12-27-2015, 01:32 PM
Yes, it was nice of Philco to not reveal the value of that resistor, wasn't it? 
Using Ohm's law, we will have to calculate it.
The five tube filaments in series will drop 101.5 volts.
The radio was, most likely, designed for a 115 volt line.
Each tube filament draws 0.15 A of current. As the filaments are in series, the entire string draws 0.15 A of current.
So the original resistor would have had to drop 13.5 volts.
R = E / I
= 13.5 / 0.15
= 90 ohms
Now we have to figure wattage.
P = I squared * R
= (0.15 x 0.15) 90
= 0.0225 x 90
= 2.025 watts
In practice, we normally double the wattage dissipation, so you need a 90 ohm, 5 watt resistor.
But hold on. If the line voltage where you are at is 240 volts, then does your radio have the extra dropping resistor plugged into the J1 socket?
Using Ohm's law, we will have to calculate it.
The five tube filaments in series will drop 101.5 volts.
The radio was, most likely, designed for a 115 volt line.
Each tube filament draws 0.15 A of current. As the filaments are in series, the entire string draws 0.15 A of current.
So the original resistor would have had to drop 13.5 volts.
R = E / I
= 13.5 / 0.15
= 90 ohms
Now we have to figure wattage.
P = I squared * R
= (0.15 x 0.15) 90
= 0.0225 x 90
= 2.025 watts
In practice, we normally double the wattage dissipation, so you need a 90 ohm, 5 watt resistor.
But hold on. If the line voltage where you are at is 240 volts, then does your radio have the extra dropping resistor plugged into the J1 socket?
--
Ron Ramirez
Ferdinand IN
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