According to my physics book, inside a resistanceless battery (it is a part of a closed circuit) the conservative field has the same magnitude but opposite direction to the non conservative field. Thus $E^* = -E$.

$E^*$ is the non-conservative field.

Anyway, I can apply the Ohm's law to a real battery because a current flows through it.

$$J = σ(E^* + E)$$

What about the current inside an ideal battery? If we try to apply the Ohm's law (we can't do this, actually) to an ideal battery, then we find an indeterminate form, because the conductibility is infinite and the net electric field is zero. I would expect that inside an ideal battery there isn't a current flow, is it?

  • $\begingroup$ yes the current still flows, without the current inside the battery flowing you would have charge buildup and the reaction would stop. The power loss is I^2R=Idelta(V) so it is zero as the internal resistance is zero $\endgroup$
    – ChemEng
    Oct 6, 2020 at 0:03
  • $\begingroup$ will a battery work without a salt bridge? NO $\endgroup$
    – ChemEng
    Oct 6, 2020 at 0:04

3 Answers 3


A battery, ideal or not, does not follow Ohm’s law. Ohm’s law is an observed behavior of a specific class of materials/devices, sometimes called a constitutive equation. It is not a universal law of nature. Not all materials/devices obey it, including batteries.

  • $\begingroup$ This is not an answer to the question, though it is a true statement. $\endgroup$ Oct 5, 2020 at 4:04
  • $\begingroup$ Why not? The question was about a problem encountered when applying Ohm’s law to an ideal battery. “It doesn’t apply” seems like the best answer to me. What are you seeing in the question that I am not? $\endgroup$
    – Dale
    Oct 5, 2020 at 4:50
  • $\begingroup$ I prefer to think of it as a definition: resistance is voltage divided by current. Now, sometimes resistance is a useful measurement, and sometimes it isn't. $\endgroup$
    – user253751
    Oct 5, 2020 at 10:24

An "ideal battery" doesn't have any structure inside. It is simply a mathematical abstraction of a device that produces a fixed voltage across its terminals. In order to do that, it absolutely must allow current to flow through it. In fact it must force a current through itself, in whatever amount is necessary to produce the required voltage across its terminals, given whatever circuit is connected to it.

But it doesn't obey the microscopic form of Ohm's Law because it doesn't have any internal structure with physical extent, and it doesn't have any material within it that could be characterized by conductivity. The materials inside physical (as opposed to ideal) batteries also don't follow Ohm's Law because for one thing the material is not uniform. There are multiple materials involved in the chemical reactions that produce the voltage across the terminals. And for another thing because those chemical reactions are producing ion concentration gradients in the electrolyte that counter any current that would be expected due to the electric field.

It also doesn't obey the macroscopic form of Ohm's Law ($V=IR$) because it isn't a resistor, and this form of Ohm's Law is essentially the definition of what it means for a device to be an ideal resistor. If your device followed this "law", it would be a resistor and not a battery.

  • 1
    $\begingroup$ Of course there is a flow of charge inside an ideal battery. Anyway, how you could define a current inside the battery if the net electric field is equal to zero? $\endgroup$
    – user248666
    Oct 4, 2020 at 16:16
  • 2
    $\begingroup$ @AngeloGiannuzzi, There is no "inside" to an ideal battery because it has no physical existence. $\endgroup$
    – The Photon
    Oct 4, 2020 at 16:37
  • $\begingroup$ In some sense, it's an anti-resistor; it produces current in the direction opposite to the voltage across it. $\endgroup$ Oct 4, 2020 at 21:53
  • 3
    $\begingroup$ @Acccumulation An ideal anti-resistor would have to follow $V = IR$ for some finite negative value of $R$, wouldn't it? $\endgroup$
    – zwol
    Oct 4, 2020 at 22:08
  • 1
    $\begingroup$ @badjohn, no. In a negative resistor, increasing the applied voltage decreases the current and vice versa. An ideal battery produces the same voltage regardless of current. The resistor (whether negative or positive) has a proportional relationship between voltage and current. The ideal battery has no such relationship. $\endgroup$
    – The Photon
    Oct 5, 2020 at 15:55

Even for an ideal battery, current conservation applies. Electrons in have to equal electrons out. An ideal battery gives a boost to the voltage regardless of the current, but current in equals current out. You can still use Kirchoff's laws to find the currents in the circuit.


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