# getting more from batteries via series connection

Let's assume I have a 100W load.

If I run this load on two 2000mAh 3.7V batteries in parallel then I should get a continuous run time of 8 min 53 seconds (given the 27 amp draw).

However if I were to take the load and modify its resistance so it would draw 100W on 7.4V then I will have a draw of 13.5 amps from the cells. This should give a continuous run time of 17 mins 46 seconds.

All this is considering the way voltage has a larger effect on wattage than current.

Are my calculations correct in this sense? (I'm aware a 100W load is rather a lot for this voltage, but its just a rounded number for an example).

Also assume the second load of higher resistance performs the same as the first, lower resistance, load e.g. light/heat output.

• The energy contained in the batteries is still the same mate.. can't get something out of nothing. Your 100W load will pull the Joules out of the battery no matter what configuration you have it. Usually, you get less losses and current related issues using higher voltages though. Commented Dec 11, 2014 at 14:18
• That's what I was thinking, which made me wonder why battery capacity is described in mAh rather than Joules, surely a Joule rating would be more accurate? Commented Dec 11, 2014 at 14:22
• mAh is a more meaningful number at first glance that Joules. Commented Dec 11, 2014 at 14:42