View Full Version : help on mah's
sratzo
10-08-2006, 10:53 PM
OK - beaar with me on this. new to electrics. Iwould like to rebuild a 9Z TX pack. While i know that 8 cells will equal the 9.6 volts. How is the mah added ? or does it stay the same for all matched cells? ex: 2700 aa cell X 8 = 9.6 volts and the 2700 mah is the amount of draw per ah ? does that sound right? Thanks for the help.
8 2700 cells in series (+ of 1 to - of the next) will result in a 9.6v 2700mah pack.
Only thing that doesn't make sense about what you wrote is "amount of draw per ah".
mah is "milliamp hours" - milliamps * hours.
If you draw 2.7 amps (2700ma) it will theoretically run for about 1 hour. - 2700ma * 1h = 2700mah.
Reduce the draw to 1350ma and it should last 2 hours.
If your Tx draws 700ma you should get 3 safe hours (~2100) before the voltage really starts to drop to unsafe levels.
T
Pinecone
10-09-2006, 05:48 PM
Yes, you are putting the cells in series, so the mAH capacity is the same as for the indivdual cells.
sratzo
10-09-2006, 07:20 PM
thanks guys. i found some good links to help out. i was confused on if and how the mah is increased. and that is in a paralle hookup but then you loose voltage. i think thats how it goes. i will keep reading. :arggg: