Sparky D wrote:Yes, in a parallel circuit, the discharge amps equals the discharge of the lowest pack.
My post was more to the thought of hooking up a 2500 mAh 8.4V pack with a 900 mAh 8.4V pack and expecting peak performance.
No on both counts. Available amps accumulate, and as a result you
can expect your motor to draw them if needed.
The amps and capacity are accumulative in parallel. Remember, mAh is just a capacity rating, a 9.6V 3000mAh battery should be able to supply 9.6V @ 1A for 3 hours. If you have a 9.6V 1000 mAh battery and a 9.6V 3000mAh battery in parallel, your new "pack" will be able to handle twice the amps if the application needs to draw it, regardless of capacity. So assuming one battery can provide 1A, 2 can provide 2A.
So for 60 minutes you could run this pack at 9.6V @ 2A. Once the small battery is empty, the bigger one can then continue with 1A @ 9.6V for another 120 minutes. These are approximate of course, resistances apply, also they never actually get to 0 during use.
The thing to remember though, the motor will only draw 2 amps if it needs it. If the motor were only pulling 1A, since the smaller battery can handle that alone (as in, if it weren't in a parallel circuit), the mAh ratings accumulate. If the motor were to draw 2A though, both batteries would be discharging full-bore simultaneously, so while the mAh ratings still accumulate, you end up not having gained any shooting time because 60 minutes @ 2A and 120 minutes @ 1A is 3 hours, which is what the 3000mAh pack can do alone at full discharge.
Remember these numbers are also completely made up and just used for reference.
Also, that's if you you're discharging at max amperage. Below twice the max amperage of the weakest battery, the cells will drain in relation to overall capacity. So, assuming one battery is adequate, adding another will increase capacity, not performance, but it doesn't decrease it either. In the hypothetical scenario that one battery weren't enough, adding another in parallel could increase performance by increasing amperage at the cost of trigger time (not capacity, that still increases). Although that's extremely theoretical because we're only at max amperage for very short periods of time.
http://www.youtube.com/watch?v=3TWk4SRoCQo#t=5m