12v conversion question

So I am re-converting my tractor to 12v as I surley screwed up the first time. Got my parts all lined up, but faile to get the resistor- parts store was out of it.

I ended up with a 12v coil, do I need a resistor for that? Or just for the 6v?

Front or side distributor?
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front distributor

Rusty......don't matter, OEM 6v squarecan coil or modern 12v squarecan coil; BOTH need the "infamous ballast resistor". NO ARGUE!!!

Iff'n you use the original 6v squarecan coil, you must also use a 12-to-6v converting resistor. But NOT for the modern 12v coil. understand??? ......Dell, yer self-appointed sparkie-meister

Dell-

'priciate the help!

So looking at the parts page, are we talking a .5ohm resistor for a 12v squarecan coil? I dunno if I can track down an OEM part, but I know Rat Shack can give me a .5 ohm resistor.

Rusty......I showed you the MANDATORY "infamous ballast resistor" for the modern 12v squarecan coil. NO ARGUE!!!

You go to radio shack and ask for a 0.5-ohm resistor, yer gonna gitta BLANK look. Radio Shack don't do 1/2-ohm 50-watt power resistors. Theys BIG CERAMIC resistors.

You may need a 0.5-ohm resistor, but it will haffta come from autoparts store, NOT tractor store. Ask for 1965? dodge ballast resistor. RU-1? .......Dell

No. We're talking the OEM resistor that's in the picture. It's a ballast resistor. That means it changes value as it gets hot. It's .3 cold & 1.7 really, really hot (just use 1 ohm).

What you will get from R/S is a fixed resistor. You might need that as well depending on the internal resistance of the coil. If it's 2.5 ohms or better, that will be ok. Less than 2.5 ohms, add another resistor. Or, keep a spare coil around.

Technology & materials being what they were in the 30's, that square coil would melt if it ran on much more than 4 amps for any length of time. (see tip # 38 for an example). In order to get a hot spark at the same time the starter was drawing max current from the battery, a ballast resistor was added in the ignition circuit. What that did was add about .3 ohms of resistance in the circuit, added to the 1.5 ohms of the coil. That got you 3.5 amps or so at start up. As the voltage increased when the engine was running to about 7.5 volts, the resistor heated up, adding more resistance in the circuit. 1.0 ohms hot, plus 1.5 ohms of the coil got you down to 3 amps or so to keep from melting the coil. The same rule (actually, Ohm's Law) applies to a 12v circuit. I= E/R. Current equals voltage divided by resistance.
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