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Ok, i think i'm getting a grasp here.
So 2 automotive batteries in a series circuit with 12V/500A = 24V/500A
but
2 automotive batteries in a parallel circuit with 12V/500A = 12V/1000A
The A will stay the same across a circut but the V will not. So if one component needs 1V/1A and another component also needs 1V/1A then the total need for the circut's power supply is 2V/1A
Alright, so now i pick a random motor off of some website just to see how much i can understand:
Motor:
Universal AC/DC Motor, Open Enclosure, 1 HP, 10,000 RPM, 115 Volts, Non Standard NEMA Frame, Service Factor 1.00, Ambient 40 C, 60 Hz, Full Load Amps 12.1, Insulation Class A, Thermal Protection None, Rotation CCW, Face with Four Tapped Size 10-32 Holes, 3 1/4 Inch OC 90 Degrees Apart Mounting, Ball Bearings, Shaft Dia 7/16 In, Shaft Length 1 1/4 In DrillSpot Price: $248.19
(like i said, just a random motor.)
Now i take a generic imaginary Auto battary of 12V/500A DC.
9.58 batteries - or 10 batteries - are needed to meet the required volts of the motor. However, that will be @ 500A (serial connection) which is too much for the motor and will mostlikely put on a pretty smoke effect and maybe some sparks. so i'll have to cut the 500A down to no more than the 12.1 amps. (currently +5v and +487.9A excess power)
To get the power down then i will need a type of resistor.
but to calculate what size then i would need apparently alot of math and a handy calculator.
But so far, I'm getting it, right?
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