Thread: Splitting The Positive View Single Post

06-24-2016, 10:08 AM
 BroMikey Platinum Member Join Date: Jan 2013 Posts: 5,266
Now it is time to calculate joules.

V x A x 3600sec. or average voltage x Average ma.

Under load figures after 1 minute connection time

1st hr = conventional discharge figures
3.85v + 3.2v / 2 = 3.525v
at 700ma average = 3.525v x .700ma x 3600sec = 8883 J

2nd hr = Split positive 3.20v x .425 x 3600sec = 5240 J

3rd hr = conventional discharge figures
same as 1st hr = 8883 J

4th & 5th hr = 3.05V x .315ma x 7200sec = 6917 J

6th hr = conventional discharge figures
3.75V X 3.15v /2 = 3.45v
at 650 ma average = 3.45V x .650ma x 3600sec = 8073 J

7th & 8th hr = split positive 2.90V x .150ma x 7200sec = 3132 J

I still have whats left in this battery that charged while the light
ran ending charge voltage only reached 3.90v with batteries
A & B now fully discharged at 3.23v each.

9th hr = conventional discharge figures
3.70V + 3.22V /2 = 3.46V
at 575ma = 3.46V x .575ma x 1200sec = 2387 J

This gives a total of 43,515 J

Conventional discharging of the same 3 batteries in parallel
only offered 22,900 Joules approx.

See previous charts above for verification.

I guess Mikey's Fluke meter must be to blame.

...........
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Last edited by BroMikey; 06-24-2016 at 11:10 AM.