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Re: adding batteries together?


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Posted by Indiana Ken on January 15, 2014 at 18:51:28 from (66.249.234.135):

In Reply to: adding batteries together? posted by NCWayne on January 14, 2014 at 18:10:58:

Quoting Removed, click Modern View to see

Lets consider 12 volt batterys, each have identical ratings of 1000 CCA. In addition for this example we will not consider, resistance of cables or length of cranking time.

Definition: Cold Cranking (CCA) rating is the discharge load in amperes which a new fully charged battery at 0 degrees F can deliver for 30 seconds and maintain a minimum voltage of 7.2 volts for a 12 volt battery or 3.6 volts for a 6 volt battery.

When a battery is loaded the voltage drops. The CCA rating is a measure of the voltage the battery can maintain at a load of 1000 ampere - it does not indicate that the battery will deliver 1000 amperes to the starter. Your starter is a known resistance therefore the current delivered to the starter will increase in proportion to voltage: I = V/R.

Electrical power: Is called Watts and defined as Volts x Amperes = Watts. For this example we will call this "Starting Potential".

From the above we can see the battery or battery combination that can maintain the highest voltage during cranking will deliver the most power to the starter and have the highest starting potential.

In order of power delivery:

1) Two 12 volt batterys in series (1000 CCA and 24 volts). Since the battery voltage is higher starter current/power will increase, typically not double.

2) Two 12 volt batterys in parallel (2000 CCA and 12 volts). Since the batterys share the starter current they see less load and will maintain a higher voltage during cranking.

3) One 12 volt battery (1000 CCA and 12 volts). Since the battery sees the full starter load the voltage will be pulled down to a greater extent.

In summary - as compared to one 12 battery:

With two 12 volt batterys wired in series, starting power increases due to the higher battery voltage. With two 12 voltage batterys wired in parallel starting power increases again because of higher voltage but in this case due to less loading.

Sorry it got a little long - hopefully someone is interested.


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