Conceptual Understanding for Voltage in a Circuit

[Big Al]

This is probably overkill, but I hope that bits of it may help to answer your questions about voltages and currents in a simple circuit.

BATTERY
The electrons in a charged-up battery are under an electric 'tension', a bit like a stretched elastic band. The chemistry in the battery is trying to push electrons OUT of the negative terminal, and at the same time trying to suck electrons back INTO the positive terminal. The size of this pushing and pulling is measured in volts, or 'electric potential'. 'Potential', because nothing happens until the battery is connected into a circuit. The 'potential difference' between the positive and negative terminals is the battery 'voltage', say 5 volts. By convention the negative terminal is regarded as being at zero volts, and the positive terminal at, say, 5 volts.

CONVENTION
Talking of convention, the pioneers of electricity (mistakenly) thought that positively charged particles flowed out of the battery, from the 'positive' terminal, re-entering the battery via the 'negative' terminal. They named the terminals and the current flow accordingly. We now know that in fact the electric current consists of negative electrons flowing from the negative terminal, round the circuit, and back into the positive terminal. But it's easier to stick with the conventional nomenclature, and imagine positively charged particles flowing in the wires -- in the opposite direction to the true electron flow. Which is what we will do in the following explanations, and which you will see on all circuit diagrams.

CIRCUIT
Suppose you connect a copper wire from the battery's positive terminal to one end of a resistor, and the negative terminal to the other end. This allows the 'tension' inside the battery to be relieved. Positive charges are pushed out of the positive terminal, through the copper wires and the resistor, and sucked back into the negative terminal of the battery. This flow of positive charges round the circuit is the 'electric current', measured in amperes. Unless there is a leak in the circuit, the number of charged particles flowing per unit time (current) at all points in the circuit is the same. A bit like water pumping round a watertight pipe. You cannot have more litres per second flowing in one part of the pipe than in another! The water would either pile up somewhere, or a vacuum pocket would form.

RESISTOR
A resistor is just something that allows electricity to flow through it, but resists this flow to some extent. The degree to which the resistor hampers current flow is measured in ohms. As the charged particles struggle through the resistor they generate heat energy in the material of the resistor. Which means the particles must be losing the same amount of energy, as the heat energy has to come from somewhere.

ENERGY OF CHARGED PARTICLES
So charged particles entering the resistor must have more energy than those exiting the resistor, to account for the heat energy generated by the flow. What is this energy that each particle possesses? It is the 'tension' provided by the battery, say 5 volts. To put it another way, it is the force driving each positive particle around the circuit. As the particles are pushed round the circuit they lose some energy whenever they encounter a resistance, that is, their 'voltage' decreases. Particles flowing through a thick copper wire lose negligible energy, since copper has a very low electrical resistance. So their electric potential (voltage) hardly decreases. In our example, the charged particles arriving at the resistor are very slightly below 5 volts, but when they leave the resistor they are at a much lower voltage, as we have seen. But what voltage is this?

VOLTAGE ON RE-ENTERING THE BATTERY
After the charged particles have got through the resistor, they then flow through a low-resistance copper wire to enter the negative battery terminal. Since very little heat is generated in this wire, the particles only lose a little energy ('voltage'). But we know that at the negative terminal of the battery the potential is zero volts, by convention. So the potential at points along the wire after the resistor must steadily decrease minutely and end up at zero volts at the negative terminal. This implies that the voltage at the far end of the resisitor is very slightly above zero. The current is able to flow along this section of wire despite the very low voltages (electromotive force) because of the correspondingly very low resistance of the copper wire.

WHY DOES 'TENSION' RELATE TO ENERGY?
An electron under tension is experiencing a force that tries to move the electron in the direction of the force. The greater this 'electromotive force' (measured in volts), the greater the potential energy the electron possesses. The more useful work it can do (heat up a resistor, drive a motor, etc.) if allowed to in a circuit. It's a bit like the elastic in a catapult. The more force (tension) you exert on the elastic, the more potential energy it stores, ready to give to the missile when you let the elastic go.

VELOCITY OF CURRENT FLOW
Negative electrons powerfully repel each other (as would positively charged particles in the conventional view) preventing them from 'piling up' anywhere in the circuit. The speed at which they flow through the wires is therefore pretty much the same anywhere in the circuit. They are actually flowing at a 'terminal velocity' in the wires regardless of the voltage changes at different points.

Big Al

 

[Biker]

I didn't know the answer to these questions before, and you can see my comments in a previous thread. What I really did to get intuitive grasp is to read Matter and interactions. I really liked the way they presented the topic, Everything was intuitive and just uses your previous knowledge of physics.
They used drude model which explains how electrons behave in metals, how you can in a sense derive ohm's law based on one assumption.. etc. You will really find circuits much easier to deal with

If you can't buy the book, You can watch videos on their website which is still useful and good.

 

[gmalcolm77]

"our ability to understand how or why they do what they do is inherently limited."

It would seem that our ability to understand anything is severely limited when you think about it. Our five senses detect only EM properties of what we suppose to be the real world. As our vision detects only a portion of the existing spectrum, Em is unlikely to be all that is out there. Until that is known for sure, what do we really know?

 

[sophiecentaur]

But, how is the magnetic field even relevant to the electrons?

How indeed, in this hybrid world in which people seem to want to live?

 

[rumborak]

Energy has been lost and that is described in terms of Potential Drop.
Work is done in the form of an interaction of the charge flow and the lattice, which causes energy loss within the lattice. It may be hard to accept that Kinetic Energy is not involved but the numbers (electron mass and speeds) give vanishingly small amounts of energy transfer. Furthermore, if KE were involved, the electron drift speed would be less (loss of KE) at the end of a wire or resistor. It is not.

I understand that, but deep inside me I feel there is some serious glossing-over going on. Or rather, that there is a good amount of circular reasoning going on. "The potential drops because work was done" and "energy is consumed because there is a potential difference between the ends of the resistor" are the same statement reworded. And as pointed out, KE of the electrons is irrelevant, so the mechanism of energy transfer is entirely unknown.

But, let me ask a different question: Would it be conceivable to shield a resistor from the external magnetic field the surrounding circuit creates? According to the Poynting vector, if there is no magnetic field in the vicinity of the resistor, there is no energy flow, which in turn would mean no Joule heating.
Is that physically not possible, or am I misunderstanding the matter?

 

[sophiecentaur]

Would it be conceivable to shield a resistor from the external magnetic field the surrounding circuit

Your magnetic field is due to the charge flow through the resistor so how could you shield from that? It's already there.
You could put the resistor in a copper box with its terminals soldered to the box at each end. Then there would be no potential across it because no work would be done in transferring the charge through the box so there would be no current through the resistor because the current path would be around the outside path.
I don't quite see why you are quite prepared to use the Poynting vector as part of an arm waving explanation which is several steps down the ladder of difficulty. That's not logical. You got as far as Poynting Vector so take it from there.

 

[rumborak]

I get the impression there is a certain deliberate unwillingness to entertain my points, but at the very least I can say I found a research paper that raises exactly my questions:

https://www.google.com/url?sa=t&sou...AuvXhdWDtJPhVmZDQ&sig2=biqYhpyUsgx4vtoA4-6wHw

I'm off to read (and hopefully understand) its conclusions.

 

[PeterO]

Sorry to be coming late to this discussion, but I was concerned with your opening sentence.
"I'm having trouble understanding circuits, particularly voltage."
I see that as something like "I am having trouble understanding the building industry, particularly the metreage."
Just because the the dimensions are measured in metres (or millimetres actually) does not mean we should refer to any of them as metreage (not even sure that is a word).
In a circuit, the electic potential at various points, or more frequently the potential difference between two points, is expressed with the unit volt - but that does not mean we should call the Potential, or Potential difference voltage - to do so is to lose the meaning of what we are talking about.
You could perhaps get away with referring to a "voltage drop" rather than a Potential Difference - but that is about the biggest stretch of terminology you should use.
We use a voltmeter to measure a Potential difference, and the answer is in volts or millivolts (depends on the meter).
We use bathroom scales to measure our weight, and the answer is in kilograms (or pounds / stones and pounds if the scales are old).
We would never refer to out weight as kilogramage, and similarly should never refer to a Potential Difference as voltage.
If you religously use the term "Potential Difference", I think you will find yourself half way to a better understanding of circuits.

 

[PeterO]

I'm having trouble understanding circuits, particularly voltage. So my question is more of a conceptual dilemma. Please correct my definitions and relations if it is incorrect. Voltage is the energy provided per charge, and this energy is converted to KE once the circuit is closed, or in other words it influences the velocity of the electrons, and in effect influences the current.

For instance, in a simple circuit made up of a battery and a resistor, the voltage drops to 0 once the electrons pass through the resistor, or the energy provided by the battery is all dissipated. Then why is the current still constant? If the voltage drops, doesn't that imply a drop in velocity of the electrons? So shouldn't it lead to a decrease in current as it leaves the resistor?

A follow up question is, why is the voltage drop the same in a resistor as the voltage provided by battery? How does a resistor "know" that it should use up the same amount of voltage provided by the battery?

Addressing your full question (again without having read all the intervening posts). I compare a circuit to the following.
Suppose you set up a running course around your nearby city. The course includes some flat ground, some gentle up slopes, some stairs going up, and of course some down slopes.
Your plan is to run around the course at constant speed.
You will not expend much energy on the down slopes, and will be working very hard going up the stairs.
Let's assume some medical/food scientist determines you will expend 1000 J of energy in one circuit of the course.
IF you were to take on food providing exactly 1000J of energy just before you started the run, you should arrive back at the Start/Finish line with exactly the same amount of energy you had just before took on the food.
(One way to ensure you ran at constant speed, would be to fill the entire circuit with people like yourself so that you were all almost touching each other. Somehow you all get moving then away you go - but remember AT CONSTANT SPEED. No slowing down on the Hills/stairs, no speeding up on the down slopes.)
If you were able to monitor your bodies energy in real time you would notice:
1. Your energy level was 1000J higher at the start (just after you ate) than at the finish. (this is like the Potential difference - sometimes foolishly called voltage - across the battery or entire circuit - a gain in Potential across the battery; a matching drop in Potential across the circuit)
2. The amount of energy you used up on the down-slopes was immeasureably small (this is like the Potential difference across the hook-up wires in a circuit - often idealised to 0 volts)
3. The greatest loss of energy occurred as you ran up the stairs (this is like the Potential Difference across the largest resistance in the circuit)

Note: if you ran too slowly, you would not use up all 1000J on a circuit. If you ran too fast you might collapse before the finish because you had no energy left. Electicity "knows" exactly how fast to run. It is not smarter than you, the Electric Field created around the circuit constantly reminds the electrons how much energy to use up at each point along the circuit (they don't go fast and slow, they just lose/use up energy at a faster rate - just as you do on the stairs compared to the flat ground).

 

[cnh1995]

I get the impression there is a certain deliberate unwillingness to entertain my points, but at the very least I can say I found a research paper that raises exactly my questions:

https://www.google.com/url?sa=t&sou...AuvXhdWDtJPhVmZDQ&sig2=biqYhpyUsgx4vtoA4-6wHw

I'm off to read (and hopefully understand) its conclusions.

It is an excellent paper! Thanks for sharing @rumborak !

 

[rumborak]

It is an excellent paper! Thanks for sharing @rumborak !

FYI, I found another piece of the puzzle in the Wikipedia article on the Poynting vector. There it says (and this is apparently derived in a Reitz book) that the inward-pointing of the Poynting vector at the resistor is due to Snell's law, i.e. due to the fact that the speed of light inside the resistor is significantly lower than the vacuum one:

https://en.wikipedia.org/wiki/Poynting_vector#Resistive_dissipation

Now I wonder whether the energy transfer is somehow facilitated through the same mechanism that changes c inside a solid.

I will try to get my hands on that Reitz book through my company library. This has become my personal mission now to understand.

 

[Biker]

Another good paper on this topic is this:
https://www.tu-braunschweig.de/Medien-DB/ifdn-physik/ajp000782.pdf

Dale showed me it a while back. The mathematics isn't hard at all and they provide good and accurate visual presentation unlike the paper provided above.

Good luck with what you are doing :D.

 

[sophiecentaur]

inward-pointing of the Poynting vector at the resistor is due to Snell's law, i.e. due to the fact that the speed of light inside the resistor is significantly lower than the vacuum one:

This bending effect occurs with mf and lf radio waves over 'real' ground. The poynting vector points a bit below the horizontal and maintains the signal level by directing power from parts of the wave that would spray tangentially to the horizon and keeps power near the (curved) ground. This effect is even strong enough to fill in the shadow that is formed when the wave passes over a city with many steel structures. A few km beyond the city, the signal strength is up where it should be.

 

[rumborak]

Interesting!

How does that behavior change for a static EM field like in a circuit? Can you still sort of say that the field travels at the speed of light (or slower), even when it doesn't change? That is, despite it not being apparent, can you treat a static EM field to be still be traveling at c? (my hunch will be yes, as any change to it would propagate with c after all)

 

[sophiecentaur]

If the field is static then there is no change to 'propagate' at c as a wave. The frequency is zero (or at least as near zero as dammit, given a long enough time to settle down)

 

[rumborak]

I was asking mostly from the viewpoint that the EM field is after all the "provider" of the energy to the resistor, and thus there seems to be a flow from the battery to the resistor, through the air. Even when not visible necessarily because it doesn't change, I guess that energy flow is still going at c.

 

[sophiecentaur]

I was asking mostly from the viewpoint that the EM field is after all the "provider" of the energy to the resistor, and thus there seems to be a flow from the battery to the resistor, through the air. Even when not visible necessarily because it doesn't change, I guess that energy flow is still going at c.

I haven't considered this very deeply but I question how relevant the speed of light is to this particular question. Certainly, if you turn off the Power switch, the interruption would only be noticed after a time d/c. But it is not valid to talk in terms of 'individual photons' making their way around the circuit.
I could ask whether you would consider the gravitational field that is providing your weight force, is traveling at c towards the centre of the earth??

 

[rumborak]

Well, it *is* a question worth asking I feel, especially when there is an energy flow associated with it like in a circuit. I mean, from a pure formula description of it, essentially nothing is happening once the EM field has been established to its steady-state value. But at the same time, we know there is energy being transported that is draining the battery and heating the resistor. *Something* seems to be traveling, likely at c, from the battery to the resistor.

 

[sophiecentaur]

I guess you could consider a wave of Zero Hz, still travels.

 

[Biker]

Well, it *is* a question worth asking I feel, especially when there is an energy flow associated with it like in a circuit. I mean, from a pure formula description of it, essentially nothing is happening once the EM field has been established to its steady-state value. But at the same time, we know there is energy being transported that is draining the battery and heating the resistor. *Something* seems to be traveling, likely at c, from the battery to the resistor.

Why does anything have to travel? Once you establish a ""Static field" (Force doesn't change over time, The field doesn't "evolve"). The field is everywhere, No propagation of anything happening. Energy doesn't propagate.

In a static field ( no radiation, no accelerated charges) , Then that means that the forces have to be instantaneous for momentum conservation to hold ( Isn't that right? )

 

[rumborak]

No propagation of anything happening. Energy doesn't propagate.

What do you mean? The energy of the battery is ending up on the other end of the circuit, heating the resistor. Energy doesn't just magically appear in places, it has to travel there.

If there is no net radiation, Then that means that the forces have to be instantaneous for momentum conservation to hold ( Isn't that right?)

If the forces were instantaneous, it would obviously violate relativity. So that's a no-go right there.

*Something* is traveling from the battery to the resistor, even in steady state.

 

[Biker]

What do you mean? The energy of the battery is ending up on the other end of the circuit, heating the resistor. Energy doesn't just magically appear in places, it has to travel there.

It is already there because the field is there. Energy is not an object. Wherever there is a field, energy is there too.(Limited knowledge of relativity just basics) First, we wouldn't care if it takes time or not, The field inside the conductor is constant. So how time can affect this?
Second, If you somehow have a charge moving with a constant velocity, no information can be obtained so it doesn't violate relativity?

 

[rumborak]

It is already there because the field is there. Energy is not an object. Wherever there is a field, energy is there too.

If that was the case, and no energy would flow since "the energy is already there", the battery would never drain. Clearly, there is a flow from the battery to the resistor (I'm starting to repeat myself here).

And yes, energy *is* an object in the sense that has to abide by the same rules of transport as regular mass does. It can't travel faster than c, it has to go from A to B like any ol' other object.

 

[davenn]

How does that behavior change for a static EM field like in a circuit?

I don't think you can have a static EM field, by nature it is always moving at the speed of light

don't confuse that with a static electric or magnetic field

If that was the case, and no energy would flow since "the energy is already there", the battery would never drain. Clearly, there is a flow from the battery to the resistor (I'm starting to repeat myself here).

At circuit switch on ( a DC circuit) there is a pulse of EM around the circuit. This pulse provides/impart6s the initial motion to the electrons aka charges
hence the light globe lights almost instantaneously. Once the electrons/charges are on the move, they are providing the ongoing energy to the light globe

This of course is different for an AC circuit where the EM field is always active and cycling.

 

[rumborak]

I don't think you can have a static EM field, by nature it is always moving at the speed of light
don't confuse that with a static electric or magnetic field

I was using "EM field" as shorthand for electric and magnetic. I wasn't implying traveling waves.
But it's an odd comment to make of yours. There is no conecptual difference between a static electric/magnetic field and an EM wave other than the rate of change, right? They are just different incarnations of the same electromagnetic field.

At circuit switch on ( a DC circuit) there is a pulse of EM around the circuit. This pulse provides/impart6s the initial motion to the electrons aka charges
hence the light globe lights almost instantaneously. Once the electrons/charges are on the move, they are providing the ongoing energy to the light globe

Errrrr ... I can only presume you have not been reading the previous posts here. The very point of this discussion is that it is NOT the electrons providing the energy, but instead the Poynting vector, I.e. ExB traveling through space. That's what was mysterious to Feynman, and to me.

 

[davenn]

I was using "EM field" as shorthand for electric and magnetic. I wasn't implying traveling waves.

then you are confusing the issue ... be clear

Errrrr ... I can only presume you have not been reading the previous posts here.

yeah I have and all I have seen is your constant argument against what everyone else is saying

The very point of this discussion is that it is NOT the electrons providing the energy, but instead the Poynting vector, I.e. ExB traveling through space.

well that's pretty vague

 

[rumborak]

Well, I can say I'm in good company with Richard Feynman, and the guys at that Israeli university who wrote the paper I linked to. All I see here is a ton of hand-waving, or plain promotion of wrong notions such as that it's the electrons transferring the energy from the battery in a "pushing down the wire" way.

To quote Feynman:

" So our “crazy” theory says that the electrons are getting their energy to generate heat because of the energy flowing into the wire from the field outside. Intuition would seem to tell us that the electrons get their energy from being pushed along the wire, so the energy should be flowing down (or up) along the wire. But the theory says that the electrons are really being pushed by an electric field, which has come from some charges very far away, and that the electrons get their energy for generating heat from these fields. The energy somehow flows from the distant charges into a wide area of space and then inward to the wire."

And my additional "wrinkle" here is that since the Poynting vector involves the static magnetic field, clearly that must be part of the mechanism of energy transfer too.

 

[sophiecentaur]

That's what was mysterious

There are many situations in Science where we have this problem. In this situation, what is actually worrisome? The 'speed' involved with the transfer of Power in a steady state situation is, I would say, not relevant because speed /velocity involves identifying how the 'location' of something is changing in time. In a DC situation, there is nothing that can be identifiable; everything is unchanging. We have already discarded the description of an EM wave as a stream of photons / little bullets and we can't approach it that way, as with water molecules in a constant flowing river. At DC, of course, the photon energy is Zero! (=hf=h0=0) so they have no meaning, in any case. The only way to approach this impasse is, I think, to consider an AC signal and see what happens as f→0. The velocity, of the wave in that case, would be c and, as there are no continually flowing DC circuits (f≠0, ever), in real life that could be good enough for us.
Perhaps it would be better to ignore the concept of any Power flowing at a 'speed' and say that the only speed involved with Power Flow from A to B is the speed of propagation of a change in Power. It may be irksome to accept that there is no 'motion' of the Power but you have to examine why we want an answer to that question. We all feel the need to leave things 'tidy' in our minds. QM is another example where this self indulgence cannot be allowed and we just have to 'Get Over It', perhaps.