My understanding of voltage is broken again - Page 1
Forum section: electronics / beginners

author: sentry7
author_rank: Regular Contributor
timestamp: January 31, 2017, 06:22:24 pm
content: So after eight years I though it made sense, but I guess not.What I think I understand:1) Suppose I had a standard AA battery -- 1.5V, right? Then the electric potential difference from A to B is +1.5 volts (ideally). The absolute electrical potentials at these points are irrelevant, the difference in electric potential is what we care about. Let's just say that B is 0V, so then A is 1.5V.2) Assuming no internal resistance and no leakage current or other madness, then if I connect a resistance across A and B, then 1.5 volts will show up across that resistance.What I do not understand:3) Suppose I had another exactly identical AA battery (which I know is impossible, but for the sake of argument...). If we connected A to D, then wouldn't it be as if we connected A to B? Why doesn't current flow from A to D by the merit that there is a potential difference between them? Maybe I also don't understand how batteries work either.

author: KD0CAC John
author_rank: Frequent Contributor
timestamp: January 31, 2017, 06:34:49 pm
content: My guess is that current only flows in one direction , maybe the drawing should look more like a diodeand or a combination of , there is a lack of circuit / load between either end of the battery / bank - adding batteries in either series or parallel - does not complete a circuit .

author: Ian.M
author_rank: Super Contributor
timestamp: January 31, 2017, 06:37:42 pm
content: Its nothing to do with direction of current flow or diodes.As the two cells have separate electrolytes, insulated from each other, when a single terminal of each cell is connected, there is no complete circuit so no current can flow#. There may be an initial potential difference between the unconnected cells e.g. due to static charges, but as soon as you connect a voltmeter with less than infinite input impedance it starts to discharge any static charge and as the maximum possible capacitance between two cylinders the size of AA 'batteries'*is very small indeed, any practical voltmeter will discharge any potential difference between the otherwise unconnected cells faster than you can read it.# That is: nosustainedcurrent. A brief transient current may flow due to static charges - see next sentence.* I'm being slightly pedantic here - abatteryis an assembly of individual cells. Common types of 1.5V 'battery' only contain one cell so don't meet the technical definition of a battery.

author: KD0CAC John
author_rank: Frequent Contributor
timestamp: January 31, 2017, 06:38:44 pm
content: And as long as I am at it , asking the question again [ asked many times in the past and never got an answer ] .When ever I click on the notify tab - to keep up with a thread .Every time I start getting random threads sent to my email , sometimes multiples , and sometimes not any from thread I linked to .

author: Ahrenp
author_rank: Contributor
timestamp: January 31, 2017, 06:41:34 pm
content: When the batteries are not connected, the voltages are floating. The only thing that is important is the relative voltage between points A and B (or between D and C). There is no absolute voltage (i.e. nocommon'ground' or 0V reference), and by connecting point D and point A all you are doing is referencing the relative '0V' of point D to the '1.5V' of point A. By doing this, you are putting point D 1.5V above point B, and point C 3.0V above B.Unless you physically connect points B and D, they are not at the same potential.

author: KD0CAC John
author_rank: Frequent Contributor
timestamp: January 31, 2017, 06:45:26 pm
content: Another thing , if you put the batteries in some type of circuit - load doing work , charging , the battery bank / cells will equalize between cells / batteries .

author: suicidaleggroll
author_rank: Super Contributor
timestamp: January 31, 2017, 07:15:19 pm
content: This is the difference between floating/isolated voltage supplies and non-isolated voltage supplies.A floating/isolated supply provides a potential difference between two points, andneitherof those two points is fixed toanyabsolute voltage. They float. A or B can be any voltage, the only driving force is that they remain 1.5V apart from each other. You could connect "A" to earth ground, putting "B" at -1.5V, you could connect "B" to earth ground, putting "A" at +1.5V, or you could connect them to any other arbitrary point in a circuit.A non-isolated supply provides a potential difference, but one of those outputs (either the + or -) is also hard-wired to one of the inputs. Its output is not floating, and cannot be arbitrarily connected wherever you want, or you might short something out.All batteries are floating until you connect them to something.

author: AG6QR
author_rank: Frequent Contributor
timestamp: January 31, 2017, 07:27:06 pm
content: Quote from: sentry7 on January 31, 2017, 06:22:24 pmElectricity flows in circuits. A circuit is a complete path (from the same Latin root as circle).A and B are connected by the internals of the battery, such that the internals of the battery do whatever they need to do (often that means supplying current) to make A be 1.5V higher in potential than B. So when you connect them with a wire, current flows through the wire and through the battery, in a circuit.When you connect A to D, you equalize their potential. But there is no path for current to flow in a complete circuit. Perhaps at the moment they touch, some static charge equalizes, but that effect is very temporary and usually negligible. This is similar to what happens when you touch two coins together -- if they had been at different static potentials before they touch, they would equalize, and a few electrons might move from one to the other, but it is not a sustained current flow, and it is usually negligible.Back to the batteries: after they have touched, then the first battery guarantees that A is 1.5V higher than B, and the second battery guarantees that C is 1.5V higher than D, so a little math shows you that C is 3V higher than A. If there were a current path for electrons to flow between C and A, there would be a circuit, and current would flow.

author: Vtile
author_rank: Super Contributor
timestamp: January 31, 2017, 07:47:23 pm
content: Interesting thing is because voltage sources are sums of each when in series connected means that if you connect A and C and measure voltage between B and D it shows zero (or the difference of the battery potentials).

author: sentry7
author_rank: Regular Contributor
timestamp: January 31, 2017, 07:48:08 pm
content: Ok, so I understand the bit about the discharge of static potential.I'm still fuzzy on what it means for something to float. So all batteries are floating until you connect them to something...you mean until you actually connect it to something, you can't know what it's voltage is? When you load the battery, the potential difference just "appears"?

author: suicidaleggroll
author_rank: Super Contributor
timestamp: January 31, 2017, 08:08:10 pm
content: Quote from: sentry7 on January 31, 2017, 07:48:08 pmAll voltages are relative. When you say "you can't know what its voltage is?", voltage relative to what? The voltage of the "B" terminal relative to the "A" terminal will be 1.5V, but the voltage of either of those relative to your foot, wall outlet, or another battery sitting on the other side of the room is undefined. There will be a voltage difference, but you can't really know what it is, it's floating (even hooking up a DMM would change it, even moving your arm to grab the DMM lead would likely change it).

author: AG6QR
author_rank: Frequent Contributor
timestamp: January 31, 2017, 08:11:02 pm
content: Quote from: sentry7 on January 31, 2017, 07:48:08 pmWhich potential difference? The voltage across the battery, or the voltage between the battery and something else? The voltage across the battery does not appear and disappear when a load is connected and disconnected.Back to your example, when the batteries are not connected to anything, you still know that A is always 1.5V higher than B. And you know that C is 1.5V higher than D. But you can't know anything for sure about the relative voltages between the two batteries, nor between any of these points and ground.The relative voltage between the two different terminals of a single battery is well-defined. But the absolute voltage between a terminal of the battery and some external point (such as ground) is not well-defined when the battery is not connected to anything. That's what it means for a voltage source to be "floating".

author: Vtile
author_rank: Super Contributor
timestamp: January 31, 2017, 08:18:01 pm
content: Yes. It does have stored energy that will create current and voltage according law of ohm when you close the circuit and allow current (electrons) to flow.V=R*I or with another typing style U=R*I, where R = resistance, I = current and V = voltage.The Float this or that, is one of the favoured terms in EE and well liked and looked after in many cases since it allows us more liberties how we set up our floating sources to make a power system. It also gives us (sometimes false) safety while working with circuits because of the properties of circuitry that you need to always have a loop to allow the current to flow.If something is floating it means it does not have any electric connection between it and the another part of the world where it is "floating from". Usually ie. Lab PSU is floating from the mains earth, so that means that it is not connected to it in any way (traditionally said to be "galvanically isolated").If you measure voltage between the floating circuit A and floating circuit B, your meter is creating a connection between the two circuits, which are now partially connected to each other, but as a 2 circuit system they are still floating from the rest of the world. Also they are only connected to only one node there is still no full circuit between the two, if you now take a wire and connect any other two places of these two circuit together, you start to see some readings in your meter, since you now have a loop of current.

author: bson
author_rank: Unknown
timestamp: January 31, 2017, 08:31:03 pm
content: The proper technical term for "voltage" is either potential energy as an EE term, or as a physics termelectromotive force(EMF). It's aforcethat wants to push charge. Like any force it's a vector relative to a reference and when you stack two batteries you add the two vectors. The force magnitude doubles, but the impedance between the terminals remains near-infinite and it's not until you form a close circuit that a path of reduced impedance is created for the force to push a charge. Electrically an open circuit is comparable to a spring-loaded rod that wants to push things sitting embedded in a block of concrete - it's staying put with nowhere to go. With two stacked batteries and no closed circuit you have two such rods back-to-back embedded in concrete. Twice the force yet nowhere to go. Electrically, vacuum, air, or other insulators is a near-complete blocker of movement of charge (this is the exact opposite of how we learn relate to physical reality through our senses).

author: sentry7
author_rank: Regular Contributor
timestamp: January 31, 2017, 08:34:40 pm
content: I understand that connecting A and D doesn't make a complete circuit, but then technically, how does connecting A and B make a complete circuit? Do electrons physically go from B to A, through the battery, and back to B? I thought there was some sort of chemical wall that stops electrons from going through the battery? If charges actually move through the cell, then it makes sense that no current can go from A to D.

author: KD0CAC John
author_rank: Frequent Contributor
timestamp: January 31, 2017, 09:36:19 pm
content: A-B would be a circuit - a short circuit .In other posts you never connected A-B = either end of a battery , you asked about connecting 2 cells in series .Now your getting all confuse , if you connect A-B a short circuited battery , your not sending voltage back to the battery - your making that connection the load - a piece of wire , the wire consumes the available potential by heating up the wire / or doing the work - making heat - depleting the battery of its potential .There is no chemical wall in the battery , the electrolyte , when charged [ basically the acid in the water goes to the lead plates ] no - to little potential , then when you charge a battery , the acid moves back into solution / electrolyte , one way to measure that is specific gravity - you are actually measuring the weight of the liquid , discharges the fluid is closer to water , charged the fluid is acid , the molecules of what makes up the acid either suspend in water = charged , or move to the plates = discharged .

author: AG6QR
author_rank: Frequent Contributor
timestamp: January 31, 2017, 11:06:13 pm
content: Quote from: sentry7 on January 31, 2017, 08:34:40 pmCharge actually moves through the cell. The details of how the chemical reaction pushes charge through the cell are best covered in a chemistry class, but the fact that charge is moving through the cell can be verified with a hall effect current clamp. It responds to the magnetic field produced by moving charge. The current clamp will show the same current flowing through the cell as through any other conductor in the circuit.Kirchhoff's current law also tells us that the same current moves through the cell as moves through the rest of the circuit.

author: timb
author_rank: Super Contributor
timestamp: January 31, 2017, 11:18:00 pm
content: Quote from: sentry7 on January 31, 2017, 08:34:40 pmIf you connect A to B with a wire, it's the same as if you connected A to B with a resistor. The wire has resistance, even if it's very small, which means (a large amount of) current flows.Now, when you try to connect A to D nothing happens. Why? Because current from A *always* has to flow back to B. If you were to connect A to D and then C to B, that would complete a circuit and current would flow (you'll also end up with 3V since the batteries are in series).Make sense?

author: KD0CAC John
author_rank: Frequent Contributor
timestamp: February 01, 2017, 12:23:55 am
content: Not sure where your loosing / confusing , maybe the concept of a capacitor - DC hits a open or wall , but AC jumps the gap ?For DC the battery is not a wall or a open - its a chemical connection , that chemical connection is potential / charge etc.Leaving the theory of AC / DC out of it for the moment , the battery can be looked at as being the same as your wall plug - 2 connections that can be applied to a load & do work , no connection until it gets back to the power plant / same as a battery - the electrolyte is the connection between the plates - you can think of the battery as a piece of wire , that has a charge / potential .The battery is not an open circuit .

author: Ian.M
author_rank: Super Contributor
timestamp: February 01, 2017, 12:47:47 am
content: ... or imagine a cell as being a conductor containing a pair of little demons batting electrons towards one end of it. The demons can only hit them so hard (i.e open circuit voltage) and cant bat more than a certain number per second (short circuit current).If you want a better understanding you have to delve into electrochemistry, specificallyredoxreactions.

author: suicidaleggroll
author_rank: Super Contributor
timestamp: February 01, 2017, 01:06:24 am
content: Quote from: Ian.M on February 01, 2017, 12:47:47 amAnd the harder and faster they hit, the more tired they get, until they finally drop dead (depleted battery).

author: sentry7
author_rank: Regular Contributor
timestamp: February 01, 2017, 02:00:31 am
content: Haven't been back to read the latest responses in depth, but I just want to thank you guys for being patient. I'm purely self taught in this stuff. It's pretty ridiculous actually but I have experience in analog, digital, and mixed signal systems. I get the complex stuff. I know what overall behavior to expect out of a given system, but when it comes to what's actually happening on the small-scale end of things it gets bumpy. I'm continuously trying to improve.

author: timb
author_rank: Super Contributor
timestamp: February 01, 2017, 04:21:01 am
content: Quote from: sentry7 on February 01, 2017, 02:00:31 amHere's an easy way to explain it for you, with images:First, let's take a battery and connect it to a 100Ohm resistor:Now, let's add a second battery in series:Notice in both of these images the current has to flow in a loop. That is, the circuit has to be complete for current to flow.Now, let's remove the load from our series battery connection:Since the circuit isn't complete, current can't flow.Think about your question in a broader context. If current still flowed even without a load then power switches wouldn't work, right?Hopefully that answers your question.

author: jamie297
author_rank: Contributor
timestamp: February 01, 2017, 06:57:53 am
content: Quote from: sentry7 on January 31, 2017, 06:22:24 pmThis is the way I visualize it:Current is the flow of charge which is a property of the electrons.Think of the electrons in the conductors as a slack chain rather than moving totally freely.As the charge drives them around from positive around to negative, in order to get electrons onto the negative side they need to be able to move off the positive side.Hope this helps

author: KD0CAC John
author_rank: Frequent Contributor
timestamp: February 01, 2017, 03:10:35 pm
content: Quote from: sentry7 on January 31, 2017, 08:34:40 pmNot sure where your at in your mind with this , but reading today your post above , another thought - it seems your thinking of the battery as a complete circuit XXXX , its only a component - not yet in a circuit , if you put a wire A-B / + to - , that would be a circuit as a short , but not a useful circuit , unless your doing a load test on the battery .

author: Brumby
author_rank: Unknown
timestamp: February 02, 2017, 05:42:11 am
content: Quote from: KD0CAC John on February 01, 2017, 03:10:35 pmYes, I was thinking the same thing.A battery IS only a component. There is no circuit involveduntilthere is an electrical path through other components and/or wires from one terminal of the battery to the other.There has to be a closed loop of some sort that includes both ends of the battery to have a circuit.

author: Galaxyrise
author_rank: Frequent Contributor
timestamp: February 02, 2017, 08:29:34 pm
content: The notion of "absolute voltage" is confusing at best when trying to understand circuits. The negative terminal of a battery is not at some fixed "absolute" potential. A battery forces charge through it, not unlike an impeller for water.Suppose an impeller can raise water 1.5m. So you get an impeller and a 1.5m pipe, fill it with water, put a - sign on the input and a + sign on the output. If you attach (water-filled) pipes at the top leading back down to the input, you can make a circuit and water will flow. (suppose water has an infinite surface tension, so won't drain out an open, downward facing pipe.) Instead of making that circuit, consider instead attaching another impeller&pipe at the top. Water would not flow. Instead, you now have a 3m point from which to source water for a circuit.

author: Vtile
author_rank: Super Contributor
timestamp: February 02, 2017, 08:33:45 pm
content: We always need to also remember that"wire" is always a component, but in basic hobbyist level of things normally so low ohm resistor that we can mark it to zero and forget its existence for sake of simplicity and just draw a line and call it as a wire, which allows potential differences to reach equilibrium through it.Basic theoretical (ideal) components are:Resistorcapacitor (behaviour changes by frequency)Inductor (behaviour changes by frequency)There is no such a thing as a wire (or spoon).All real components (or mathematical models of them) are combinations of these 3 theoretical components and/or two basic sources.Which are:Voltage sourceCurrent sourceWhere "behaviour changes by frequency" means that its reactive resistance (known as reactance) changes when frequency changes. This simplification keeps the ohms law usable tool.

author: sentry7
author_rank: Regular Contributor
timestamp: February 02, 2017, 09:36:57 pm
content: So after many long walks and thoughts, I think I'm satisfied with the reasons everyone has provided as well as my own. The things I'm expecting to happen can't happen for reasons that I'm perfectly aware of. I've been low on sleep, energy, and a lot is going on with me, so I'm just not going to let this hang me up. Thanks all.

author: rs20
author_rank: Super Contributor
timestamp: February 03, 2017, 12:23:44 am
content: I think one of the most important points, one that was hinted at in other replies, is to think very carefully about the fact thatall voltages are relative(except,maybe, static electricity, which is of no interesting to electronic circuit design and can be completely ignored). When you measure a voltage in real life, you put the red lead in one place, and you must always put the black lead somewhere else. When someone speaks of "the voltage on X", there is an implicit ground that a measurement is being made relative to -- almost always the ground symbol on a circuit. The "voltage of a battery" is, obviously, the voltage of the positive lead relative to the negative lead. With this in mind, it is clear to see that discussing the "voltage of the positive terminal of the battery" is dangerously vague and ill-defined, which I think is the root of your confusion here. If you were to expand your wording to be explicit about what you're measuring relative to, the contradictions simply evaporate as you're no longer comparing apples and oranges.This extends into simulation, even. When you do a simulation in LTSpice, you must provide a ground symbol, so all "absolute" / "single-point" voltage measurements are relative to that ground. In many respects, that's theonly purposeof the ground symbol in LTSpice or any other simulator.

author: IanB
author_rank: Super Contributor
timestamp: February 03, 2017, 12:27:00 am
content: Quote from: rs20 on February 03, 2017, 12:23:44 amExcept when it damages what you are working on through inadequate protective measures

author: KL27x
author_rank: Super Contributor
timestamp: February 03, 2017, 07:54:24 pm
content: Even if you replace the batteries with charged capacitors, the result is the same.If you connect resistor between A and D, you have no current flow and no voltage drop. The A side of cap #1 is super electron deficient. The D side of capacitor #2 has an excess of electrons. So why no flow?The electrons can't actually be physically removed from the donor material by very far, without an enormous amount of energy added to the system. The electrons piled up on side D are only "borrowed" from the very nearby material on the other side of the dielectric barrier on side C. They are never physically moved very far. Even though the leads and wires and circuitry that were involved created a "route" that was perhaps inches or even miles in length, the electrons on the C side of the wire were "traded" for ones that were already on the B side. The electrons were never separated from their donor by more than a some few microns of physical space.In order to move electrons from D to A without completing the circuit between B and C, the electons/holes in each capacitor are no longer matched. There are now electrons in B with corresponding holes in D. Since B and D are not separated by a few microns thick dielectric barrier, but they are physically separated by a significant amount of space, this would make the universe explode. Well, no, not really. I'm not sure what happens to the capacitor if this occurs (but I have suspicion to follow), but I know it would require a net input of (significant*) energy to move even the first electron over.*The wire/lead itself, is not a willing donor/acceptor of electrons. The wire is a conductor that will trade electrons, not lend or borrow them. The only lending/borrowing going on happens over very tiny distance, only. Your example is trying to circumvent this. Since force of attraction is inverse of distance squared, removing the electron any significant distance beyond "very very small" will require a finite amount of energy. When far enough away, the attraction will essentially be lost. This quantum amount of energy perhaps represents the energy barrier of a phase change. Also, the resulting charge of the material will drastically change material property. Perhaps, either the materal will suck/dump electrons out of the air or else must undergo phase change to plasma state. And the capacitor will molecularly rearrange (evaporate/sublimate and recondense/resolidify) causing bulging or material creep, if done at a modest rate, while air completes the circuit. Or if excessive rate, it will explode. It will literally vaporize, temporarily, and explode (or at least crack) due to rapid expansion of solid/liquid into plasma.

author: Ratch
author_rank: Regular Contributor
timestamp: February 06, 2017, 07:14:34 am
content: Quote from: sentry7 on January 31, 2017, 06:22:24 pmAre you asking if connecting two voltage sources (VS) in series is the same as shorting out one of the VS? Why would you even consider that it would be the same? The connections you specify are at different locations.QuoteThere is no voltage difference between A and D. Measure it with a voltmeter if you don't believe me. There is a voltage difference between A and B, however. Do you know why?QuoteI am sure you don't, but it doesn't matter. Any voltage source would act the same way.I read some pretty wild stuff by folks answering the OP's question. I believe many of you do not really have a good grasp of what voltage is.Voltage is a density. Density is a ratio between a quantity of a substance and one of its properties. An example is mass per volume of a substance. For voltage, the density is energy per charge quantity. This density is measured in volts, whose units are joules/coulomb for the MKS system. In other words, it is the energy contained in an amount of charge. It takes energy to bring an amount of charge together. It takes more energy to bring the charge carriers closer together. It takes even more energy to bring additional charge carriers together. So just knowing the amount of charge does not give you the voltage. You also have to know the state of togetherness.The charge carriers don't like to be together, so if there is a conduction path to a lower energy density (lower voltage), the charge carriers will move to the lower voltage. Along the conduction path there will be resistance. By quantum action, the resistance will convert some the the electrical energy to heat. The charge carriers will arrive at the end of the conduction path with less energy than they started with due to heat loss, and therefore have less energy density and a lower voltage. Since voltage is a density, it is not a vector quantity. It is a scalar quantity. I invite discussion of this subject.Ratch

author: KL27x
author_rank: Super Contributor
timestamp: February 06, 2017, 09:06:52 pm
content: QuoteQuoteWhen I first started educating myself about electronic circuits, I had same internal incongruity as OP. Lots of negative charges on the negative side of cap #1. Lots of positive holes on the positive side of cap #2. Since all these charges, as you say "don't like to be together" why not some of the electrons move from negative side of cap #1 to positive side of cap #2 when you connect them with a conductor?You have explained all around it without explaining it. Saying "there is no voltage, check it and see" puts the explanation down to a DMM. LOL. That's not an explanation.I think maybe we are not understanding the OP's question in the same way. Cuz I see no way in which you even attempted to address it.QuoteThe way I read OP's question, I consider only my answer as a direct and complete explanation that doesn't include a "just because!" (I can fill in any gaps to anyone that can't connect the dots... they all connect! It's essentially chemistry/physics!) LOL. Go figure. We are all looking at the same thing but seeing something different as the OP's fundamental incongruity of understanding/logic.

author: Ratch
author_rank: Regular Contributor
timestamp: February 07, 2017, 12:00:29 am
content: Quote from: KL27x on February 06, 2017, 09:06:52 pm

author: helius
author_rank: Super Contributor
timestamp: February 07, 2017, 01:53:08 am
content: It would help tremendously if you would just learn how to quote.Voltage is defined as apotential differenceof electrical energy between two points. That it is adifferenceinpotential, makes it exactly analogous to gravitational potential (height). When you raise a mass of 1kg to a height of 1 meter above its previous position, the gravitational field does work on that mass. When you move a charge of 1einto an electric field of 1 V/m, the electric field does work on the charge. As a result, the gravitational potential energy (in the first case), or the electrical potential energy (in the second case), of the object has been changed.Your (Ratch) repeated incorrect assertions werealready debunked elsewhere.In electrical engineering, one of the points is often defined as a circuit node, such as "ground". This makes it convenient to say "the voltage at this other node is 1.5V" but this is really not correct. Voltage is always a differencebetweentwo points (remember that it is apotential difference). If we take it as a given that the electric field infinitely far away is knowable, we could also define the "absolute potential" as the potential of a charge that has come from infinitely far away to its present position. That is the normal approach for statics.The OP example of two dry cells is interesting, because it shows how this potential difference idea is critically important. If you thought that the voltage was a simple scalar value, then having the voltage at B be 0V, A be 1.5V, D be 0V, and C be 1.5V, might appear to make sense. So the question of why there isn't a "voltage difference" between A and D would be a puzzle. In fact voltage is not defined this way after all. Thepotential differencebetween A and D is zero, because no work is being done on a charge moving from A to D.

author: sentry7
author_rank: Regular Contributor
timestamp: February 07, 2017, 02:00:13 am
content: Quote from: Ratch on February 07, 2017, 12:00:29 amYou somehow managed to read over my responding questions and my concluding thread. This demonstrates that you did not thoroughly read my responses, but rather the "pretty wild stuff" in everyone else's.Quote from: Ratch on February 06, 2017, 07:14:34 amAnd this is a explanation?Quote from: Ratch on February 06, 2017, 07:14:34 amYou answered my question with another question.Regurgitating material from a physics book does not help me understand the subject in question.I am a computer programming tutor, and there is a way to induce people to think for themselves, and presenting their ignorance about something to them is not how to do it. You have done this not only in this thread but multiple others. Instead of the attacking the people who are trying to help, try contributing to the topic. So far, counter-intuitively, your post was the least helpful.Also, you should really learn how to properly use quotes.

author: sentry7
author_rank: Regular Contributor
timestamp: February 07, 2017, 03:30:59 am
content: Also, during one of my walks, I think it finally came in a way that makes sense to me.Voltage really is a game of "relatives".Hypothetically, I have a brother, a sister, a mother, and a father. If I am the reference point, then this person (relative to me) is my brother, or my sister. If a person is not connected to me through my family, then the question "Relative to me, who are you?" doesn't make sense. However, if my sister married this stranger, then I can say that person is my brother-in-law because he is connected to my sister, who is connected to me.Numbers work well too.1 is less than 2, which is less than 3. We can say that 5 is greater than 3 because 5 is greater than 4 which is greater than 3. We can get to 3 from 5 because they are connected through 4. A question like "Is 5 greater than blue?" makes no sense because blue is not connected to any numbers.But without any reference point, "sister" has no meaning. It even applies to directions. To the left.........to the left of what?Back to the example, A and B are "connected" because they are two parts of the same cell, but A and D are not. Without physically connecting A to D, asking what is the voltage of D w/respect to A is undefined.

author: helius
author_rank: Super Contributor
timestamp: February 07, 2017, 04:20:48 am
content: Quote from: sentry7 on February 07, 2017, 03:30:59 amYou make the same mistake over again! "voltage of D w/respect to A" implies that voltage is a property of a single terminal (relative to some reference). It isn't: Voltage is an affair of two terminals, or in circuit theory, one port. This is because its definition requires us to move a charge from one place to another!The voltage of the port D:A is not undefined, it is zero. Voltage is defined betweenanytwo points, because you could bring a test charge from any initial point, to any final point, measuring the work required. TheEfield (without moving magnetic fields) is conservative, meaning that the path taken doesn't matter. And this field exists everywhere in space: between unconnected cells, inside the cells, on the surface of the cells. Voltage is everywhere.

author: rs20
author_rank: Super Contributor
timestamp: February 07, 2017, 04:58:55 am
content: Quote from: helius on February 07, 2017, 04:20:48 amHelius, please tell me what these "port voltages" are:?D:A = 0V (as you've already mentioned)D:BC:AC:BI'm curious to see how you arrive at consistent, well-defined numbers for all of these without resorting to asymmetric assumptions, breaking Kirchoff's Voltage Law (e.g. D:A + A:B + B:D = 0V; if D:A and B:D are both zero, then you've got 0 + 1.5 + 0 = 0 which is a probably a problem) or describing some/all of these voltages as undefined. Also, it seems that if you're so insistent on defining the voltage of port D:A as zero volts, you've just assumed without justification that neither of the batteries are carrying any static charge. But more than willing to be corrected.At the very least, sentry7 is ~100% correct if he/she substitutes "high-impedance" for "undefined", because describing it as "high-impedance" is far more valuable, useful andrelevantthan describing it as "0V".

author: suicidaleggroll
author_rank: Super Contributor
timestamp: February 07, 2017, 05:02:38 am
content: Quote from: helius on February 07, 2017, 04:20:48 amZero? Not at all, it will besomethingfor sure, with a very high input impedance DMM you could attempt to measure it, but it almost certainly won't be zero if the points aren't connected. Why would you think two unconnected points would have a voltage difference of 0? What about static charge, ESD, etc?

author: Ratch
author_rank: Regular Contributor
timestamp: February 07, 2017, 05:28:20 am
content: Quote from: helius on February 07, 2017, 01:53:08 amWhat does "potential difference of electrical energy" mean? Does it mean that the positive terminal of a 5 volt voltage source is 5 joules more that the negative terminal is? Voltage is not energy.QuoteI don't think they were debunked. I answered every one of those challenges in that thread.QuoteYes, voltage needs a reference point to measure from. Unless otherwise stated, infinity is assumed to be the reference.QuoteIf A and D were connected, there still would be no work done because the resistance is negligible. But the electric field at A and D would be equal, so the voltage at C could be measured with reference to B.Ratch

author: rs20
author_rank: Super Contributor
timestamp: February 07, 2017, 05:39:51 am
content: Quote from: Ratch on February 07, 2017, 05:28:20 amWell, in any electronics context, the reference is assumed to be the ground symbol on the schematic (if available); failing that it'd be Earth, or the negative terminal of the power supply. I don't think any electronics engineer has ever resorted to the edge of the universe as his/her reference.

author: Ratch
author_rank: Regular Contributor
timestamp: February 07, 2017, 06:03:22 am
content: Quote from: sentry7 on February 07, 2017, 02:00:13 am

author: gnif
author_rank: Unknown
timestamp: February 07, 2017, 06:09:44 am
content: Quote from: KD0CAC John on January 31, 2017, 06:38:44 pmAsk it in a thread that is specifically about the problem and you may get an answer, it is completely off topic here. If you had done this, either I would have noticed or someone else would have and then sent me a PM, at which point I would have investigated and performed some testing to try to isolate the cause.

author: Ratch
author_rank: Regular Contributor
timestamp: February 07, 2017, 06:15:29 am
content: Quote from: rs20 on February 07, 2017, 05:39:51 am

author: sentry7
author_rank: Regular Contributor
timestamp: February 07, 2017, 06:47:35 am
content: Quote from: Ratch on February 07, 2017, 06:03:22 amThis is a public online forum; there are no "localized conversations" in a public thread. The level of conversation between me, them, and you is the same.Quote from: Ratch on February 07, 2017, 06:03:22 amDeductive reasoning is only effective when one has the correct premises. Since you yourself have made it clear that I don't have the correct factual establishment, it is illogical to expect deduction to be effective.Quote from: Ratch on February 07, 2017, 06:03:22 amFirstly, a student cannot be aware of what he (or she) doesn't know; If he knew, then the question would be moot. If he didn't know, then how does he know what to look for?Secondly, I would never consider you to be a teacher, especially not mine.Quote from: Ratch on February 07, 2017, 06:03:22 amIf I could relate physics concepts to the problem, this thread would not exist. Again, a failed observation on your part.Quote from: Ratch on February 07, 2017, 06:03:22 amSince you are not the offended party, you are unqualified to specify what is and is not an attack. I think you mean to say that you did not intend to attack anyone. You do know what it means to be challenged because you are constantly defending. One does not defend what is not an attack.Quote from: sentry7 on February 07, 2017, 02:00:13 amQuote from: Ratch on February 07, 2017, 06:03:22 amKeep hoping.

author: Ian.M
author_rank: Super Contributor
timestamp: February 07, 2017, 07:16:26 am
content: @Sentry7: I strongly recommend looking at users' posting history if their replies seem in any way 'off'. Its then your choice whether you have sufficient moral fortitude to ignore trolls or whether you need to go to your profile and add them to your 'Ignore List' or whether they must be debunked for the common good.

author: helius
author_rank: Super Contributor
timestamp: February 07, 2017, 07:27:20 am
content: So youdoknow how to quote!Quote from: Ratch on February 07, 2017, 05:28:20 amPotential difference is a difference of energy due to the repositioning of an object in a field. The amount of energy is scaled based on the parameter of the force equation that corresponds to that object. For instance, Coulomb's Law is \$ F = k {q_1 q_2 \over r^2} \$. The parameter corresponding to our object is \$ q_1 \$, the object's charge. In your example, the potential difference between the terminals is 5 joules per coulomb, or 5 volts. Volts are a derived unit, not a base unit.This would have been obvious if you had bothered to read the rest of the paragraph, which contains the gravitation analogy. The potential difference of an object raised 1m is the increased potential energy scaled by the object's mass: 9.81 m2/s2. We can say that this is the potential difference of thevertical distance 1m, or equivalently, that the gravitational field is the gradient of this potential.Quote from: suicidaleggroll on February 07, 2017, 05:02:38 amSorry. I should have said that the voltage would be 0 in the case where there is no external field besides the field caused by the dry cell's chemical reaction. If the cells are in any arbitrary position in space, or do not have neutral net charge, you would not know that. So the voltage may be (depending on conditions) unknown, but never undefined. If the points were to be connected by a wire, the voltage becomes known as a result of the equipotential surface of the wire.

author: KL27x
author_rank: Super Contributor
timestamp: February 07, 2017, 09:19:33 am
content: ^Now that is a very detailed reponse. But suffice to say, "zero" is pretty much accurate, lol. Yeah, you can take into account possible static surface charge of the component lead, haha.QuoteGoing back to OP: this is wrong. There is a meaningful and measurable electrical potential (voltage) between A and D. And as has been discussed, it is determined solely by the existing static charge on the part leads (which we can call perhaps totally random and which might go in EITHER direction and has nothing to do with the polarity of the battery terminals) and/or presence of electrical field, etc.Your revelation is thus totally unsatisfying for me, personally. Because I do not feel this is a correct understanding.it's funny how some answers don't seem to account (from my chair) for the fundamental missing logical leap that must exist for the OP to have this question in the first place. Maybe I'm not understanding other common fault/fallacy/assumption/trap that other people might have?To address this, one must first ask the question: why should anyone think there ought to be current flow between A and D to begin with?I recall from my own first question over this very same issue, I was working with a simplified view to the effect of, "hey... lots of electrons over here on this negative terminal of cap/battery #1. And lack of electrons over here on this positive terminal of cap/battery #2. So when you connect them, would not the electrons be pulled from the negatively "charged" terminal to the positively "charged" terminal?In case this helps OP to find a better understanding:The stored energy only exists between the halves of the cell/cap. In the cap, the electrons on the one side want to go only one place, and that is back to the other side of that cap. They have never completely "left" the donor molecules on the other side. They can do this only in solution, and still only in close proximity to the respective/opposite ion. In a battery, it's a chemical reaction that stores the energy. But is can be looked at much the same way. The reaction can't progress until a connection is made which allows electrons to flow from the one side to the other. The negative terminal of the cell can't "give out" electrons to anywhere else. The only place they can go which is a downhill reaction, is to the other half of the same cell. (Or in series, they can play musical chairs, cathode to anode of another battery, but in a loop.) Anywhere else you want to make them go but the other side of the cell, you would have to add energy (and you would turn some of the electrolytes into plasma in the process).This might sound dumb, but here goes my super simplified view: The negative terminal isn't bursting with electrons waiting to shoot out at soon as something is connected to it. It has a net negative charge but it is internal, not on the leads. And the dielectric is desperately holding this charge from being sucked IN to the other side of the cell/cap. The external connection just happens to be the only way these charges can get from one side to the other. For each microcoulomb of electrons that passes thru the circuit, just imagine instead that they are pouring across the dielectric in the cell/cap, directly. But with a little waterfall generator in there. That's what's powering the circuit. The electrons moving from one side of this divide to the other, just a hop/skip away. You can make the elecrons take connecting flights (as long as the total voltage drop doesn't exceed the willingness of these electrons to get home), but if the last flight doesn't end on the other side of the cell, they aren't gonna get on the plane. Add cells in series, and same thing. The electrons in each cell take a circuitous route, but they are only really going to the other side of their own cell, with a detour. But now you have two pumps on the same line, adding their voltages. You see that with multiple cells, each electron that passes through the circuit sends one electron "home" in EACH of the cells/caps which are in series.

author: Sigmoid
author_rank: Frequent Contributor
timestamp: February 07, 2017, 12:16:21 pm
content: Do we REALLY need to bring static charges into this, when someone is apparently having trouble grasping voltage?I think the real question is, how can voltage be explained in a way that's intuitively understandable. There are some examples here of how, and some, markedly, of hownot. (And "technically more precise" is not always the best way to go about it.)Quote from: sentry7 on February 07, 2017, 03:30:59 amI'm not certain that the part about the relatives and numbers is entirely on the right track, but ultimately, it's safe to think that without a connection, the voltage difference between A and D is undefined.For the specific case where you connect cells in series, you can think about it in terms of circuits (I believe you said you understand circuits in a practical way)... If A and B are connected, according to Ohm's law, there is no potential difference between them. Uad=R*I, R=0 (we assume an ideal conductor connecting the two points), therefore Uad=0*I=0Sorry probably scratch that, I don't want to confuse you further.

author: Sigmoid
author_rank: Frequent Contributor
timestamp: February 07, 2017, 12:41:50 pm
content: Following your example with relatives, think of it like this...There is Adam, who has a son, Bob: Bob = Adam + 1There is Claire, who has a daughter, Denise: Denise = Claire + 1Let's set Claire as 0 (the reference) completely arbitrarily.If Adam and Denise marry (Adam = Denise), then Bob is the grandson-in-law of Claire:Claire = 0 (reference)Denise = Adam = Claire + 1 = 1Bob = Adam + 1 = 2If Bob and Claire marry, then Adam is the grandfather-in-law of Denise:Adam = Bob - 1 = -1Bob = Claire = 0 (reference)Denise = Claire + 1 = 1

author: Ratch
author_rank: Regular Contributor
timestamp: February 07, 2017, 03:12:59 pm
content: Sentry7,I am sorry you feel so discombobulated about my response to your question. Hopefully, others in this forum can give you the help you seek in understanding this circuit senario.Ratch

author: Ratch
author_rank: Regular Contributor
timestamp: February 07, 2017, 03:29:02 pm
content: Quote from: helius on February 07, 2017, 07:27:20 amIt appears that you only want see the last response of what I am answering. If you made that clear earlier, I could have complied sooner. I hope I "got it" now.QuoteThe potential difference of energy due to charge movement within an electric field depends on the amount of charge and the change in electric field strength. That is what voltage defines, joules/coulomb.Ratch

author: sentry7
author_rank: Regular Contributor
timestamp: February 07, 2017, 03:46:41 pm
content: Quote from: Ratch on February 07, 2017, 03:12:59 pmYou have already stated this in post #43; there is no need to restate it.Quote from: Ratch on February 07, 2017, 03:12:59 pmI believe you mean "scenario".

author: Ratch
author_rank: Regular Contributor
timestamp: February 07, 2017, 03:55:58 pm
content: Sentry7,Sorry for the typo.Ratch