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I Voltage in a Circuit

  1. Oct 5, 2015 #1
    I feel that I don't understand how voltage works in a circuit. I understand voltage to be electric potential energy per unit charge (kq/r). In the case of a circuit, electrons flow from low potential to high potential. But I don't understand how resistors cause a voltage drop. Isn't voltage based on position? How can the resistors cause a drop in potential energy? I can understand the resistor causing a drop in kinetic energy slowing the electrons down, but how does it lower the voltage? And if the total voltage is equal to the sum of the voltages of the resistors, then the voltage difference in wire after the last resistor and the positive terminal would be zero, right? Then how would the electrons be able to flow back to the positive terminal? Wouldn't they just stop?

    The best analogy I can think of is a river flowing downhill that turns a turbine. But in this case the turbine doesn't cause a drop in the water's potential energy. It only takes some of the water's kinetic energy.
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  3. Oct 5, 2015 #2


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    If you like the water analogy think of voltage as pressure. A resistor is then like a narrow restriction in a pipe. You get a pressure (voltage) drop across the resistor when current is flowing through it. The water analogy isn't perfect though.

    One way out of that problem is to realise that there is no such thing as an "ideal wire". Even a wire has some resistance so there is some voltage drop down the wire from the last resistor to the positive terminal (even if it's very small).
  4. Oct 5, 2015 #3
    What about the question about kinetic and potential energy? The original definition of voltage potential energy per unit charge doesn't seem to make sense here as explained in my initial question.
  5. Oct 6, 2015 #4
    Voltage is defined as kq/r. It should vary as charge moves across the circuit regardless of whether there is resistance or not. I don't understand why resistance causes voltage drop and no resistance doesn't.
  6. Oct 6, 2015 #5
    There is potential drop even without resistor.
    For ideal battery, the potential difference between the terminals is the same with and without resistor. So there is the same voltage drop.
    Without resistance the drop is all on the connecting wires. With the resistor, most of the drop is on resistor and just a little on the wires. The total is the same (again, for ideal source).

    Voltage is not defined like that. It is a formula which gives the potential for a point charge only. Not relevant for circuit analysis.

    And don't assume that the KE drops in a resistor. The drift velocity of the electrons may increase or decrease in the resistor, as compared with the wires. It depends on the material and geometry of the resistor.
  7. Oct 6, 2015 #6
    A voltage drop in an ideal circuit is equal to the resistance times the current. Lumped-element circuit analysis is all about the ideal case in which we can make useful approximations instead of Maxwell's laws. When you deal with circuits of higher frequencies (as in, in the MHz/GHz range), those approximations no longer work, and you're forced to use Maxwell's equations.
  8. Oct 7, 2015 #7
    Can't you use the formula (kq/r) for finding the difference in potentials of a single test charge at different points on the circuit though? Why is most of the drop on the resistor? Yes I know there is energy lost to the resistor..but how can potential energy possibly lost? Potential energy by definition should only change with position and voltage is potential energy per unit charge. In the case of the river flowing downhill, the potential energy of the river is only based on its height. The energy lost to the turbine (resistor) doesn't cause a loss in potential energy.
  9. Oct 7, 2015 #8
    No, that formula only works for a the potential in the field of a point charge. The field in the wires is not created by a point charge.

    A potential "drop" (or voltage drop) means that there is a potential difference between two points.
    The potential energy of the charge decreases when going through the wires or resistors. It is a matter of interpretation to say that the charge loses PE or just that some of its PE is converted in other forms (like heat or radiation).
  10. Oct 7, 2015 #9


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    That depends on the turbine design - consider a waterwheel for example. It lowers the water, reducing its potential energy while using the energy to do work.
  11. Oct 7, 2015 #10


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    It is essential to realise that the KE of electrons plays no part in the energy transfers in a wired electric circuit. The mass and RMS velocity imply that KE is insignificant.
  12. Oct 8, 2015 #11
    But why can't we calculate the voltage at each point on the circuit with kq/r? Why doesn't that work? Why is the voltage drop more across areas with more resistivity? Why isn't voltage difference based solely on position as it was in electrostatics? If the resistor does negative work on the charge, why should that affect the potential energy? Isn't it like rubbing a box on a hill against the ground? It loses energy to friction but it's potential energy stays constant.
  13. Oct 8, 2015 #12
    Which part of this (below) you don't understand?
    It will be useful to see your thinking before trying to answer more of your questions.
  14. Oct 8, 2015 #13


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    If it helps you visualize things, try the hydraulic analogy. A resister is like a narrow pipe full of baffles. The pressure at the inlet end has to be higher than the pressure at the outlet end if the flow rate is to match the rate at which water comes in from the upstream pipe and goes out to the downstream pipe.

    If you imagine the power supply as a source of fixed pressure, the flow rate increases or decreases to whatever the pressure can drive.
    If you imagine the power supply as a source of fixed flow, the pressure increases or decreases to whatever is needed to match the flow.

    Tiny build-ups of charge are adequate to create large potential differences. From that perspective, the electric field in the resistor is still a function of charge position, just like with electrostatics.
  15. Oct 8, 2015 #14
    But as I understand it, the current is uniform throughout the circuit, correct?

    And from what my teacher taught us, kq/r still works when its not just a point charge..you just have to apply kq/r to the multiple point charges creating the field and add the voltages. So why can't you do that when dealing with a circuit?

    I also feel stuck on how voltage is POTENTIAL energy per unit charge. How can friction play a role on potential energy which is based soley only location? If it was losing kinetic energy across the resistor I would understand because friction and resistance slow things down.
  16. Oct 8, 2015 #15
    Oh, yes you can use it that way, in principle. If you knew the position of each point charge in the circuit. I thought you are asking if you actually can, not if it would be possible in principle. :smile:

    The charges do "loose" KE energy every time they interact with the lattice. And then their KE energy increases again due to the field. As the particle moves to lower and lower potential it gains small amounts of KE and these are transferred to the lattice. At the second end of the resistor, the average KE is still the same and the PE is decreased as compared to the PE at he first ("entry") end.
    So the net effect is a decrease in PE of the charges and increase in thermal energy of the lattice.

    A little similar to a body moving down a ramp with constant speed (imagine that you throw it down along the ramp with some initial speed). The tangential component of gravity is equal to the friction force. So at the bottom the KE is same as at the top but the PE is decreased and both ramp and body are hotter.
  17. Oct 8, 2015 #16


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    Yes, and in a loop of pipes, the flow rate of the water is uniform throughout the circuit as well.
  18. Oct 8, 2015 #17


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    Friction plays no part in the Potential Energy. Friction would, however, reduce the amount of useful energy to be gained by using the GPE of a falling mass as the source. The Electrical Potential Energy 'used' (or transferred) around a circuit will be mainly in the Load (Heater. Motor, Audio Amp etc etc.) and a small amount in heating up the wires.
    It is strange that people always seem to question what happens in the wires of a circuit, rather than the Electrical Appliances and Devices. It requires very little energy to get the electrical power from source to load - so the majority of the Voltage Drop is across the Load. If it's not, then you need to be re-designing your circuit.
    Have you calculated the amount of actual KE that your model would involve? To transfer 1W of energy into a resistive load, using KE, how much Mass of electrons would you need, if the mean drift speed is just a few mm/s? Would you say that it is the Kinetic Energy of a bicycle chain that accounts for the power transferred from the cyclist's legs to the wheel? How long would the bike keep accelerating if you took your feet of the pedals of a fixed wheel bike? KE doesn't need to be involved in Electrical power transmission or a chain drive.
    This KE can be considerable in an electron beam through a discharge tube, when the electron speeds can be a significant fraction of c - but not in a solid conductor.
  19. Oct 8, 2015 #18
    We already had this discussion and I calculated and showed you that it fits. I suppose you forgot about it. It was at least one year ago. :smile:
    I had the same impression before doing the calculation, that it won't work.

    But the bottom line is that you don't need huge mass of electrons because they repeat this process very fast. And I mean very. The relaxation time in metals is of the order of 10^(-14) I believe. I have to look it up again.

    PS. This is not my model. I did not make it up myself. It started with Drude and was refined later.
  20. Oct 8, 2015 #19


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    Oh boy. With my memory span for recent events, a year represents many half lives. :rolleyes:
    So, fair enough, you have used the idea of KE to account for an energy transfer mechanism from PE to thermal on a microscopic level. How does this apply to very low resistivities? Why doesn't this imply higher and higher electron drift speeds as resistance is reduced? (i.e. less and less KE is taken from the drifting electrons) The drift velocity only involves carrier density and cross sectional area (afair) so that implies that carrier density should be inversely proportional to resistivity (??). Can you give me a ref to your argument of last year? I find it hard to accept - but I could be convinced. I also have a problem in applying your idea to a non-resistive load, where there are no 'collisions' as such but the energy is transferred in the form of mechanical motion.
  21. Oct 8, 2015 #20
    I am sorry, I am not sure what is the part you don't agree with. The scattering of electrons in a lattice is treated in all solid state books. Do you think that the energy is transferred to lattice by a different mechanism, which does not involve electron-lattice interaction?

    Your favorite analogy with the bicycle chain is just this, an analogy. It does not explain how the energy gets to the lattice vibrations that correspond to thermal energy.
    Do you have another mechanism in mind?

    Of course, I was talking only about ohmic heating, not other forms of energy transfer. I never said otherwise.
    I am not sure what is your problem with low resistivities. The drift velocity is not a material parameter. It depends on the current density.
    The estimate we discussed last time was for a metal. Do you mean something with even lower resistivity?
    A metal has about 10^22 free electrons per cubic cm^2 and relaxation times of the order of 10^(-14) s.
    So in order to transfer 1W, each individual interaction will average to 10^(-36) J. (for a 1cm^3 resistor).
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