Sorry to question your explanation, but that just doesn't add up to me. Every Mopar ignition "Bypasses" the ballast resistor completely during start up. Which according to your theory would result in hard starting. Why would the factory design it that way?
It's because the stock system doesn't use a 3 ohm coil. What I was saying was that, in theory, you could reduce the ballast resistance to zero and maintain overall resistance value by using a 3 ohm coil which ought to improve spark output (due to more volts across the coil). However, that would probably cause hard starting because, as with the stock system, you would be dropping the available voltage due to powering the starter, but, unlike the stock system, you would be also trying to overcome 3 ohms in the primary ignition circuit (rather than about 1.7 ohms on the stock system). As noted by AussieChallanger, you would be operating at about 9 volts, but you would have a circuit designed to operate at 12 volts.
mojavered,
So it seems you have a coil rated at 7 ohms and a ballast resistor rated at 8 ohms. When the engine's running your coil will be operating at less than 6 volts which is not really making the best of that coil, though it will probably still run OK. Starting should be fine. The problem is that your total resistance will be only 1.5 ohms. So at 12 volts that's drawing 8 amps.
When you are driving it's not a constant 8 amps because the system is constantly switching on and off, but I think it is still in excess of the design load for the ECU (although for a chrome box I don't know what that is supposed to be, I doubt it is as high as 8 amps).
If you left the ignition on without the engine running it would be a constant 8 amps.
This is why people often find they cooked their ECU parked up listening to the radio.
So basically it might be OK depending on how tough the chrome box is supposed to be, but I didn't think they were built to handle that sort of current except maybe for short periods (like racing).
