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Here's one mental picture of a battery:

The electro-chemical reactions inside the battery happen only when there's a closed circuit. When you place a voltmeter across the poles, then you create a closed circuit (with very low current) - the electro-chemical reactions start, and you now have a voltage between the poles. The placement of the voltmeter changes the thing you are trying to measure. Without the voltmeter, there's no voltage between the poles, because there's no closed circuit.

But the above seems completely inconsistent with how people speak of open-circuit voltage, as existing independent of actually measuring it.

My reasoning must be wrong somewhere. But where?

(I know this is a very simple question, but I'm genuinely confused by it.)

Here's an example of the typical use of open-circuit voltage (see 24:00), from an MIT lecture.

Edit: replaced voltage-drop with voltage.

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    \$\begingroup\$ The ions are pushing all the time (and creating an electrical potential difference). They just can't go anywhere so there is no flow of charge and I is zero so the chemical reaction is, in effect, stopped. Think of a water pump on a closed loop of filled pipe. Close a valve and you still have pressure but no flow. \$\endgroup\$ – Transistor Feb 4 at 12:52
  • \$\begingroup\$ Interesting thoughts... Think of the battery of as a tensioned spring... or a car tire full of compressed air... or just like any container that stores potential energy... \$\endgroup\$ – Circuit fantasist Feb 4 at 12:53
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    \$\begingroup\$ Bill Beaty discusses a facet of what you may want here. Bill is active in the NNTP part of the internet and is a research engineer at the University of Washington's Chemistry Department in Bagley Hall. You probably could just call him up and discuss your thoughts and questions. \$\endgroup\$ – jonk Feb 4 at 18:55
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    \$\begingroup\$ Thought experiment: If potential difference requires a closed circuit how would current ever flow when switching from an open-circuit to a closed circuit? There would be no EMF (electro-motive force) to generate a current to boot the chemical reaction in the battery. \$\endgroup\$ – Transistor Feb 4 at 19:46
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    \$\begingroup\$ "If a battery is in the forest, and there is nobody around to measure it, does it have a voltage?" \$\endgroup\$ – Michael Feb 7 at 18:28
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The electro-chemical reactions inside the battery happen only when there's a closed circuit.

For an ideal battery, this is true. However, all batteries have some, hopefully very small, leakage current, and will eventually discharge. Hopefully it takes years for this to happen, but that is not always the case.

When you place a voltmeter across the poles, then you create a closed circuit (with very low current)- the electro-chemical reactions start, and you now have a voltage drop between the poles.

The voltage between the poles existed prior to connecting the voltmeter or closing the circuit. It is the voltage between the poles that keeps the electro-chemical reactions from proceeding. If the battery is not connected to anything, the voltage between its poles exactly matches the electro-chemical potential of the reaction.

The placement of the voltmeter changes the thing you are trying to measure.

Yes.

Without the voltmeter, there's no voltage-drop between the poles, because there's no closed circuit.

This is mistaken. There is a voltage difference between the poles whenever the battery is holding a charge. There is no current when there is no closed circuit.

But the above seems completely inconsistent with how people speak of open-circuit voltage, as existing independent of actually measuring it.

Open circuits may have voltage, and this voltage exists independently of attempts to measure it.

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    \$\begingroup\$ Maxwell-Faraday law says that in the absence of a varying magnetic field, the curl of the electric field is 0, thus the electric field is conservative, thus, is equal to the gradient of a potential. That potential we call voltage. Nothing about an electric field across a battery contradicts this. If you take any loop, the sum of the voltage differences between points around the loop will algebraically sum to 0. That is Kirchhoffs Voltage Law, which is observed by a voltage across a batteries terminals. \$\endgroup\$ – Math Keeps Me Busy Feb 4 at 14:54
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    \$\begingroup\$ If you cannot understand that a voltage may exist in the absence of a chemical reaction (or magnetic induction) you will have difficulty understanding pn junctions in semiconductors. Learn about diffusion current and drift current, and the voltage across a disconnected pn junction. \$\endgroup\$ – Math Keeps Me Busy Feb 4 at 15:08
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    \$\begingroup\$ choose any loop, say from the anode of the battery, through the battery to the cathode, then through the air back to the anode. The voltage between the anode and cathode is the same when the path is through the battery as it is when the path is through the air. That is what KVL says. \$\endgroup\$ – Math Keeps Me Busy Feb 4 at 15:19
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    \$\begingroup\$ Yes, there is an electric field in the air, and yes if you shake a charged battery in a vacuum it will induce magnetic fields. \$\endgroup\$ – Math Keeps Me Busy Feb 4 at 15:29
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    \$\begingroup\$ @JohnO when they put the chemical together to manufacture the battery, a small amount of them react and create a voltage drop. The voltage drop then stops the chemicals from reacting any more. \$\endgroup\$ – user253751 Feb 5 at 3:49
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The chemical reactions inside the battery cause opposite charges to build up on the terminals.

The buildup of opposite charges creates a voltage difference between the terminals, whether the batter is connected or not.

When the battery is disconnected, the charge builds up until it's sufficient to stop those chemical reactions. That's why the reactions only happen when the battery is in a circuit.

When a load is attached to the battery, it drains those built-up charges away, allowing the chemical reactions inside the battery to proceed until they're built back up.

The voltage on the terminals is greatest when the battery is disconnected, because it's sufficient to stop the chemical reactions. When the battery is being drained by a circuit, the voltage is a little lower, allowing reactions to proceed. The higher the drain, the lower the voltage, the faster the chemical reactions, and the greater the current. That effect creates what we can measure as the battery's internal resistance.

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Without the current loop there's a balance state generated inside the battery. As much electrons have taken the new places in the forming new compound as possible before the generated internal charge imbalance compensated the attraction caused by the free lower energy electron states. But some amount of electrons have taken the new places, so they are off from their original places. We see it as +charge at the +pole of the battery. Respectively there's some extra electrons in the minus pole. Thus the voltage exists without any load such as a voltmeter.

The questioner suspects that no voltage should exist with no load because no electro-chemical reaction should happen with no current. I said that it happens, but stops due the built balancing electric field. The balance occurs as no-load voltage.

BTW. I must admit that I have no idea what could cause the claimed current before there's any voltage.

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  • \$\begingroup\$ If you place an electroscope near the poles of the battery, does it show evidence of charge? \$\endgroup\$ – John Feb 4 at 14:46
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    \$\begingroup\$ Yes if you have sensitive enough electroscope and you can place it near enough the gap between the plus pole and the minus pole. But inserting an external device causes current, no matter it's only capacitance charging. That spoils the idea of "NO load!" \$\endgroup\$ – user287001 Feb 4 at 14:57
  • \$\begingroup\$ Do you know of any video demos of that effect? \$\endgroup\$ – John Feb 4 at 15:12
  • \$\begingroup\$ I have no idea how to construct such small electromechanical electroscope. But a mosfet hopefully could get you confirmed that the voltage exists. Connect the battery between G and S. Have high enough battery voltage and use an ohmmeter to show does the fet conduct. For more confirmation you can put capacitors to the wires - say 10 nF plastic insulated - only remember to discharge them by shorting them momentarily before connecting. The gate of the fet must also be discharged beforehand without recharging it accidentally (=difficult, but possible) \$\endgroup\$ – user287001 Feb 4 at 15:58
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It is always Ohm's law and voltage deviders.

Say you have two battery terminals. Between the terminals there is air, which have a conductivity, very low though. In theory if you take a multimeter and probe from one battery terminal to the air in the middle of the 2 terminals you should have seen half the battery voltage. But you do not. In that case the resistance of the air is so incredibly high that not enough current can flow into the multimeter.

In the case you describe you can see the battery, as an ideal battery with a series resistance internally. It could be very low, say in lithium-ion batteries, much less than 100 mohm. In other batteries it can be much higher like 3 ohm.

But the multimeter which will complete the circuit is often 1 megaohm. Sometimes it has its own internal voltage divider with 9 megaohm to 1 megaohm (10:1).

On a 3.6 volt battery that would equate to 3.6 V/(1 megaohm + 3O ohm) = 3.6 uA. Then taking this current in relation to the internal resistance of 3 ohm, 3.6 uA * 3 ohm = 11 mV.

So by completing this particular circuit you measure 11 mV less than the actual battery voltage.

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Ideal voltage sources have 0 resistance. Yet all “Practical” ones have some low value. So the voltage is always present in this open circuit and only reduces from current passing thru it’s internal resistance to an external resistance in a current loop.

Batteries are voltage sources with an effective series resistance, ESR often neglected if ESR is <1% of load R, yet you expect battery voltage will drop by this ratio of ESR / (ESR + Rload).

Do not think of electrochemical battery cells as 0V in open circuit. This only happens if shorted or expired from previous loads for a long time. The short circuit currents can be very high if ESR is very low. Voc/ESR= I short cct.

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There is always a voltage-difference V between the terminals, regardless of being connected to a load or not.

If you stick a piece of metal in a solvent, a voltage develops between the metal and the solvent. It's just as simple as that. The voltage depends only on the chemistry, and not on the geometry (the size of the plate, for example).

Apparently the voltage-gradient is confined to a small area near the surface of the metal, of the order of a few molecules thick.

I'm not sure, but the constancy of the voltage looks like it might be an instance of Le Chatelier's principle.

Your original error may have been the assertion that the chemistry is "turned off" when there's no circuit. I think that may be false. It may be a case of mistaking equilibrium for nothing happening.

From General Physics, by Landau, Akhiezer, and Lifshitz:

Thus chemical equilibrium (and in fact other types of thermal equilibrium) is dynamic on the molecular scale; the reactions do not actually cease, but the forward and reverse reactions occur at equal rates and therefore produce no overall effect.

That is, the chemistry is happening even in the disconnected cell, with forward and backward reactions in balance, and keeping a steady state (neglecting long-term effects). That would explain a lot.

To understand this better, it helps to focus on the lead-acid battery. These references were helpful:

http://ecee.colorado.edu/~ecen4517/materials/Battery.pdf

http://amasci.com/elect/battvolt.html

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  • \$\begingroup\$ The first version of this answer was goofy. I have re-written it. \$\endgroup\$ – John Feb 5 at 15:39
  • \$\begingroup\$ "If you stick a piece of metal in a solvent, a delta-V develops between the metal and the solvent." This doesn't seem quite right. I think you need a second, dissimilar electrode. (The downvote isn't mine.) \$\endgroup\$ – Transistor Feb 5 at 15:42
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    \$\begingroup\$ I personally find your usage of Voltage "drop" a bit confusing. I'd say there is Voltage across the terminals of a battery (or differential and so on), but I'd only use "drop" when referring to voltage losses related to internal resistances or connection resistances and so on. Point being, there doesnt seem to make sense that there is a "drop" in the voltage of an unconnected battery. Perhaps this is just weird to be because I'm not a native english speaker and your usage might be 100% ok. I'd like to hear from other members if possible. \$\endgroup\$ – Wesley Lee Feb 5 at 15:46
  • \$\begingroup\$ I will change it to "difference" instead of drop. \$\endgroup\$ – John Feb 5 at 15:51
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    \$\begingroup\$ @WesleyLee, your English is very good if you can detect that. You are correct that we usually use "voltage drop" in the context of voltage losses due to resistance (anywhere) in a current carrying loop. \$\endgroup\$ – Transistor Feb 5 at 16:21
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... how people speak of open-circuit voltage, as existing independent of actually measuring it.

Or mass: how people speak of mass, as existing independent of actually measuring it.

Pressure makes a better analogy than mass, though. Are you familiar with the pneumatic tools mechanics use in automobile-repair shops? Suppose that the shop's air compressor has charged the pressure tank with compressed air to a pressure of 1000 kPa. Suppose that the mechanics have stopped work to take a ten-minute break. No air is flowing from the tank. Does the tank's pressure still exist?

Voltage is much the same.

If unsure, consider that

  • pneumatic power equals pressure times volumetric flow rate, whereas
  • electric power equals voltage times electric current (that is, times electric-charge flow rate).

The analogy is fairly precise.

VENTING

It's confusing until you realize that venting the pressure tank lets the pneumatic charge escape, whereas venting the battery does not let the electric charge escape. Venting the battery resembles stopping the pressure tank by stuffing a plug into the vent, for air is an electric insulator.

If you wish to plug a battery so that the electric charge cannot escape, plug it with air!

If you want electric charge to escape, though, then you must remove the air that is plugging the battery, so to speak. Vacuum won't do that, but a copper wire will, because the copper wire is a medium through which electric charge can escape.

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