Theoretical Question for Dell, souNdguy, Bruce(VA), etc...

But if I can get the correct answer to this, it may allow me to help others with some of the most commonly asked questions.

I know the orginal Ford ignition system dissiapated about eighteen watts in the circuit of the coil primary.

P = I x E
P = 3 x 6
P = 18

Shouldn't a properly-done twelve volt conversion dissipate the same amount (surely it couldn't dissipate much less!)?

In that case, what should the dashboard ammeter read? Since we orginally had six volts at three amps, and we now have 12 volts, shouldn't the ammeter read an amp and a half to give us our eighteen watts?

18 = I x 12
18/12 = I
I = 1.5

• 12v Conversion

In the original 6-volt ignition system, TOTAL resistance sets the total current. I=E/R (Ohms Law) Iff'n ohms stay constant, and you increase volts, AMPS gotta increase (and POWER gotta increase too)

The ignition points (switch contacts) actually set the 3-amp requirement, called "area rule". More amps, bigger points required. This is true for mega-volt electrical power systems too. They gotts some really big contacts switches.

So to take advantage of the 3-amp rule, 12-volt ignition coils have higher primary winding resistance than 6-volt coils. The greater 12-volt winding creates a stronger internal magnetic field that creates HOTTER SPARKIES.

Yes, you can use a 12-to-6 volt converting resistor with a 6-volt coil and get acceptable 12-volt ignition operation (3-amps), but why? You LOOSE the 12-volt HOTTER SPARKIE advantage for better ignition operation.

The "only advantage" of using a 12-to-6-volt converting resistor ...is... its cheaper than buying a new 12-volt coil. Me? I like hotter sparkies..... ..Dell, a 12-volt advocate for the right reasons

That said, I haven't found the right reason for my eazy starting 6-volt 52-8N, and I know how to do it right the first time. Infact I know 8-ways to do it and they all work.

P=I*E=3*3=9 watts. the coil does not see the full six volts, the balast resistor drops the voltage to about 3 volts.

Joe

A.Bohem made mention of how much energy the ignition disipated. that would take into account any series primary resistances AND the coil.. not just the coil.

Bob gave a good breakdown.. etc.

soundguy

Quite right. I was thinking of the SYSTEM current which is of course displayed by the ammeter, including the ballast resistor.

To approach the same question from a different angle, what does a dashboard ammeter read with a correctly designed twelve-volt conversion?

Couple that with a non precision ammeter movement.. and it's gonne be hard to tell if your ammeter is reading between 1.5-4a in most cases... especially the old loop meters..e tc.

All said.. your choice of coil and any resistors inline, and condition of all electrical connections, exact votlage of the charge system, etc.. all those will be variables for the ohms law equation.. etc.

Soundguy

The current draw by the ignition system is dictated by the battery voltage divided by the resistance of the coil, ballast resistor(s), wiring, and electrical connections. If you double the battery voltage then the resistance will need to double to keep the voltage constant thru the coil. This is why most 12v systems add another ballast resistor. The current draw shown on the meter would be the same.

Ike

ASSUMUMING the original 6-Volt coil is retained for the 12-Volt conversion, the coil is "seeing" the same voltage and current as it did in the 6-Volt system, and the "converting resistor" is dissipating about the same wattage as the coil and orignal ballast resistor together.