Short circuit = no power?

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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.

Answer 5

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).

Answer 6

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.

^{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.

^{simulate this circuit}

Answer 7

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.