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Let's say I have three points: A, B and C. I measure the voltage between A and B: it's 0. I then measure the voltage between A and C, and it's again 0. So, the voltage between B and C must also be 0, right? Well, not quite so.

Let's say I have two batteries, and I call the + terminal of the first battery A, the + terminal of the second battery B and the - terminal of the second battery C. Here the previous conclusion fails: the voltage between A and B as well as between A and C is 0 (both measured with a voltmeter), but the voltage between B and C is naturally the voltage of the battery. What's wrong here?

I guess the answer must be related to the absence of a closed circuit, but I'm feeling that the conclusion of my first paragraph should apply regardless of that.

Edit:

My question was motivated by this other popular question. There it is stated (accepted answer) that: for current to flow, a circuit does not need to make a physically closed loop. If point A is fixed at \$ 0V \$ and point B is fixed at \$V_b\$, they don't need to be physically connected for this property to be true. When we connect a resistor R between point A and point B, the current from B to A will be \$ I_{BA}=V_b/R \$.

But the answer I've got so far is telling me that I need an a priori closed circuit in order to be able to measure a voltage difference.

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LGenzelis
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    Bla bla bla.. nope, sorry, my brain refuses to translate this into a schematic, if you want anything close to an answer **you** will have to draw a schematic. Imagine a TV repair technician not having a schematic but only: collector of T1 is connected to R1, R1 is also connected to C2 which is connected to the emitter of T2 which is connected to the base of L3... you get that ? – Bimpelrekkie Jan 12 '17 at 12:54
  • The most important part here is the 1st paragraph. I didn't draw an schematic since I'm talking of points between which there is a voltage difference. I'm not necessarily speaking about a circuit. This would be the "schematic": A--------0v-----B (measured voltage between A and B) A--------0v-----C (measured voltage between A and C) What would then be the measured voltage between B and C? – LGenzelis Jan 12 '17 at 22:38
  • Voltages are not absolute, they require a reference. If you have two batteries that are completely separate, you call the - of one 0V and the + of the other Vb, then the voltage between those two points is not necessarily Vb, it is unknown. – Tom Carpenter Jan 12 '17 at 23:29
  • @TomCarpenter, so are you saying that the text I quoted in my edit is wrong? – LGenzelis Jan 13 '17 at 02:13
  • Yes, very. In fact I'm quite convinced the accepted answer on the question you quoted from is utter nonsense. – Tom Carpenter Jan 13 '17 at 08:09
  • *I'm not necessarily speaking about a circuit.* Well yes you are, it involves devices which have electrical connections so you have a circuit and you should draw it so that it is clear what you mean. – Bimpelrekkie Jan 13 '17 at 08:27
  • @tom capacitance can sustain the voltage. That is my point. – Andy aka Jan 13 '17 at 21:08
  • @Andyaka except, going back to the original thing of two batteries - if you move them nearer or further apart, that voltage will change. If you touch it, the voltage will change (you have a charge). If you want to talk about femto farads or less, then yes, everything it capacitively coupled with everything else, but in any practical sense the voltage between them cannot be quantified because there is simply no way of measuring it. – Tom Carpenter Jan 13 '17 at 22:30
  • @tom my answer doesn't disagree with what you have said. In fact that is what I've tried to hint at. – Andy aka Jan 13 '17 at 23:16

2 Answers2

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What's wrong here?

Battery A isn't galvanically connected in any way to battery B hence when you measure between A and B your voltmeter is measuring an open circuit with no current flow. When you measure between A and C it's the same story.

Your meter measures 0V because there is no galvanic connection and your meter's input impedance (although in the mega ohms range) acts like a short and rapidly equalizes the potentials by discharging the floating battery called A.

EDIT for clarification

Note the point I made above "equalizes the potentials by discharging the floating battery called A". This was an important hint.

The linked Q and A (subsequently added to the question) is talking about celestial bodies and it has been shown in other questions that the capacitance that exists between these bodies is in the realms of tens if not hundreds of uF. This makes a big difference.

With (say) batteries placed on an insulating table, the capacitance between battery A and battery B might be a pico farad.

Let's now say that battery A (12 volts) is charged up to a 100 volts above battery B (9 volts). Let's say this 100 volts exists between the two negative terminals. This means that: -

  • A negative to B positive is 109 volts
  • B negative to A positive is 112 volts
  • A positive to B positive is 103 volts
  • A negative to B negative is 100 volts

If you took your high impedance voltmeter and measured these voltages you would immediately start to discharge the 100 volts but at what rate. Well, it's a simple formula; it's called the RC time constant.

For "large bodies" at different voltages, C is about 100 uF, the meter input impedance is about 10 Mohm and the RC constant (seconds) is 1000 seconds. In other words, after 1000 seconds of measuring, the 100 volts that previously existed between the two bodies decays to about 37% of what it was i.e. 37 volts.

Taking a quick 2 second measurement you wouldn't notice the small drop in voltage and all the voltages in the bullet points above would largely remain the same.

But, you probably have only about 1 pico farad so the RC time constant is 10 micro seconds. In other words, by the time your meter registered anything the charged voltage would, for all practical purposes become zero and you would have no current flow.

Andy aka
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    As you can see Andy has a much better text-to-circuit translator than I do, well at least not such a lazy one ;-) – Bimpelrekkie Jan 12 '17 at 13:13
  • Thanks @Andy, but that doesn't really answer my question. Please look at the edit I've just made. – LGenzelis Jan 12 '17 at 22:51
  • @LGenzelis And you could have avoided that confusion by making a drawing, yes a schematic. – Bimpelrekkie Jan 13 '17 at 08:25
  • @LGenzelis OK so the linked Q and A is talking about celestial bodies and it has been shown in other questions that the capacitance that exists between these bodies is in the realm of tens if not hundreds of uF. This makes a big difference. I shall add this to my answer. – Andy aka Jan 13 '17 at 09:21
  • "is charged up to a 100 volts above ". - in order for that to be the case there must be some connection between the two batteries. If they are not connected in any way you cannot quantify a voltage between them because there are no common nodes to reference a voltage to. – Tom Carpenter Jan 13 '17 at 19:35
  • @tom I'm talking capacitive connections and then pointed out how flimsy they are likely to be. – Andy aka Jan 13 '17 at 21:06
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But the answer I've got so far is telling me that I need an a priori closed circuit in order to be able to measure a voltage difference.

At the moment that you connect a physical voltmeter, a current must flow and there must be a loop to measure properly.

There are Electrostatic voltmeters

but these are for measuring static charges like charge that is trapped in some material (I might be incorrect, I have no experience with this). So not for a voltage (potential) difference.

Batteries do not have charge trapped inside them but provide a voltage difference by means of some chemical reaction. For a potential difference (because that is what a voltage is) you need a common reference. That is why the return wire is needed, to align the reference between two separate batteries.

Bimpelrekkie
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