# Couple questions regarding batteries

1. Jun 3, 2009

### DocZaius

Hello there!

Couple questions:

1. When a battery's terminals are connected by a copper wire, why is that called a circuit? Aren't the electrons moving, as a result of voltage difference, from the anion compartment of the battery to the cation compartment in a one-way direction? The fact they don't loop back around seems to disqualify the setup as a circuit.

2. Speaking of those electrons, wouldn't the voltage for a battery immediately drop the second the terminals are connected, and do so at a slower and slower rate until each compartment is equal in charge? Because of the highest charge difference at the beginning, the electrons would be pushed harder from the negative to positively charged areas. Then as time goes on and the potential difference is reduced (due to the electrons having brought their negative charge over to the other side) the total net force would be reduced and so on until there would be no charge difference.

The above is how I would imagine a battery to work, by pushing electrons from one side to the other due to the voltage difference. Yet, my model requires the voltage to be constantly decreasing from the moment the battery's terminals are connected, and for this rate of decrease to be at its highest at the beginning of the connection. The problem with this, is that I don't observe such behavior. Batteries seem to have a fairly constant voltage early in their lives. Could someone explain the problem with my description of a battery?

Thanks!

2. Jun 3, 2009

### Mapes

1. It's a circuit because ions are completing the loop inside the battery.

2. Charge is replenished at the electrodes by a chemical reaction.

3. Jun 3, 2009

### mikelepore

http://en.wikipedia.org/wiki/Alkaline_battery

In the chemical reaction written out on this page, notice one reaction putting out electrons and another reaction taking in electrons.

At the negative side, zinc hydroxide is converted into zinc oxide and water and a couple of free electrons. At the positive side, electrons taken in, plus something else, yields something else.

4. Jun 3, 2009

### chroot

Staff Emeritus
The battery does not have "compartments" of charge. Electrons are not piled up at one terminal.

If this were so, an absolutely enormous attractive force would exist between those compartments, and the battery would instantly implode.

Instead, electrons are distributed throughout the wire, the terminals, and the fluids inside the battery. All of the parts of the battery are electrically neutral.

The battery is just a "pump" which pushes electrons around the loop of wire connected to it. It is analogous to a loop of pipe filled with water, with a pump pushing the water around the loop.

- Warren

5. Jun 3, 2009

### DocZaius

So if I'm understanding this correctly:

Connecting one terminal to another in a battery allows the chemical reactions that were "paused," or rather brought to an equilibirum, in each compartment to "resume". One chemical reaction in a compartment was "paused" because the solution needed to get rid of electrons to resume and it now can, and the other chemical reaction in the other compartment was "paused" because the solution needed to capture electrons and it now can. Is that accurate?

6. Jun 3, 2009

### Mapes

This sounds good, but note that the chemical reaction doesn't pause. When no wires are attached, the opposite reaction runs at the same rate, and the battery is in equilibrium. It's two ways of looking at the system, and you expressed it both ways, but the first is really too simplistic for a technical discussion. (Think grade school / high school definition vs. college / grad school definition.)

7. Jun 3, 2009

### Bob S

The electrons reach the electrodes of the battery and are carried across the electrolyte as charges on ions.

The no load voltage of a battery is based on the electrode voltage of each half-cell in the battery. See
http://en.wikipedia.org/wiki/Standard_electrode_potential_(data_page [Broken])
Work out the battery potential (two half cells) for a lead-acid battery.

In the real situation, as the battery is discharged, the internal resistance R increases, and the IR drop is responsible for the measured voltage drop,

Last edited by a moderator: May 4, 2017
8. Jun 4, 2009

### dE_logics

1)
I'm not good at chemistry so I keep away form what's happening within the battery...just assume it as a capacitor.

Point is since charge can't 'appear' from nowhere, and there's no accumulation of charge within that circuit, what can be revolving in the whole of the circuit are the same electrons present within that wire.

They make a way through the battery, and into the wire, then opposite.

2)
Yeah, that actually happens in a real battery, its called a 'short'.

No the charge difference is maintained...that's why we call the battery a constant voltage source.

This is the exact problem, this does not happen.

9. Jun 4, 2009

### dE_logics

If I'm getting it right...yes you're right.

For the reaction to continue, the 2 terminals need to be connected.

10. Jun 6, 2009

### Naty1

On the other hand, when a load, such as a wire, is connected if the current of electrons is high enough (that is if the resistance wire is low enough) electrons are depleted from the battery solution and the voltage drops. In other words, when a battry puts out a lot of power the voltage drops almost immediatetly because the chemical reaction isn't fast enough to provide all the electrons required. In a diesel engine for example, the starter current might be well over 500 amps and for the brief second or two that is being provided to the starter the battery voltage will be substantially lower than when idle.

Note I am not saying that the electrons move through the wire at a high rate of speed, say anywhere near the speed of light, for example; that is NOT the case, in fact electron drift is what causes the current flow and that is rather slow...on the order of a few meters per second....and the ion movement in the battery is likewise quite slow..

11. Jun 6, 2009

### mikelepore

I believe that's why it's standard for police cars to have two batteries in parallel. If they turn off the engine and keeps all the lights flashing, radio turned on, etc. there's a danger that one battery wouldn't be able to maintain a constant 12 volts.

12. Jun 6, 2009

### mikelepore

I wouldn't recommend visualizing the battery as a capacitor. That picture might give you the right answer in just a few problems if the only thing a person is interested in is the currents and voltages at only one infinitely-short moment in time, like a snapshot of the conditions about a millionth of a second right after a switch has been thrown. But for general use, imagining the battery as capacitor will give all the wrong answers.

13. Jun 7, 2009

### dE_logics

Another explanation of the voltage drop can be by the maximum power theory.

A lead acid battery is known to give good surge voltages, that's why it's prefered (apart from the lower initial cost and very stable P.D).

Not practically, but just assume a very powerful capacitor, the only difference would be that the P.D will be completely defined by a mathematical function, where as in a battery, it depends on they type of battery, and very much dependent on the surrounds also (capacitor, I think, is not that much dependent on the surroundings.)

14. Jun 22, 2009

### jayeshb.work

hii..frnds,as we knw dt Li ion battery having lifecycle arround 1000 cycles charge nd discharge. along with as no of cycle increases its internal resistance also increases, so current decreases. do u have ne sort of solution for dis,so dat we can get a constant current arround 700 mA,as long as the battery works.ya..we need to add a additional circuit with it. which type of circuit we sd use....can u suggest me ne proper design .