Why does a real DC circuit with 1 resistor have uniform current?

Question

Imagine a DC circuit with small but non-zero resistance in wires and large resistance in a single resistor in series with the battery, all ohmic.

Connecting the battery I imagine a surge of electrons from the - plate due to high concentration of negative charges. Using Coulombic concepts I intuit a movement on average of charge away from the - plate which is uniformly through the conducting wire, loosely a brownian motion of valent electrons in the atomic lattice down a gradient of electron concentration.

When they enter the resistance, by the definition of resistance from resistivity as R=pl/A=me*l/(ne^2tA) where n is # electrons per volume and t the average time in between electron collision, I suppose that (l/A) rises and n and t increase. I then intuit that the current in the resistor falls. This is not where the problem is, because as a result the current coming out of the - plate of the battery will fall, so that the (electron) current coming out of the battery is equal to the electron current in the resistor.

However, when the electrons come out of the resistor, I intuit that current again increases. Why? Because l/A falls and t increases (although perhaps n falls). The first surge of electrons continues out of the resistor at what I see to be a higher average velocity than inside the resistor. Their motion is driven by their mutual repulsion which on average is in the direction of the + plate.

If the rate of electrons entering the sinks in the + plate is equal to the rate of electrons leaving the - plate, and the electrons move from + to - in the battery, then I can explain why the current leaving the resistor is equal to the rest. However, if this is not necessarily the case then I cannot. Can anyone clarify and point out where my intuitive image is wrong?

Answer 1

I found it a bit difficult to follow you, but I think your basic problem is with the image that something changes regarding the flow of electrons (the magnitude of current) because the electrons "start out" by encountering the very low resistance of the wire and the encounter the large resistor also connected to the battery. There is no difference in the current in the wire and resistor. If there were, electrons would "pile up" in the circuit. The following is offered by way of a simplified explanation.

Prior to connecting the battery, the mobile electrons have various speeds in random directions due to thermal energy (what you call brownian motion), but no particular average net speed (a.k.a, drift velocity) along the conductors in the circuit.

When the battery is connected an electric field is simultaneously (near the speed of light) established all along the low resistance wire and the larger resistor connected to the battery. The electrons don't start out entering the wire and then enter and exit the large resistor as separate events. The electrons simultaneously move through both the wire and the resistor at all locations.

On the other hand, the potential differences (voltages) across the wire and resistor will not be the same. The potential difference V between any two points in the circuit will equal the work per unit charge required to move the charge between the points. The greater the resistance between the points, the greater is the amount of work per unit charge (voltage) needed to move the charge between the points. So relatively little work is needed to be done by the battery to move the electrons through the low resistance wire compared to the work required to move them through the large resistor.

At the microscopic level, electrons are continually losing kinetic energy when they collide with particles in the resistance material (in the form of heat) while simultaneously regaining that kinetic energy from the energy of the electric field, so that the average drift velocity of the electrons (and the current) remains constant throughout the wire and the resistor.

But Bob, I see that point. What I’m asking, however, is what specifically, explicitly makes it impossible for the rate at which electron current moves out of the resistor to be different from the rate at which current moves through the wire.

Have you ever seen Newtons Cradle? If not, Google it up. When the steel ball at one end impacts the ball next to it, the ball at the other end of the series of balls seems to instantaneously move out. It's a chain reaction transfer of momentum. Similarly, when an electric field is applied to a conductor, the electric field that propagates at nearly the speed of light causes an electron at one end to move and simultaneously the electron at the other end moves. It is somewhat like a chain reaction.

Hope this helps.

Answer 2

The moment you connect your batterie to the circuit, you have almost immediately (light speed) an electric field along the circuit, so all electrons start to move, according to the resistance given, since the resistor inhibits the movement, all e move slower than without the resistor. But to keep the same amount of e per second passing any area A, the speed in the resistor with its smaller area ist greater, (thats why the resistor gets warm or hot. Compare it withe the flow of water in a closed pipe, same amount of water per sec through any area, so it flows faster in parts with smaller area. So in same respect your intuition ist missed. (and of cause the same amount of e leaving one side arrive at the other side.)