## Help!!!!! Voltage drop is driving me insane!!!!

if you have water at one pressure connected by a hose to another container at a lower pressure water will flow from one to the other. water will continue to accelerate through the hose until the pressure drop from friction equals the difference in pressure.

 Quote by granpa if you have water at one pressure connected by a hose to another container at a lower pressure water will flow from one to the other. water will continue to accelerate through the hose until the pressure drop from friction equals the difference in pressure.
Yea I guess that makes sense. Thanks. I'm not really used to thinking of water pressure, I'm more used to thinking of it the way I described earlier.

I think when it comes to electrical engineering, its easier to not question why voltage drops, but to just accept that it does and to solve for it correctly.
 As far as anaogies go, I don't want to know what "It" is like. I want to know what "It" is......And I also want to know anything about "It's" WHYs..and HOWs I know that voltage is a measurment that is valid when refering to OPEN terminals, and not in effect when part of a closed circuit........In an OPEN supply measurement there is no generated EMF other than that of the supply. I also know that when you measure the voltage across a device with a closed circuit you are NOT reading the supply voltage anymore, but instead VOLTAGE DROP, in relation to the given current and known resistance of your meter..... you are measuring the suppy EMF acting on the rest of the circuit, or the "generated" EMF, known as Voltage drop. again....spare me the lame analogies, Why does the energy of electrons or charge carriers diminishes when it goes through a circuit element? please refrain from explaining things just simply because of OHMs law....OHMs law is described or "DEFIJNED" BY what happens....it does NOT DEFINE WHAT IS HAPPENING.

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 Quote by eugenius So as far as direct current from a battery is concerned, Each terminal of the battery has so many electrons missing on the positive side, and the same amount of electrons added to the negative side. Each terminal has the same exact electric field but with opposite charge.
A cell does not function the way you have described it. Your description is more suitable for that of a parallel plate capacitor than a battery. As I mentioned (and provided a link to) earlier, just because separation of charge causes an electric field which in turn induce current flow doesn't mean that current can flow only in the existence of a charge separation. The two half-cells of a battery do not contain either a surplus or a deficiency of electrons. Instead, what we have here are electrochemical galvanic potentials, and one half-cell undergoes oxidation while the other undergoes reduction. Oxidation in one half-cell (anode) causes the electrons to be liberated and they travel along the circuit wires to the other half-cell (cathode), where electrons are required for reduction to take place. It is these electrochemical potentials which induce a voltage across a circuit, causing current flow (locally I believe, but which spread out over the entire circuit loop after a while).

 Quote by eugenius I think when it comes to electrical engineering, its easier to not question why voltage drops, but to just accept that it does and to solve for it correctly.
I'm an EE major as well, and I think it's always better to understand the fundamentals than accept it without question, provided it doesn't confuse you too much.

A current is basically a flow of electric charge induced by an electric field associated with a voltage. When an electron moves in a conductor under an electric field, it will accelerate. However it will inevitably be scattered by collisions with the lattice, and its speed decreases to zero. But the electric field is still present, and it induces the electron to accelerate (and move again) until the next collision occurs. Each one of these collisions cause the electrons to lose its kinetic energy, and this KE is dissipated as heat. It is for this reason that wires feel warm after use. The repeated collisions ensure that the average KE of the electrons is very low. Hence the total energy of the electrons is nearly equivalent to its potential energy (PE). It is evident that the potential energy of the electron decreases when it gets closer to the positive terminal, just as we might say that an electron situated closer to a positively charged plate has lower PE.

This is where voltage comes in. The energy provided to the electron is given by the potential difference or voltage. A resistor with high resistance causes the electrons to lose more energy due to more electron scattering. This is why we have R=rho l/A. The longer the wire (or conductor) the more scattering events occur throughout the length and the electrons lose more energy. Similarly the larger the cross-sectional of the conductor, the more "space" there is for electrons to move throughout the conductor length. The actual picture is really a lot more complicated than this. You'll have to use the Kronig-Penney model as approximation to the electron wells in the lattice, then introduce something known as k-space (reciprocal vectors) to come up with something known as the dispersion relation (E-k diagram) and then from there you can derive the carrier transport equation for charge flow under some conditions. But that is solid-state physics and I'm not familiar with it.

 Quote by XPTPCREWX I know that voltage is a measurment that is valid when refering to OPEN terminals, and not in effect when part of a closed circuit........In an OPEN supply measurement there is no generated EMF other than that of the supply.
What do you mean by "open supply"?

 Quote by XPTPCREWX I also know that when you measure the voltage across a device with a closed circuit you are NOT reading the supply voltage anymore, but instead VOLTAGE DROP, in relation to the given current and known resistance of your meter..... you are measuring the suppy EMF acting on the rest of the circuit, or the "generated" EMF, known as Voltage drop.
Kindly explain what is "supply voltage".

 Quote by XPTPCREWX again....spare me the lame analogies,
There is nothing "lame" about the analogies granpa or anyone else has provided in this thread. We don't know your level of expertise and knowledge and it's impossible to teach solid state physics to high school students just so we can derive Ohm's law and the concept of resistance and voltage drops. Hence analogies are necessary so that educators can teach students the basics of circuit theory before delving more into the fundamentals.

 Quote by XPTPCREWX Why does the energy of electrons or charge carriers diminishes when it goes through a circuit element? please refrain from explaining things just simply because of OHMs law....OHMs law is described or "DEFIJNED" BY what happens....it does NOT DEFINE WHAT IS HAPPENING.
See above.
 I think you need to calm down a bit. If you take a step back, along with a deep breath and then read the posts by Defennder, you could probably answer your own questions or at the least, clarify them for us so we could answer them better. As regards EMF, the term is misleading, there is no "force" such. What happens is that the chemicals inside the cell could become more stable by losing energy and they could facilitate this loss of losing energy by gaining or losing electrons (gibbs free energy), and you could read up on a bit of basic thermodynamics and electro chemistry if you want to understand that process better. Now there is a REASON for the chemicals to lose or gain electrons because it helps them attain a lower energy state and hence become more stable. This energy that they lose to attain stability is the energy available to us to utilize as we see fit in a circuit. Now, naturally, we cannot use more energy than is available to us and as we are more concerned with the practical application of such principles, we call this tendency to trade electrons a potential difference. The flow of electrons through a path between the two chemicals is called electric current. The opposition the electrons face when traveling through the path between the two terminals is called resistance.

This makes sense, and actually raises more questions.
 Quote by Defennder When an electron moves in a conductor under an electric field, it will accelerate. However it will inevitably be scattered by collisions with the lattice, and its speed decreases to zero. But the electric field is still present, and it induces the electron to accelerate (and move again) until the next collision occurs. Each one of these collisions cause the electrons to lose its kinetic energy, and this KE is dissipated as heat.
This is very interesting but..
1. If these events are pretty much instantaneous, meaning that action and reaction coexist …accelerating and deceleration and colliding practically simultaneously and inseparably….then repeats itself inherently, influenced due to the presence of an electric field and other electrons…in a chain of events that constitute electron flow, and kinetic energy dissipation…..then voltage drop or kinetic energy loss must be a continuous chaotic, somewhat uncertain ongoing and changing value in a negligible way…..How is this “dissipation of kinetic energy” or “Voltage drop” calculated if it is occurring in a million different ways, when electrons are taking the form of many different paths of least resistance inside a resistor………..I hope I’m asking this right….
 Quote by Defennder A current is basically a flow of electric charge induced by an electric field associated with a voltage. It is for this reason that wires feel warm after use. The repeated collisions ensure that the average KE of the electrons is very low. Hence the total energy of the electrons is nearly equivalent to its potential energy (PE). It is evident that the potential energy of the electron decreases when it gets closer to the positive terminal, just as we might say that an electron situated closer to a positively charged plate has lower PE.
2. I understand that the “EMF” generated by the “potential energy” is in the form of kinetic energy as soon as you have electron current flow…or power loss (dissipation)…… and that this kinetic energy is proportional to the potential energy that imposed the current flow in the first place, However…..what I do not understand is that… If the kinetic energy is constantly dissipating its energy in the form heat due to electrons colliding into each other then the PE…and KE are no longer equivalent only in the form of electric energy, but rather equivalent in electric and thermal energy output (dissipation). Why is it that in a circuit the voltage drop..or KE is always equal to the PE when the KE should be less due to heat dissipation.
 Sigh.... No wonder my professor said on day 1 that the concept of voltage is really hard to grasp. But according to its definition (integral of the electric field over distance between the 2 charged bodies causing the electric field) it makes perfect sense to me. http://img525.imageshack.us/img525/7...rcuit56wc7.jpg What pulls charged bodies? What causes attraction? The electric field..... I'm not talking about alternating current from a generator. That’s a totally different reason for current and its expressed through Faraday's law. Voltage is the pressure, the electromotive force. Voltage DESCRIBES what causes the electrons to start moving in the first place. Unless you want to be an atomic physicist it doesn’t need to be any more complex than that for the average electrical engineer. And as far as a battery is concerned, my description earlier comes from many books, in which its all the same. I had a small taste of the Oxidation and Reduction reactions in chemistry. That is going way too deep into chemical reactions to understand the basic concept. Its all simple. In general, the negative terminal has an excess of electrons. The positive terminal has a deficiency. The cause of this and the actual details of the reactions don't need to matter to grasp the basic concept of why a battery causes electrons to move. When you connect the path from negative to positive, electrons have a means to move towards the electric field which is pulling them. You added a conductor path. So they move. Now the part which is UNLIKE a capacitor. When the electron reaches the positive terminal, the reactions within the battery force the electron back onto the negative plate. And so the process starts all over again, until the battery has no more reactant, or whatever is the official chemical reason, sorry I'm not a chemist. Basically until the battery dies.

 Quote by eugenius Sigh.... No wonder my professor said on day 1 that the concept of voltage is really hard to grasp. But according to its definition (integral of the electric field over distance between the 2 charged bodies causing the electric field) it makes perfect sense to me. http://img525.imageshack.us/img525/7...rcuit56wc7.jpg What pulls charged bodies? What causes attraction? The electric field..... I'm not talking about alternating current from a generator. That’s a totally different reason for current and its expressed through Faraday's law. Voltage is the pressure, the pull, the electromotive force. Voltage is what causes the electrons to start moving in the first place. Unless you want to be an atomic physicist it doesn’t need to be any more complex than that for the average electrical engineer. And as far as a battery is concerned, my description earlier comes from many books, in which its all the same. I had a small taste of the Oxidation and Reduction reactions in chemistry. That is going way too deep into chemical reactions to understand the basic concept. Its all simple. In general, the negative terminal has an excess of electrons. The positive terminal has a deficiency. The cause of this and the actual details of the reactions don't need to matter to grasp the basic concept of why a battery causes electrons to move. When you connect the path from negative to positive, electrons have a means to move towards the electric field which is pulling them. So they move. Now the part which is UNLIKE a capacitor. When the electron reaches the positive terminal, the reactions within the battery force the electron back onto the negative plate. And so the process starts all over again, until the battery has no more reactant, or whatever is the official chemical reason, sorry I'm not a chemist.
First you state that the electric field pulls charged bodies. Then you state that voltage is the pull. Actually, you had it right the first time. The Coulomb force between 2 charged bodies is an axiom based on empirical observation. What cause this force is unknown, but from Coulomb's force law, other laws are derived. After Coulomb's law, it was observed that the action is not instantaneous, but propogated with finite speed. This gave rise to the field concept. Thus a charged body propogates an E field, which results in a force being observed when it reaches another charged body. Voltage is then derived from the integral of E field and path. What "causes" electrons to move in the first place is the inherent nature of charged bodies. They exert forces upon one another with fields propogating at speeds near that of light. Just thought I'd chime in.

 Quote by cabraham First you state that the electric field pulls charged bodies. Then you state that voltage is the pull. Actually, you had it right the first time. The Coulomb force between 2 charged bodies is an axiom based on empirical observation. What cause this force is unknown, but from Coulomb's force law, other laws are derived. After Coulomb's law, it was observed that the action is not instantaneous, but propogated with finite speed. This gave rise to the field concept. Thus a charged body propogates an E field, which results in a force being observed when it reaches another charged body. Voltage is then derived from the integral of E field and path. What "causes" electrons to move in the first place is the inherent nature of charged bodies. They exert forces upon one another with fields propogating at speeds near that of light. Just thought I'd chime in.
you make sense, you seem like you can answer my bloody question. see above, (my latest post).
 Well voltage is not exactly the pull itself. I mean if a ball is on top of a roof, what is pulling it? Gravity of course. But the ball still has potential energy, which is due to gravity and the height above the ground. Voltage is potential energy. The electron is just like that ball. Electric field is like gravity. The only reason I said that voltage is the pull itself, well because people have a really difficult time understanding what energy is. It seems so vague. Energy is the ability to do work. Energy is not the work itself, but it DESCRIBES the work. So voltage describes the electric field. So then its closely related to the electric field. I mean Voltage sorta predicts the future. Its potential energy. That means the electron has the potential to move at a certain current, when the conductive path to the electric field is completed. Yea pull is the wrong word to use here. Describes the pull is better. I will edit my previous post.

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 Quote by XPTPCREWX 1. If these events are pretty much instantaneous, meaning that action and reaction coexist …accelerating and deceleration and colliding practically simultaneously and inseparably….then repeats itself inherently, influenced due to the presence of an electric field and other electrons…in a chain of events that constitute electron flow, and kinetic energy dissipation…..then voltage drop or kinetic energy loss must be a continuous chaotic, somewhat uncertain ongoing and changing value in a negligible way…..How is this “dissipation of kinetic energy” or “Voltage drop” calculated if it is occurring in a million different ways, when electrons are taking the form of many different paths of least resistance inside a resistor………..I hope I’m asking this right….
We cannot predict whether a coin will land heads up or tails, but we can say that out of a thousand or million drops, half will be on heads and half and on tails. Similarly, though we cannot predict the path of an electron in the conductor, the drift velocity of the electron is in the general direction opposite the electric field. Current isn't just the movement of one charge, two charges etc. It's the flux of the vector sum of the current density in a conductor through some specified surface: $$I = \int_S \mathbf{J} \cdot d\mathbf{S}$$. So what this means to us is that though we don't know what which possible paths an electron would take in a conductor, we can understand the behaviour of a multitude of electrons under the influence of an electric field. And that is sufficient for us to understand how current flows.

If you're interested as to how to derive ohm's law $$\mathbf{J} = \sigma \mathbf{E}$$, read up on the Drude model. Unfortunately I can't elaborate here because I don't know the details.
 Quote by XPTPCREWX 2. I understand that the “EMF” generated by the “potential energy” is in the form of kinetic energy as soon as you have electron current flow…or power loss (dissipation)…… and that this kinetic energy is proportional to the potential energy that imposed the current flow in the first place, However…..what I do not understand is that… If the kinetic energy is constantly dissipating its energy in the form heat due to electrons colliding into each other then the PE…and KE are no longer equivalent only in the form of electric energy, but rather equivalent in electric and thermal energy output (dissipation). Why is it that in a circuit the voltage drop..or KE is always equal to the PE when the KE should be less due to heat dissipation.
The drift velocity of an electron through a conductor under E-field is usually very low. And we know from statistical physics (I hope this is the correct field) that the electrons don't have a fixed velocity but instead a normal distribution (normal as in the statistical sense) of velocities. The drift velocity is something like the root-mean-square-speed of all the possible velocity values. As said before KE is continually dissipated because of lattice collisions. If we didn't have lattice collisions and instead had perfect electron transport, the electrons in the conductor would accelerate without stopping and the current would keep increasing. So in high school textbooks, we assume that the electrons in the wire are travelling at some fixed velocity (the root-mean-square speed) and there is no lattice collision in the wires which are assumed to be perfect conductors.

In semiconductor band theory, we draw the band diagram of the electron as sloping downwards across a conductor when a field is applied. The electrons travel at the bottom-most part of the conduction band because we assume the KE is very close to zero. The voltage drop across a segment of conductor is given by the drop in PE and the lost in KE. So we know this as qV where V is the potential difference across the circuit element. This energy is manifested as both dissipated heat and loss of PE.

 Quote by cabraham They exert forces upon one another with fields propogating at speeds near that of light.
Hi cabraham, did you really mean to say near light-speed? Isn't it supposed to be at the speed of light? Or is there some transimission line considerations here?

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 Quote by XPTPCREWX Why does the energy of electrons or charge carriers diminishes when it goes through a circuit element?
The electric potential energy "drops" because it is transformed into another form of energy (i.e. light, heat).

CS

 Quote by Defennder We cannot predict whether a coin will land heads up or tails, but we can say that out of a thousand or million drops, half will be on heads and half and on tails. Similarly, though we cannot predict the path of an electron in the conductor, the drift velocity of the electron is in the general direction opposite the electric field. Current isn't just the movement of one charge, two charges etc. It's the flux of the vector sum of the current density in a conductor through some specified surface: $$I = \int_S \mathbf{J} \cdot d\mathbf{S}$$. So what this means to us is that though we don't know what which possible paths an electron would take in a conductor, we can understand the behaviour of a multitude of electrons under the influence of an electric field. And that is sufficient for us to understand how current flows. If you're interested as to how to derive ohm's law $$\mathbf{J} = \sigma \mathbf{E}$$, read up on the Drude model. Unfortunately I can't elaborate here because I don't know the details. The drift velocity of an electron through a conductor under E-field is usually very low. And we know from statistical physics (I hope this is the correct field) that the electrons don't have a fixed velocity but instead a normal distribution (normal as in the statistical sense) of velocities. The drift velocity is something like the root-mean-square-speed of all the possible velocity values. As said before KE is continually dissipated because of lattice collisions. If we didn't have lattice collisions and instead had perfect electron transport, the electrons in the conductor would accelerate without stopping and the current would keep increasing. So in high school textbooks, we assume that the electrons in the wire are travelling at some fixed velocity (the root-mean-square speed) and there is no lattice collision in the wires which are assumed to be perfect conductors. In semiconductor band theory, we draw the band diagram of the electron as sloping downwards across a conductor when a field is applied. The electrons travel at the bottom-most part of the conduction band because we assume the KE is very close to zero. The voltage drop across a segment of conductor is given by the drop in PE and the lost in KE. So we know this as qV where V is the potential difference across the circuit element. This energy is manifested as both dissipated heat and loss of PE. Hi cabraham, did you really mean to say near light-speed? Isn't it supposed to be at the speed of light? Or is there some transimission line considerations here?

Now is it safe to assume voltage drop is really kinetic energy drop? (in reference to the energy expended but reflecting the potential)

 Quote by XPTPCREWX Ok, this is helpfull... Now is it safe to assume voltage drop is really kinetic energy drop? (in reference to the energy expended but reflecting the potential)
Oh yes that really is the question here. Voltage is potential energy, but it seems like the following

 The electric potential energy "drops" because it is transformed into another form of energy (i.e. light, heat).
is really a kinetic energy drop.

I mean if a ball falls through the air and air resistance begins to affect it, only its velocity drops which is kinetic energy. The ball's potential energy only depends on gravity (which doesnt change) and on its height above the ground. Sure the height changes, but not due to the air resistance.

You see what I mean? Air resistance has no effect on the potential energy of the ball.

Now is an electron the same as the ball, or does it work differently when it comes to electrons? I could have probably summed up all my previous posts into this one question.
 Now....if voltage drop, is kinetic energy drop...then the voltage across a resistor should be the PE...right?

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The voltage drop is due to both KE and PE. PE of the electrons is converted to KE, which is then dissipated as heat loss. Remember that the energy of the electrons is the potential energy it has when it is accelerated by the E-field. Upon application of the E-field, the electron accelerates due to Newton's second law. We then have the conversion of PE to KE. This KE is then lost later due to lattice collisions.

 Quote by eugenius I mean if a ball falls through the air and air resistance begins to affect it, only its velocity drops which is kinetic energy. The ball's potential energy only depends on gravity (which doesnt change) and on its height above the ground. Sure the height changes, but not due to the air resistance.
This is inaccurate. At no point in time does the velocity of the ball decrease. It increases all the time. Only the rate at which it increases varies. The drag force due to air resistance increases until it balances out the weight of the object, at that point the ball attains terminal velocity. When the drag force increases, the net force acting on the ball downwards decreases, and this means that the ball's downward acceleration decreases. But this does not imply that the ball decelerates. Deceleration $$\neq$$ Decreasing acceleration.

The only time when the ball loses velocity is when it hits the ground and comes to a standstill, or if a parachute which is attached to it suddenly opens.

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 Quote by Defennder The drift velocity of an electron through a conductor under E-field is usually very low. And we know from statistical physics (I hope this is the correct field) that the electrons don't have a fixed velocity but instead a normal distribution (normal as in the statistical sense) of velocities. The drift velocity is something like the root-mean-square-speed of all the possible velocity values. As said before KE is continually dissipated because of lattice collisions. If we didn't have lattice collisions and instead had perfect electron transport, the electrons in the conductor would accelerate without stopping and the current would keep increasing. So in high school textbooks, we assume that the electrons in the wire are travelling at some fixed velocity (the root-mean-square speed) and there is no lattice collision in the wires which are assumed to be perfect conductors.
I need to correct something here. The electrons are continually scattered while traveling in the wires. Based on the concept of the mean relaxation time (the average amount of time an electron spends in between collisions), we can derive a mean drift velocity using some statistical formulation and assumptions. My original post had the impression that the mean drift velocity is just the RMS speed of the electrons without taking into account lattice collisions. This is mistaken.

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