Electric circuit, drift velocity, battery

[edington]

Hello,

This is a problem that is buzzing me around for quite sometime since I took fundamental physics course. Consider an electric circuit with one battery and one resistor. And ignore the internal resistance of the battery and the resistance of the wires.

1) Does battery supply electrons to the circuit?
When we connect the battery, it generates a potential difference (en electric field) which causes the electrons to move. I learned that even electrons move very fast, the net motion of electron is very slow (drift velocity). So, an electron can reach to the positive terminal hours later it left the negative terminal. Well, when it reaches the positive terminal what happens? Does that electron go through the battery and pumped back again from the negative terminal?

2) What I learned was that even free electrons are not that "free" to move. They have region to move. So, an electron located at point A initially can only go little far and can hardly reach the other end. Through collisions the energy is transferred. This is confusing little bit, I thought the electrons are moving with Vd (drift vel.) from one terminal to the next. Could you enlighten me about this?

3) This is the part the buzzes me the most;
Consider the simplest electric circuit. One battery, connected to a resistor (wires are resistanceless). All the potential energy that is gained from the battery will be lost across the resistor. So that for any closed path, the DeltaV is zero. Now, the electrons in the wire (before they cross the resistor) have a higher potential (thus higher kinetic energy), but when they cross the resistor they will lose the energy they gained from the battery. So, the electrons in the wire on the other side of the resistor have less energy (less kinetic energy).
We know that because the current (which is defined as the amount of charge that passes through a cross section per unit time) is the same in the wire on the both side of the resistor, then their drift velocity must be the same. How this can happen? am I wrong with my reasoning?

 

[Quinzio]

1) Does battery supply electrons to the circuit?
When we connect the battery, it generates a potential difference (en electric field) which causes the electrons to move. I learned that even electrons move very fast, the net motion of electron is very slow (drift velocity). So, an electron can reach to the positive terminal hours later it left the negative terminal. Well, when it reaches the positive terminal what happens? Does that electron go through the battery and pumped back again from the negative terminal?

No, a battery is not like a water pump in an aquarium, that pumps the water from the tank to a filter and then back in the pool.
I'm not really an expert, but a battery basically creates ions through electrolysis. Electrons are accumulated in the cathode, the -ve pole.
When electrons reach the +ve (passing through the resistor) they "annihilate" with the +ve charges created by the same electrolytic process.
There is no really current flow inside the battery from the two poles.

2) What I learned was that even free electrons are not that "free" to move. They have region to move. So, an electron located at point A initially can only go little far and can hardly reach the other end. Through collisions the energy is transferred. This is confusing little bit, I thought the electrons are moving with Vd (drift vel.) from one terminal to the next. Could you enlighten me about this?

Again, I?m not an expert, but electrons do really move in a conductor (e.g. a copper wire). They do jump from an atom to the next.

Another example: in a vacuum tube, like the "old" CRT monitor tube, electron did fly into the vacuum inside the tube.

3) This is the part the buzzes me the most;
Consider the simplest electric circuit. One battery, connected to a resistor (wires are resistanceless). All the potential energy that is gained from the battery will be lost across the resistor. So that for any closed path, the DeltaV is zero. Now, the electrons in the wire (before they cross the resistor) have a higher potential (thus higher kinetic energy), but when they cross the resistor they will lose the energy they gained from the battery. So, the electrons in the wire on the other side of the resistor have less energy (less kinetic energy).
We know that because the current (which is defined as the amount of charge that passes through a cross section per unit time) is the same in the wire on the both side of the resistor, then their drift velocity must be the same. How this can happen? am I wrong with my reasoning?

An electron itself doesn't have really a "potential" energy. It does have energy only if it is related to another charge in the space.
It makes sense to talk of potential energy only between two charges, not about one charge alone.
The electrons flowing in the +ve wire of your circuit has the same "importance" of the electrons on the -ve.
We are used to see a ground path in all electronic circuit, as if it was a "waste pipe" for "exhaust" electrons, but it's not really the case. Electrons in the "return" path have the same energy or whatever than the electrons in the +ve parts of the circuit.
In the -ve pole of battery there is a bigger "density" of electrons respect the +ve, so they tend to reach places where there is a smaller density, if there is a conductive path.

I was not really rigorous, but I hope I shed some light.