what does voltage is dropped across the resistor mean?

Thread Starter

elara

Joined Jan 12, 2016
15
Here's an analogy which may help. Think of an old-fashioned egg-timer with a volume of sand above a narrow passage. The number of sand grains going into the passage in unit time equals the number of grains coming out (the sand flow rate, analogous to current). There is a big pile of sand above the passage but not at the exit of the passage (analogous to potential difference).
Thanks for responding.
I can understand that the big pile of sand is a "reserve" of sand a.k.a potential energy. The potential difference here is measured in terms of how much sand is in "reserve" and below surface of the passage (two points) - this is voltage. But once the flow starts (assuming the rate is restricted because of the narrow passage), and you want to measure the potential difference "just" above and "just" below, the rate of current should be the same (according to Kirchoffs law of current). In your analogy you are still measuring the potential difference from the "reserve" point and the surface point below the narrow passage even after the flow has started, and not during the "actual" flow i.e. at the point where the "sand" grains are going into the passage and at the point where the sand grains are coming out of it. Pardon me if I understood it wrong.

Here is your fundametnal flaw. Voltage does not necessarily mean an excess of electrons on one side. Yes, when you have excess electrons on one side you have an electric field, which turns into voltage difference, but the implication doesn´t necessarily work the other way. Electric field is what constitues voltage, and you can have two plates with same amounts of electrons that are at different voltage, just beacuse they are placed at different places in the the external electric field.

Battery does not have an excess of electrons on one pole. It has excess of electrons with higher potential that are willing to go to the other side, but the absolute amount on both poles is the same.
Could you please elaborate on what you mean by this? If the number of electrons on each pole is the same, what does electrons on one pole has higher potential means? I am asking cause I couldn't find the explanation anywhere.

A review of the chemistry of batteries might be helpful. In chemistry the use of square brackets implies "the concentration of". Concentrations in a solvent are usually expressed in moles per liter. One mole of any substance, or element, or ion is equal to Avagaro's number, which 6.02E23, which as you may be aware is bigger than a boatload!

http://www.science.uwaterloo.ca/~cchieh/cact/c123/battery.html

All of the ions in a DC circuit involving a battery and a resistive load are contained within the battery and the electrolyte. They do not leave the battery. All of the atoms in the wire and in the resistor are neutral atoms, there are no ions anywhere in sight. All of the electrons bound to atoms of the conductor have discrete energy levels. Above the highest energy level for the atoms of the conductor there is a band gap where no electrons have an energy level within the band gap. Above that energy level are numerous electrons from the battery that move along in the conductor with energy levels above the band gap. These energy levels are not discrete and form a continuum of energy levels. Most conductors have a very small band gap or none at all because the valence band and the conduction band overlap. Quantum mechanics also allows electrons to drop through the band gap to lower levels and tunnel through the bad gap (potential barrier) to higher levels. On a macro scale it is impossible to observe or measure this behavior.

As @kubeek has pointed out the battery contains an excess of electrons with enough potential energy to enter the wire, make it through the resistor, and proceed to the other electrode of the battery where chemistry dictates what happens next.
Thanks for this. I was able to grasp some of what you mean. But the question remains, the flow of ions (or electrons) in the battery only starts when there is difference in potential between positive and negative terminal. What does higher potential on one side mean when the number of electrons on both poles are the same? Regardless, there is a clear potential difference between the poles in the battery, but when you are measuring the potential difference above and below the resistor (through voltmeter - as potential difference is only measured between two points) you also observe there is "potential difference" between these two points? Now what would this be or mean? I could understand if someone said that the "potential energy" is dropped at this point as energy is released due to the involvement of resistor i.e. the potential difference (from my previous understanding) created between the battery poles is used up.

A charge (electron) accelerates in response to an electric field. In a resistor (which means in most materials) a charge (electron) moving through the material wants to slow down and just stay put (wiggling around its current position due to thermal energy). This is much like friction. So to get the electron to move from one end of the resistor to the other, we need an electric field that is constantly accelerating the electrons which are then constantly giving up that energy to the friction-like effects and slowing back down. Recall that a voltage drop is the result of an electric field. So the bottom line is that we need an electric field across the resistor to give the electrons enough energy to make it through the resistance (with that energy being converted to heat in the process).

As a mechanical analogy, imagine a block of wood sliding on a long wooden board. The "voltage" across the board is the difference in height from one end of the board to the other (which is, in fact, a measure of the gravitational potential difference across the board). The "resistance" of the board is the degree of roughness between the block and the board. The "current" is the block of wood sliding down the board. The goal is to raise one end of the board just the right amount so that the block slides down the board at a constant speed. If the board and block are highly polished and lubricated, the high end of the board doesn't have to be lifted much at all. But if the two surfaces are very rough (think fine sandpaper) then one end of the board has to lifted a lot to find the angle that will result in the block sliding down at a constant speed.
Thank you! My understanding is that voltage drop refers to energy release. It is the terminology maybe that has got me confused. In the context of Ohms law, voltage is affected by resistance, and voltage is always discussed as being potential difference between two points - so any time there is resistance, there is change in voltage - so potential difference is affected. So when referring to potential difference before and after the resistor points - I don't see "where" is the potential difference as there is no difference in rate of flow.
I think it would make sense to say that "energy is released" as a result of resistance but voltage remains constant throughout. In other words potential energy is the only thing that is affected not the potential difference.
 

hp1729

Joined Nov 23, 2015
2,304
Thanks for responding.
I can understand that the big pile of sand is a "reserve" of sand a.k.a potential energy. The potential difference here is measured in terms of how much sand is in "reserve" and below surface of the passage (two points) - this is voltage. But once the flow starts (assuming the rate is restricted because of the narrow passage), and you want to measure the potential difference "just" above and "just" below, the rate of current should be the same (according to Kirchoffs law of current). In your analogy you are still measuring the potential difference from the "reserve" point and the surface point below the narrow passage even after the flow has started, and not during the "actual" flow i.e. at the point where the "sand" grains are going into the passage and at the point where the sand grains are coming out of it. Pardon me if I understood it wrong.



Could you please elaborate on what you mean by this? If the number of electrons on each pole is the same, what does electrons on one pole has higher potential means? I am asking cause I couldn't find the explanation anywhere.



Thanks for this. I was able to grasp some of what you mean. But the question remains, the flow of ions (or electrons) in the battery only starts when there is difference in potential between positive and negative terminal. What does higher potential on one side mean when the number of electrons on both poles are the same? Regardless, there is a clear potential difference between the poles in the battery, but when you are measuring the potential difference above and below the resistor (through voltmeter - as potential difference is only measured between two points) you also observe there is "potential difference" between these two points? Now what would this be or mean? I could understand if someone said that the "potential energy" is dropped at this point as energy is released due to the involvement of resistor i.e. the potential difference (from my previous understanding) created between the battery poles is used up.



Thank you! My understanding is that voltage drop refers to energy release. It is the terminology maybe that has got me confused. In the context of Ohms law, voltage is affected by resistance, and voltage is always discussed as being potential difference between two points - so any time there is resistance, there is change in voltage - so potential difference is affected. So when referring to potential difference before and after the resistor points - I don't see "where" is the potential difference as there is no difference in rate of flow.
I think it would make sense to say that "energy is released" as a result of resistance but voltage remains constant throughout. In other words potential energy is the only thing that is affected not the potential difference.
The potential difference is not "used up" until the battery is completely dead. As long as there is a potential difference current will flow. Same current out of the battery as there in into the other end of the battery. As long as there is current flow the voltage is dropped across the resistor.
Would it be easier to see if we had two equal resistors in series? Current is the same at any point in the circuit. Half the voltage is dropped across each resistor.
 

Thread Starter

elara

Joined Jan 12, 2016
15
The potential difference is not "used up" until the battery is completely dead. As long as there is a potential difference current will flow. Same current out of the battery as there in into the other end of the battery. As long as there is current flow the voltage is dropped across the resistor.
Would it be easier to see if we had two equal resistors in series? Current is the same at any point in the circuit. Half the voltage is dropped across each resistor.
Sure I agree. Potential difference is what produces the electric field which causes the electrons to flow. As I have stated in my previous post, According to Ohms law, voltage is affected by resistance, i.e. the potential difference is also affected as voltage is the potential difference between two points. But in my mind perhaps, when you measure the potential difference "above" and "below" the resistor, or any point in the circuit when there is a resistor, the potential difference is not affected as the "rate of flow of current" above and below is the same. The potential energy however is affected as this is released in collision with resistor intervention. Now maybe I have got the definition of what "potential difference" wrong and I suppose this is the answer I am seeking. I had thought potential difference is the difference in the excess electrons between two points, but @kubeek and @Papabravo told me that the it is not the difference in electrons but the "high" potential between two points that is different. I am not sure what this means. So to conclude, for now I am of the understanding that "the potential difference remains constant as the flow starts but potential energy is what is affected".
 

profbuxton

Joined Feb 21, 2014
421
Let me see If I can explain "voltage drop across resistance" in my simple way.
Consider the following: a battery(or some DC power supply). This battery (or supply) has at its terminals a voltage(we wont worry about how its generated at this stage).
Now connect a resistance across the terminals(we can chose a suitable one which wont take too much current).
A 12v bulb would be ideal (if battery is 12v or supply set to 12v)
Now the bulb should light up agreed?. Now we take a meter and measure from one bulb terminal to another.
The meter will read 12v(ignore wiring resistance). You have now measured the "volt drop" across a resistance.
Note current through the bulb will be some value, same through every bit of wire and bulb coming out of positive terminal through wires ,bulb and back into negative terminal( don't care about what flowing and what direction at this point).
Now for the fun part! We now connect another same bulb(12v) in SERIES with the original one. From plus terminal to bulb 1 terminal A, bulb 1 terminal B to bulb 2 terminal A, bulb 2 terminal B to supply negative.
Now measure across bulb 1. Meter will read 6 volts. Measure across bulb 2. Meter will read 6 volts(assume both bulbs equal resistance). Volts across supply is still 12 volts. Current through SERIES circuit is still the same at all points BUT will be half of one bulb current since we have doubled resistance.
We now say that we have 6 volts "drop" across each bulb and TOTAL volts across both still equals 12v(supply volts). So volts across bulb 1(6) and volts across bulb 2(6) equal supply volts.
I hope that answers your query about "VOLT DROP" across resistance
 

WBahn

Joined Mar 31, 2012
30,296
Thank you! My understanding is that voltage drop refers to energy release. It is the terminology maybe that has got me confused.
Voltage drop does not refer to energy release, no more than a height difference does. Just as there is no energy release associated with a brick being held over a 12 ft drop, so too there is no energy release associated with a disconnected capacitor that is charged to 12 V. Height and voltage are measures of the potential for energy conversion. When you drop the brick and it actually falls through the 12 ft, you have conversion of energy from gravitational potential energy to some other form. When you connect the capacitor so that charge can flow through the 12 V drop you have conversion of energy from electrical potential energy to some other form.

voltage - so potential difference is affected. So when referring to potential difference before and after the resistor points - I don't see "where" is the potential difference as there is no difference in rate of flow.
I think it would make sense to say that "energy is released" as a result of resistance but voltage remains constant throughout. In other words potential energy is the only thing that is affected not the potential difference.
If potential energy is affected, then the voltage is affected precisely because voltage is a measure of potential energy.
 

Thread Starter

elara

Joined Jan 12, 2016
15
I re-read all the responses and other sources to understand the fundamentals of atom and electrons and I think I understand what high potential might mean now. The "potential" here is basically describing at what energy levels the electrons are orbiting in an atom. On the negative terminal of the battery - I think the electrons are in their highest energy levels (i.e.on a higher orbit around the nucleus) and starts flowing when connected to circuit. Once they hit the resistor, they go back to their "normal" energy level emitting heat/radiation (photon?)- i.e. an EM wave..

The potential difference is difference in where the electrons are orbiting - the higher the level the more the potential.

Is this right? Could someone respond?
 

DC_Kid

Joined Feb 25, 2008
1,072
so, from a physics view, the electrons have energy associated with them (work), the harder they push (V/R=amps) the more energy the resistance takes on (usually converting to heat). the resistance strips energy from the etrons. their potential continues to decreases until they get back home. if its one bridge (R) between + and - then it is that one bridge that converts all the energy. if its many bridges (R's) then each R converts a portion. so, the more amps the more etrons per unit time cross the bridge. so now you see the direct relationship between a pool of electrons that have potential, and the amount of flow the bridge (R) allows per unit time. work in (move electrons to the other side) = work out (remove the energy on their way home).

a direct analogy is heavy block on rough surface. kinematic friction, you put in a force to have v (velocity, not volts) and that force over distance = work in = work out (the heat generated by the friction)
 

Thread Starter

elara

Joined Jan 12, 2016
15
so, from a physics view, the electrons have energy associated with them (work), the harder they push (V/R=amps) the more energy the resistance takes on (usually converting to heat). the resistance strips energy from the etrons. their potential continues to decreases until they get back home. if its one bridge (R) between + and - then it is that one bridge that converts all the energy. if its many bridges (R's) then each R converts a portion. so, the more amps the more etrons per unit time cross the bridge. so now you see the direct relationship between a pool of electrons that have potential, and the amount of flow the bridge (R) allows per unit time. work in (move electrons to the other side) = work out (remove the energy on their way home).

a direct analogy is heavy block on rough surface. kinematic friction, you put in a force to keep contant v (not volts) and that force over distance = work in = work out (the heat generated by the friction)
So the higher potential that the electrons start with gets somewhat "used up". Right? In other words, the electrons have moved to a lower potential (lower level). So, I am still confused with the term potential difference between the point before the electrons hit the resistor and the point after it. What would be the potential difference here?

If someone could explain how this is measured and why is it this way, It would really help.
 
Last edited:

nsaspook

Joined Aug 27, 2009
13,553
so, from a physics view, the electrons have energy associated with them (work), the harder they push (V/R=amps) the more energy the resistance takes on (usually converting to heat). the resistance strips energy from the etrons. their potential continues to decreases until they get back home. if its one bridge (R) between + and - then it is that one bridge that converts all the energy. if its many bridges (R's) then each R converts a portion. so, the more amps the more etrons per unit time cross the bridge. so now you see the direct relationship between a pool of electrons that have potential, and the amount of flow the bridge (R) allows per unit time. work in (move electrons to the other side) = work out (remove the energy on their way home).
Don't confuse the current loop with energy flow. The electrons are a necessary component of the circuit as a system but the actual energy flow process should be understood in a way that's compatible with AC and later RF energy flow.

http://amasci.com/elect/poynt/poynt.html
 

bwilliams60

Joined Nov 18, 2012
1,450
If you have two objects exerting the exact same amount of energy directed towards each other, there would be no resultant movement. So lets say two cars were bumper to bumper facing each other and they exerted the same energy towards each other, There is no movement in either direction. In order for current to flow, there has to be an area of high pressure and an area of low pressure (potential) for electrons to flow in one direction. Does not matter conventional or electron theory. This difference in pressure is your potential difference.
When electrons move through an area of resistance, current flow will remain the same through the circuit (series) but with each resistance, potential will be lost.
 

Thread Starter

elara

Joined Jan 12, 2016
15
Voltage drop does not refer to energy release, no more than a height difference does. Just as there is no energy release associated with a brick being held over a 12 ft drop, so too there is no energy release associated with a disconnected capacitor that is charged to 12 V. Height and voltage are measures of the potential for energy conversion. When you drop the brick and it actually falls through the 12 ft, you have conversion of energy from gravitational potential energy to some other form. When you connect the capacitor so that charge can flow through the 12 V drop you have conversion of energy from electrical potential energy to some other form.



If potential energy is affected, then the voltage is affected precisely because voltage is a measure of potential energy.
Thanks @WBahn, I kind of understand, the only question I then have - Does the "potential energy" (e.g. 1.5 V)of the battery ends up across the resistor in the form of heat? If so how is the electric field created at this point to "push" the electrons back to the battery to make them energised again?

Does this question make sense?

EDIT: Perhaps the reason for this is that the electrons are forced out of the resistor because there are "more electrons" piled up in front of the resistor - In other words, a secondary source of electric field is established across the ends of the resister to drive them through. So this suggests that "density of electrons" at each end of the resistor is different (at least until a steady state is reached). This also explains why the rate of flow of current is same across all points once steady state is reached...

EDIT#2: Or perhaps that the battery is introducing different electrons each loop? In other words, the potential difference (with a 1.5V battery) on one end of the resistor is 1.5V, and the other end of the resistor is 0V (in a series circuit with only one load). This implies that the current is what is determining the exact drop of voltage in the resistor. This would also make sense why voltage is divided if different resistors are present in a series circuit.

Hoping to get some answers...
 
Last edited:

Thread Starter

elara

Joined Jan 12, 2016
15
Reiterating my thoughts of the comment I made above:

Does the potential difference from the battery (once connected to an external circuit) ends up around the resistor? That is translated into other form of energy(conserved) to heat. What is the driving force at this point that is promoting flow of electrons?

My thoughts:
When the potential energy is realised at the resistor (in a series circuit) - there is a secondary electric field formed around the resistor because of the accumulation of electrons at the negative polarity of the resistor. Hence, causing a new "potential difference" between two ends of the resistor (as per Ohms law - lesser than the source voltage).This electric field promotes electron movement and controls the rate of flow (steady state). This is why the current is same at all points in the circuit as a result of steady state..

Excuse me for stating the obvious, I feel like I cannot move forward unless I understand the fundamentals and it is easier to remember it this way. So I would really appreciate if someone could comment on this?
 

WBahn

Joined Mar 31, 2012
30,296
Any path that you take from the positive terminal of the battery to the negative terminal of the battery will exhibit an electric field that, if integrated along that path, will result in whatever the voltage potential is between the terminals. At any point along that path an electron will be accelerated by the net force acting on it. One of those forces is the electric field due to the battery. Other forces exist including gravity and the electron's interactions with the materials that the electron is a part of.

Have you read the blog entry I recommended regarding what a battery is and isn't?
 

WBahn

Joined Mar 31, 2012
30,296
Excuse me for stating the obvious, I feel like I cannot move forward unless I understand the fundamentals and it is easier to remember it this way. So I would really appreciate if someone could comment on this?
If you want the fundamentals, then forget about resistors and take a step back and start with the forces between charged particles and, from there, the notion of the work to move one charged particle in the vicinity of another charged particle and, from there, the notion of an electric field to serve as a measure of the net influence of all of the other charged particles in the universe at a particular point in space, and from there, the notion of a voltage potential to serve as a measure of the energy required to move a charge from one point in space to another in the presence of an electric field.

In short, study the first chapter or so of a second semester physics course (usually called E&M I).
 

Thread Starter

elara

Joined Jan 12, 2016
15
Have you read the blog entry I recommended regarding what a battery is and isn't?
I will look at it now!

If you want the fundamentals, then forget about resistors and take a step back and start with the forces between charged particles and, from there, the notion of the work to move one charged particle in the vicinity of another charged particle and, from there, the notion of an electric field to serve as the net influence of all of the other charged particles in the universe at a particular point in space, and from there, the notion of a voltage potential to serve as a measure of the energy required to move a charge from one point in space to another in the presence of an electric field.

In short, study the first chapter or so of a second semester physics course (usually called E&M I).
Thanks for this, I will do that. However, I must point out that not many sources give a detail explanation (microscopic view) about what exactly happens at each point in the circuit, i.e. how resistors interact with electrons, what happens to the electric field and how exactly is voltage dropped etc. I thought a forum post would make things easier but my lack of understanding of fundamentals perhaps is the source of confusion. Thanks for your help so far!
 

WBahn

Joined Mar 31, 2012
30,296
The detailed microscopic view you are looking for involves quantum mechanics, wave functions, spatial and temporal probability distributions, partial differential equations, complex analysis, and a few more things. Are you up for that?
 

Thread Starter

elara

Joined Jan 12, 2016
15
The detailed microscopic view you are looking for involves quantum mechanics, wave functions, spatial and temporal probability distributions, partial differential equations, complex analysis, and a few more things. Are you up for that?
"Microscopic" In a sense the part that explains the effect of electric field in a circuit at different points. For example, analysis of collision of electrons inside the resistor, the reason which constitutes to "accumulation", which promotes the formation of a secondary electric field.

Your response here
"Any path that you take from the positive terminal of the battery to the negative terminal of the battery will exhibit an electric field that, if integrated along that path, will result in whatever the voltage potential is between the terminals. At any point along that path an electron will be accelerated by the net force acting on it. One of those forces is the electric field due to the battery. Other forces exist including gravity and the electron's interactions with the materials that the electron is a part of"

Answers my questions - perhaps not at a microscopic level but enough for me (for now) to move forward.

Thank you.
 

bwilliams60

Joined Nov 18, 2012
1,450
Keep in mind also that conductors also have some resistance in them in which case a small amount of voltage is still being used up in each section of wire. This in itself will allow electrons to be returned to their origin. On a very basic level :)
 
Top