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?

  • $\begingroup$ If the current is higher in one place in the circuit and lower in another, that means more electrons are going in than are coming out. What happens to the extra electrons? Does the wire get full of electrons and swell up and pop like a balloon? $\endgroup$
    – user253751
    Sep 15, 2020 at 19:14
  • $\begingroup$ Would you be satisified if I just told you that electrons repel each other so they can influence electrons ahead of them and behind them thereby keeping the "space" between them even? There is no good mechanical analog because a group of mechanical objects are unable to maintain their spacing equally unless they are crowded up right against each other (like a bumper-to-bumper traffic jam). Electrons don't need to be up against each other. $\endgroup$
    – DKNguyen
    Sep 15, 2020 at 19:40
  • $\begingroup$ @DKNguyen, To some extent, you could liken it to a railroad train. The amount of space between wagons can vary by a little bit, but not by much. When the engine changes speed, the "news" of that speed change travels very quickly down the length of the train until, after a fraction of a second (actually, can be two or three seconds for a very long freight train) all of the wagons are moving at the same new speed. $\endgroup$ Sep 15, 2020 at 20:08

2 Answers 2


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.

  • $\begingroup$ Bob D, I taught high school physics for 13 years, and I don't know where this concept comes from that electrons travel at different speeds in different circuit elements of a series circuit. However, I do know one thing. Students can be taught the correct concepts during daily lessons, and 2 weeks later when they take the test, many of them revert right back to their old and incorrect concepts. For some students, you literally can't beat the invalid concepts out of them. $\endgroup$ Sep 15, 2020 at 19:29
  • $\begingroup$ @DavidWhite Indeed. It comes up time and time again. $\endgroup$
    – Bob D
    Sep 15, 2020 at 19:31
  • $\begingroup$ Bob D, yes it does, and if I knew how to permanently get the correct concepts into students' heads, I would have done it by now. I'm afraid that this same type of question will appear again in 1-2 months. $\endgroup$ Sep 15, 2020 at 19:33
  • $\begingroup$ Bob D, this might be a bit pedantic but then again this entire question is pedantic: do all electrons start to move simultaneously? Surely it takes time for forces from the electrons on the - plate to propagate (like it takes time for a compression force to travel all the way through a light year long stick)? $\endgroup$ Sep 15, 2020 at 19:51
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    $\begingroup$ @AdriaanBerger I updated my answer to respond to our follow up question. Hope it helps. $\endgroup$
    – Bob D
    Sep 15, 2020 at 21:31

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.)


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