# Help me understand runtimes based on battery capacity

#### thermal guy

##### Flashaholic
If the current draw is 2000 ma at 7.4v

Then how is the run time only 1 hr 15 minutes minutes with two 3400 mah 18650's.

This is an example that I have taken from a site because it falls inline with my question

People keep telling me it's just simply math Divide draw by battery capacity but there's something I'm missing because that would put runtime at over 3 hours.HELP! and please go slow 😄

#### archimedes

##### Flashaholic
2x (3.7V nominal, 3400mAh nominal, in serial configuration) 18650 = ~ 7.4V and ~ 3400mAh

3400mAh / 2000mA = 1.7h

Reduce 1.7h ... because 2000mAh is a pretty high current draw, and capacity is generally max at low draw, and less at higher draw ... so probably < 90 minutes

Yeah 75 minutes sounds in the ballpark to me, if my math is correct

And that's ignoring factors like regulation, converter efficiency, voltage or output cutoff, etc, etc

Last edited:

#### raggie33

##### *the raggedier*
2x (3.7V nominal, 3400mAh nominal, in serial configuration) 18650 = ~ 7.4V and ~ 3400mAh

3400mAh / 2000mA = 1.7h

Reduce 1.7h ... because 2000mAh is a pretty high current draw, and capacity is generally max at low draw, and less at higher draw ... so probably < 90 minutes

Yeah 75 minutes sounds in the ballpark to me, if my math is correct

And that's ignoring factors like regulation, converter efficiency, voltage or output cutoff, etc, etc
can you say that again so even me can understand think of me as homer simpson and peter had a child while they drank during my pregnancy

#### thermal guy

##### Flashaholic
3400+3400=6800 don't it?

#### parametrek

##### Enlightened
No, 3400+3400 does not equal 6800 right now. Either the volts add together or the mAh add together but not both. Not allowed to have the cake and eat it.

Since the cells are in series the voltages add and the mAh does not. If the 2x18650 were in parallel then you'd have 6.8Ah and 3.7 volts. Since they are in series it is 7.4V and 3.4 Ah.

Also think of it as a total amount of watt hours that must be conserved. An 18650 has 3.7V * 3.4Ah = 12.58 watt hours. It doesn't matter how the batteries are arranged. Parallel or series they will have to 2*12.58. Adding both volts and mAh would give 50 watt hours!

Last edited:

#### thermal guy

##### Flashaholic
Ok so if the batteries were in parallel and you get the volts up then you get more run time yes? Is this the way they configure high output 4X18650 lights. The batteries are always in parallel ?

#### thermal guy

##### Flashaholic
Now that helps. Thank you my friend it really is pretty easy to understand now. Granted the pictures helped 😄

#### archimedes

##### Flashaholic
Yes, some people are visual learners