12 volt system on a Model A

K

Ken Ehrenhofer

Well-known member
Hi All, I am not familiar with the 12 volt conversions for the Model A so I have a question. I thought the output of the ceramic mounted resistor would be 6volts going to the ignition switch and then on to the distributor points. The one I have in the shop is 12volt in and also 12 volts out??? I do not understand thanks for the help, Ken
 
I think this is what you are seeing:
With the ignition switch closed and the points OPEN you will see "static" voltage (12 volts)at the output of the ceramic resistor. With the ignition switch closed and the points CLOSED you will see 6 volts at the output of the ceramic resistor. A function of current flow.
 
Yup. There is a formula for that.
E (Voltage) = I (Current) x R (Resistance)

What happens is 6 volts are 'dropped' across the resistor when the points are closed, so that the coil sees 6 volts rather than 12 volts.
E (6 volts) = I (4 amp Current Thru Coil) x R (1.5 ohms of ballast resistor)
 
With a three OHM coil the ignition will also operate on 12 volt . If a resistor and six volt coil are used , the ignition will be running on a weaker six volts and won't run as well .
 
Purdy Swoft said:
With a three OHM coil the ignition will also operate on 12 volt . If a resistor and six volt coil are used , the ignition will be running on a weaker six volts and won't run as well .
Click to expand...

Except they are two different coils. The "weaker" voltage drives the "stronger" coil, which the car has had all along. The two arrangements should be equivalent. The resistor just wastes >30W of power in heat.
 
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