What you write is correct for DC or constant current AC circuit. However, the coil circuit is reactive, changing with time under the influence of a switched inductance. A meter, especially a digital meter, will time-average a changing voltage using its internal algorithm for such measurements. While the voltage across the resistor will drop, the purpose of the resistor is to limit current. There is a big transistor in the module that takes the place of the points. Following the signal from the distributor, the transistor opens and closes to switch the coil. There will be a big inductive kick when the coil switches - here's a typical trace of the primary and secondary voltage: The primary is the blue trace, and the red is the secondary (shown in thousands of volts). The points open at zero. Notice the big inductive kick at this point which trails off to where the spark ends at 1.1 ms. The inductor (coil) then recharges, oscillating back to the a voltage near 12V. Now, with the points closed and the engine off, the resistance of the primary is in series with the ballast resistor, and there will be a voltage drop. That big transistor that takes the place of the points should be open if there is no signal from the distributor ... maybe not, I don't know. Ideally, it would be open if there's no switching signal from the distributor. That big transistor also has an internal resistance which you can calculate from the transistor's characteristics. That resistance will always be in the circuit, and it's what makes heat when the module switches. The amount of heat is dependent on how much current flows from emitter to collector (through the transistor). The ballast resistor is in the circuit to limit the amount of current in the circuit ... Ohm's law! Sorry if I'm flogging this... but I get started on an explanation and have to complete it.