Can someone explain why? Brightness has anything to do with pot value?
When a pot is fully on (10) then the ac current will take the path of the least resistance, which is from pick-up and out to the amp (as the resistance at that junction (setting ten) is nil.
When the pot is turned back to five, then in the case of a 250K pot, the resistance will be 125K to earth - hence half the signal will be drained away.
If a 500K pot is set half-way, then the resistance is 250k - but it's still split 50-50.
I suppose it's something to do with the total resistance facing the pot (250K or 500k), but at full volume, when the pot resistance is virtually nil - all the pick-up current is sent to the amp.
So are they both as bright when at full output?
lee
When a pot is fully on (10) then the ac current will take the path of the least resistance, which is from pick-up and out to the amp (as the resistance at that junction (setting ten) is nil.
When the pot is turned back to five, then in the case of a 250K pot, the resistance will be 125K to earth - hence half the signal will be drained away.
If a 500K pot is set half-way, then the resistance is 250k - but it's still split 50-50.
I suppose it's something to do with the total resistance facing the pot (250K or 500k), but at full volume, when the pot resistance is virtually nil - all the pick-up current is sent to the amp.
So are they both as bright when at full output?
lee
