I am testing a Omni Powe 12V 260Ah battery.  C100 = 260AH and C10=200AH

 

On a inverter using avg 375W for an hour an 25 minutes before the inverter cut of on 11V.   What is the condition of the battery and how long should it last?

375W at 12V? 375W on the output? Lets say around 90% efficiency, so 375/12/0.9=35A. Nice round number.

That's more than the C10 rate, so you will not get the full 200Ah. Assuming a Peukert factor of 1.1. I can work it out from the 260Ah and 200Ah times, but I'm too lazy to look up the formula and the book where I know it is explained (BMV manual) is at home. Anyway, plugging into an online calculator, says that will reduce the capacity to 140Ah and you should get around 4 hours out of it.

So sounds to me like that battery is gone. If you can only do 35 amps for about half an hour, then by simplistic math you have about 10% of the original capacity left.

What say you other gentlemen?

Edit: Oh hang on! An HOUR and 25 minutes! Okay that changes things a bit. Sounds like it has about 40% of the original capacity left then. Not completely gone, but definitely a sign that it is not well.

Edit 2: Yup. By trial and error on the linked calculator, it looks like you have around 60Ah left. Or 30%.

From the same battery big bank, I tested another battery the same way.

The first round I got 52 minutes out of it and after discharging it slowly (with 2 car headlamps)  with again a last monetary fall to below 8V (hard to check the last part, but will build a low voltage cutoff switch at 9.5V)  and recharging it to full capacity, I got 2 hours 45 minutes out off it.

The difference between the discharge of this and previous (where I only got a 1h 30 min after recycling) discharges, is that I only used the"dimms", so slower discharge. 

I think I will redo it, but with a cutoff switch doing the monitoring.

http://www.instructables.com/id/9v-battery-status-indicator-circuit/

(Using 1K below green Led instead of 330R works.  (I have tested it. Previous circuit doe not work.)