Let's assume that the resistance of a wire is zero. Now, suppose the wire has a length of 10 m and is connected to a battery with an emf of 10 V. According to my physics textbook, the electric field should be constant across the wire. Using the equation ΔV = -ʃE·ds, the voltage drop for a path Δs in the wire should be -EΔs. Also, the electric field should be of 1 V.m. Why then doesn't the electric potential decrease of 1 V for every meter in the wire? When solving problems with Kirchoff's laws, we assume that the voltage is constant until there is a resistor in the circuit and I don't understand why. Even if the resistance of the wire is zero, shouldn't the potential in the wire decrease gradually according to ΔV = -ʃE·ds?

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    $\begingroup$ I don't understand your reasoning. If we assume (as you have asked) that the resistance of the wire is zero, isn't the $\Delta V$ zero for any value of current through the wire? Put another way, if it were a zero ohm resistor across the emf source, doesn't Ohm's law demand that the voltage across the resistor be $I\cdot 0\Omega = 0V$ for any value of current $I$ through the resistor? $\endgroup$ Commented Jun 12, 2020 at 17:22
  • $\begingroup$ If there is a battery across the wire, then there will be a voltage. It is the potential difference that dictates the current and not the opposite. Therefore if the resistance was close to zero the current would approach infinity. But it's not the value of the current that interests me, it's how the voltage changes in the wire. Regardless of what ohms law says, according to ΔV = -ʃE·ds, the voltage should drop gradually through the wire and not only when there is a resistance. The definition of voltage is ΔU/q = -ʃE·ds and I'm trying to make sense of this equation in the case of a circuit. $\endgroup$
    – Pierre
    Commented Jun 12, 2020 at 17:48
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    $\begingroup$ Well ... $\Delta V=0$ and $E=0$ does not violate your integral. Nonetheless, you should always be suspicious of conclusions reached by analyzing non-physical situations. Our theories are not designed to describe phenomena which are physically possible. The analsis is invalid, and conclusions are meaningless. How fast can a unicorn run? $\endgroup$
    – garyp
    Commented Jun 12, 2020 at 17:51
  • $\begingroup$ All the voltage drop takes place inside the battery when you "short" the terminals. $\endgroup$ Commented Jun 12, 2020 at 17:54
  • $\begingroup$ @garyp, ChemiCalChems provided an answer similar to yours. I hadn't thought of E=0, but if the electrons aren't accelerating in a wire with zero resistance than E has to be equal to 0 and thus ΔV=0. Anyways thank you and perhaps you meant "physically (im)possible"? $\endgroup$
    – Pierre
    Commented Jun 12, 2020 at 18:07

2 Answers 2


You are assuming the electric field within a perfect conductor in a steady state is non-zero. Let's go through what this would mean.

For starters if it really is a perfect conductor, it would mean charges would be accelerating within your conductor. In a steady state, this does not happen, for the current inside the circuit is constant.

It could also mean your charges are meeting some resistance, and thus doing work, causing a voltage drop, in which case, they would indeed need some electric field to push them past the resistance. But then, this means your conductor is not $0 \,\Omega$.

Thus, the electric field within the conductor not being zero implies either the conductor not being perfect, or charges being accelerated continuously to infinity. We can thus conclude the electric field within a perfect conductor is 0 in a steady state.

Thus, no, you would not get any sort of voltage drop across it. Current comes in, current comes out, and it has to fight nothing, because it's a perfect conductor, so no voltage drop across it, or across any segment of it.

  • $\begingroup$ I hadn't thought of the possibility of an electric field of zero. Your answer makes perfect sense. Thank you! $\endgroup$
    – Pierre
    Commented Jun 12, 2020 at 18:02
  • $\begingroup$ Glad it helped @PierreChamoun ! $\endgroup$ Commented Jun 12, 2020 at 18:03
  • $\begingroup$ To be sure, the electric field within a perfect conductor is zero regardless, i.e., there is no steady state requirement. For example, a perfect conductor has zero skin depth. $\endgroup$ Commented Jun 12, 2020 at 18:11
  • $\begingroup$ @AlfredCentauri I mean, sure, else charges would accelerate to infinity rather quickly. Was just trying to illustrate why that can't be. $\endgroup$ Commented Jun 12, 2020 at 18:17

If there is a battery across the wire, then there will be a voltage.

If the battery is physical, it has non-zero internal resistance, and then the emf will be dropped across this internal resistance when the zero resistance wire is connected across it, i.e., the battery terminal voltage will be zero. The resulting current through the zero resistance wire has a specific name: the short circuit current. See, for example, Battery Internal Resistance & Short Circuit Current

If the context is ideal circuit elements where the battery has zero internal resistance (is an ideal voltage source), then the KVL equation for the circuit you propose is invalid. For example, if the battery fixes the voltage across its terminals at $1.5V$, and the zero resistance wire (ideal short circuit) fixes the voltage across its ends to be $0V$, the resulting KVL equation is

$$1.5V = 0V$$

So, in the context of ideal circuit theory, this type of circuit is 'not allowed'.

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    $\begingroup$ I was not assuming that there is no resistor in the circuit, only that the wire has zero resistance. I was only wondering why there is no gradual voltage drop between the resistors. Of course, a circuit with zero resistance would end up in a short circuit. Thanks for your help! $\endgroup$
    – Pierre
    Commented Jun 12, 2020 at 18:15

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