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The difficulty of understanding voltage

by PhysE
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davenn
#19
Mar6-13, 03:25 AM
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I was under ... and obviously the mistaken idea ... going by your comments, that electron drift was relatively constant, rather there was just more electrons going past a given point ... not because they were moving faster, but because the "power" source was supplying more of them

so does this mean that all the comments by various people that the electron drift velocity is a few mm / sec
isnt correct ?

I'm always willing to learn :)

D
e.chaniotakis
#20
Mar6-13, 04:41 AM
P: 72
Quote Quote by davenn View Post
I was under ... and obviously the mistaken idea ... going by your comments, that electron drift was relatively constant, rather there was just more electrons going past a given point ... not because they were moving faster, but because the "power" source was supplying more of them

so does this mean that all the comments by various people that the electron drift velocity is a few mm / sec
isnt correct ?

I'm always willing to learn :)

D

There is some misunderstanding here. The power source DOESN'T supply the circuit with electrons.The electrons are already in the wire . In a first crude approach you can depict it as an ideal Fermi gas of electrons moving almost free of electric forces.
When you increase the voltage each individual electron will increase its drift velocity by very little ( the order of magnitude is ~mm/s) . But the voltage affects many electrons. So you have many electrons , each of them affected infinitessimaly by the voltage.
Now, the current is :

I = (charge per unit volume)*(drift velocity)*(wire cross sectional area)
I = (electron charge)*(nr of electrons per unit volume)*(drift velocity)*(wire cross sectional area)

the drift velocity per electron increases with voltage so, (electrons)*(drift velocity) increases therefore current increases with voltage
Ratch
#21
Mar6-13, 08:19 AM
P: 315
davenn,

so does this mean that all the comments by various people that the electron drift velocity is a few mm / sec
isnt correct ?
e.chaniotakis has it correct. Electron drift velocity is very, very slow. Increasing it several times is still very slow. A voltage source takes in as many electron as it expells, so the net electron gain/loss of the source is zero. There is an uncountable sea of electrons in a good conductor like Cu, so the drift velocity does not have to be very much to supply large currents. And the finite density limit of free electrons in a conductor (8.4E28 electrons/meter for Cu) insures that the drift velocity has to speed up in order to support larger currents.

Ratch
cabraham
#22
Mar6-13, 09:16 AM
P: 1,041
Quote Quote by davenn View Post
I was under ... and obviously the mistaken idea ... going by your comments, that electron drift was relatively constant, rather there was just more electrons going past a given point ... not because they were moving faster, but because the "power" source was supplying more of them

so does this mean that all the comments by various people that the electron drift velocity is a few mm / sec
isnt correct ?

I'm always willing to learn :)

D
We must be sure to understand which quantities are held constant and which vary. If a conductor and/or fixed load resistance is connected across a CVS (constant voltage source), then the source, assuming it has high compliance (low internal series resistance & adequate energy conversion ability), will maintain a fixed voltage. If the resistance is fixed as well, which it appears to be in this example, then we can compute current per Ohm's law.

If the source voltage is increased to a new fixed value, but the resistance is the same as before, than the current must increase per Ohm's law. So we arrived at the question "does the increase in current incur a higher drift speed, or the same drift speed with increased charge carrier count in motion?"

I believe we've come to a consensus here that the drift speed would have to increase in order to increase the current, which must happen since voltage was increased with resistance held constant. To understand why at a mIcro scale involves solid state physics. When I have free time (hardly ever) I can refer to my Kittel text on said subject. But the model of a Cu conductor and a sea of electrons appears correct.

The no. of available carriers (electrons) in Cu is a fixed physical constant. These electrons will drift as soon as an external source provides an E field. All available electrons are in conduction, since Cu is a metal. An increase in source voltage incurs an increase in E field with constant resistance. The increase in current can only come about via an increase in drift speed. The percentage of free electrons participating in conduction is already at its maximum value, a fixed quantity.

Thus the drift speed must increase to realize an increase in J the current density. I believe the explanations above are correct. I initially pondered the thought that a material with current density J1 can undergo an increase in current to a new value J2, by 2 means. But in this example the R value does not change, so the only way to increase J is by increasing drift speed. We seem to have consensus.

Claude
PhysE
#23
Mar6-13, 01:35 PM
P: 10
Thanks for all the replys, they certainly made for some interesting, at least the parts I could get my head around did. I have another question regarding voltage that confuses me, I have tried to use a practical example to explain it, hopefully you can follow my logic and spot any places I have slipped up.

There is a component of a circuit, it has a P.D of 12V. This means that every coloumb of charge that is moved through the component does 12J of work. The current through this component is 2 Amps = 2 coloumbs per second. The resistance of this component could be worked out with R=V/I. The resistance with the above valies would be 12/2 = 6 ohms.
The P.D is changed from 12v to 10v and the resistance is kept the same. This leads to an decrease in current. Why exactly does this happen? If the P.D across this component falls does it mean (excuse my rather crude terminology) that the "driving force" of the current has decreased, yet the charge still has to face the same amount of resistance, meaning the current that gets through decreases? This is how I see it at the moment, but I do not really understand why the P.D across the component ( the work that is done when charge moves through it), affects the flow of current. If it was the EMF I would understand it, but does the P.D across a component also equate to the "driving force" across the component?
cabraham
#24
Mar6-13, 02:02 PM
P: 1,041
Quote Quote by PhysE View Post
Thanks for all the replys, they certainly made for some interesting, at least the parts I could get my head around did. I have another question regarding voltage that confuses me, I have tried to use a practical example to explain it, hopefully you can follow my logic and spot any places I have slipped up.

There is a component of a circuit, it has a P.D of 12V. This means that every coloumb of charge that is moved through the component does 12J of work. The current through this component is 2 Amps = 2 coloumbs per second. The resistance of this component could be worked out with R=V/I. The resistance with the above valies would be 12/2 = 6 ohms.
The P.D is changed from 12v to 10v and the resistance is kept the same. This leads to an decrease in current. Why exactly does this happen? If the P.D across this component falls does it mean (excuse my rather crude terminology) that the "driving force" of the current has decreased, yet the charge still has to face the same amount of resistance, meaning the current that gets through decreases? This is how I see it at the moment, but I do not really understand why the P.D across the component ( the work that is done when charge moves through it), affects the flow of current. If it was the EMF I would understand it, but does the P.D across a component also equate to the "driving force" across the component?
Instead of "driving force", the terminology in use is "Lorentz force". If a charge carrier is in the region where there are E & B fields present, and it is moving at a speed "u", then the 2 forces incurred by the charge carrier are "Fe & Fm" (electric & magnetic.

Fe = q*E, and Fm = q*(uXB). F = Fe + Fm.

Of course charged particles have their own E field, so that when collisions occur, and a layer of charges builds up in a region, then incoming charges feel this force due to accumulated charges. Likewise, a circuit possesses self inductance so that when charges move, this is current, a magnetic field exists which tends to oppose a change in current.

All of these characteristics enter in when circuits are studied. In your example, decreasing the source from 12V to 10V results in a decrease in E field, and a decrease in Lorentz force as well as decreases in current. J = σE. so since σ is constant for the material, a decrease in E results in a decrease in J the current density.

How this decrease occurs was already discussed. The drift speed decreases.

Claude
PhysE
#25
Mar6-13, 02:14 PM
P: 10
Thank you for your reply. However, I think I worded my question incorrectly. I am wondering about lowering the PD of a component, for example using a 10V bulb as opposed to a 12v bulb. If the resistance of the two bulbs are the same the current through the bulb must decrease, why does the P.D across a component affect the current through it. I feel I now have an understanding of the rlationship between the the current and voltage of the source but still lack information on why reducing the PD of a component would change the current across a component?
davenn
#26
Mar6-13, 03:15 PM
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Quote Quote by e.chaniotakis View Post
There is some misunderstanding here. The power source DOESN'T supply the circuit with electrons.
well it has to else there is no current flow ... see Ratch's post following yours
# in = # out = flow

The electrons are already in the wire .
Yes I realise the wire is already full of electrons that can move "freely"

In a first crude approach you can depict it as an ideal Fermi gas of electrons moving almost free of electric forces.
When you increase the voltage each individual electron will increase its drift velocity by very little ( the order of magnitude is ~mm/s) . But the voltage affects many electrons. So you have many electrons , each of them affected infinitessimaly by the voltage.
Now, the current is :

I = (charge per unit volume)*(drift velocity)*(wire cross sectional area)
I = (electron charge)*(nr of electrons per unit volume)*(drift velocity)*(wire cross sectional area)

the drift velocity per electron increases with voltage so, (electrons)*(drift velocity) increases therefore current increases with voltage
can you actually put some numbers into those formula

so for example if at 1A current, the electron drift velocity is say 1mm/sec and we increase
the current to 10 Amps or 100 Amps. Is there a 10 fold or 100 fold respectively increase in the electron drift velocity ?

Just trying to get a feel for how much the drift velocity changes ... ie. is it significant for significant currents ?

Dave

PS... NONE of this ( the above) was I taught in electronics classes during my training. Had never heard of electron drift velocity till maybe a year ago, when it became prominent in discussions on this forum
e.chaniotakis
#27
Mar6-13, 04:53 PM
P: 72
Drift velocity changes, but the nr of electrons passing per unit volume remains constant. Their product (current density)increases as voltage increases and thus the current (which is only an integration with respect to cross sectional area).

Now, about numerics, check that out to see a derivation of electron drift velocity in copper:
http://www.physicsforums.com/showthr...=674062&page=2

You will see that u = (constant)*E = (constant)*V (1)
I = (constant)*u (2)
thus I =(constant)*V (3)

if you change voltage, say by 10% , drift velocity changes by 10% (equation 1).
Thus, current changes by 10% too (equation 2).
- This in order to keep the constant term in equations actually constant:-)
Ratch
#28
Mar6-13, 05:24 PM
P: 315
davenn,

... Just trying to get a feel for how much the drift velocity changes ... ie. is it significant for significant currents ? ...
What you need is a good physics book. See the attachment. Use the example to fiddle and diddle with the wire area, current, etc. As you will note, doubling the current will double the drift velocity.

Ratch
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