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Old 10-24-2012, 08:28 PM   #1
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a little help battery math

i need a little math help. i've got 4 -6volt golf cart batteries. rated at 20amp hr. - 230 amp. 5 amp hr.- rate 174 amps they are wired 12 volt. here's the math question the batteries will charge to 14.6 volts. intenverter shuts of the 120 volts at 10 volts ?. so will someone build me the formula on the usable watts in this equation ????
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Old 10-24-2012, 08:46 PM   #2
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Having four 6 volts in a 12 volt system doubles the volts and doubles the amps so you have 12 volts (2 X 6) and 460 amp hours (2 X 230 at 12 volts) to draw against. BUT you only want to draw them down 50% so you can use 230 amp hours at the 20 amp rate. Taking them to 10 volts will destroy the batteries in short order.
Our OEM Interstate U-2200's lasted just under 10 years, but I almost never went below 11.0 - 10.9 volts.
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Old 10-24-2012, 08:54 PM   #3
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Old 10-25-2012, 03:52 AM   #4
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Old 10-25-2012, 06:03 AM   #5
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Need to consider how many amps you are drawing when you decide how far to take them down to.

For instance - if you were drawing 2 amps and wanted to stop when they were 50% discharged, you would need to stop at around 12Volts. If indeed you kept going until 11V at this small current, the battery would be very close to flat.
Draw 50 amps (small inverter) and you should stop at (say) 11V - because there is higher internal voltage drop to take into account, plus other factors - and you will find that if you then drop that load off, the voltage will rise rapidly and if you have judged it right, will settle out at around 12V, so roughly 50 % discharged.
Take out 200 amps - as some do to drive a large inverter - and the voltage will immediately drop well below 11V and perhaps closer to 10 volts and for that load, you can stop at 10V to end up with a 50% discharged battery.


Note that I'm not saying you should drag 200 amps out of your battery because that too will reduce the lifespan of the battery, just as much as excessive discharge will.
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