Originally posted by mglavin
Interesting, but. If the charger is rated for 200mA output @ 15V and the cell count is five or 7.2V static. Presuming the combined cells would consume 7.2V under charge is unrealistic, IMO. The math or formula is accurate, but the V rate at which the battery pack will be charged is unknown. Most likely is it will be much higher than 7.2V as the charger is capable and rated to deliver 15V. It knows not what's at the other end. A simple exercise with a DVM will reveal the actual V under load.
I am not convinced that the charger will charge at a higher mA output simply because the V is well below the output rating. There is no load other than the cells IR...
As Jazz mentioned math and theory are often worlds apart from reality.
I'm sorry but you've completely missed the point. If you connect a 6V battery to a charger rated for a maximum of 15V the output of the charger is dragged down to 6V. There's no way you can pull a 6V battery up to 15V. Batteries are voltage sources, they decide what voltage they take. You can't apply Ohm's Law to them as though they were resistors. In your terms the charger certainly does know what is at the other end and the output voltage is forced down by it. Your simple exercise with the DMM will clearly show this.
Because of this and the way the wall wart chargers are constructed at the lower voltage the output CURRENT of the charger will increase. A simple measurement with an ammeter on any wall wart will confirm this too.
Steve