If a DC multimeter is attached to the plus pole of a battery and the other pin to ground, then in my understanding there is a positive potential on the positive battery pole and a zero potential at ground. Shouldn't the multimeter show the difference in potential between those two points of contact?

What is the difference between connecting the negative terminal of my multimeter to the negative terminal of the battery vs connecting it to earth ground?

  • 5
    \$\begingroup\$ Ground is not "the ground". Ground is a reference point in the circuit. \$\endgroup\$ Commented Jul 3, 2017 at 19:05
  • \$\begingroup\$ See this question, plus the many others it links to. \$\endgroup\$ Commented Jul 3, 2017 at 19:11
  • \$\begingroup\$ Then replace my reference to ground by some zero potential... maybe a piece of rubber or anything \$\endgroup\$
    – Alon
    Commented Jul 3, 2017 at 19:11
  • 3
    \$\begingroup\$ Voltmeters must draw a tiny bit of current to work. At the voltages used, you can't get enough current through a sheet of rubber (or the air.). An electrometer can show voltage differences with out a current flow. \$\endgroup\$
    – JRE
    Commented Jul 3, 2017 at 19:14
  • 2
    \$\begingroup\$ Note: Masse is used for "ground" in the field of electronics in Germany. Bezugsmasse translates as "reference/ground" and Gerätemasse translates as "device ground." So it's easy to see the usage of 'masse' as it applies in electronics. The OP did his preparatory work, except that he probably didn't realize how it sounds to an English-only speaker to add "mass" in the question. \$\endgroup\$
    – jonk
    Commented Jul 3, 2017 at 19:59

2 Answers 2


This illustration may help:

enter image description here

Figure 1. Without a ground reference the car screw-jack is unable to provide any lift.

In a similar manner to the car-jack analogy, a battery without a ground reference cannot provide any lift / thrust / force / current. We need to close the circuit.

So you have one DC multimeter and attach one pin to lets say the plus pole of a battery and the other pin to ground/mass (<- is there a difference between those btw.? haven't found anything googeling).


simulate this circuit – Schematic created using CircuitLab

Figure 2. In (a) the battery is not referenced to ground other than through the meter. No current will flow and the voltmeter will read zero. (b) In this case the circuit is completed through the ground and current can flow around the loop through the meter.

In electrical circuits ground is some reference point and, by definition is the zero volts reference. This is analogous to a surveyor picking a zero reference from which to reference all his/her other measurements. Note that voltages can be positive or negative with respect to the reference point.

In my understanding there is a positive potential on the positive battery pole.

It is only positive with respect to the other pole. Again, a height can only be positive with respect to a reference height.

Shouldn't the multimeter show the difference in potential between those two points of contact?

The fact that there is no circuit, the act of connecting the multimeter will connect the battery + to ground. In practice you will see a stray voltage reading due to capacitance between the battery and the earth but there is no power behind it and it couldn't, for example, light an LED.

Have a look at my answer to Few questions about basic concepts in electronics for more on this.

  • 1
    \$\begingroup\$ Germans refer to ground as "Masse." It has been incorrectly translated as "mass." \$\endgroup\$
    – JRE
    Commented Jul 3, 2017 at 19:48
  • \$\begingroup\$ Haven't got the hang of it entirely yet but your concept of the open circuit gave me a basic idea that's pretty solid yet to me, thanks \$\endgroup\$
    – Alon
    Commented Jul 4, 2017 at 20:22
  • 1
    \$\begingroup\$ 4 years later, I'm coming back to this and I finally understood. Thank you again! \$\endgroup\$
    – Alon
    Commented Dec 7, 2021 at 20:37
  • 1
    \$\begingroup\$ @Alon, thanks for letting me know. You might like this - and it may help you if you ever have teenage sons: "When I was a boy of fourteen, my father was so ignorant I could hardly stand to have the old man around. But when I got to be twenty-one, I was astonished at how much he had learned in seven years." --Mark Twain \$\endgroup\$
    – Transistor
    Commented Dec 7, 2021 at 20:44

If you have a flashlight battery in your hand or sitting on the table, then usually the entire battery will be charged with respect to ground. In theory, the voltage on either battery terminal can be measured.

For example, one battery terminal may be at +715.5V with respect to earth, and the other terminal at +714.0V wrt earth (but still showing the expected 1.5V difference between the battery terminals.) Or, if it's sitting on a portable plastic table, and someone had earlier brushed their skin across that table, then the battery terminals may be at -18,000 and -18001.5 volts! If your hand then approaches the battery, those voltages may change drastically because of e-fields and capacitive-divider effects, yet the 1.5V between battery terminals remains constant.

In other words, the battery is actively maintaining the 1.5V between terminals. But also, the entire battery as a whole is acting as a capacitor-plate with a few picofarads, and the surface of nearby ground is supplying the second capacitor plate. Scuff your shoes on the carpet while you stand near the battery, and the battery voltage wrt earth will vary enormously. (However, standard DVM voltmeters cannot measure it. That 18KV which had been on the battery will vanish as soon as the grounded DVM touches one battery terminal.)

An ideal voltmeter can show these large voltages on the battery if one meter terminal is connected to earth. But a standard DVM voltmeter with 10meg input impedance will behave more like a short circuit, and instantly drain out the charge on the few-picofarads of battery surface-capacitance. Even though the battery terminals differ by 1.5V, the meter reads zero for either terminal. This occurs because the terminal-voltage is instantly altered by the 10meg-ohms of the meter, wrt ground. The meter discharges each battery terminal that it touches. (The other "un-touched" terminal then changes too, maintaining the constant 1.5V between terminals.)

How "instantly" does the meter alter the battery terminal's voltage?

Suppose the entire battery acts as a 2pF capacitor plate, with ground being the other plate. If we try to measure the voltage on a single battery terminal, our DVM connects a 10M resistor between that terminal and ground. The measured voltage will fall to zero, with a time constant of:

Tc = RC
Tc = 10^7 ohms * 2pF
Tc = 20 microseconds

The moral of the story: in circuits, voltage is measured between two points on the circuit, not one. If you try to measure the voltage at a single point wrt ground, you'll end up with crazy enormous values which are strongly influenced by nearby static-charged objects. (But also you'll need a specialized voltmeter to measure these values at all. 10M ohms is far too small. Instead use an "electrometer" voltmeter with 10^20 ohms input impedance.)

  • \$\begingroup\$ lots of good input thanks I'll have to meditate over this once or twice \$\endgroup\$
    – Alon
    Commented Jul 4, 2017 at 20:26
  • 1
    \$\begingroup\$ if this answer wasn't in existence, I'd have written it. \$\endgroup\$
    – Neil_UK
    Commented Sep 23, 2021 at 13:17
  • \$\begingroup\$ hi @wbeaty, i had 2 12V supply adapters. i tried to measure the voltage between positive terminal of one supply and negative terminal of another. I thought I'll get 12V reading in my multimeter. to my surprise it stayed in zero. theoretically we can say the circuit is not closing, but i thought, since there is an imbalance in the charges between these two terminals, so a current would flow, can you explain what's wrong with my approach ?? \$\endgroup\$
    – Vignesh C
    Commented Jun 20, 2022 at 17:42
  • \$\begingroup\$ @VigneshC an imbalance of charges, that's a capacitor, might be 100pF, maybe lots more, maybe lots less. Your voltmeter is actually a 10Meg resistor. If you connect a 10M resistor across a 100pF capacitor, it discharges the capacitor in roughly 1/1000th of a second. (To measure the 12V without drawing any current, your voltmeter can't have ten megohms, it needs to have BILLIONS of megohms. Not 10^7 ohms, instead try 10^20 ohms. That way you can measure the voltage across a 100pF capacitor (without discharging it instantly!) \$\endgroup\$
    – wbeaty
    Commented Jun 20, 2022 at 23:28
  • \$\begingroup\$ Thanks @wbeaty, I can understand the concept you explained. I have one more doubt, if we have a power supply we have a positive terminal. it will pull the electrons from negative terminal. so the negative terminal is in lower potential so the electrons are flowing. if i connect the same positive terminal to negative terminal of another supply adapter, the same should happen right ?? \$\endgroup\$
    – Vignesh C
    Commented Jun 21, 2022 at 5:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.