Hello, I am working on a 41-290. It was left in some old lady's attic and is very dry and crispy. I will need to rewire and recap. One thing of concern is the main power resistor. It has taps for 15,31,146 ohms (I'm guessing from memory). If I wanted to replace that, what is the wattage on this one? I want to use 3 separate metal oxides or wire wounds. Any suggestions?
Does each section still measure close to schematic value ?
Wattage consumed can be determined with Ohms law: voltage squared divided by resistance.
If sections are bad....
some folks are good at calculating voltage drop by following the schematic. I'm unable to determine this one....
I didn't measure the values yet but the taps are crumbling. My best bet is to replace it. I don't want to assume this will hold up and have it fry the transformer after all the work I am doing to restore it. I will guess and use 10 watt wire wounds. If they get too hot, I will try something of a higher wattage. Any better suggestions?
You won't be able to find resistors that exactly match the values of the original resistor strip, but you can come pretty close using two common-value resistors in series. I think you can use 1/2 watt resistors for all, but I would probably use 1 watt resistors for one of them. See my calculations and explanations below. 10 watt power resistors are way overkill for this circuit.
The calculation is pretty simple. It is not absolutely accurate, because it ignores the fact that not all of the B current goes through all of the resistors in the strip, but it is very close, and it squares very well with measurements I have made in other radios. This calculation errs on the safe side for each of the resistors in the strip except for the 146 ohm, for which the calculation is absolutely accurate, since all the current goes through that resistor.
Including the field coil resistance, the total resistance in the circuit (from the schematic) is 1100+146+31+18=1295 ohms. The voltage drop is 265-180=85 volts (from the schematic). Therefore the current, I=E/R, is 85 volts /1295 ohms = 0.065637 Amps. The total power is I squared R. I squared is .06537 X .06537 = .004356, so I squared R for the whole circuit = .004356 X 1295, or 5.6 Watts.
Since the current at all points in the circuit is the same (ignoring the small amount of current that bypasses the 31 and 18 watt resistors in the voltage divider), you can use the same value of I squared to calculate the power for all of the replacement resistors.
Most of the 5.6 watts dissipated in the circuit is dissipated by the 1100 ohm field coil, leaving less than one watt to be dissipated by the resistor strip. .004356 X 1100 = 4.7916 watts. 5.6 - 4.8 = 0.8 watt. (Rounding fractional values up.)
The 146 ohm resistor dissipates .004356 X 146 = 0.635976 watts. You can approximate this by using a 100 ohm resistor (.44 watt) in series with a 47 ohm resistor (.2 watt). I would use a 1 watt for the 100 ohm and a 1/2 watt for the 47 ohm. That gives you a better than 50% margin in both cases. They will get hot, but no hotter than they are designed to run.
The 31 ohm resistor dissipates .004356 X 31 = 0.135 watts. A 30 ohm resistor (0.131 watts) approximates this value pretty well. I would use a 1/2 watt resistor for this one, although a 1/4 watt resistor would give better than 50% margin.
The 18 ohm resistor dissipates .0784 watts. You can approximate this value by using a 10 ohm resistor (.04356 watt) and an 8.2 ohm resistor (.036 watt) in series. I would use 1/2 watt resistors for these, although 1/4 watt resistors would be perfectly adequate for each.
Before they started using resistor strips in (I hink) 1941, Philco used discrete resistors for the voltage dividers. They used 1 watt and 1/2 watt resistors for this purpose, so this squares with my calculations.