How many electrons does one Ohm prevent/resist?

[Justinmcg67]

Hello All! I am having a rather difficult time locating the answer that I am seeking. What I am trying to determine is a little more understanding about resistance. I know that 1 Amp is coulombs per second, and that is 6.241×10¹⁸ electrons. I know that 1Ω is 1 Amp across 1 Volt. However what I am trying to make sense of is how many less electrons are flowing through more than one ohm? I am not the best when it comes to explanation so I will try to add some clarity...

1 Amp = 1 Coulomb per Second
1 Coulomb = 6.241×10¹⁸ electrons
1Ω = 1 Amp across 1 Volt

I am going to say therefore that 1Ω is still passing 6.241×10¹⁸ electrons...?

How would I set up an equation to determine how many less electrons are flowing as the value in Ohms increases? This question is for nothing more than curiosity and any help would be greatly appreciated.

 

[Drakkith]

Just find the amperage. That itself tells you how many charges are flowing through a point in the circuit.
For example, if you double the resistance to two ohms, then the amperage falls to 0.5 amps, which means you also have half a coulomb per second.

Edit: Note that you could have a hundred volts instead of one, but if you also increase the resistance to 100 ohms then the current is still one amp, and therefor you still have one coulomb per second.

 

[Justinmcg67]

Just find the amperage. That itself tells you how many charges are flowing through a point in the circuit.
For example, if you double the resistance to two ohms, then the amperage falls to 0.5 amps, which means you also have half a coulomb per second.

Well I was certainly expecting a much more elaborate explanation but you cleared my ignorance for this question! To make sure I am able to figure this out on my own from here on out...

1. Difference in Amperage = Difference in Charge
2. Subtract the initial Electrons from resulting electrons
3. Identifies how many less electrons are flowing

Am I correctly understanding this?

 

[dauto]

I don't think that description makes sense. The current doesn't tell you the number of electrons moving. That's given by the number of free electrons in the conduction band of the material. If you reduce the current you will have pretty much the same number of electrons moving, but they will be moving more slowly (smaller drift velocity).

 

[Vanadium 50]

I don't think that description makes sense. The current doesn't tell you the number of electrons moving. That's given by the number of free electrons in the conduction band of the material. If you reduce the current you will have pretty much the same number of electrons moving, but they will be moving more slowly (smaller drift velocity).

I agree. The answer to the question "How many electrons does one Ohm prevent/resist?" is "all of them".

 

[sophiecentaur]

I agree. The answer to the question "How many electrons does one Ohm prevent/resist?" is "all of them".

Yes. And a proportion of the energy of every electron will be transferred to heat in the resistor.
But, as usual, it is much easier (and better) to talk in terms of charges, currents etc. than introducing the nature of the electrons that are involved in the charge transport. The added complication does not, imo, add anything at all to the understanding of the situation - unless you happen to know an awful lot about electrons. (I suspect you do not, Justin)

 

[phinds]

Yes. And a proportion of the energy of every electron will be transferred to heat in the resistor.
But, as usual, it is much easier (and better) to talk in terms of charges, currents etc. than introducing the nature of the electrons that are involved in the charge transport. The added complication does not, imo, add anything at all to the understanding of the situation - unless you happen to know an awful lot about electrons. (I suspect you do not, Justin)

I not only agree w/ that, I would take it a step further and say that since the kind of thing the OP is talking about is basically electronic circuits with passive components, the introduction of electrons into the discussion is a total waste of time even if you DO know a lot about electrons. Just use ohms law and be done with it.

 

[sophiecentaur]

You have to sympathise with the poor devils, though. For the last twenty five years, teachers have been told to give the electron model of electricity and students all believe that it's the way to understand it. Students all think that any macroscopic description of a phenomenon is somehow incomplete. I give thanks, almost daily, that we did Electricity in school without any reference to electrons (MUCH TOO HARD - we were told, and something that you will come to later.)

 

[Dale]

I don't think that description makes sense. The current doesn't tell you the number of electrons moving. That's given by the number of free electrons in the conduction band of the material. If you reduce the current you will have pretty much the same number of electrons moving, but they will be moving more slowly (smaller drift velocity).

That is true if you are considering all of the electrons in the entire circuit. But if you simply consider the number of electrons that pass a given location in a given amount of time then it is clear that doubling the resistance halves that number. You are correct that it halves it by reducing the velocity rather than by reducing the density, but either way, doubling the resistance halves the number of electrons drifting past a given location in a given amount of time.

Personally, I have no problem with the OP's question. You can convert 1 A = 1 C/s = -6.2E18 e/s. So if you double your resistance you half the current and therefore half the number of electrons passing any point in the circuit.

 

[sophiecentaur]

Perhaps you should include the word "net" in the second sentence.

 

[dauto]

That is true if you are considering all of the electrons in the entire circuit. But if you simply consider the number of electrons that pass a given location in a given amount of time then it is clear that doubling the resistance halves that number. You are correct that it halves it by reducing the velocity rather than by reducing the density, but either way, doubling the resistance halves the number of electrons drifting past a given location in a given amount of time.

Personally, I have no problem with the OP's question. You can convert 1 A = 1 C/s = -6.2E18 e/s. So if you double your resistance you half the current and therefore half the number of electrons passing any point in the circuit.

The problem with the question (as stated) is that the OP insists on talking about counting electrons instead of counting electrons per second which would have been fine. There is a natural tendency to interpret statements in a way that make them more correct than they really are. The reader fills in the gaps. That tendency sometimes prevents us from actually noticing when the person doesn't understand some basic concept such as the difference between charge and current. I'm not saying that's the case here, but it could be and there is no way to make sure that it isn't except by pointing out that the question doesn't make sense as stated. That may make me sound a bit pedant but the point really is to figure out whether or not the OP (a person I know nothing about) understand the basic concepts necessary to make sense of the answers we're providing.

 

[sophiecentaur]

The OP should just do some reading and then ask a question, based on what he has read. The post was clearly a 'fishing' exercise and they never work well.

 

[russ_watters]

So, the OP's question was poorly conceived, which led to a fairly useless answer if answered exactly as the question is worded. I consider it part of our job to correct the question before answering it -- indeed I think that's more important than the answer. Which is what the first few posters did.