outcome of voltage in series and amp/h

2 Posts
Sep 12, 2011 09:11 pm
outcome of voltage in series and amp/h

 I'd like to know the result  connecting  batteries with different voltage(V) and different currents(AH) in series Let's add up ,in series,= Rolls battery 2V/1050AH + Rolls 4V/1350AH + Rolls 6V/683AH!!!! If I get 12V,what will be the Amp Hour? 3083AH? Whatever the result, can it be applied to solar panels setup? I need clearification.
« Last Edit: Sep 12, 2011 09:17 pm by kay andy »
97 Posts
Sep 14, 2011 04:08 pm
Re: outcome of voltage in series and amp/h

Hi Kay.  Looks like no one is anxious to jump on this one, but I will take a stab at it.  I am not an expert on lead/acid batteries, but common sense tells me that this series combination would be a very bad one.  The useful capacity of the string would be that of the smallest amp/hr battery, the Rolls 6v 683amp/hr.........it would be the first one to discharge, since the current is the same in all components of a series configuration.  After it discharged, the other batteries, being larger, would still have charge capacity, and they would be forcing a current through the 6v Rolls that would ruin it for any further use  other than lead ballast weight.  If you are going to wire batteries in series, they should be the same voltage, amp/hr rating, and preferably the same age, in order to avoid a situation where the oldest battery ruins the string and renders the newer batteries useless as well.  Jon C.
2 Posts
Sep 14, 2011 06:56 pm
Re: outcome of voltage in series and amp/h

Hi Jon c, thank you for taking the pain to treat my question.I'm so novice in this things.Actually I thought it's just a question of adding up in series and get the voltage you want.I took note of that now.Bye Kay.
99 Posts
Oct 14, 2011 01:01 pm
Re: outcome of voltage in series and amp/h

I'm not convinced that's true.  Why would a discharged battery receiving current from larger batteries ruin it?  Seems to me that it would tend to keep it charged.  Imagine a big tank of water pouring into a 5-gallon bucket pouring into a glass which pours into your mouth.  The glass has very small capacity but is being constantly refilled by the bucket.  The bucket likewise has smaller capacity than the tank, but is also being refilled as quickly as it is discharged.  If anything, the largest capacity battery would have the most volatility in charge level with the smaller ones only discharging very much when the biggest has drawn down significantly.  This of course is all conjecture and should actually be tested.  I would recommend trying the setup for a few full charge/discharge cycles and measure the level of each battery throughout the process.  If you graph the three batteries % of full charge throughout the cycle, I suspect that the largest capacity battery will show the biggest swings.  I'd be very interested in seeing the result of this mission, if you choose to accept it.
97 Posts
Oct 15, 2011 11:28 pm
Re: outcome of voltage in series and amp/h

Thomas, discharge current flows externally from negative to positive terminal of a battery.  Charge current flows just the opposite, internally from negative to positive in the battery.  In a series string of batteries, if the smallest battery loses charge capacity first, then the other batteries will force a current through it that is reverse to a charging current.  I.E. from positive to negative, internally.  Current is the same in all parts of a series circuit and would have to flow this way through the small battery.  This would, in fact ruin the smaller battery by forcing a current that is in reverse to a charging current.   Jon C.
97 Posts
Oct 16, 2011 09:27 am
Re: outcome of voltage in series and amp/h

A simple bench experiment would prove this concept.  If you series wired a C cell and an AA cell (battery neg to pos), and a flashlight bulb, you would have a 3 volt battery pack using the bulb as a load device.  (sub C and AA nicad battery would be a better example of a rechargeable power source and would yield better result).  Assuming a fresh charge on the batteries, the bulb would burn brightly at first, and then, after some time, the bulb would dim as the batteries run out of capacity.  The first battery to go flat would be the smaller AA battery, and the C battery would not serve to charge it further, but rather, would discharge it further.  Eventually, the smaller AA battery would totally run out of capacity and would go resistive, exhibiting a reverse polarity voltage across its terminals.  The battery may or may not recover from this discharge state, and may be permanently damaged by the polarity reversal.  A lead acid battery most certainly would be permanently damaged.  From this test, one can conclude that the capacity of the battery pack is that of the smallest battery, and that the batteries should have been the same capacity rating for best results.  All multiple cell battery packs are, or should be, critically balanced for cell capacity to get the best performance from the battery pack.   If you have had experience with rechargeable battery packs and banks, as I have, you should know this concept. 

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