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Voltage loss w/ resistor scaling with current

  1. Mar 11, 2015 #1
    So according to the ideal voltage loss equation V = IR if I double I (per se) I will double the loss in voltage. This is a bit odd and I just want to make sure my intuitive explanation is correct.

    I am assuming voltage loss is mainly due to some electrons bouncing off the resisting material and thereby lowering the kinetic energy of the electrons that come later from the collisions that ensue (this would lower current again but less so until it reaches a sort of equilibrium). If I increase the current, the speed with which an electron bounces off the material will increase and therefore the KE loss of electrons further down the line (related to voltage loss) would increase as well. This doesn't seem exactly proportional but close enough where I hope it is the right explanation.

    I am only now worried about the connection with KE loss and both current and voltage, so I may have to rethink some things (sorry I may have posted too soon).
    Last edited: Mar 11, 2015
  2. jcsd
  3. Mar 11, 2015 #2


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    Your intuitive explanation is not correct. You have to remember that voltage is a measure of electric potential energy. That is, if I place a particle in an electric field that particle will have potential energy. The amount of potential energy the particle has between any two locations within this field is defined as voltage. When you place a resistive component into a circuit, some of this potential energy is converted into thermal energy and is then lost from the circuit. This lowers the potential energy available to the rest of the circuit. In other words, the voltage (electric potential energy) is lost, leaving less voltage available to the rest of the circuit. Hence, voltage drop.

    Current is a flow of charges, but it itself is not a measure of how fast they flow, it is a measure of numbers. You could have a few electrons traveling very quickly, or a lot of electrons traveling very slowly and you'd get the same number passing a point in the circuit over time. In reality the electrons are moving about randomly simply due to thermal motion and all their different speeds and directions add up to a net of zero over time. When you apply a voltage you have slightly more motion in one direction than the other, which manifests as current flow.

    There's also the fact that kinetic energy scales with 1/2 of the velocity squared. So when you double something's speed you quadruple its kinetic energy. But in a circuit if you double the current all you've done is double the energy delivered. You can see this from the power equation: P=IV. If you halve the resistance of the circuit, which allows you to double the current but keep the voltage the same, you get twice the power (energy over time), not quadruple.

    You can't just change the values of the equation willy nilly, you have to think of how the values in a real circuit can physically be changed. For example, if you want to double I then you either have to double V or halve R. If we have a simple ideal circuit with a voltage source and a single resistor, all of the voltage drop is across this resistor. If you then replace the resistor with one that has half the resistance the current doubles but the voltage drop remains exactly the same.
  4. Mar 11, 2015 #3
    Assume with an approssimation that V is costant, how is usually for a battery which doesn't immediatly discharge.
    So if I double I, R must become half. The current I is due to R.
    If R is infinite there isn't current and if R=0 the current get its maximum.
    The electric potential energy is due to V and the kinetic energy transferred from it is bringing by charge in motion.
    If I increase R, keeping costant the potential energy, less electrons can move and so the current decreases.
    Obviously V can't hold costant for a long time because of the Joule effect, due to electrons bouncing the lattice.
  5. Mar 11, 2015 #4
    Ok. So let's pretend I place a bigger battery in a circuit instead of a smaller battery (I'm now doubling voltage). There is twice as much potential energy per charge across the circuit, and twice as much potential energy per charge is lost across a certain resistor. This is just to get the situation straight.


    I need to find a way to explain how twice as much energy (per charge) is lost in terms of thermal when the potential energy (per charge) is doubled.

    Beyond this section I'm a little in the dark so this may be a bit of a repeat of what was said before only because I figure if I can't propose another answer I can at least clarify my older attempt.

    Previous thoughts perhaps explained a little more clearly (if it helps clarify what I'm confused over):

    The reason I was referring to kinetic energy is because my assumption is that in order for potential energy to convert to thermal it must first convert into kinetic energy and then that KE must bump into things (presumably the wire et al.) and thereby lose the expected thermal energy. My initial explanation was in terms of the electrons bouncing back and interfering with the others because this would at least scale with an increase in potential energy (per charge); this is because an increase in PE over the same distance would imply an increase in the gain in KE over that distance and thereby an increase in the speed with which the electron would bounce back after a collision with the wire (et al.) and impede the later electrons. Otherwise I'm sort of in the dark as to how the loss in potential energy (per charge) across a resistor would increase as the potential energy (per charge) increased in such a nicely proportional manner.

    The weird part about current:

    I actually knew that current wasn't purely speed but rate of charge (which can increase with cross section area of the wire). This was really only an afterthought that electrons bouncing back would affect the current of the electrons before as it impeded their motion (assuming they are not so spread out as to not interfere). (This is realizing that motion is only one factor of the current equation.) So if I increased the voltage I would increase the voltage loss, but afterwards slightly decrease the current (and therefore the voltage) due to the increased voltage loss.


    I guess what I'm really scouting for is how that potential energy is turned into thermal so I can know how an increase in that potential energy would increase the amount turned into thermal so proportionally.

    Again sorry if this is a repeat it's just that I'm a bit confused. It's possible that there was some implication from the power equation provided that I did not use.
    Last edited: Mar 11, 2015
  6. Mar 12, 2015 #5
    Sorry I waaay muddled around with that. I think I now see where this is going, but I'm going to have to write it out more properly. Sorry if I caused a headache. What I said wasn't thought through clearly.
    Last edited: Mar 12, 2015
  7. Mar 15, 2015 #6
    Ok problem resolved. Thanks for the early responses btw.
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