Help Voltage drop is driving me insane

[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).