Help Voltage drop is driving me insane

[eugenius]

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 doesn't 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.

 

[XPTPCREWX]

Now...if voltage drop, is kinetic energy drop...then the voltage across a resistor should be the PE...right?

 

[Defennder]

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.

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 doesn't 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.

 

[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 traveling 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.

 

[cabraham]

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?

Greetings defennder. Yes, I should have said "*at* or near the speed of light". In air or vacuum, the speed would be that of light. In a t-line, the speed approaches light, but not quite equal to light. BR.

Claude

 

[eugenius]

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.

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.

Ah yes of course. I know that. Forgot about terminal velocity. Maybe air resistance is a bad example to compare to electrical resistance, but what about friction eh?

 

[Defennder]

Greetings defennder. Yes, I should have said "*at* or near the speed of light". In air or vacuum, the speed would be that of light. In a t-line, the speed approaches light, but not quite equal to light. BR.

Claude

I see, thanks for clarifying.

Ah yes of course. I know that. Forgot about terminal velocity. Maybe air resistance is a bad example to compare to electrical resistance, but what about friction eh?

Hmm, I believe it's conceptually distinct, though they bear certain resemblances. For the electrons in the wire, they undergo repeated acceleration and collision, whereas in the case of friction (assuming the force provided isn't constant otherwise the moving mass would still accelerate due to net force on it or it moves at a constant velocity), the mass simply gets slowed down if the frictional force exceeds the applied force on it. But in both cases, we don't have like what happens to the electrons in the wire since the moving mass doesn't stop, accelerate, stop repeatedly.

 

[XPTPCREWX]

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.

If I have a "Potential Energy" of 100v across a battery terminal, this means that the voltage drop AKA the "Kinetic Energy" across the total resistance will also be 100v difference when under load or under influence of the existing PE of the 100v Electric field...this reflects the conservation of energy.
But if I measure ONLY a segment of the resistance of that same circuit…measuring a voltage drop or "Kinetic energy drop" of 50v across half, what I seem to understand is that I will get a 50v Potential Energy….BUT of what?
a. The PE of the ENTIRE Electric Field of the circuit which we already established had a PE of 10v
b. The PE of that one electron, or group of electrons.
c. The PE “ASSUMED” to cause that Kinetic energy drop or loss.

 

[XPTPCREWX]

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.

If I have a "Potential Energy" of 100v across a battery terminal, this means that the voltage drop AKA the "Kinetic Energy" across the total resistance will also be 100v difference when under load or under influence of the existing PE of the 100v Electric field...this reflects the conservation of energy.
But if I measure ONLY a segment of the resistance of that same circuit…measuring a voltage drop or "Kinetic energy drop" of 50v across half, what I seem to understand is that I will get a 50v Potential Energy….BUT of what?
a. The PE of the ENTIRE Electric Field of the circuit which we already established had a PE of 10v
b. The PE of that one electron, or group of electrons.
c. The PE “ASSUMED” to cause that Kinetic energy drop or loss.

 

[Defennder]

A voltage drop of 10V means to say that the 10J of energy was supplied to each coulomb of charge as it passes through that circuit element; 10J worth of energy was dissipated per unit coloumb of charge passing through that circuit element. By the way, by Kirchoff's voltage law, the sum of voltage drop around any closed loop in the circuit is equivalent to the supplied emf for that loop. This can be proven by the conservation of energy and conservation of charge (Kirchoff's current law).