Why is the current the same at points A and B in a resistor?

[Joker93]

Yes,the resistor tends to slow the charges down(the current),and thus initially(before the system reaches steady state conditions) you have charge accumulation near the start of the resistor.As the charge accumulates it creates its own electric field and thus gives some extra "push" to the charge that will tend to also accumulate with them and finally there will not be any more accumulation of charge but a steady state will be reached and you will also have steady current in the resistor.The same mechanism is what drives current when the wire bends within a circuit.So yes,the resistor DOES actually slow down the current,but it does so until the accumulated charge caused by it produce an equal amount of electric force to keep the current steady

 

[Drakkith]

As the charge accumulates it creates its own electric field and thus gives some extra "push" to the charge that will tend to also accumulate with them and finally there will not be any more accumulation of charge but a steady state will be reached and you will also have steady current in the resistor.The same mechanism is what drives current when the wire bends within a circuit.So yes,the resistor DOES actually slow down the current,but it does so until the accumulated charge caused by it produce an equal amount of electric force to keep the current steady

Does this imply that there are negative charges built up on one side of resistive elements?

 

[nasu]

Yes,the resistor tends to slow the charges down(the current),and thus initially(before the system reaches steady state conditions) you have charge accumulation near the start of the resistor.As the charge accumulates it creates its own electric field and thus gives some extra "push" to the charge that will tend to also accumulate with them and finally there will not be any more accumulation of charge but a steady state will be reached and you will also have steady current in the resistor.The same mechanism is what drives current when the wire bends within a circuit.So yes,the resistor DOES actually slow down the current,but it does so until the accumulated charge caused by it produce an equal amount of electric force to keep the current steady

Slow them in respect to what? It seems that you are talking about the transient time, when the circuit is established but still don't see what could be the meaning of this "slowing down". It seem to imply that is takes some time for the electrons to reach the resistor.
Indeed there is surface charge creating the field, in the steady state but this field is what gives the electrons their drift velocity. Is not slowing them down, whatever than means. So if you think about the initial state, before the field is established, the drift velocity is zero and then increases to the steady state value. Where would a slowing down will fit in this?

We can compare the circuit with and without rezistor (just wires) and in this case is clear that the drift velocity in the conductors is lower in the case with resistor than in the case without, given that the same voltage source is used..
If we compare the drift velocity in resistor and in wires for the case with resistor, it can be either way. There is no reason to assume that the drift is slower in resistor.

 

[OmCheeto]

Does this imply that there are negative charges built up on one side of resistive elements?

/me scratches head...
I think it does.

/me thinks about it for another 2.3 seconds
hmm... It makes total sense to me.

Never really thought about it before.

Somebody should build a new analogy model, as I don't think any of the previous ones work here.

Unless of course, all the fans going to the football game have really bad B.O. (You require a mutual repulsive force that fits the square law)
F = k * (Stench₁ * Stench₂)/d²

Science!

Seriously. Why else would they show a voltage drop across a resistor?
Of course, I could be wrong.

 

[Dale]

Does this imply that there are negative charges built up on one side of resistive elements?

Yes. This is an interesting detailed calculation.

EDIT: corrected link
http://depa.fquim.unam.mx/amyd/arch...ia_a_otros_elementos_de_un_circuito_20867.pdf

 

[OmCheeto]

Yes. This is an interesting detailed calculation.

http://mathpages.com/rr/s2-04/2-04.htm

Relativistic doppler shift? Now I'm really confused.

 

[Dale]

Relativistic doppler shift? Now I'm really confused.

Oops. Copy paste error.

Here is what I meant to post.

http://depa.fquim.unam.mx/amyd/arch...ia_a_otros_elementos_de_un_circuito_20867.pdf

 

[OmCheeto]

Oops. Copy paste error.

Here is what I meant to post.

http://depa.fquim.unam.mx/amyd/arch...ia_a_otros_elementos_de_un_circuito_20867.pdf

Delightfully serendipitous mistake though. We've been discussing doppler shift navigation over on the Ceres thread. I think you may have just answered my question. Thanks!

 

[skepticwulf]

Regarding an analogy, from the textbook I'm studying: "
A common misconception is that the electrons are “used up” by the lightbulb. A good analogy
would be water flowing across a water wheel in a flour mill. The water flows onto the wheel at
the top (high potential) and causes the wheel to rotate as the water descends along the wheel. The
amount of water that leaves the bottom of the wheel is the same as the amount that entered at the
top, but it does so at a lower point (low potential). The change in potential energy goes into work
in the wheel. In a lightbulb, electrons at higher potential energy enter the lightbulb and give off
that energy as they pass through the bulb. As with the water on the wheel, the number of
electrons exiting and entering the bulb is the same"