# Battery question

Discussion in 'General Electronics Chat' started by CanElec, Jan 2, 2009.

1. ### CanElec Thread Starter Member

Nov 23, 2008
24
0
I was just wondering why it isn't possible for current to flow from the -ve of one battery to the +ve of another? And why when you measure the voltage between the two, you get zero? The -ve of one battery should still be attracted to the +ve of another I would think.

Even if no current can flow between the two due to the chemical reaction process that takes place(though I'm not sure why this reaction still couldn't take place), I would think you could still at least measure the potential difference between the two batteries. I mean voltage is still voltage right? The 0v neg. of battery A should still be attracted to the 9v pos. of battery B.

Obviously I'm wrong about this because I just tried to measure with my multimeter and ended up with 0v between the two 9v batteries (voltage between neg. term. of one and pos. term. of the other), but I can't figure out why.

2. ### Wendy Moderator

Mar 24, 2008
20,735
2,498
Try drawing a schematic, it will make it more obvious. Current only exists if there is a path, and voltage requires a connection between two batteries, no connection and they are isolated from each other.

3. ### italo New Member

Nov 20, 2005
205
1
Batteries are current sources tie two together and one will equalize the other and viceversa. If one is 8v the other is 9v by themselves then the voltage will be 8v and the 9v will supply all the current to make it so.
Batteries should never be tie together for this exact reasons.One will drain the other. To share the load a diode or a small resistor should be put in series to insure load equalization.

4. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
513
Most decidedly not

Current sources have (theoretically) infinite internal resistance and maintain the fixed current in an external circuit, regardless of that extenal circuit's resistance, adjusting their own voltage to cause this to happen.

Voltages sources have (theoretically) zero internal resistance and maintain a fixed voltage across an external circuit, regardless of that extenal circuit's resistance, adjusting the current to cause this to happen.

Batteries are voltage sources.

5. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
513
It's not to do with attraction.

It's that you haven't made a circuit.

A circuit is a complete path starting at one terminal of a battery and passing through a collection of components, external to the battery, to the other terminal.

If one of these other components is a second battery current (electrons) will indeed flow from the -ve terminal of one battery to the +ve terminal of the other.

But you must have a circuit for this to happen.

Remember also that conventionally we think of current as passing from +ve to negative. this is called 'conventional current flow'.

6. ### CanElec Thread Starter Member

Nov 23, 2008
24
0

I have attached an image. So if I have a circuit, then current can flow from the neg of one battery to the pos of another?

In the case of (A), current would flow through the load?

In the case of (B), current could flow from the negative terminal of (1) to the positive terminal of (2), going only through 2's load?

• ###### Common2.JPG
File size:
10.8 KB
Views:
29
Last edited: Jan 2, 2009
7. ### CanElec Thread Starter Member

Nov 23, 2008
24
0
I am still confused as to why I cannot measure a voltage across the negative terminal of one battery and the positive terminal of another battery.

As far as I know, you don't need any current to flow to measure a voltage. So when you measure 9 volts across a (single) battery, no current is flowing, thus no chemical reaction is happening inside the battery.

If there is no change inside the battery, then why can't I read a voltage when I connect one probe to the negative terminal of one battery and the other probe to the positive terminal of another battery.

When I touch one probe of my meter to the negative terminal of a battery, all of the electrons start pushing against the probe. Now when I connect the other probe to the positive terminal (of the same battery), it has an electron deficiency, so it is attracting electrons. Yet no current is flowing, so it is simply this push or pull that is being measured by the meter. So why can't it detect this push or pull (now between two different batteries) between the negative terminal of one battery and the positive terminal of another?

8. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
513

A is not a circuit.

B is two separate circuits. Each battery / resistor comprises one circuit. there is also a link from one circuit to the other but this is not a complete circuit.

9. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
513
Yes, you can have a voltage, without any current flowing. You can also measure this voltage.

That is exactly what an isolated battery is. No current flows, because there is no circuit.

Connect one terminal of this battery to one terminal of another battery with a wire link.

You still have no circuit

You can now measure the voltage from the link to the free terminal on either battery. Or you can measure the combined voltage across both free terminals.

You cannot do this without the link

Connect the two free terminals of the joint battery together through a load resistor.

You now have a circuit and current flows.

10. ### syed abdullah New Member

Nov 26, 2008
4
0
I need some details about deep cycle batteries as assignment

11. ### khusmann Member

Jan 2, 2009
21
0
CanElec, I understand where you're coming from - I've had the same question myself, and was never satisfied by the "it isn't a complete circuit" argument. If you connect something more positive to something more negative, current should flow - its a circuit! Looking at your diagrams with this in mind, one would expect current to flow from the positive side of one battery, through the resistor and then into the negative side of the other battery. But, as you've found, it doesn't. I've never seen a satisfying answer for this in a textbook or heard one from anyone else officially, but I've come up with a way to rationalize the behavior to myself, which may be correct for all I know.

Here's what I figure: Batteries are constant voltage sources, as was mentioned earlier. This means that they try to keep a constant voltage relative to the two terminals. So they always try to make the plus terminal always 1.5V (or whatever) higher than the minus terminal.

So, if you two batteries sitting on the table, all you know is that the voltages between the two terminals on a given battery is 1.5V (unless they're dead ). You do NOT know the voltage difference between two terminals of two different batteries. Let me give you an example - say a terminal of one of the batteries has a TON of extra electrons somehow, while a terminal of the other battery has very few extra electrons. In this case, there is a large voltage difference between those two terminals. If you connected these two terminals with a wire, current would flow to equally distribute the charge on those two terminals, making them have the same voltage. (and once they're at the same voltage, current would stop flowing).

The important part of this is that when you connect the negative of one battery to the positive of another battery, they become the same voltage. So, no current will flow except the small amount of current for a quick moment (basically instantanious) to equalize the voltage at the terminals. Even try to measure the voltage difference between two terminals on different batteries using a multimeter, the terminals equalize to the same voltage because your multimeter actually allows a small amount of current to go through it (it has an internal resistance). If you had a magic multimeter that actually didn't draw current, the value you would measure would depend on the difference in charge of the two terminals (Not 1.5V!).

So say you connect the positive of one terminal to the negative of a different battery's terminal. So now, the negative terminal and positive terminals you connected together are at the same voltage, and the remaining negative and positive terminals are at +1.5V and -1.5V relative to this common voltage. (Because remember the batteries try to maintain that constant voltage difference) This you can actually measure with your multimeter. And if you measure the voltage between the two remaining neg and pos terminals, you should get 3V as expected.

Your problem is that you are seeing the positive and negative terminals of batteries as absolute voltages. But that's the problem - there's no such thing as an absolute voltage. Voltage is always measured relative to something else.

Anyway, there's my two bits and I hope it helps.

12. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
513
Not quite a circuit.

Something cannot be 'more positive or negative without reference to a third point. Both items have to be connected to this third point to say they are more positive (negative).

And whadda ya know if you now connect these two points together you will have a circuit.

Then current will flow.

13. ### khusmann Member

Jan 2, 2009
21
0
Yeah, technically I agree, but I can understand how CanElect feels like it is a circuit. Sorry, bad wording on my part.

Don't you only need two points to have a potential difference? For example, if I rub my shoes on the carpet, I collect a bunch of extra electrons. The potential difference between me and a doorknob (that lets say has a close to equal number of electrons and protons) is now very large - I can say that I am a lot more negative than the doorknob, independent of any third point.

14. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
513
No you need a third point.

If you connect any two points together, with no other connection,
they will be at the same potential.

This is true whether you connect through a resistor or a plain wire.

Each battery terminal is only positive (negative) with respect to the other battery terminal on the same battery.

Its potential (voltage) is indeterminate with respect to any other point in the universe, including the terminal of another battery.

If you connect it to the terminal of another battery it will have tha same potential as that terminal, as I outlined above.

As a matter of interest we usually call the third point earth or ground and refer everything in the circuit to it.

Another way of putting this is to realise

All voltages are voltage differences between two points.

So we connect together all the second points and call the result earth or ground.

Last edited: Jan 3, 2009
15. ### khusmann Member

Jan 2, 2009
21
0
I agree. All I'm arguing is that this is the steady state response.

Say you have 2 bodies in space - one with many more electrons than protons (net negative charge) and another with equal electrons and protons (net neutral charge). You would agree that if you connect these 2 bodies, current flows (a small amount of current, for a split second). After the current flows, they are now at the same voltage. Before the current flowed, they were at different voltages right? Or am I missing something here?

16. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
513
Yes your example is true, but batteries have equal numbers of +ve and negative charges.

Since you cannot have excess charge at a point, let us give these bodies real dimensions.

In your example, if for instance they are spheres, the one with excess negative charge has these distributed evenly over its surface, by mutual repulsion.

Now connect the charged and neutral shpheres by a sturdy wire.

As you say the excess charge now redistributes itself over the surface of the newly formed dumbell shape.

You now have a single body with excess negative charge.

So ?

17. ### khusmann Member

Jan 2, 2009
21
0
So before you connected the two bodies, they were at different voltages right? 3rd reference point not needed.

18. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
513
How were they at different voltages?

With respect to what?

What is the voltage of a single isolated body in space, with or without excess negative charges?

There is no answer as the question is meaningless.

Yes both are at different voltages, with respect to a third refernce point, which may be inside one of the bodies, but still exists.

19. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
513

Take a 9volt battery.
Connect the negative terminal with croc clips and a flying lead to the +60 rail of my power amplifier.

What is now the 'voltage' of the battery negative terminal?
What is now the voltage of its positive terminal?

Have I 'charged up the negative terminal' to this new voltage?
In other words has any current passed from the amplifier power rail to the battery?

If so what would happen if I disconnect the croc clip from the power rail ( but not from the battery) and touched it to the negative power rail? would I see sparks? (I would certainly see sparks if I touched a lead connected to the positive rail to the negative one.)

If I don't see sparks, what happened to the charge, if any, that passed to the battery when I connected it to the positive rail?

20. ### khusmann Member

Jan 2, 2009
21
0
Great, I think we're closing in on my misunderstanding here - thanks for being patient with me!

Here's what I was thinking - to find the voltage difference between the two charged bodies, integrate the electric field on the shortest path from one body to the other body. Is this incorrect thinking?