Using Ohm's Law,I=E/R where I is current, E is voltage (no, I don't know why. Some people use V) and R is resistance in ohms: In the case of a 12 volt conversion: I=12/1000 I= .012 Or 12 mA. This is a typical grid current in a vacuum tube amplifier, but not good for a Kettering ignition system. Extrapolating with the original six volt system, I=.006 A, or 6 mA. Not a whole lot of current by any standard. Using P=IE for power consumption, P=.144 watts at twelve volts and .036 watts at six. Many Kettering systems on this class of tractor (and many cars and trucks, for that matter) work well dissipating about 25 watts. There is a BIG difference between .036 watts and twenty watts. So what, you ask? Remember the ammeter on your dashboard? Well, let's assume for the sake of argument that the standard ballast resistor on your dashboard develops a high-resistance fault that gives it a value of 1k ohm (not uncommon in sixty-year-old resistors). In this case, the meter mught wiggle a little at the instant the switch is thrown, but would in all probablity not show any legible deflection from zero, as the current flow is very tiny and the meter is designed to measure accurately more twenty times as much current. Simple, neni pravda?
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