Let´s say I have a 2 AA 1,5V 1000mAh Alkaline battery cells in series.
The circuit it´s connected to consumes 500mA, works at 3V and it´s powered on always, and taking into account other electrical specs as ideal, the time it would take to last would be:
1000mAh / 500mA = 2h
Is this correct?
I´ve checked this question and it confuses me.
1000mAh / 500mA = 2hIdeally yes, however the voltage will change (rapidly) as the battery capacity decreases, making current consumption go up (if there is a regulator there demanding 3V). Even if you don't have a regulator, you won't be able to drain the battery all the way to 0, your actual runtime will be closer to 1 hour. Check this other question for details. – Ron Beyer May 12 '20 at 13:49