Switch set to off yet there is still voltage present

    
As you can see in the picture, the switch is set to the off position.
When I test the continuity, there is no beep.
However, when I test the outgoing voltage, it is 18 Volts D.c.
What gives?
"https://www.dropbox.com/s/sngn1wyod4ni19i/Switch.jpg?dl=0 "
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snipped-for-privacy@gmail.com says...

You did not specify the input voltage or if you had a load on the ouput.
Let me make a wild assumption. YOu are using a digital voltmeter and there is no load on the output. The digital voltmeters draw such little current you are probably picking up a leakage voltage due to the insulation being damp, an arc over that is leaving a carbon trace, or just weak insulation. If you put a very small load on it,the voltage should go away.
People run into this all the time with the 120 volts AC in houses. You can almst sitck one end of the meter in the wall socket and hold the other lead in your hand and show a voltage.
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On Fri, 5 Apr 2019 14:39:42 -0700 (PDT), A K

Without some kind of load on it, you can't trust voltage readings on a digital meter. For reference, even a "good" analog meter imposed about 20,000 ohms per volt on the scale, (10v scale was 200,000 ohms).
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On Friday, April 5, 2019 at 6:20:05 PM UTC-4, snipped-for-privacy@aol.com wrote:

My digital meter is much more convenient to use. But whenever I get weird voltages, I go get the Simpson. If they agree then it's real. Mostly it's a phantom voltage from somewhere though.
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snipped-for-privacy@aol.com says...

I worked in a very large plant and there was so much wire in the conduits the digitals were almost worthless for electrical power trouble shooting. Used the Simpson most of the time. I think Fluke has came out with some meters that will place a load on the circuits now to help eliminate the problem.
I knew I had the breaker pulled on a circuit but was getting a shock. Even the simpson was showing about 70 volts . That was even with the wires disconnected from the breaker. I discovered if I started off on the highest Simpson range and stepped down the meter would stay in the same relative position if it was the induced voltage.
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wrote:

Rig up an adapter that lets you put about 10 meg ohm across the leads and it will work fine. For 99.99% of the users, it should be inside the meter anyway. It is only when you are probing CMOS or similar circuit that inserting a load that small into the circuit will give you a bad answer. Even then all CMOS pins are supposed to be driven or terminated.
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On Friday, April 5, 2019 at 8:01:48 PM UTC-4, snipped-for-privacy@aol.com wrote:

As I recall, Fluke and probably others have an adapter that you can buy and plug into the meter terminals before the leads that does that.
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wrote:

Yes https://www.fluke.com/en-au/product/accessories/adapters/fluke-sv225
Makes more sense to have a switch that does that,
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On 4/6/2019 1:39 PM, Rod Speed wrote:

...

But then they wouldn't be able to sell the added accessory at Fluke prices... :)
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More likely that its not that easy to do with a switch and retain the full voltage isolation at the highest ranges etc.
Fark, $70
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On 4/6/2019 3:02 PM, Rod Speed wrote: ...

Missed the smiley, eh? :)
--


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snipped-for-privacy@aol.com says...

Some companies have adapters like that.
However 10 megs is not nearly low enough. That is the impedance of most better quality meters. The adapter should be much lower, probably a few thousand ohms at the most.
Where I worked we had so much induced voltage in the wiring that I made a tester out of a 7 watt small light bulb as we had thousands of them on the equiipment as indicators , so had lots of spares.
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On Saturday, April 6, 2019 at 9:39:40 AM UTC-5, Ralph Mowery wrote:

I found this.
This is an old-style unregulated "wall wart" adapter. The internal circuit is probably this, and nothing more:
schematic
simulate this circuit – Schematic created using CircuitLab
The transformer is designed to fail in a safe manner (no external flames, n ot too much smoke and isolation between mains and output maintained) if it develops a shorted turn or is fatally overloaded.
The reverse leakage of the diodes in BR1 is quite low at room temperature a nd the electrolytic capacitor (perhaps 10,000uF rated at 16VDC) itself typi cally has not much leakage, so the charge can be maintained for minutes or more with nothing connected.
More modern switching supplies (as well as linear regulated adapters) typic ally drain down much faster.
If you're connecting such an adapter to a circuit of your own making, it's worth remembering this as the adapter can easily hold enough charge to dest roy something if it's connected incorrectly. With it pulled out of the wall socket, you can short the output with a wire and most of the charge will b e removed, but a bit of voltage will creep back due to the way capacitors b ehave (dielectric absorption effect).
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On Fri, 5 Apr 2019 14:39:42 -0700 (PDT), A K

The switch has a light in it???
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On Friday, April 5, 2019 at 8:37:10 PM UTC-5, Clare Snyder wrote:

Switch has no light.
Andy
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On Saturday, April 6, 2019 at 5:38:04 AM UTC-4, A K wrote:

How are you measuring "the outgoing voltage"? And what is the normal system voltage? What is the load?
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On Saturday, April 6, 2019 at 8:22:58 AM UTC-5, trader_4 wrote:

I test the voltage coming out of the transformer.
Normal house voltage is about 118 Volts A.C.
Output of transformer is 18 volts D.C 700 mA.
Andy
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On Saturday, April 6, 2019 at 10:08:41 AM UTC-4, A K wrote:

So to summarize from your last two posts:
The switch is on the primary side. You're measuring the 18V at a power supply output, with the switch turned off and it's apparently the charge left inside the power supply capacitor.
Correct?
If you had included that important info initially, the responses would have likely been different.
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On Saturday, April 6, 2019 at 1:30:52 PM UTC-5, trader_4 wrote:

I do my best, but am not perfect.
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