Source with short circuit, 0 potential difference?

[The_Lobster]

In an example in my book (Uni. physics), there's a circuit with emf = 12V, internal resistance, r = 2 Ohm, and no external resistance, R = 0, short circuit.

What I don't get, is that it says that across the terminals, the potential difference between these points must be zero since there is no resistance in the outer loop. (And I can see from V = IR that this _must_ be the case). But the current now will be I = emf / r = 6 A, so there is a current, and the emf does work to move charges from negative terminal to the positive terminal? So how can there still be no potential difference? Isn't this contradictory?...

Edit: Can't this be rewritten as an ideal source with external resistance?

 

[vk6kro]

No, it is OK.

A battery is assumed to be a perfect EMF with a resistor in series with it. This is a bit of fiction as there is no actual resistor and it may be the result of the battery chemistry or connecting wires, but it is convenient to think of it as a resistor.

So, even if you short-circuit the terminals of the battery, there will only be a current flowing that is limited by the internal resistance and driven by the EMF.

If you have an external resistor, then the current is limited by both this resistor and the internal resistance, in series.

If the battery is open circuited, you could measure just the EMF at the terminals, because there is no drop across the internal resistance due to there being no current flowing.

 

[The_Lobster]

Thanks for your reply!

I still don't quite get why we say there is no potential difference across the terminals if we short the circuit. How would this relate to the water fountain analogy? If there is nothing limiting the water coming down from the top of the fountain, it still has potential when it is pumped to the top?

 

[vk6kro]

I don't like water analogies, because they always cause confusion.

I'll try, though.

The shorted battery might be like a pump with a small outlet which limits the water flow. If you connect the outlet of the pump to the inlet, the small outlet will still limit the total water flow.
The small outlet is like the internal resistance.

Electrical circuits are always closed circuits while water analogies are not usually closed. Garden hoses and fountains etc are not comparable with electrical circuits, because they are not closed circuits.

 

[schip666!]

If current is flowing in some conductor with exactly zero resistance you will not be able to measure a voltage difference. But in practical terms there _has_ to be a tiny resistance in whatever is used to short the battery terminals, so you should see some _small_ voltage across the short. You may not actually be able to measure it, but there it is.

The battery internal resistance is a way to describe the current supply limit of the battery. If it had absolutely ZERO internal resistance it would have to be able to supply infinite current. If the battery has a very small resistance then you might actually see a voltage across the terminals even with a dead short -- at least until the shorting object's resistance caused it to vaporize.

The water analogy works fine for me if you include evaporation, rain, and all the apparatus of hydraulic delivery systems in the circuit. Physicists hate it because "Electrons don't work that way." But most physicists don't build and fix electrical equipment for a living...