Continue to Site

Welcome to our site!

Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

  • Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

Voltage Drop Concept

Status
Not open for further replies.

Wayneg

New Member
- Given that opposition to current flow is known as resistance
- If I have a series circuit with a 12 volt source and 2 resisters, R1 & R2 at 2Ω and 4Ω respectively
- I will measure 2 amps at any point in the circuit
- Why are the resisters, which oppose current flow, not reducing the current individually?
- Is the voltage drop across each resister the force "consumed" (my word) by the resistor as the "force" tries to push a constant amount of electrons (2 amps in this case) against the resistance?
- If that were the case why wouldn't I measure 8 volts at the far side of R1 (E=2amps X 2Ω= 4 volts = voltage drop at R1)?
- If I did measure 8 volts after R1 why wouldn't I use that value to calculate the voltage drop at R2 instead of the applied voltage of 12 volts?

This stuff is really confusing. I'm sure I'm missing something obvious.
 
Last edited:
Hi,

The resistors *do* reduce the current. If you replace them with s short, you get an even higher current.
If you replace them with nothing (open circuit) you get no current at all.

Well, the voltage is not consumed, it simply appears across the resistor. The current is not consumed either, it simply flows through the resistor. The power (or energy) is consumed and that is the product of the voltage and current:
Power=voltage times current=volts*amps

Voltage is also polarized, so when you add the voltages in a circuit you have to add them algebraically (ie pay attention to their sign). So 12v-6v=6v, and 12v-2v=10v, and 12v-4v=8v. If you connect the 4 ohm resistor to the positive terminal of the 12v source (battery or whatever) and the 2 ohm resistor to the 4 ohm resistor, then the other end of the 2 ohm resistor to the negative terminal of the source, then you'll measure 12v-8v=4v at the junction of the two resistors. That's the same 4v you get when you multiply 2 ohms times 2 amps.

Dont hesitate to ask more questions if this still isnt clear. It becomes clearer as time goes on and you inquire more often and think about it a while. It also helps to do more experiments like this and try to figure out the results and why they make sense.
 
First. the flow of water analogy is bad, but works initially.
Volts = pressure
Amps = flow (electrons or holes don't flow in a wire)
Resistance = inverse diameter of a pipe (smaller d. More R)

Picture a 4' diameter and then a 2' pipe main with a tiny hole in it. Measure the flow.
So a 1 ohm resistor in series with 1 M doesn't do anything.

Warning: Conventional current is from + to negative (holes in semiconductor physics) and helectrons go the other way.
 
Thanks for your quick response.

I guess I accept that resistance decreases current given I=E/R. But I am confused about why in the series circuit it happens across the whole circuit at once and not in a stepwise fashion at each resistor.

Clearly, my vocabulary is not as well tuned as it needs to be. Notwithstanding, it seems you answered my voltage question. That was a huge help. Based on your answer I assume that the voltage between the 2 ohm resister and the negative terminal would measure 0. Is that correct?

Thanks,
Wayne
 
KISS,
You are right about one thing, the water analogy is bad. I find it more confusing than electron flows. And that's another issue. Please don't tell me that electrons don't flow in a wire. That's the basis of my limited understanding. I thought I finally understood what electricity was. What about valence rings and free electrons and all that kinda stuff? Anyway, you say conventional current flows from positive to negative but electrons go the other way. If they "go the other way" how are they getting there if not flowing through the wire?

My reading tells me that "Negative" charge results from too many electrons and "Positive" charge from not enough (or fewer) and the electrons from the "Negative" flow to the "Positive" to attempt to balance the charge. Is this not correct?

Regards,
Wayne
 
resistors add in series just like hairballs do, more clog less flow! you are right that electrons go from negative to positive, just like water flows downhill,

dont concern too much with valence, the more electrons are loose in a wire the less resistance and better conductance, (on average)
if you are looking at the valence in the battery you will see more in to how "opposite" "valences" attract, the pressure they create which form the volts, an where its balanced off in a dead battery.

conventional just says how we like to look at where positive is: the point of suction which wants to pull all those electrons in. (which factually is short a few electrons)
usually the electron source is ground, which is what we consider neutral since it is is like the sea level, and where there are abundance of the electrons, (not to be confused with negative voltages)




sry iv always liked the water method
 
Conventional vs. electron current: https://www.electro-tech-online.com/custompdfs/2013/01/elect_flow_vs_conv_I.pdf

Electrons don't flow in a wire. At least in the sense, that the same one emerges at the other end. The electrons get bumped out of orbit (that's wrong too, but close enough) and another takes it's place. So there is an apparent flow, not a real flow.

In a Vacuum tube or Valve, electrons do flow. In an electron microscope, they flow. In a wire or semiconductor material, they don't.

An Ampere is defined as 1 Coulomb/sec. https://en.wikipedia.org/wiki/Ampere A Coulomb is a measure of charge.

Teaching depends a lot on your level of understanding and as time goes on, the simple concepts change, In grade school, we though of electrons in circular orbits. Later, they are probabilities that you will find an electron.

Early, I was taught about convention current, but when you do a chemistry lab experiment and have to KNOW where the electrons are, you have to switch gears a bit. Same for solid state physics. In that case, they invented the concept of "holes" to represent conventional current flow.

Early, you learn about the slope of a line, Later you learn differentiation which is the slope of a curve at any point.

Electricity is the flow of charge and Ben Franklin was wrong so we have conventional current because we can't go back and rewrite history.

At the same time Power is defined as the power dissipated. Negative power is power generated, BUT we never say we have -16 MW power plant. But it has to be negative to make the math work out assuming an equation of the form (something)=0. If I say (Power Generated) = (Power Consumed), but it's not in the proper form for solving.

Also in electrical engineering j is substituted for i for √-1 because i is generally reserved for current.

I/we don't know where your level of understanding is at, but I can tell you that in College, I challenged the teacher many times and won. His answer was "Your not supposed to know that yet".

Go with what your taught, but it will change over time.

Shaking a wire in mid-air generates a current, but for all intents and purposes it does not. I had to worry about it though. Paper is an insulator, but I could measure it's resistance (actually Conductance = 1/R) when the values are high. A short piece of wire has inductance, but in the world which most of us live it doesn't. At high frequencies it does have appreciable inductance.

So, know your surroundings and simplify.
 
Last edited:
Hi,

Im not sure it is a good idea to state that electrons dont flow in a wire. That is only correct if we consider the 'flow' to be of a certain character such as water where usually most of the molecules travel more or less in straight lines or something like that. But flow doesnt have to follow a certain definition really, as long as something moves in a regular way (laminar flow or turbulent flow for example) and we do have electrons moving in a regular way on average which would probably be better likened to turbulent flow. In DC current, we also have electrons that enter in one end exit at the other end of a short length of wire. We can get very technical here and bring in some definitions from quantum mechanics, but i dont think that's necessary in every discussion. Maybe just to make the reader aware of the complexities of electron flow while maintaining some user level degree of simplicity that can help get the major facts across.

To be more specific, the electrons 'flow' or 'move' in various directions sometimes along the wire and sometimes not, but their average movement is along the wire and that is usually referred to as 'drift'. This drift is what we usually call current flow. Taken on average, we can say that the electrons flow through the wire from one end to the other in DC current, but it is much slower than we might guess offhand. In a #10 gauge wire a 1 amp current might flow a few centimeters per second or slower just to quote a simple example. That is far, far slower than the speed of light for example.

So i think it is ok to say that electrons flow as long as we remember that it is the drift speed not the full electron movement speed side to side or back and forth.
 
Last edited:
So i think it is ok to say that electrons flow as long as we remember that it is the drift speed not the full electron movement speed side to side or back and forth.

hi Al,
I agree with your statement, electrons do move/drift in a conductor as the result of a potential difference between the ends of the conductor.

E.
 
I want to thank you all for your responses to my question. You have helped immensely. But just to be sure, allow me to state what I think I understand at this point. First, since my study of the topic is at a very elementary level some hugely complicated notions are being "dumbed down" for the purposes of illustrative simplicity. OK - I'll just go along for the ride at this point and leave the complexities to later study. Second, I shouldn't worry about which way whatever is "flowing" since it will not impact my analysis of a circuit as long as I keep my signs straight. Third, and this was an eye opener (thanks for the PDF link), there is current and conventional current which are two ways of discussing the same thing.

Thanks again to you all.

Regards,
Wayne
 
Hi again Wayneg,

It sounds like you have it right. And dont worry too much about the technical details because they are really only necessary when you study this deeply and for most circuits you never need to know more than 'conventional' current flow. To give you an example, i've been working with circuits and doing quite a bit of in depth analysis of the circuits themselves for the past 30 or more years and the only time i have to discuss "electron drift velocity" is when someone asks about more information about the true nature of electrical current. Otherwise i would probably not talk about it at all :)
It's a subject that comes up in a university study of physics as well, but that isnt necessary for a good understanding of circuit analysis which is really just about how the circuits work. For that conventional current flow (and some times the so called 'electron current flow' which is just conventional in the other direction) usually works just fine.

So to sum up, if you are interested in circuit analysis a brief knowledge of the true nature of electrical current is ok without going too deep. If you are interested in the true physics (as in a class on physics) then you would want to look deeper. Alas, alas, alas, the deeper we look the more we find out that there is no one yet alive that understands the ultimate truth about particle physics, so dont fret too much if you find that asking questions leads to still even more questions :)
 
Last edited:
Nicely said!

PS: At RF frequencies, silver plated copper tubes are sometimes used as conductors. Turns out the middle of the conductor doesn't contribute. So, make them hollow to save material costs. Some instruments that work in the GHz range might look like a plumbing contraption rather than electrical wires.
 
Wayneg,

Given that opposition to current flow is known as resistance

Reactance also reduces current.

- If I have a series circuit with a 12 volt source and 2 resisters, R1 & R2 at 2Ω and 4Ω respectively
- I will measure 2 amps at any point in the circuit

That is correct. It has to be that way. You cannot have 4 amps at one point and 1 amp at another. In a series circuit, charge flow has to be the same at all points, otherwise charge will accumulate or deplete.

- Why are the resisters, which oppose current flow, not reducing the current individually?

They do so. Each resistor is dissipating the energy of the charge flow, which in turn reduces the amount of charge flow, also called current.

- Is the voltage drop across each resister the force "consumed" (my word) by the resistor as the "force" tries to push a constant amount of electrons (2 amps in this case) against the resistance?

Voltage is not "consumed" by the resistors. Resistors consume energy. Charges move from a higher energy point to a lower energy point. Voltage is the energy density of the charge. At one end of the resistor, the voltage or energy density is higher, so the charges move to the opposite side of the resistor to the lower voltage or energy density. Charges lose energy as heat when they move through a resistor, so the voltage is less at the end of the resistor.

- If that were the case why wouldn't I measure 8 volts at the far side of R1 (E=2amps X 2Ω= 4 volts = voltage drop at R1)?

You will measure the voltage difference or energy density loss of the moving charge across each resistor.

- If I did measure 8 volts after R1 why wouldn't I use that value to calculate the voltage drop at R2 instead of the applied voltage of 12 volts?

The energy density of the charge at the source (battery) will equal the energy density of the charge lost at the resistors. That is one of Kirchoff's laws.

This stuff is really confusing. I'm sure I'm missing something obvious.

Just remember that voltage in not force, it is the energy density of the charge. The energy density decreases as the charge travels through resistance because its energy is dissipated into heat by the resistors. This loss of energy also causes its energy density to decrease also because the same number of charges are moving.

Ratch
 
Status
Not open for further replies.

Latest threads

New Articles From Microcontroller Tips

Back
Top