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I came across a rather simple looking circuit of a voltage divider circuit. It has 5V at one terminal and 0V at the other terminal. Let's start by connecting two 1K resistors in series between these two terminals. The voltage at the center is found to be 2.5V(using, say, a Multimeter). Increasing the resistor values to 1M gave the same result. I keep on increasing the resistance values till I get the series resistance equivalent to air and then I remove the physical resistances, leaving the air itself as a resistance. If I now measure the voltage midway between the two terminals I should be getting 2.5V but I do get nothing(0V) instead! Why? What other factors are in play here?

enter image description here

EDIT: Thanks for pointing out the incapability of meter to measure this voltage. My meter might be incapable of measuring this voltage, but is it really 2.5V in the mid over there?

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    \$\begingroup\$ Your meter has resistance. Perhaps as much as from \$1 \:\text{M}\Omega\$ to \$10 \:\text{M}\Omega\$. So when you connect one lead to \$0\:\text{V}\$ and the other lead to "air", you are using your "low" impedance meter to "ground" the air. So it reads 0. (The impedance of air is very much higher, with any appreciable gap.) \$\endgroup\$
    – jonk
    Commented Feb 5, 2020 at 6:13
  • \$\begingroup\$ a electrosope will read half voltage at the midpoint, but I've not seen one that gives a reading at 5V \$\endgroup\$
    – Jasen Слава Україні
    Commented Feb 5, 2020 at 7:39

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Your meter needs current to make a measurement. The objective is for your meter to sink or source so little current compared to the circuit that it does not significantly affect what is being measured. This lets you approximate the current from your meter as zero. It's the whole quantum mechanics thing where you can't observe/measure something without somehow influencing that thing thus disturbing the original state that you were trying to measure.

But if the current that the meter sinks/sources is truly zero (or below that required by the meter itself which, in a way is determined by how high an input impedance it is relative to what you are measuring so that it has minimal effect as stated by @jonk in the comments. The air resistance is so high compared to the meeter input impedance the meter is like a short circuit across the air), then your meter can't sink/source this current, then no measurement.

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  • \$\begingroup\$ So can I assume that the voltage would be 2.5V over there despite the meter's inability to measure it? Is it really 2.5V in the mid? \$\endgroup\$
    – Bhuvnesh
    Commented Feb 5, 2020 at 6:11
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    \$\begingroup\$ @Bhuvnesh No, because the currents could be so low that they are comparable with stray charges moving in the surrounding area. It's not a nice neat straight path from A to B anymore. \$\endgroup\$
    – DKNguyen
    Commented Feb 5, 2020 at 6:23
  • \$\begingroup\$ @Bhuvnesh Consider heat. Think how heat doesn't flow in a straight line from point A to B. It kind of spreads around and smears everywhere as it goes. That's because materials don't have enough of a difference between thermal conductors and insulation for us to confine the heat neatly with well defined paths and boundaries. But we can with electricity because the difference between conductors and insulators is so high we can...but in this case your air is an insulator and the surrounding in air is also an insulator. So the nice neat approximation of flow doesn't work anymore. \$\endgroup\$
    – DKNguyen
    Commented Feb 5, 2020 at 6:25
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    \$\begingroup\$ +1 nice analogy with heat. I would think the center point would in fact be 2.5V ideally, because of symmetry. But it would not change linearly between the two points. And if one side was a sphere and the other a pointed tip or whatever it surely wouldn't be 2.5V at the midpoint. \$\endgroup\$
    – Spehro 'speff' Pefhany
    Commented Feb 5, 2020 at 14:47
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    \$\begingroup\$ @SpehroPefhany if it was symmetrical i would think so. but schematics dont show geometry. I had not thought about linearity. That is another can of worms \$\endgroup\$
    – DKNguyen
    Commented Feb 5, 2020 at 15:41
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Excellent question... and not so simple... If only there were a lot of questions like that, I wouldn't be leaving the site... I also often (not to say constantly) do such thought experiments and think about them. For example, "What will happen in the arrangement below, if the resistanse is increased up to infinite?"

Floating resistor

(https://electronics.stackexchange.com/a/479174/61398)

Also, what will happen with the well-known definition of an ideal current source - an electrical source with infinite high internal resistance? Its resistance will melt into the environment...

In this area (EE) we deal with circuits that are systems of separate (lumped) elements connected with conductors... and all these components differ from the environment (insulator) with some resistance. It can be high but still resistance. Otherwise, we should work environments having distributed parameters. For example, KCL will be about a sphere, not a node. The "linear" voltage divider will be 3D... here is an example:

3D voltage divider.

A colleague of mine stabs electrodes in a flowerpot to illustrate what "step voltage" is on the subject of electrical safety.

In electronics, we solve the problem of the impossible static infinite high resistance with differential (dynamic) infinite high resistance. So, a transistor current source can have low static resistance but extremely high dynamic resistance.

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    \$\begingroup\$ Thank you for your appreciation and explanation! Curiosity is what keeps one alive \$\endgroup\$
    – Bhuvnesh
    Commented Feb 5, 2020 at 18:04
  • \$\begingroup\$ The ideal current source is not "defined" as "an electrical source with infinite high internal resistance". An ideal current source is a two-terminal element that constrains the current flowing through it. It is an indivisible element. I don't know what an "electrical source" is. \$\endgroup\$
    – Elliot Alderson
    Commented Feb 5, 2020 at 18:10
  • \$\begingroup\$ @ Elliot Alderson, and how does the ideal current source from the subject on EE constrain the current? It has not "dynamic resistance"; it is not a transistor, tube, etc. Then what is it? They model it by a combination of an ideal voltage source and infinite high resistance.. "Electrical source" is "real voltage source"... neither voltage nor current one... something between them. Both we and our students are human beings that understand electric phenomena not by strict definitions but rather by such mental experiments and free thinking... if we were machines (computers), we would need them... \$\endgroup\$
    – Circuit fantasist
    Commented Feb 5, 2020 at 18:42
  • \$\begingroup\$ We do not care how an ideal current source works. It is, by definition, an elementary circuit element. There is no need to model it...it is its own fundamental model. I think you are mistaken about the students. They are taking classes where they learn to analyze circuits that are composed of ideal circuit elements, and the definitions of those elements are formalized and strict for good reasons. The ideal resistor, the ideal voltage source, the ideal current source...you can't just change their definitions on a whim without causing confusion. \$\endgroup\$
    – Elliot Alderson
    Commented Feb 5, 2020 at 18:51
  • \$\begingroup\$ @Elliot Alderson, I have said in the other comment what beginners need above all - imagination, intuition, associations, thought and real experiments, human attitude to them with many explanations of the phenomena from different perspectives, richly illustrated but not ascetic white&black circuit diagrams... a rich environment in which notions about invisible electrical phenomena mature... This is best seen in the questions they ask here... If everything was the way you present it, they wouldn't ask such questions... \$\endgroup\$
    – Circuit fantasist
    Commented Feb 5, 2020 at 18:56

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