0
\$\begingroup\$

Assuming that the internal resistance of the battery is zero, assuming that the internal resistance of the wire is zero and last but not the least, assuming that the measuring device is ideal and does not draw any current.

In a scenario where there is only one resistor between source and the ground. Depending on the value of the resistor the current’s value is decided. The voltage across the resistor is same as that of the source voltage. If the resistor’s value is changed the voltage across it still equal to the source voltage. But, the resistors control the number of electrons that get blocked and these blocked (accumulated) electrons creates forces what we refer to as a voltage with respect to ground. If a larger resistor blocks more electrons and a smaller resistor blocks less, then in a simple series circuit with only one resistor between the source and the ground, the voltages across different resistors should be different. The question is why is it the same ?

My understanding is, the source sends electrons to make the voltage equal to its own level. This is current. These electrons come and settle down on resistor’s top terminal. Before the next batch of electrons arrive, the present electrons would have passed to the ground. The measuring device is about to show that there are no electrons hence no voltage but before that happens, a new batch of electrons come and the measuring device always shows constant voltage equal to source voltage across the resistor, no matter what the value of resistor is.

So, Can I say that the measuring device is not fast enough to sense that little bit of drop that was present between the two batches of current that the source sent in order to achieve equilibrium.

Is my understanding correct or have I completely misunderstood it ?

\$\endgroup\$
5
  • \$\begingroup\$ Ah, Let me see. (1) I don't agree with you saying the following: "These electrons come and settle down on resistor’s top terminal.". I think a resistor is not like a capacitor, therefore no charges are crowded at any terminal, (2) Electrons do not travel in batches: When one electron comes out of the negative terminal of a battery, another electron, goes into the positive terminal, after a very, very short time (called relaxation time, of the order of 10**-8 second, to be precise). So the electrons "seem" lying on the surface of a moving belt, all moving "at the same time". \$\endgroup\$
    – tlfong01
    Dec 28 '20 at 2:05
  • 1
    \$\begingroup\$ Thank you for your answer. Okay, so this relaxation time is so short that the multi-meter or any other measuring device would not be able to take it into consideration and hence it always gives a constant reading equal to the source voltage across the resistor. \$\endgroup\$
    – Shai
    Dec 28 '20 at 2:33
  • \$\begingroup\$ Ah, let me see. The multi-meter's operation time is about 300 ms, so the electrons would LOL when they know that any stupid human uses a multimeter to measure their speed. To electrons, even the fast oscilloscope is too slow, much much slower than even making slow motion pictures. My US$300, 50MHz digital storage scope can do 1 Giga samples per second, that is, 10 to -9 second per sample. I heard that rich guys do have 1GHz scope, but I am a poor guy, so I can only image. On the other hand, electrons can pass a 100 nano meter long tunnel of a tunnel diode, in no time! /to continue, ... \$\endgroup\$
    – tlfong01
    Dec 28 '20 at 3:27
  • 1
    \$\begingroup\$ @Shai Suppose a wire forks into two parallel paths where two resistors are located downstream from the wired fork. Suppose the resistors have different values. The pre-fork end of the wire goes to the battery (-) and the opposite ends of the two resistors are brought to the (+) end of the battery with each separately touching the (+) battery terminal. Ask yourself this, "How do the electrons know that one amount of them should 'go left' while a different amount of them should 'go right?'" Keep in mind that right at the fork, it's just an all-copper fork with resistors still a long ways away. \$\endgroup\$
    – jonk
    Dec 28 '20 at 3:49
  • \$\begingroup\$ Ah, elementary: there are two relaxation controllers! :) \$\endgroup\$
    – tlfong01
    Dec 28 '20 at 7:47
1
\$\begingroup\$

It sounds like you are groping your way to a transmission line model understanding of what's happening.

With models, we like to use the simplest model that captures the performance of the circuit we're working on. For most people, for most of the time, that's the circuit theory model. In it, wires have zero resistance, constant voltage everywhere, they transmit effects instantaneously, and electrons don't exist, just current and voltage.

However, you are worrying about time, so it's time to take one step up, and consider the transmission line model.

Take a long line, connected at the far end to a resistor to ground. There's a fast DMM at both ends of the line. At t=0, you connect a voltage source with a certain output impedance to the line.

The input end of the line jumps up to some voltage. A voltage wave sets off along the line, and a current wave with it, the current wave charging the capacitance of the line up as it goes. It's not a result of the voltage wave, it doesn't cause the voltage wave, but the two go hand-in-hand along the line. Their ratio depends on the geometry of the line, the square root of the ratio of the line's inductance and capacitance, and is called the line characteristic impedance. This has nothing to do with the line's resistance, which can be assumed to be zero without affecting this first order description of what happens.

The initial voltage that the near end jumped up to was set by the ratio of the source output impedance and the line impedance. It has nothing to do with the resistor at the far end. At t=0, the far DMM is still reading 0 volts. The source end of the line doesn't 'know' anything about the far end.

After the wave has propagated down the line at the speed of light in the line (often around 0.6c for plastic insulated lines), it reaches the resistor. The far DMM jumps up to some voltage. The voltage it jumps to is determined by the line impedance, and the value of the far resistor. If these are not equal, a return wave is generated, and this sets off back down the line again, until it reaches the source. A further reflection may or may not happen, depending on the value of the source impedance and the line impedance.

Each time a reflection is generated, it's smaller than the previous one, and the wave will have a new voltage and current amplitude. After sufficient number of reflections, the line settles down, and the source and resistor have at last 'agreed on' a voltage and current that suits them both. This means that the voltage and current are only determined by the source impedance and the load resistance, and not the characteristic impedance of the line.

For short lines and slow instruments, you can ignore all this detail, and just use circuit theory as a quick, and perfectly adequate, way of figuring out what's going to happen when you connect a battery to a bulb, or a transistor to a biassing circuit.

I've not mentioned electrons in this newer model. They don't really help, not with understanding anyway. If you must, consider them as the mechanism that underlies charge in metal conductors (other charge carriers exist in other conductors, don't get chauvanistic about electrons). Positive charge is a local deficit of them, positive charge a local buildup, and current the net rate they move from one place to another.

If you really, really, want to worry about electrons, look up the Drude model. Be warned that it's not very accurate, has little predictive power, and is a hand-waving classical approximation to what really needs to be a quantum mechanical description, but some people find it helps. Working electronic engineers can safely ignore electrons, they are more the domain of physicists.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.