I'm having trouble conceptualizing why the voltage drop between two points of an ideal wire (i.e. no resistance) is $0~V$. Using Ohm's Law, the equation is such:

$$ V = IR \\ V = I(0~\Omega) \\ V = 0$$

However, conceptually I can't see how there is no change in energy between these two points.

It is my understanding that the electrical field of this circuit produces a force running counterclockwise and parallel to the wire which acts continuously on the electrons as they move through the wire. As such, I expect there to be a change in energy equal to the work.

Voltage drop is the difference in electric potential energy per coulomb, so it should be greater than $0~V$:

$$ \Delta V = \frac{\Delta J}C \\ \Delta J > 0 \\ \therefore \Delta V > 0 $$

For example, suppose I have a simple circuit consisting of a $9~V$ battery in series with a $3~k\Omega$ resistor:

Simple electric circuit of a 3 kilo-ohm resistor in series with a 9 volt battery

If the length from point 4 to point 3 is $5~m$, I would expect the following:

$$ W = F \cdot d \\ W = \Delta E \\ F > 0 \\ d = 5 > 0 \\ \therefore W > 0 \\ \therefore \Delta E > 0$$

Since work is positive for any given charge, the change in energy for any given charge is positive -- therefore the voltage drop must be positive. Yet, according to Ohm's Law it is $0~V$ since the wire has negligible resistance.

Where is the fault in my logic?

  • 1
    $\begingroup$ Don't forget that an electric field accelerates electric charge. If there were, in fact, an electric field acting continuously on the electrons in the ideal conductor, the electric current would continuously increase since the electrons would be continuously accelerating. $\endgroup$ Commented Oct 11, 2013 at 16:33
  • $\begingroup$ Does this mean work is $0$ throughout the entire circuit since the velocity of the electrons remain constant? $\endgroup$ Commented Oct 11, 2013 at 16:47
  • $\begingroup$ No, and I'm not sure how you could infer that from my comment. Electric power (time rate of change of work) is the product of voltage and current. Since the voltage across an ideal wire is zero, there is zero power associated with the wire for any current. But the voltage isn't zero across the resistor or the battery so there is power associated with these circuit elements. $\endgroup$ Commented Oct 11, 2013 at 17:20
  • $\begingroup$ Possible duplicate: physics.stackexchange.com/q/8675/2451 and links therein. $\endgroup$
    – Qmechanic
    Commented Jan 11, 2014 at 16:08
  • $\begingroup$ If there is no friction on the electrons moving through the wire, then you don't need to perform any work on the electrons to keep them moving at a constant velocity (Newtons first law of motion). Thus a constant electric current can flow through an ideal conductor with no voltage drop. $\endgroup$
    – jabirali
    Commented Jan 15, 2015 at 3:47

2 Answers 2


The key thing is that there is NO electric field within the perfect wire. So, there is no force acting on the electron, and thus no work done on it (while it's in the perfect wire).

This goes back to the definition of a perfect conductor (which the perfect wire is). Within a perfect conductor, there is no electric field. Instead, the charges (which have infinite mobility) rearrange themselves on the surfaces of the conductor in such a way as to perfectly cancel out any internal field.

So, the only fields in your circuit would be 1) in the battery, and 2) in the resistor.

I should also add that this is due to the approximation of the wire as 'perfect'. A real wire has some resistance, or equivalently, its charges don't perfectly reorder so as to perfectly cancel an internal field.

  • $\begingroup$ I wasn't aware that the electrical fields cancel out. Your answer seems to make sense, but I'm having trouble visualizing the charges. Would you be able to expound on how the arrangement of the charges differ in the presence and absence of resistance. $\endgroup$ Commented Oct 11, 2013 at 16:46
  • $\begingroup$ Think of the wire as a horizontal cylinder. If you apply an electric field pointing to the left, the electrons in the wire will move to the right, so that eventually they collect on the right side, and there is a deficit of electrons on the left. This distribution of charge (positive on the left, negative on the right) produces a field of its own, pointing to the right, which works against your applied field. This process will continue, until there is no net field left inside the conductor; the equilibrium is reached once there is no more field and thus the electrons experience no net force. $\endgroup$ Commented Oct 11, 2013 at 16:50
  • 1
    $\begingroup$ The cause of resistance in materials is a very complex topic (that's why it's just glossed over by either calling a wire 'perfect', or sweeping everything into the one value of 'resistance'). Resistance in a wire isn't due to the electrons bouncing off the nuclei; quantum mechanics makes things more interesting. I'd like to explain more, but don't have space in a comment. Search for "impurity scattering" and "phonons"; those are the two main culprits for electrical resistance in a normal wire. $\endgroup$ Commented Oct 11, 2013 at 17:08
  • 1
    $\begingroup$ I think that's the best way to think about it. The difficulty is that when you start asking "what actually IS resistance", you need to use quantum mechanics, and "force" just isn't a very good concept there. So, my summary would be: "in an ideal conductor, there is no net electric field, and in a non-ideal conductor, there is a net electric field but the force from that field is countered by a 'force' due to the resistance." Does that help? $\endgroup$ Commented Oct 11, 2013 at 17:21
  • 1
    $\begingroup$ @Zane Beckwith even I was having the same question you have answered very nicely but I have another question.If there is an electric field in a resistor then when an electron moves across it 1.Potenial Energy of an electron is lost 2.Energy is lost because the electrons collide etc.Now when I was studying I saw that the heat dissipation of a resistor is simply the loss of potential energy of an electron accros the resistor.How can we equate these two? $\endgroup$
    – user264098
    Commented Jun 18, 2020 at 17:31

$F$ is not greater than 0 in an ideal wire, think 'frictionless surface' if it helps. In this idealisation, the electrons are considered to move from 4 to 3 without effort... Therefore there is no need to invoke any energetic loss.

If this doesn't appeal then you need to drop the idealisation and consider resistivity and then you can the case more physical...

If you're feeling philisophical: "In an idealisation, there is no wire."

  • $\begingroup$ but if there is no wire there is potential difference across two terminal $\endgroup$
    – banned
    Commented Apr 7, 2022 at 8:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.