Question regarding voltage and current in batteries

Hopefully I'm posting this in the correct area, I figured this being the "power" section this may be the best lace for this. Apologies in advance if this is a rookie question, I'm still new to the hardware side of things as my background is mainly on the software side.

First of all my understanding is this:- A battery has a voltage (e.g. 1.5v) and "capacity" (e.g. 2000mA/h). When you connect this in a circuit the battery will output 1.5v but the current is not fixed to 2A necessarily. The current in can pretty much be anything as it's determined by the voltage and resistance in the circuit (so it can be 500mA or 4A depending on what's in the circuit).

Assuming the 1.5V is adequate for the circuit components, if the current in the circuit is 500mA in this case, the battery will last 4 hours. If it's 2A it lasts 1 hour, 4A 30 mins and 20A 6 minutes.

First of all, can you confirm that understanding is correct or have I got this absolutely wrong?

As you use a battery, does the voltage remain constant but the capacity is drops as it is used, or will the voltage also drop over time?

For instance, if I have a 1.5v battery that is 2,000mA/h, if the circuit draws 1A and it is used for 30 mins, does that mean it will be 0.75V or will the voltage remain the same? Or does the current that can be drawn from the battery start to drop?

Thanks in advance for looking at my newbie post!

First of all, can you confirm that understanding is correct or have I got this absolutely wrong?

Not absolutely wrong but a bit naive. The current you can get out of the battery has nothing to do with the batteries capacity, it is a totally different thing. A 2000mA / hour source will not deliver that in a usable form you are better off halving that for any practical purpose. That capacity will quickly go down as you use the battery.

As you use a battery, does the voltage remain constant

No the voltage drops slightly until it is nearly exhausted then drops rapidly.

does that mean it will be 0.75V or will the voltage remain the same

No not any of that, the voltage drop with dropping capacity is different on different types of battery, any attempt to guess the remaining capacity based on the voltage is a bit hit and miss as you will know if you have ever used a laptop.

All batteries have some internal resistance so that the voltage will drop as current increases in a reversible way - this limits the useful current from the battery, and can cause excessive heating in large battery packs. As well as the voltage depending somewhat on state of charge this internal resistance will change with state of charge and temperature (each battery chemistry has its own peculiarities).

The capacity will depend on the discharge rate - high discharge rates can reduce the apparent capacity markedly - though what counts as a high discharge rate depends on the chemistry. Because of this battery capacities are usually quoted at a specific rate (often the "10 hour rate"). Capacity of rechargeables will decrease with age and use too.

All of this imperfection and variability means that if you require 2Ah capacity you'd better go and get a 3 or 4Ah battery, and check what discharge rate it can cope with.

MarkT: The capacity will depend on the discharge rate - high discharge rates can reduce the apparent capacity markedly - though what counts as a high discharge rate depends on the chemistry. Because of this battery capacities are usually quoted at a specific rate (often the "10 hour rate").

Since OP is describing 1.5V batteries I thought I'd point out that Duracell has datasheets that show the Ah ratings of their batteries given different discharge rates. What MarkT said is absolutely correct, but, y'know, it's helpful to see a picture too ;)

http://www.duracell.com/en-US/Global-Technical-Content-Library/Product-Data-Sheets.jspx?icn=AdLob/ProductDataSheets&cc=AdLob

Cheers guys, I think I have a bit of a better understanding of all this. When my new parts for my circuit arrives, I'll have some things to play around with to get a better understanding of all this. I appreciate you passing on some of your knowledge :)