# The difficulty of understanding voltage

davenn
Gold Member
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

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:

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:-)

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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