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Now I want to build stuff and I'm really interested in learning things (consider I'm starting from scratch).
So I'm reading all of this website and the following line in this article got me scratching my head for some time:

[about the power rating of a circuit]
Likewise, if we have a short-circuit condition, current flow is present but there is no voltage V = 0, therefore 0 x I = 0 so again the power dissipated within the circuit is 0.

I'm quite sure that you can melt stuff when connecting it to both ends of a battery. Not that I tried it myself but even touching both ends of an AAA battery with a metal wire produces sparkles and heat. Is it really correct that there is no power dissipated within the circuit in a short-circuit condition?

Also, I remember that there couldn't be an electron flow in a circuit if there was no voltage drop between both ends of the circuit. Then, isn't the line I quoted kind of contradictory?

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    \$\begingroup\$ A short circuit dissipates zero power but a 1 milli ohm wire connected across a 12V car battery isn't a short circuit and it will glow and melt. \$\endgroup\$ – Andy aka Mar 6 '18 at 18:00
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    \$\begingroup\$ That's a rather misleading statement you quote. No power will be dissipated in the short itself but it certainly will be dissipated in the internal resistance of the battery, which is part of the circuit. \$\endgroup\$ – Finbarr Mar 6 '18 at 18:08
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    \$\begingroup\$ unless it is a "low temp superconductor" everything has resistance .. everything.. batteries, caps , inductors, transistors . If not a conductor, it is a dielectric which also has resistance in series and parallel \$\endgroup\$ – Sunnyskyguy EE75 Mar 6 '18 at 18:18
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    \$\begingroup\$ @finbarr and in the wires. House wiring may be 12AWG but the 20A breakers are rated to interrupt 10kA. \$\endgroup\$ – Harper Mar 7 '18 at 7:50
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    \$\begingroup\$ @Alexander according to Wikipedia superconductors can maintain current without voltage so I would assume no power dissipation for this case as it would. For applied voltage it looks like superconductor just stops having 0 resistance: 'there is another important concept in superconductivity: that of a critical current. This is the largest current (density) that a particular superconductor can carry without becoming resistive.'. \$\endgroup\$ – Maciej Piechotka Mar 8 '18 at 8:58
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You should not be so hard on your professor.

Much of the confusion newcomers to EE struggle with is that we talk about theoretical IDEAL circuits as part of the teaching process. In ideal circuits things often act rather contrary to your intuitive and experimental notions of how things actually work.

Things like short circuits, transformers, diodes, and pretty much everything else we work with, have ideal models we use to describe and understand them within the scope of how we try to use them. The reality is far more complicated and much harder, if not impossible, to define entirely.

As such the definition of a "short circuit" is in fact an "ideal component". It is a resistance with zero resistance, that is \$0\Omega\$. That is, the force of the battery will act through it with no opposing force. Pushing on nothing, you do no work, and no power is dissipated.

In real life of course, the wire you use to short out the battery has some small resistance. The battery itself also has some internal resistance. Since both of those are small, the resultant current is very large. That means lots of power is dissipated in the wire, and in the battery and things quickly get rather warm.

As I said, do not be so hard on your professor. A lot of EE is accepting the ideals at face value while realizing that reality is rather different. The ideal models give us a base point to work from which allow us to design things to a working level of accuracy without getting lost in the chaos of real world effects.

However, we always have to be mindful that the ideals are a myth.

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    \$\begingroup\$ Good answer. From the chemistry side, what is happening is that you are basically letting the chemicals at the positive and the negative electrode react with each other freely. Since these chemicals are selected for their great reaction energy (among other things) letting this run amok will quickly generate heat and destroy the battery and possibly the surrounding container, items, people, buildings... the severity of this only depends on the size of the batteries. \$\endgroup\$ – Stian Yttervik Mar 7 '18 at 5:52
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    \$\begingroup\$ At half of practical engineering is knowing when the simple models of components break down, and when you are in the (95+%) of cases where they are good enough to be useful. For most things, much of the time, a practical resistor can be modelled as a theoretical one and the results will be about the same, sometimes however the stray capacitance, series inductance, tempco, tolerance, voltage breakdown or some other such thing starts to really matter, the art is knowing when the physics of the real thing is likely to stuff you. \$\endgroup\$ – Dan Mills Mar 7 '18 at 12:25
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    \$\begingroup\$ I actually wish courses would take a moment around this point to talk about real hardware. This is the perfect place to point out that our models don't capture all of the effects. And while you may feel foolish for finding this confusing, I can say quite assertively that you are not alone. If you look at this site long enough, you can found countless new EEs which are confused by the behaviors of ideal components in pathological constructions (like those involving shorts). \$\endgroup\$ – Cort Ammon Mar 7 '18 at 16:15
  • \$\begingroup\$ I feel like the professor's claim is "more wrong" than what you're saying here. If we're going to allow idealizations, I is infinite and 0*I is not 0 but an indeterminate form. \$\endgroup\$ – R.. Mar 8 '18 at 0:09
  • \$\begingroup\$ @R.. yes well.. it's a mistake to overthink the theory. It will drive you mad. This example is bad enough, transformer theory will definitely drive you over the edge. \$\endgroup\$ – Trevor_G Mar 8 '18 at 13:23
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even touching both ends of an AAA battery with a metal wire produces sparkles and heat

To analyze this circuit, you have to consider both the internal resistance of the battery and the actual resistance of the wire.

Since a real wire has non-zero resistance, some power will indeed be delivered to the wire and turned into heat.

But also, since a real battery has internal resistance, some power will be converted to heat inside the battery where it doesn't do any good and may damage the battery.

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The statement (from the website) is correct only in a purely theoretical sense, as there really is no such thing as a 0 ohm short. All wires have some resistance, and a battery itself has internal resistance. Your professor was indeed correct - if there is current flow, then there is a voltage drop, though it may be very small.

In fact, one way of measuring current in a circuit is to place a small calibrated resistance (called a shunt resistor) of typically 0.01 ohm in series with the load, and measuring the voltage drop (usually in millivolts) of the shunt.

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A zero voltage with a short circuit is only true if there is zero resistance. That is a theoretical statement.

In reality (at least for us at room temperature) there will always be some resistance and thus a short circuit will have some voltage and thus power.

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  • \$\begingroup\$ Do you mean that when there would theoretically be zero resistance, electric potentials of both ends in the circuit would instantly "cancel each other out" ? \$\endgroup\$ – qreon Mar 6 '18 at 18:03
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    \$\begingroup\$ Zero resistance is when you have super-conductivity. I have no knowledge in that area. (You probably need to get to the physics stack exchange :-) \$\endgroup\$ – Oldfart Mar 6 '18 at 18:06
  • \$\begingroup\$ @qreon: If you can "instantly" reduce the resistance between both ends of the circuit, then yes, "the electric potentials of both ends in the circuit would instantly cancel each other out"! \$\endgroup\$ – Guill Mar 9 '18 at 7:11
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Consider the ideal circuit (a) below. There is a 2 A current flowing through the circuit. It goes from A to B, through the resistor to C, then back to D, and through the voltage source to A, completing the circuit.

Now, what is the voltage drop in the A-B wire, and how much power is dissipated there? That's an ideal wire, so its resistance is zero, and therefore the voltage drop and power are also zero. Regardless of the fact that there's a 2 A current flowing through it. An ideal wire is a short-circuit, and here's one that dissipates no power, just like your teacher said.

schematic

simulate this circuit – Schematic created using CircuitLab

What about circuit (b) then? We can calculate the current as \$ V_2\ /\ R_2 \$, except that \$ 10\ \mathrm{V}\ /\ 0\ \Omega \$ is a division by zero, and we can't calculate that. But if we assume \$ R_2\$ has some positive value and see what happens when it gets smaller and smaller we'll quickly see the current approaching infinity. (We could write that as \$ \lim_{R_2\to0} {V_2\over R_2} = \infty \$ if we wanted to play mathematicians.)

Obviously, there's also a voltage drop, since the net voltage around the circuit must be zero. A non-zero voltage times infinite current gives infinite power. This is different from (a), since here, the whole voltage source was short-circuited.

schematic

simulate this circuit

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When a number of resistive elements are connected in series and driven by a voltage source, the total amount of power will be inversely proportional to the total resistance (to be precise, it's voltage squared divided by resistance), but the fraction of power received by each individual resistive element will be proportional to its resistance.

If one has a wire with a resistance of 1 ohms connected in series with a light bulb whose resistance is 99 ohms, and that combination is driven with a 100 volt source, then the total power will be 100 volts squared, divided by the 100 ohms total resistance, i.e. 100 watts. Of that power, 99% would be dissipated in the light bulb and 1% would be dissipated in the wire.

If the resistance of the light bulb were to fall to 0.001 ohms, then the total power dissipated would be 100 volts squared divided by the 1.001 ohms total resistance, i.e. 9,9990 watts. Of that power, about 0.1% (10 watts) would be dissipated in the shorted light bulb and 99.9% (9980 watts) in the wire. Note that the maximum power dissipation in the light bulb would occur if its resistance were equal to that of the wire. In that case, 5,000 watts would be divided equally between the wire and the bulb (each receiving 2,500 watts).

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This seems to stem from an assumption that, even in the idealization, the current through the circuit is still finite, and thus V=IR implies V=0.

A more reasonable model a real-world short would be that voltage remains nonzero; in the ideal case of zero resistance you'd therefore have infinite current. The power P=IV would likewise be infinite.

Your question made me curious, so I posted my own. The comment tere by Nick Alexeev, I think, basically answers your question — the model of the short-circuit you are reading about is meant for modeling more benign circuits, not ones that melt.

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  • \$\begingroup\$ "The power P=IV would likewise be infinite." -- if current is infinite, then the voltage would be infinitessimal; infinity * an infinitessimal is an undefined result: it could be anything, but infinite power isn't really a realistic result. \$\endgroup\$ – Jules Mar 7 '18 at 9:25
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    \$\begingroup\$ @Jules: Infinite power is a much better approximation to the phenomena the OP observes than zero power, namely a rapid discharge of energy. \$\endgroup\$ – Hurkyl Mar 7 '18 at 9:37
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    \$\begingroup\$ If we are using ideal components, the battery provides exactly 12V. So current is 12V/0Ω = ± ∞ Amps, and likewise power is ±∞ watts. This is all a bit silly, particularly the ± part. Calculus uses limits to avoid infinitesimals. If we define a short circuit as one where resistance approaches 0, power also increases without limit (until components cease to resemble ideal ones). \$\endgroup\$ – gmatht Mar 7 '18 at 17:21

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