I have a jet ski with a 19 AH battery, I measured a constant drain of 3.6 ma when it is turned off.
How can I figure out how long it will take for that to drain the battery enough where the jet ski will not start? I am guessing the battery needs about 75% capacity to crank it sufficiently to start.

Thanks

 

Your battery is 19,000 mAH, so draining 3.6mA is approx 219 days.

@ 25% = 219/4 = 54.75 days.

 

19AH = 19000mAH
75% = 19000mAH * 0.75 = 14250mAH
25% = 19000mAH * 0.25 = 4750mAH
4750mAH / 3.6mAH = 1319H
1319H / 24H = 55 days

 

Thank you for these answers. It appears that as long as I use it at least once per month, I shouldn't have a problem.

Best regards!

 

Hi,

There is another factor involved in determining the run time of a given application.
For a battery as big as 20AHr where that 20AHr was measured with a 4ma drain current the run time would be:
20000/4=5000 hours
but because battery capacity is usually measured at around C/20 that would mean the test current would normally be:
20000mAHr/20Hr=1 amp
and this would lead to a 20 hour run time.
But with a current that is 1000/4=250 times lower than the standard test current, another factor would kick in. This factor
makes the battery look like it has more capacity than at the standard discharge current level.
For that small of a current you might see as much as 2x the run time. So the max run time could be as high as:
20000mAHr*2/4m=40000mAHr/4m=10000 hours.
So we went from 5000 hours to 10000 hours. Maybe more reasonable would be 7500 hours but to get the more accurate
time we'd have to know a lot more about the battery. One thing is certain though, with a very small drain current like that we should see a much longer run time than normal.
The other thing to account for though is self discharge. That could be as high or even higher than the drain of 4ma you talk about.
We'd have to know this too of course.

 

Thanks, That should mean it has even a better chance of being ok as long as I use it for a few hours at least once a month.