I was performing some 12 volt DC voltage drop tests and I got a reading that I did not understand. Hopefully someone here can satisfy my curiosity.

I had the positive lead on a 12 volt battery positive. When I touched the tip of the common lead with my fingers, I got an 11 volt reading. I was not grounded to the system. Out of curiosity I then touched a small piece of metal that was not grounded to anything...it had a 1 volt reading. How is that possible?

I thought the common lead had to return to battery ground to measure the difference between the leads. I know now that is not the case, but I don't understand what is happening to get that reading. Can a human complete a circuit and power a load without returning to battery ground?

  • Get an analog meter. They aren't as sensitive to the stray fields that are everywhere now... JMHO.
    – JACK
    Oct 20, 2020 at 14:44
  • I am not sure how an analog meter would answer my question.
    – MAL
    Oct 21, 2020 at 3:20
  • I'm not trying to answer your question or I would have posted it as an answer. Just a suggestion so you don't get all the crazy readings .
    – JACK
    Oct 21, 2020 at 12:25

1 Answer 1


On some ranges, a digital voltmeter (DVM) may have very high input impedance. A small static charge builds up on any non-grounded conductor. Atmospheric electricity is always present. The potential averages about 100 V/m height difference, so there is might be a 200 V difference between a 2 m tall person's head and toes. Of course, the wet, salty body is a fairly good conductor, so it effectively short-circuits most of the charge to ground... unless, perhaps, you're wearing rubber-soled shoes. So the DVM could have been measuring charge buildup.

A DVM also may be subject to self-rectification of AC pickup. Even though the scale is set to DC voltage measurement, RF pickup from radio stations and 50 or 60 Hz pickup from the power mains can still show as an erroneous DC reading.

As an experiment, I just set an inexpensive (US$6) DVM on the 200 mV scale, separated the leads at arms length, but with no contact to the probe tips, and walked around the room. Motion on the carpeted floor generated a large static charge, that bled off in a few seconds. Placing the leads near a WiFi router picked up a small signal, i.e., accidentally rectified RF.

You can make an even higher impedance voltmeter, an electroscope, with household materials. BTW, a spark jumps an air-gap at, very roughly, 1,000 V/mm, so a 1 cm spark made by walking across the floor and reaching for a grounded object might be 10,000 V. Shocking!

But you ask could you power something. It depends on the current (I), i.e., amount of electricity (compare with amount of water, whether a drop or a bucket full), as well as the voltage (compare with water pressure). Wet human skin, in series with the body, may have a resistance as low as ~500 ohms, and dry skin as high as 100,000 ohms. Resistance impedes, or partially blocks electricity, so you would not have much current, measured in amperes (A), milliamperes (mA, 1/1,000 A) and microamperes (µA, 1/1,000,000 A). Current = voltage / resistance (I=V/R), so at 12 V DC with 100,000 ohms, that would yield a current of 0.00012 A (0.12 mA), which could light a LED (light emitting diode) dimly. It would be safe to hold one lead of the LED, touch a finger to one side of the battery and the other LED to the other side of the battery. If it doesn't light (dimly, so this won't be visible in daylight or a brightly-lit garage), reverse the leads of the LED.

There would not be enough current to power an incandescent lamp, motor, or almost any other device than a lone LED, or perhaps a LCD clock, though.

  • Thanks DrMoishe, I should have prefaced this with I am not an electrician, and other than continuity I have limited understanding of electricity, I just ran the tests again with the same results...I held the leads longer this time, it dropped to 10V and .8V on the metal part. It was damp outside for both tests if that makes a difference. Here is what I don't understand If the difference between the leads is 12 volts at the battery terminals...there are 12 volts available, do I have 11 volts available to power a load when connected with battery+?
    – MAL
    Oct 20, 2020 at 3:26
  • There are two factors: voltage and current. See modified answer. Oct 20, 2020 at 17:50
  • I think I am starting to understand. It sounds like if I would have applied a load to my human/battery positive circuit, I would have seen a very large voltage drop in the circuit. Can a circuit be completed and power a load using the battery positive from 1 battery and the battery negative from a different battery or any other outside ground?
    – MAL
    Oct 21, 2020 at 15:52
  • The word "circuit" is just what the name implies: the current must make a full circle through al devices. If one lead of a battery are not connected, no DC current flows. That said, AC current can flow due to capacitive effects, and the impedance of free space -- see en.wikipedia.org/wiki/Impedance_of_free_space Oct 21, 2020 at 16:46

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