# Why does current still flow with no voltage?

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## Main Question or Discussion Point

I need this clarified. I've been told that if you have a circuit with a load resistance of 0, but you have a power supply of internal resistance, when you have no voltage coming from the power pack you will still get a current. I don't understand how this works. I've been told its because the internal resistance creates a potential difference but I do not understand how it does this.

Could anyone help me out? :)

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You are talking of a situation such as a shorted out battery.
If the wire (or whatever you use to make the short circuit), has zero resistance, then placing a volt meter across the battery terminals will show no voltage, although a current flows through the wire.
(It's not actually possible for the short circuit to have a resistance of zero though unless it's a superconductor.)
It practice it will have some small amount of resistance and it will get hot as a result of the current, and a very small voltage should be detectable despite the short circuit.

You are talking of a situation such as a shorted out battery.
If the wire (or whatever you use to make the short circuit), has zero resistance, then placing a volt meter across the battery terminals will show no voltage, although a current flows through the wire.
(It's not actually possible for the short circuit to have a resistance of zero though unless it's a superconductor.)
It practice it will have some small amount of resistance and it will get hot as a result of the current, and a very small voltage should be detectable despite the short circuit.
So since there is short circuit, there is excessive current flow with no voltage, and this is caused by no load resistance? Ok, well, doesn't the internal resistance contradict this?

berkeman
Mentor
So since there is short circuit, there is excessive current flow with no voltage, and this is caused by no load resistance? Ok, well, doesn't the internal resistance contradict this?
The point is that in the real world, all connection wires do have a (very small) resistance. V=IR is true for all connecting wires. So the current in the shorting wire is determined by the total of the internal resistance of the battery and the (much smaller) resistance of the shorting wire. You will measure a non-zero (but very small) voltage across the shoring wire. It may take a very accurate DVM to measure the small value, but it is there.

The point is that in the real world, all connection wires do have a (very small) resistance. V=IR is true for all connecting wires. So the current in the shorting wire is determined by the total of the internal resistance of the battery and the (much smaller) resistance of the shorting wire. You will measure a non-zero (but very small) voltage across the shoring wire. It may take a very accurate DVM to measure the small value, but it is there.
Okay so, the current in the wire would be I = V/R, where V is that very small voltage and R is the internal resistance + the very small resistance in the wire? So current still flows because the electrons still want to get to the positive charge and the zero-load resistance allows them to do so?

berkeman
Mentor
Okay so, the current in the wire would be I = V/R, where V is that very small voltage and R is the internal resistance + the very small resistance in the wire? So current still flows because the electrons still want to get to the positive charge and the zero-load resistance allows them to do so?
Um, not quite right. The current is I = V/R, where V is the source EMF of the battery or power supply, and R is the sum of the internal and external resistances. The V you measure across the wire is the small voltage that you get with the overall current I flowing through the small resistance of the shorting wire.

Um, not quite right. The current is I = V/R, where V is the source EMF of the battery or power supply, and R is the sum of the internal and external resistances. The V you measure across the wire is the small voltage that you get with the overall current I flowing through the small resistance of the shorting wire.
I don't understand then. This is when the power pack is set to 0V, so I guess essentially turned off, so why would you put the EMF of the battery into the equation if its not on?

If you send a current through a zero resistance conductor, there will be no voltage drop.
That is what V=IR says.

I don't understand then. This is when the power pack is set to 0V, so I guess essentially turned off, so why would you put the EMF of the battery into the equation if its not on?
You can not turn a battery off.
In order to get zero voltage across its terminals, you either short it with a zero resistance wire or you discharge it completely.

berkeman
Mentor
If you send a current through a zero resistance conductor,
How do you "send" a current?

There's still a voltage, it just doesn't register on the meter - it's a measurement anomaly.

Think of it in terms of Isaac Asimov's water analogy. Voltage is the pressure of the water, which is supplied by the source. Current is the amount of water being moved. Resistance is the size of the pipe. If pressure exists between the input and output of the source, water will flow, so you'll measure a current as long as the connection exists. To actually have no flow you need to turn off the source - i.e. actually break the connection to the battery, turn off the power supply's switch, etc. Then there's no pressure on the water in the pipe, If there's pressure across the pipe, water will flow.

When a short circuit exists it's the same as an infinitely (in theory) large water pipe, so your meter, which has limited sensitivity, doesn't detect any pressure difference. However since there's no such thing as an infinitely large pipe there's still flow, but your meter can't detect it because the pressure difference across the pipe is close enough to the internal pressure difference of the source that the meter, which detects a pressure differential between the two, can't see it. But there's still a lot of water moving.