Re: Voltage Readings
From a recent Bubba post:
Ignition “ System” divided into 6 volts of electrical balance…….
In a perfect world using ohms law. The ignition coil winding would offer .6 ohms, the resistor would offer .6 ohms .
This would equal 1.2 ohms of resistance allowing a current flow of 5 amps.
The wattage ( volts x amps) of the circuit would be divided equally dividing the heat build up caused by the resistance.
This is fine until we start the engine , then the voltage becomes higher ( like 78 volts), lets use 7 volts.
7 volts divided by 1.2 ohms would be 5.8 amps .
This increase in amps must be taken up somewhere! The increase flow would increase the heat in each resistance and the small gauge wire to the distributor would take up the rest. This lead is a special resistance lead designed to take up the surges allowing good point life!!!
Testing with volt drop the drop across the coil would be .6 ohms x 5.8 amps= volt drop of 3.4 volts , the same would work for the resistor with a voltage drop of 3.4 volts…….the increase of resistor heat caused by flow would drop another 1 volts.
Using 7 as a spec , the distributor lead and the contacts could see approx. .2.4 of a volt in this final voltage drop….
Moral of the story is use the factory designed set up and live happily ever after. A regular coil will supply 2030,000 volts. A flathead with low compression and a richer than normal air fuel mixture would never need this much voltage from the coil.
