The W is probably Wire Wound. Colors should be brown-black-black-silver for 1 ohm 5%. Then the 3 is 3W.
Probably an audio amp? WW is usually common and they normally break to infinity.
Might be a lead resistance problem.
Reading low value resistors can be problematic. Check with probes shorted and subtract reading.
The W is probably Wire Wound. Colors should be brown-black-black-silver for 1 ohm 5%. Then the 3 is 3W.
Probably an audio amp? WW is usually common and they normally break to infinity.
Might be a lead resistance problem.
Those kind of resistors commonly go high in value when they have been overloaded, which that one obviously has (by it's colour).
It's really a matter of feedback - the resistor is overloaded so gets too hot, this makes it's resistance increase, which makes it get hotter still - back to start and repeat
Quite often they get so hot it seriously damages the copper tracks, and you can't even solder to it properly any more - so you need to run wires to further back along the PCB tracks.
If it;s wire wound, they usually open completely (e.g. Fuseable resistors). carbon tend to get larger in vlaue. Metal film puddles. Metal oxide usually breaks completely.
If it;s wire wound, they usually open completely (e.g. Fuseable resistors). carbon tend to get larger in vlaue. Metal film puddles. Metal oxide usually breaks completely.
The type of resistor shown in the picture commonly goes higher in value, due to the reasons in my previous post. I'm making no claims as to what type of resistor it is - just observations of MANY such failed resistors across a wide range of units that used them. I would presume they aren't wirewound though, as it's often easily visible.
It's really a matter of feedback - the resistor is overloaded so gets too hot, this makes it's resistance increase, which makes it get hotter still - back to start and repeat
It depends on if the voltage across the resistor tends to stay constant or the current through it tends to remain constant. You would need to know where it was in the circuit and the nature of the initial fault that caused the resistor to overheat. My guess (As we have been given no information on it's function in the circuit.) is that is sensing current and something else has failed causing the current to increase above the normal level and something else (Other than this resistor) is now limiting the current. so I think P=I² R is more likely apply in this case.
It depends on if the voltage across the resistor tends to stay constant or the current through it tends to remain constant. You would need to know where it was in the circuit and the nature of the initial fault that caused the resistor to overheat. My guess (As we have been given no information on it's function in the circuit.) is that is sensing current and something else has failed causing the current to increase above the normal level and something else (Other than this resistor) is now limiting the current. so I think P=I² R is more likely apply in this case.
The voltage (and dissipation) increases because the resistor has increased value - but the voltage is squared, where the resistance is linear.
A classic example would be emitter resistors in audio amplifiers (or current sharing resistors) - if you're only using 0.1 ohm resistors you can get away with low wattage ones, but if you move up to 1 ohm, then you're in wire wound 4 or 5 watt territory.
As always, you can combine the various formulas to get the result you want, from which ever two variables you have.