05-03-2013, 11:58 AM
Measure the voltage across the resistor and you can calculate the power (wattage) dissipated. You can use the nominal resistance for the calculation--it's close enough. If you want to be super accurate you can measure the actual resistance after desoldering one or the other lead.
E=IR, so I=E/R, and Power = I squared R. Power should be half of the rated value or less. If you are super concerned about the heat, you could buy a power resistor from Mouser or Allied that bolts to the side of the chassis and uses the chassis metal as a heat sink. It will run cooler than your power resistor, but the metal of the chassis will get a little warm. If you buy one of those, check out the derating specs of the resistor. They will often advise derating at 4 times the rated power or more, so if your resistor dissipates 3 watts, buy a 10 or 15 watt power resistor.
I think that's a better solution than drilling a large hole. But my guess is, Philco engineers did their homework and your resistor, assuming it is not defective or degraded, will be fine.
E=IR, so I=E/R, and Power = I squared R. Power should be half of the rated value or less. If you are super concerned about the heat, you could buy a power resistor from Mouser or Allied that bolts to the side of the chassis and uses the chassis metal as a heat sink. It will run cooler than your power resistor, but the metal of the chassis will get a little warm. If you buy one of those, check out the derating specs of the resistor. They will often advise derating at 4 times the rated power or more, so if your resistor dissipates 3 watts, buy a 10 or 15 watt power resistor.
I think that's a better solution than drilling a large hole. But my guess is, Philco engineers did their homework and your resistor, assuming it is not defective or degraded, will be fine.
John Honeycutt
