Here is a Lucas tutorial on the original mechanical points style ignition system, which gives a good explanation of the electrical principles of traditional coil based ignition systems.
http://www.da7c.co.uk/History%20Section/LUCAS%20ACHIVE/lucas%20coil%20course%203.pdfLuddite, I think your explanation is entirely accurate. My reading of the Unilite wiring diagram above is that the 12V ignition switch supplies 12V to the ballast resistor whenever the ignition switch is in the ON position. This would result in maybe 6-9V being applied to the coil, depending on ballast resistor value. If you move the ignition switch to the sprung START position then 12V is applied directly to the coil (and the starter solenoid) thus ensuring that 12V is briefly applied to the + terminal of the coil until the ignition switch is released from the START position and returned to the ON position. The diagram, like so many auto-wiring diagrams is not very clear to my mind.
Generally, there is 3 to 4A of DC current going through the coil if the ignition is on. So as power is I*I*R then the coil will act as heater dissipating 20 to 50 W depending on the coil resistance (1.5 ohms approx if you do have an ballast resistor installed or 3 ohms approx if you don't). So the coil will get hot. But there is also the effect of the current in the coil being turned on and off by the points or the electronic ignition - I'm not sure how much this affects the power dissipated in the coil as AC currents flow into and out of the magnetic fields in the coil. That probably makes the coil hotter in normal use - are some coils oil filled now to help cooling?
If you choose to use a coil that is intended to have a series ballast resistor without the ballast resistor (ie directly connect the + connector on the coil to 12V, then twice the intended current will flow through the coil. In that case (as power is proportional to the square of the current but linearly proportional to the resistance,) then you will increase the power dissipated in the coil by 2 - and this is why the coil is more likely to burn out in such circumstances. Again this is a DC argument, not quite sure whether it is made worse if you consider the changing ignition current.
At times, I have run my 4/4 Ford crossflow with with conventional points - well, actually an optical points sensor but essentially the same thing, or the full Accuspark electronic ignition, with or without a ballast resistor. In all cases it worked fine, (except when I wired it up wrong) provided I used a ballast resistor with a coil intended for a ballast resistor (1.5 ohm coil approx) or did not use a ballast resistor and used a coil (3 ohm) intended for use without a ballast resistor.
I imagine once you get rid of the mechanical distributor (as in a modern car) and use a sensor on the crankshaft and have a bunch of electronics and computers driving separate ignition coils for each spark plug as in a modern car, the amps and ohms numbers quoted above go completely out of the window. I don't understand modern cars at all I'm afraid so can't really comment.
