Hi again,
Capt Pit Bull: The question clearly states ampere hour. So my reasoning remains correct.
Keith Williams: I strongly suggest that you revise a bit of basic theory of electrical circuits. You insist in including resistances in your arguments although there is no information whatsoever about that in the question. You can’t talk about a load until you know what load it is. Of course, a battery is of no use if you do not have a load but when you do, you know how much current is being drawn because you can measure it. In addition, in the capacity tests you mentioned, the load is known.
I agree with you when you say that the value of the internal resistance may be neglected for most of the cases. But I doubt that it is 24ohm as you suggested. Just think that 40 A are being drawn from a battery (due to a load equivalent to 0.3ohm) . If the internal resistance of the battery was about 20ohm then, a voltage of 20 x 40 = 800 V would develop across the internal resistance. As this is not possible, the output voltage of the battery would just drop to zero and the current would stop. Therefore, the battery could not meet the specifications! The internal resistance is very low and much less then 1 ohm! Ideally, it should be zero but unfortunately, that is not the case.
Cwatters: You may connect two batteries in parallel provided that both are serviceable. If there is a difference in the output voltage of the two batteries, then a current will flow from one to the other and charge it until both are at the same voltage. This is what happens when you charge a battery or jump-start a car with a dead battery. Also, an “alternator” charges a battery with a higher voltage. The process of charging a battery always involves the connection of a higher voltage source to the battery. “Intelligent” chargers may control electronically the amount of current delivered to extend battery life.
This is not a very difficult subject but it is important that we discuss it until everybody agrees!!