Does Electronic Ignition Need a Resistor?

My 2N has been converted to 12V with electronic ignition. My dad, who did the conversion, felt that the coil was running to hot after the conversion. Question..does the electronic ignition need a ballast resistor on the coil wire?

A resistor in the ignition circuit is not there for the points or EI. It's there for the coil.

Front coil or side coil conversion?

If it's a front coil, it needs the oem ballast resistor no matter if it's a 6v or 12v coil.

If it's a 6v coil, you will need an additional resistor......or just get a 12v coil.

See tip # 30.
75 Tips

Yep...#30 plain as day. It is a 12 V coil, with a front mount distributor. I'll add the resistor. I assume it the standard one if I look up 2N parts. Thanks Bruce.

I found this in the archives, think this would be a good add the the 75 tips also Bruce. It is an excellent explanation of the "why" we need the resistor for those folks that have an interest.

An old post from Bruce.

"....as to whether the resistor should be removed or not when replacing a 6v to 12v coil. "
What resistor are you talking about? If you are referring to the OEM ballast resistor on top of the terminal block, the answer is no. If there is another resistor in the circuit in order to use a 6v coil, the answer is yes, remove it.
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.
It used to be before the "Land of Almost Right" started making coils that you could count on a 12v frontmount coil as having 3 ohms of internal resistance & the 6v coils as having 1.5 ohms or less. Thus, thanks to Ohm's Law, you could calculate what additional resistance you needed in the circuit to limit coil current to 3.5 amps. So, you can assume that your 12v coil has 2.5 ohms or more internal resistance or you can measure it. A digital multi-meter has two probes & a switch. Set the switch on resistance. Put one probe on the top of the coil & the other on the pigtail at the bottom. It will give you a reading in ohms. Lets just say it reads 3.0 ohms. Your OEM ballast resistor (which you must use) is about 1 ohm hot. A coil a 3.0 ohms, plus the ballast resistor at 1 ohm (hot) gives you 4.0 ohms resistance in the circuit. Your 12 volt alternator puts out 14.5 volts. You need to determine current (amps). 14.5 v divided by 4.0 ohms gets you 3.6 amps; that's ok. But, and this is the problem......what if the coil is only 2 ohms? Do the math. 14.5 volts divided by 3.0 ohms gets you 4.8 amps! Not good! And, if the coil is less than 2 ohms (and some are) it will fry quickly. So, to get it to 3.5 amps, you need another resistor in the circuit. Either that, or keep spare $30 coil around.