brass Rat \'57 EE)

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• posted on June 12, 2006, 4:08 am
On 6/11/06 8:20 AM, in article snipped-for-privacy@4ax.com,

You may note that as you change the voltage setting by a factor of 10, the measured voltage is also changed by a factor of 10. That is, the needle position is the same for the two settings.
What this suggests to me is that you are measuring capacitive coupling from the hot side of the line across to the disconnect hot side past the switch. The small capacitance is high impedance compared to the resistance of the meter. In the end, you are getting constant capacitive current flowing through your meter.
If you know what your meter resistance is, you can place a resistor of that value in parallel with the meter. The "measured" voltage will be cut in half because equal currents are flowing in the meter and in the shunting resistor.
Meter resistance is easily determined from the meter's rating in ohms per volt. Just multiply the ohms per volt by the full scale voltage setting.
Bill -- Ferme le Bush

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• posted on June 10, 2006, 11:41 pm
Steve wrote:

Hi, Then try to hook up a small flash light bulb to see if it lights up.

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• posted on June 10, 2006, 11:39 pm
Ralph Mowery wrote:

Hi, Having high impedance input, DVM will pick up anything. Try old fashion analog meter such as Simpson 260, an old work horse. I still keep it handy besides my Fluke DVM.

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• posted on June 10, 2006, 6:22 pm
Basically, your meter is far better than meters of yesteryear. It has extremely high input impedence, so it doesn't 'labor down' the circuit it's measuring, even slightly.
What you're measuring (almost certainly) is a small AC voltage that's induced by the lengths of wiring being in close proximity to each other, much like a transformer. This voltage has no current delivering capability, as can be seen by shorting the wire to ground.
This perplexing phenomenon gives people fits until they understand it.
Steve wrote:

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