Electronic – getting more from batteries via series connection

amperagebatteriesparallelserieswatts

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.

Best Answer

Your calculation for the series case is incorrect. You forgot that since the batteries are in series, you can only use the 2 Ah of each cell. 100 W / 7.4 V = 13.5 A. 2 Ah / 13.5 A = 9 minutes, just like your answer for the parallel case. You can't cheat conservation of energy.

Also, calculating battery runtime to better than 1 part in 500 (one second out of 8 minutes 53 seconds) is absurd. Battery capacity is a rough number at some set of nominal conditions. Various factors, like temperature, drain rate (current), age, what you consider "empty", number of charge/recharge cycles for rechargables, and part to part variation change the total charge you will get from the battery. Take a look at a battery datasheet. Batteries are complicated.