[Raleigh]

I don't think you really need 5 watt resistors to replace the candohm in this radio. You can calculate the current through the resistors from data in the schematic, and then you can calculate the power dissipation. Of course, there's nothing wrong with using 5 watt resistors as long as they fit.

From the schematic, the voltage drop across the field coil is 265 - 180 = 85. Using ohm's law, the B current is about 77 mA. Since all of the DC in the radio goes through the field coil, 77 mA is the return current to the transformer center tap through the candohm.

Most, if not all, of the return current goes through the 146 ohm section of the candohm. For simplicity, let's say it all does. The power calculates out to .87 watts. I suspect that very little current goes through the 31 and 15 ohm sections, but if we say all 77 mA does, they'd dissipate .18 and .09 watts, respectively.

You could probably use a 150 ohm 1-watt resistor and half-watters for the other two. For derating, I'd probably double up two 1-watt 75 ohm resistors in series to make the equivalent of a 2-watt 150 ohm, since I don't have many 2-watt resistors lying around. Most likely I'd use 1-watters for the other two, but that's really overkill.

I bought a 42-350 that someone had partly recapped, and he replaced the three-section candohm with a 20 watt and 2 10-watt resistors. Way overkill.

If you still have the chassis out of the cabinet, it would be interesting for you to measure the voltage drop across the 3 sections and see what you get, to see if my calculations are bogus or not. I'm not a EE so I could easily be mistaken.