Reading this question and its answers (as well as other questions), it seems that in an idealized short-circuit with zero resistance, one concludes the voltage is zero.

This seems completely wrong.

The justification is given by V=IR. Assuming current is finite you would indeed conclude that V=0. But why would you assume finite current?

Yes, real-world currents must be finite, but real-world resistances must be nonzero. This is an idealization; the idealized values don't have to be physically attainable.

And, in a real-world approximation of an ideal short circuit, one sees very large current; nonzero voltage, infinite current, and infinite power seems like a much more accurate idealization than the finite current, zero voltage, zero power idealization.

Thus my question. Is this idealization of finite current and zero voltage really the common one to make? And why?

Edit: to make it explicitly clear, in this idealization, the parameters of the ideal circuit are allowed to attain idealized values — specifically, a priori, a literally infinite for current is allowed (for mathematical precision, I mean the extended real number ∞). With R=0 and I=∞, Ohm's law puts no constraints on the voltage; every extended real number value for V is consistent.

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    \$\begingroup\$ "What happens when an unstoppable force meets an immovable object?" In the real world we have neither unstoppable forces nor immovable objects. In theory, you can approximate strong forces as unstoppable and heavy objects as immovable, but if you find yourself in a situation with an unstoppable force hitting an immovable object, your approximation has failed. \$\endgroup\$ – user253751 Mar 7 '18 at 0:50
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    \$\begingroup\$ An ideal Voltage source cannot be shorted because the source and the short both have no resistance which is a violation of OHm's law. You can short the source if it has resistance or an ideal source with a low resistance but not both. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Mar 7 '18 at 0:57
  • \$\begingroup\$ "why would you assume finite current?" - why wouldn't you? "real-world resistances must be nonzero" - not true. Superconductors have exactly zero electrical resistance en.wikipedia.org/wiki/Superconductivity \$\endgroup\$ – Bruce Abbott Mar 7 '18 at 1:26
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    \$\begingroup\$ @BruceAbbott: I wouldn't make that assumption for the usual reason: doing so gives a model that doesn't approximate a case of interest, but the alternative does give a good approximation. (i.e. the real-world kind of short circuit that melts) \$\endgroup\$ – user90235 Mar 7 '18 at 1:29
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    \$\begingroup\$ A 'short' is presumed to have much lower resistance than the rest of the circuit, implying that there is some resistance which limits current to a defined value. Short circuit:- "A short circuit is an abnormal connection between two nodes of an electric circuit intended to be at different voltages. This results in an electric current limited only by the Thévenin equivalent resistance of the rest of the network" \$\endgroup\$ – Bruce Abbott Mar 7 '18 at 4:06

No resistance. Finite current. No voltage across. These are the assumptions for an ideal conductor. That makes the short circuit look like an ideal conductor. When doing benign [small signal] circuit analysis, the ideal conductor assumption is useful. When analyzing something less benign that can glow and melt, ideal conductor assumptions might no longer be useful.

Different kinds of assumptions for different kinds of problems.


If you assume ideal components in a circuit you will get contradictions - you can't have A because B.

An ideal voltage source has no internal resistance and will deliver a constant voltage regardless of current.

An ideal short circuit will have zero resistance, hence must have zero voltage across it regardless of current.

If you connect an ideal short circuit across an ideal voltage source, you have an impossible situation - both a fixed voltage (from the voltage source) and zero voltage (due to the ideal short circuit) between the same two points.

In the Real World, voltage sources do have some internal series resistance (for batteries) or limited current capacity (for power supplies), and any conductor will have some resistance, all of which will limit the maximum current that can flow, and the resulting voltage across the voltage source/short circuit.

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    \$\begingroup\$ What forces zero voltage regardless of current? I know the V=IR argument, but that only forces zero voltage in the case of finite current; it no longer allows that conclusion in the idealization of infinite current. \$\endgroup\$ – user90235 Mar 7 '18 at 0:42
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    \$\begingroup\$ What ordinary arithmetic calculation are you referring to? Plugging in R=0 and I=∞ does not imply V=0; every value of voltage is consistent with Ohm's law in this edge case. Do you have some other calculation in mind? \$\endgroup\$ – user90235 Mar 7 '18 at 0:59
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    \$\begingroup\$ While I'm sure this is a good answer to some question, I don't think this is an answer to this question. In particular, I feel like I've asked "Why would we make assumption X? It leads to bad result Y" and this answer says "Y." and when pressed, "because we assumed X." \$\endgroup\$ – user90235 Mar 7 '18 at 2:15
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    \$\begingroup\$ @NickAlexeev: No, "P=0" (as a consequence of "V=0") is the bad result Y I'm referring to -- by "bad result" I mean that the hypothesis leads to a model that doesn't at all resemble the real-world situation by the same name that I was considering. I think your comment that an ideal short-circuit isn't supposed to have any relation to a real world short-circuit is probably the best answer to my question. \$\endgroup\$ – user90235 Mar 7 '18 at 2:24
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    \$\begingroup\$ @Hurkyl You seem to be working from the assumption that since V/R=I, and 1/0=∞, we must have infinite current. However, this is not true. Division by zero is undefined. Moreover, infinity is not a number. It's a concept, and you cannot use it like numbers. This answer is correct: using idealized components, there is no meaningful answer. It's like asking, what angle should a unit vector have to reach the point (2,3). There's no solution as that point does not lie on the unit circle. \$\endgroup\$ – Sanchises Mar 7 '18 at 8:35

in an idealized short-circuit with zero resistance, one concludes the voltage is zero.

Don't forget about the inductivity of the shortcut. If you also idealize the inductivity you really have infinite currents.

but real-world resistances must be nonzero

Even this is not true: Superconductors have zero resistance but a nonzero inductivity.

And there are even electrical circuits in the real world where a nonzero voltage is applied to a "shortcut" (if you define "shortcut" as \$R=0\$): Superconducting magnetic energy storages

As long as a nonzero voltage is applied to the shortcut (the SMES' coil) the current is rising according to the formula \$\frac{di}{dt} = \frac u L\$.

As soon as no voltage (zero volts) is applied to the shortcut you have a constant current flowing in the SMES' coil. This current represents the energy stored.


The idealisation of a short circuit is not "finite current and zero voltage", the idealisation is "zero resistance". How much current is going to flow depends on the rest of the circuit. If calculations for the whole circuit show that infinite current will flow through the short circuit in that situation, it means that you cannot use the idealisation of the short circuit, and you need to use its real resistance.

  • \$\begingroup\$ If calculations for the whole circuit show that infinite current will flow through the short circuit in that situation, I would imagine a better response would be "This circuit will self-destruct" and consider the analysis complete (modulo error checking), unless I had a good prior reason to expect otherwise (or reason for it to be worth knowing exactly how it would self-destruct). But I'm not an EE, so my instincts may well be wrong. \$\endgroup\$ – user90235 Mar 9 '18 at 4:24

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