Posted 08 June 2007 - 09:53 PM
Hmm, Ive thought about this a bit more, my Electrical knowledge is a bit rusty, but i will do my best and try and keep this simple. It really is a lot more complex, you have to take into account the internal resistances of the bateries, the resistance of the load, weather you would wore the batteries up in series or parallel.
Any Electronic grads that can correct me or explain it simpiler please do.
The 2500mAh is its Capacity, or maximum drain. So if you had a 1.5V 2.5A light bulb it would work for 1 hour, or a 1.5V 1.25A bulb would work for 2 hours.
Watts = Amps x Volts
So your single AA battery (assuming 1.5 V)
1.5V x 2500mA = 3750 mW
This means at maximum drain it could supply 3750mW for 1 hour.
When connecting two batteries in Series you are doubling the voltage while maintaining the same capacity rating (amp hours). ie 3V at 2500mAh
When connecting two batteries in Parallel you are doubling the capacity (amp hours) of the battery while maintaining the voltage of one of the individual batteries. ie 1.5V at 5000mAh.
Next you need to know how many Volts your 1KWh device will be running at.
As an example I will say 12V. So to supply 12 V you will need to have.
12/1.5 = 8
So you will need 8 batteries in series, however these 8 batteries still only give you 12V at 2500mAh, but
12 x 2500 = 30Wh
But you are now 970Wh short.
To find out how many banks of batteries you need, devide the total by the amount per bank. 1Kw =1000W
1000/30 = 33.3333
So you would need 34 banks of 12. 33 would leave you slightly underpowered.
So in total you would need 12 x 34 = 408 batteries
Hmmm, this doesnt seem like a lot, maybe i messed up with the maths.
Anyways, this is very basic as the are still lots of other things to take into account, and is only a rough estimate.
I hope it gives you some insight into the complexities of electrical theory.