So this is a 2 watt resistor. I can't seem to get the color codes to work out.

A wire shorted to ground burning it, and a pot, up. I am going to have to replace the pot by trial and error. Not having a full resistor set in 2 watts, and having 2 variables... I need to figure this out.

 

Try another picture, with your light source originating from the un-burned side... Your colors are indecipherable.

 

I see 47 ohms 5%.

 

Sorry colors are
Yellow gold green violet Yellow

Thanks, I wanted to be sure. Doesn’t seem to add up exactly

 

Sorry colors are
Yellow gold green violet Yellow

Thanks, I wanted to be sure. Doesn’t seem to add up exactly

I think you are reading it backwards. So, yellow-violet-green would be 4.7 Mohms. Not likely to fry under almost any condition unless dielectric breakdown occurred. Are you sure the middle band is green and not black?

 

Sorry colors are
Yellow gold green violet Yellow

Thanks, I wanted to be sure. Doesn’t seem to add up exactly

I think you are reading it backwards. So, yellow-violet-green would be 4.7 Mohms. Not likely to fry under almost any condition unless dielectric breakdown occurred. Are you sure the middle band is green and not black?

If green, and a 5 band resistor, then
Yellow. 4
Violet. 7
Green. 5
Gold. X0.1 multiplier
Yellow +/- 4% tolerance.

I would say 47.5 ohms.



If black, it would be 47.0 ohms.


Yellow is not a common tolerance band but, it is old looking and may have been popular at one time.

 

Well I feel a lot better now that it is a little confusing. The middle band has a green tent and looks like it cooked darker. But .5 Ohms seems unlikely and pointless so maybe the heat, or age added the green tent.

It is from the 70s I think. Someone replaced a wire on a pot, and allowed it to short to ground. I will go with 47 ohms and then start with a large value pot and see what happens... No schematics or parts are available for it.

Thanks
Cory

 

How many ohms in the pot?
(Placing 47 ohms in series with a 100k pot simply doesn't make sense.)
With a 2 watt rating, we would assume the resistor uses up most of the voltage because most pots are about 1/2 watt.
Assuming the resistor really needed to dissipate one watt, the pot should be vaguely 20% of the total resistance in that circuit.

Not a real answer, but a method to "ballpark" the value.

 

Well the pot is burnt up and is old enough that the pot manufacturer doesn't have records of it. There are several things tied together at the control panel, and honestly the resistor being in line with the pot was just a guess.

I knew I needed those 2 parts just to give it a try, will be looking into it more next week.