The internal balast resistor is only used on the primary side to regulate current it does change the current in the secondary by cutting it in half. The ignition coil itself is a basic step up tranformer. These transformers work off a turn to turn ratio. So let's say we need 20,000 volts on the secondary going to the engine. In order to find the turns ratio we would divide 20,000 by 6(volts). This calculates to 1(primary) to 3,333(secondary) turns of wire. Now if we had a 12 system we would need the same 20,000 volts. This coil would need to be coil 1(primary) to 1,166(secondary) since this is 20,000/12. So your 6 volt system requiring 20,000 would be 6v/20,000=amps of the secondary/4amps primary. You would end up with .0012 amps and 20,000 volts. If we use the 12 volt coil with the 6 volt system. Now, we would have 6 volts multiplied by 1,166 from the 12 volt secondary. This would equal 9,996 volts. Your current will also change. To find the current :6v/9,996=amps sec/4amps primary. By the way amps is found by dividing 6v by the resistance 1.5 ohms. The secondary current would be .0024 amps compared to the original .0012. So just as wrench bender stated your current would double. Hot rod Lincoln was on the right track with your voltage being thrown off which is now 9,996 volts. The gap of your spark plugs depends on this high voltage to jump the gap. These numbers aren't exact. I just used 20,000 as an example. I hope this is some what clear.
