Ohm`s Law:- R(esistance) = Volts(V)/Current(I). Transpose:- I = V/R. Then, 12/1.5 = 8Amps :: 12/3 = 4 Amps :: 6/1.5 = 4Amps.
Feeding the 6V, 1.5 Ohm Coil at 12V would virtually double the H-T and burn out over time. However when fed the voltage, as dropped during cranking, which can be as low as 9-10V, for just that short period it produces an enhanced spark but only during startup. During the running phase, The Resistance feed (ballast resister) reduces the 12V to circa 6V at all other times.
Which IMV, is the point Pete was making before.
Please accept my apologies if this is "granny sucks eggs" to many. But was how it was spelt out to me back when I was the "recipient" of this information!.!
Pete