Reason for my electronics 101 question

Meanwhile, here's where it all started. I am replacing the resistor in a 3-position generator control/light switch on a 1940's tractor. The old open (uncovered) wound-wire resistor was burned out like many of them are. I picked up a 3.3 ohm ceramic resistor (with a 10 watt rating) from an electronics supply house to wire into the switch. The original resistors were "about" 3 ohms on most of those switches for a 6 volt system. I don't know the watt rating for those original resistors. I am wondering just how hot that resistor will get when I wire it into the switch. I can't check it out right now since a running tractor all hooked up and operating won't happen for a while yet, but I am getting stuff ready and prepared.

This ceramic brick resistor is about 1/4" square and about 2" long if that means anything, and i assume is a wire-wound type inside the ceramic brick.

Any idea as to whether this resistor will do the trick without exploding or vaporizing? The system is a 6 Volt system and the resistor circuit grounds the generator's field circuit which a manual says puts out about 3.8 - 4 amps but maybe that is meaningless and is determined by ohms law or something. Once again, your help is appreciated.

The real solution here is to measure the current through the field winding, or at least measure the resistance of a field winding & estimate the current. I'd walk out to the shop and measure one, but we had a foot or so of snow here today & I don't feel like getting all bundled up.

Someone post measurements, OK. Otherwise, I'll check the next time I get out that way.

Thanks,

Keith

At 7 volts and 5 ohms, the current through the series combination of the external resistor and the field winding is about 1.4 amps. 1.4 amps squared x 3.3 ohms gives about 6.5 watts being dissipated in the series resistor.

6.5 watts is pretty consistent with the size of the orignal wire wound resistor. A 10 watt resistor can't dissipate the full 10 watts inside the switch housing (they're rated in free air at a particular temperature) without overheating, but should be able to dissipate 6 or 7 watts just fine.

I'd go ahead and use the 10 watt resistor. If you're concnered, go to a slightly higher value, say 5 ohms. Under these same conditions, the current through the resistor would be a shade over an amp, and the resistor would dissipate about 5 watts. Since the output of the generator is directly proportional to the current through the field winding, the output current from the generator would be reduced by about 40% compared to a 3.3 ohm resistor, but should still maintain the battery without any problems.

Keith
(Electrical Engineer, 30+ years experience)

With the resistor IN the circuit, at the generator's low output, likely the field current will be only a couple of Amps.

Watts = I2 X R.

2 Amps squared = 4 X the resistance of 3.3 Ohms = a power dissipation of 13.2 Watts.

Even at a field current of only 1.5 Amps, the resistor would have to dissipate nearly 7.5 Watts.

Even that would likely be too much for all-day sustained operation, with the resistor in the enclosed switchbox.

Better go back to the electronics supply and get a 20 or 25 Watt unit.