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DaveGrima

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You can't. Not directly, anyway. However, you can calculate it. Get a schematic diagram. Set the selector switch to the pickup you want to measure. Volume all the way up. Measure the DC resistance. Then, looking at the diagram, figure out the parallel paths and their resistances. Using those, calculate the DC resistance of the pickup.

For instance, IIRC, a Les Paul typically has a 250K volume pot. The tone pot shouldn't matter, as it has a capacitor in series with it. So, if you were to measure 9.6K ohms, you could calculate that the DC resistance of the pickup is 10K, because 250K in parallel with 10K gives you 9.6K. If the pickup is 9K, you'll get 8.7K. And so on.

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Hmm. I just put one of the multimeters tips to the pickup's lead wire at the volume pot and the other tip to the lug right next to it. I got 8.05 for the bridge, and 7.78 for the neck. The pickups are a 490t and 490r so that's about what they should be. Are you saying theyre a little higher? BTW, typically humbuckers use 500k pots don't they?

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Yes, you're right about the 500K pot. I have no idea what your pickups' DC resistance ought to be. What I am saying is that whenever two resistances are in parallel, the total resistance will be less than either resistance alone. The resistance of the pickup will be much less than that of the parallel resistance if the volume is full up. 500K will have less of an effect than my 250K example. If you got 8.05K with the pickup in parallel with 500K, then the pickup's DC resistance is 8.18K. For 7.78K, 7.90K However, I don't understand your measurement points. How is your tone pot wired?

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I cant explain. Its just the way Gibson wired it at the factory. Pretty standard probably. I dont know exactly what points i touched. I know one of the tips was on the contact point where the pickup was wired to the pot, the middle lug. The other point I just kinda poked around on the pot till I got a reading.

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