So, we’re playing around with a little app for smartphones. Trivial, not even a joke but a jokule, perhaps a jokette.
And electric current is electrons moving along, right?
So, how many electrons makes how much current?
As all know, my relationship with engineering is akin to that of Polly’s with logic. But there’s two (?) ways of measuring current, volt and amp. Being, as I understand it, the rate and volume (err, perhaps pressure?)? Umm, roughly?
If that be true then again as I understand it, then 1 volt at 1 amp is going to be half as many electrons as 2 volts at one amp, or one volt at two amps.
Are people laughing at me already?
But if that is so then, say, how many electrons is one volt at one amp for one second? And is it then just simple multiplication after that? 10 volts at 10 amps is 100 times as many electrons?
It’s the actual number of electrons on the move that I want to find out.
An ampere is the current flow of one coulomb per second. One electron carries a charge of -1.6*10^-19 coulombs so one coulomb has 6.25*10^18 electrons. So one ampere is 6.25*10^18 electrons per second
“Are people laughing at me already?”
Erm… yes. ;-D
But we’re here to help. Have a read:
Electrickery explained so that even a 16 year old RAF apprentice could understand it. Well, I did when I was. 😉
Think of amps as flow rate and volts as pressure – so I don’t think volts effects how many electrons are moving, just how hard each one is pushing.
A useful analogy is Volts is the height of the hill and Amps (current) is the amount of water flowing per second. Increase the height and, other things being equal, the current will increase proportionally. Widen the channel and you get more current for the same voltage, equivalent to reducing the electrical resistance.
By the way, yes we are laughing at you.
Flatcap’s answer looks correct.
In the flow-of-water analogy, you might think of the voltage as being the height where the electrons go into your gadget, compared to where they come out. 1A at 1V delivers 1 Watt of power.
A (perfect) kettle drawing 10A at 10V (i.e. consuming 100W) will take about 43 minutes to boil a litre of water. A more powerful one drawing 10A at 100V has ten times the power (1kW) and boils the litre of water in 4.3 min. However the same number of electrons will pass down the wire in each second, because that’s exactly what a current of 10A means, but each one carries more energy into the kettle.
Amp(ere) is to current as Volt is to voltage, as Watt is to power, as Joule (or kJ) is to energy… as Pound is to weight.
So it’s only ampage that I need to think about for the number of electrons? Volts don’t make any difference?
probably better stick at economics!
I think Ritchie might be about to start a “Wibbles by Worstall” section.
I seem to remember you had to do *this* (points fingers and thumb in different directions) and that meant something about amps and volts or something.
Does that help?
Tim must be an arts graduate…
actually this link is relatively helpful
Not *number of electrons* – Number of electrons *per second*
(current = Amps = charge per unit time = coulombs/second, and each electron carries about one weenieth of a coulomb charge).
If you are only interested in the number of electrons on the move then voltage does not matter. If however you are interested in how much work they are doing, it does.
Probably Wikipedia is your friend here 🙂
Why not ask Murphy (under a pseudonym obviously given it’s you), he know all about physics – did he not reconcile general relativity and quantum mechanics once in a lunch break on an unfolded napkin?
I was going to make a similar point.
Although Murphy paid no attention during physics lessons because he know what he was being taught was wrong, I am sure he could nevertheless knock up a nuclear power plant in his shed if he wanted to. Orthodox neophysiceral thinking is just a plot to reduce worker power.
The number of charge-carrying electrons is vastly different in different conductors. The current is the number of charge-carrying electrons times their drift (i.e. average) velocity.
Incidentally, in metals the drift velocity is a tiny fraction of the thermal velocity. If you were a sitting in the crystal lattice watching the electrons go by, you wouldn’t be able to tell which direction the current was flowing in.
In semiconductors, the drift velocity is much higher but the number of charge carriers is much lower. (I say ‘charge carriers, because they may be ‘holes’ rather than electrons. Holes are missing electrons with negative effective mass. But I digress.)
For those on here clearly far more knowledgeable about electricity than me and having to put up with the wibblings of those, however well meaning, who clearly know nothing about the subject, now you know what it’s like with those of us such as myself with a career in tax having to read the inane meanderings of Murphy on tax.
If I suddenly start explaining that HEAVY electricity must be carried in wires but that it is perfectly feasible to carry LIGHT electricity around in buckets and I refuse to back down on my assertions no matter how patiently you explain things to me only then to find that I claim that of course I knew all along that there are not two weight categories of electricity and that what I was saying was a metaphor and how could you have thought otherwise, well that’s what arguing with Murphy on tax matters is like.
Ah yes, heavy electricity
You need to think about resistance to those electrons. All conductors have resistance, even super conductors.
If the wire your electrons are flowing down has a resistance of 1 ohm then it will need. A voltage of 1 volt to generate current of 1 amp. Halve the resistance and you only need half the voltage.
Unless your talking about alternating currents as in the mains, then it gets in to realms you probably don’t want to know about.
Just to make it worse, the actual electrons flow in the ‘wrong’ direction and so often in electronics people talk about the flow of ‘holes’, i.e. spaces where electrons used to be.
“I think Ritchie might be about to start a “Wibbles by Worstall” section.”
If Murphy faced the problem he would have announced to the worl din a great fanfare exactly how many electrons were flowing.
Anyone who then challenged him would be denounced as neo-physicist and he would have received support from the new physics foundation. Any further argument would be shrugged off as sophistry and the commenter banned.
I had you down as more a Karla the elephant sort of person.
It’s not the elections moving that causes the current it’s the passing of charge, I seem to recall that electrons rather nonchalantly drift…
BiND, I=V/R, right? Ohm’s Law.
But if I’m reading Tim right, he knows the current, just wants to know the number of electrons involved. so resistance doesn’t matter.
Though, given SJW’s comment, it depends a bit on the detail. It’s the rate of electrons falling out of the end of the wire, doesn’t tell you as much about what they’re doing on the inside. I think…
Think of there being a gate in your conductor (wire). Current (amps) is a measure of how many electrons (what charge) pass through that gate in a given time.
Voltage is how hard you had to push to make that amount of charge flow. So it depends on the resistance of the conductor (yes, yes, V=IR and all that).
It all gets a bit fuzzier when you remember that the electrons are thermally excited and wobble about lots, so that these things are actually about averages and tendencies.
Don’t confuse the poor boy. Flatcap Army’s answer is correct; everything else is persiflage. A wire carrying one amp of current is moving 1 coulomb’s worth of electricity past a given point in one second. The charge carriers in wires are electrons. Resistance, voltage, capacitance etc.: none of that makes any difference. 1A ≝ 1 C·s⁻¹. An electron has charge 1.6021766 10⁻¹⁹ C. The End.
Robert Clayton is also right. Electron drift velocity in copper wires is typically of the order of a few tens of microns to a few mm per second. It’s an easy calculation: each copper atom donates a single electron to the conduction band, we know the density of copper (8960 kg·m⁻³) and its atomic weight (63.546 g·mol⁻¹) so we know how many moles of copper and hence conduction electrons there are per metre of a conductor of given cross section. To get a given current a certain number of these must flow past a given point in a given time.
Drift velocity of electrons is very leisurely. What travels fast is the electrical field when applied, which travels at the speed of light.
It’s the electrical field that causes the electrons to drift.
I think the difference between Tim and Dick is that Tim acknowledges his ignorance.
“the electrons are thermally excited”
Nice explanation of trying to debate with Murphy. Sounds like he and Socialist “Economics” are made for each other.
It’s the light electricity that you get in wires, heavy electricity requires a vacuum 😉
Explanation: The mass of an electron increases as its velocity becomes a significant fraction of the speed of light, but you can only get them that quick without air or the atoms inside of a wire to bump into. Yer bog standard wire-electricity only moves a few mm per hour on average. Electrons in an old fashioned TV tube are going maybe a quarter of the speed of light and “weigh” about 6% more than their rest mass.
Doesn’t quantum theory suggest you can’t know?
By definition, we know where they are. In your bit of wire. So that’s their position sorted. Therefore, their vector must be unknowable.
When spouting bollocks, always helps to use flash words.
(Following economics taught me that. Clever, innit?)
Imagine a clear day in a pub garden, steady sunshine and no wind (so minimal external variances).
The barman is pulling your pint.
I’ve forgotten what point I was going to make.
abacab: the change in the electric field travels at the speed of light in the conductor. In a bare copper conductor that’s about 2/3 c although the presence of a dielectric material round the wire (e.g. a coax cable) will alter this.
One thing that’s quite interesting is working out how long the wavelength of the electric field amplitude is in a wire carrying AC. At 50Hz it’s about 3800 kilometres.
Finally, as a fun aside: in a wire (assumed to have non-zero resistance) carrying a current, what is the direction of energy flow? In other words, what is the Poynting vector of the E and B fields? This is very counterintuitive but physically sound. Feynman II.27 goes into detail.
I like the answers from “Flatcap Army”, “Social Justice Warrior” and “Bloke in Costa Rica”.
If I add any more details myself I’ll just confuse things.
Doesn’t quantum theory suggest you can’t know? By definition, we know where they are.
Not as I understand it. By definition we don’t know where the electrons are (if Heisenberg is right).
My understanding is that we are better to treat an electron as a smear than a point object.
However, while quantum theory is indeterminate, statistics gets us past that issue. For sufficiently large quantities, and there’s a lot of electrons in a Coulomb, we can measure to quite unbelievable accuracy now (mm to the moon type accuracy).
If you just want a very, very basic level of understanding then
(A famous cartoon, no idea where it is from originally, of a “Volt” pushing an “Amp” while an “Ohm” tries to stop it.)
“It’s the actual number of electrons on the move that I want to find out.”
There’s a difference between the number of electrons on the move, and the number of electrons passing a given point.
The number of electrons moving is just the total number of free electrons in the conductor. It makes no difference what current or voltage is flowing, so long as it is non-zero. Either they all move slow, or they all move fast, but the number of them is the same.
If you want to use the hydraulic analogy, which helps intuition, then the number of electrons on the move is like the number of cubic metres of water in the river. The number of electrons passing a given point (the current in Amps = Coulombs per second) is like the flow rate of the river – the number of cubic metres per second passing a given point. The voltage is the potential energy per unit charge (in Joules per Coulomb) which can be thought of as like the height of the river (potential energy increases with height). surface. You need a height difference to get water to flow along the river, and the steeper the slope the faster the water flows.
You can, of course, generate another economic analogy using flows of money. The charge corresponds to the total number of pound notes, the current corresponds to the number of pounds passing through an account per second. The voltage corresponds to the utility (human happiness) obtainable per pound, that causes the flow.
So when you ask “how many electrons are on the move?” it’s like asking “how much money is moving through the economy?” If you mean “how many are moving past each point per second?” then Flatcap Army’s answer is what you want.
If you really do mean “how many are moving?” then each mole (64 g) of copper contains Avogadro’s number (6E23) free conduction electrons. Work out how much copper you’ve got in grams, divide by 64, then multiply by 6E23.
No wonder you accept the Globalwarmmongering frauds.
Shoulda gone to Ampereforth.
if this is really about electrons moving where are they now? All over the floor? Shouldn’t you put them back.
“if this is really about electrons moving where are they now? All over the floor? Shouldn’t you put them back.”
I’m sure you’re just joking, but…
If you’re talking AC then they just jiggle backwards and forwards a little.
If you’re talking DC then (usually) they’re back in the battery, having gone round in a loop.
The need to avoid a big pool of electrons building up somewhere is why you generally need a circuit. There are of course exceptions, like lightning strikes. But these exceptions do generally involve big pools of electrons having built up somewhere. In the case of lightning bolts, they do, indeed, end up all over the floor! 😉
… I can’t resist a bit of physics education. 🙂
A wave moves quickly through the water; the water itself moves very much more slowly. The same sort of thing occurs when electricity passes through a wire; the electricity moves very much more quickly than the electrons themselves.
Imagine a 100ft long 1inch tube full of 1inch balls. You shove a ball into one end of the tube and one pops out of the other end instantaneously. All the balls move on one place down the tube. The motion (or electricity) has traveled 100 feet but the balls (or electrons) have traveled 1 inch. This is not an exact analogy but I hope it maybe useful nonetheless.
It’s just P=IV Power, P in watts is the actual energy consumption, I is the current and V is the voltage. So if the current goes up by a factor or 10, with the same voltage, then the power consumed goes up by 10. If the current remains the same and the voltage goes up by 2 then the consumption doubles.
200V at 10 amps = 2000 watts. 100V at 20 amps = 2000 watts.
I am concerned murphy may wish to implement an electron tax once he discovers how many there are.
Electrons actually move really slowly. It’s the effect of their movement that travels really fast, like people in a crowd shuffling forward and, um, *analogy breaks down at this point*.
Anyway, “orthodox” physics is just neoliberal sophistry.
Grace Hopper used to use a piece of wire to demonstrate the speed of information (and therefore presumably electricity) in a computer. The wire was about, as I recall, 12 or 15 inches long and represented a nanosecond.
Have I muddied the water enough?