Relationship between potential in batteries and zero potential at infinity - Page 1
Forum section: electronics / general

author: Banned User
author_rank: Newbie
timestamp: July 15, 2017, 03:11:03 am
content: I have been wondering really hard about what is the relationship between the defined zero potential at infinity as we do in physics, and the potential differences in circuits using batteries or other voltage sources.Suppose we have a circuit with a battery and a resistor, and we measure the voltage across the resistor, what is the relationship between this potential and the zero defined at infinity?As we know, the potential at a point due to an electric field is the integral from infinity to that point of E.dl, so how do you use this to calculate the potential across the resistor?I know the zero of the circuit is in the circuit not at infinity as that would be dumb, but still this question hurts my brain. I want to know how do you calculate the potential across the resistor using the zero at infinity.Can I define the potential across the resistor in relation to the zero at infinity ?Can I calculate the potential across the resistor by taking a path integral from one end of it to the other ? Can I do it by taking the integral from one end of it to infinity, and then from infinity to the other end, and will the answer be the same as the voltage source reading?I have tried doing this by considering a situation with two charged spheres, with charge on the surface. So there is a potential difference from one sphere to another since there is a charge difference +Q and -Q. So if I take an integral from the +Q sphere to infinity, I will have to take the -Q sphere into account too as it has an electric field. Then take the integral from infinity to the -Q sphere as well, and subtract the two with the correct signs. This seems to be equivalent to a battery or voltage source situation. Would anyone like to help ?Another slightly complicated question I have is about the shape of the electric field from the + end of a 9V battery to the - end. Would it go straight from the + end to the - end on top of the battery? I know it has electric fields inside it as well, but on the outside, does it go from + to - in a straight line? Then however if I connect a wire from one end to the other, and assuming it's a wiggly wire, how does the electric field arrange itself inside the wire? Does it follow the wire in small linear elements?Thanks!

author: IanB
author_rank: Super Contributor
timestamp: July 15, 2017, 03:29:44 am
content: When you consider the circuit with the battery and the resistor, what you care about are the potentialdifferencesaround that circuit. As such, the absolute potential of any part of the circuit referenced to infinity is irrelevant. It is like a constant of integration that cancels out.Suppose you have some difference, D = A - B. Now lets suppose that you change the reference potential of the system so that you have D = (A + 1000) - (B + 1000). The result is still D = A - B. So if you isolate the system from the surroundings, any difference in potential between the system and the surroundings doesn't matter.

author: IanB
author_rank: Super Contributor
timestamp: July 15, 2017, 03:35:42 am
content: Quote from: Analog86 on July 15, 2017, 03:32:24 amYou are asking how to calculate the potential across a resistor using the zero at infinity. I don't think you can. Therefore there is no answer to your question.

author: Alex Eisenhut
author_rank: Super Contributor
timestamp: July 15, 2017, 03:50:03 am
content: Go away, Electronicsgeek.

author: ConKbot
author_rank: Super Contributor
timestamp: July 15, 2017, 04:10:20 am
content: If the zero point is at infinity, it will take forever for the potential difference to be realized, so you take the limit as time goes to infinity, get the inevitable heat death of the universe, take the derivative of that, and get zero, because it's the heat death of the universe.

author: Alex Eisenhut
author_rank: Super Contributor
timestamp: July 15, 2017, 04:54:53 am
content: Quote from: blueskull on July 15, 2017, 04:44:39 amToo bad. I was hoping he'd destroy my computer from the inside, I'm looking for an excuse to upgrade...

author: rs20
author_rank: Super Contributor
timestamp: July 15, 2017, 04:56:19 am
content: Quote from: Analog86 on July 15, 2017, 03:11:03 amYes. The electric field is a gradient field, and as such its path integral is dependentonlyon the start and end of the path, not the journey inbetween. This includes journeys that travel via a point at infinity.Now as for what you see if you compute the path integral from either node of the circuit to a point at infinity, well, that depends entirely on the static charge on the circuit; a concept that is completely irrelevant to everyday electronic engineering concerns.Quote from: Analog86 on July 15, 2017, 03:11:03 amThis makes no sense -- speaking of a potential across a resistor is already asking for a potential difference, but "in relation to" means taking another difference. You can take a difference of differences and a difference of absolute values, but you can't meaningfully take a difference of a difference and an absolute value. It's like asking what the differences between our heights are, "in relation to Yao Ming". What would that even mean?

author: ConKbot
author_rank: Super Contributor
timestamp: July 15, 2017, 04:59:30 am
content: Forget knowing how to answer the question, I'm not even reading the garbage. Im just here for some reddit style shitposting now.

author: Banned User
author_rank: Newbie
timestamp: July 15, 2017, 05:02:08 am
content: Quote from: rs20 on July 15, 2017, 04:56:19 amFinally someone normal replying.I meant to say the voltage at one of the nodes of the resistor not across it when using the infinity point. So I gather that the infinity point is only relevant for electrostatics and not for electrodynamics of the circuits. Hmm.. That is a very interesting point which I totally missed. I will think about this thank you!

author: Banned User
author_rank: Newbie
timestamp: July 15, 2017, 05:04:36 am
content: Quote from: ConKbot on July 15, 2017, 04:59:30 amConKbot ? Sounds like you are a TA player haha

author: Zero999
author_rank: Super Contributor
timestamp: July 15, 2017, 05:34:04 pm
content: Quote from: Alex Eisenhut on July 15, 2017, 03:50:03 amAlthough likely, we've no proof.Quote from: blueskull on July 15, 2017, 04:44:39 amAs far as we know, the original poster hasn't broken any rules (at least not in public: the allegations of threatening private message(s) should be raised with the mod's via PM, no in public) so leave them alone for now.EDIT:I've just emailed Simon and he's been banned.