Measuring pickups at the input Jack, how much off?

[rjb]

You only know the value at the Jack 7k.
If the reading is 7k at the jack and you presume the value of the pot is 500k, but it could be 300k.
What is the value of the pickup?

Short answer: A little over 7k. (Assuming the pot is turned to 10.)

Long answer: R_pup = R_jack / (1 - R_jack/R_vol)

I guess your scenario is that you have a Gibson guitar and aren't sure if the volume pot is 500k or 300k.
Let's look at the extremes. A 500k pot with +/-20% tolerance could be 500k+20%=600k and a 300k pot could be 300k-20%=240k

For a 600k pot:
R_pup = 7k / (1 - 7k/600k) = 7.08k

For a 240k pot:
R_pup = 7k / (1 - 7k/240k) = 7.21k

Rounding off, I would say the pickup is in the range of 7.1k to 7.2k.
Or thereabouts.

[big_teee]

I like
7000+5%=7350
7.35K close enough.
T

[rjb]

OK, 5% worst case. But even with a 240K volume pot, the error is only 3%. As you say, close enough.

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Beating a dead horse:
The error will be higher for an overwound pickup and low value volume pot than for an underwound pickup and a high value volume pot.
Example: My MIM Tele originally had a 14K humbucker in the neck(!!!), and a 250K volume pot.
R_jack = (14K x 250K) / (14K + 250K) = 13.26K
R_jack + 5% = 1.05 x 13.26K = 13.9K
So, in this extreme case, the estimate of R_pup = R_jack + 5% would be a tiny bit low.
But it didn't take any measurements at all to tell that the pickup sounded like mud!

[big_teee]

I checked my Les Paul neck pickup yesterday.
It is 7.3k I wound it myself. 500k pot dimed.
It measures 6.95k in the guitar.
6.95+5%=7.2975
Close enough.
T

[Dave H]

You only know the value at the Jack 7k.
If the reading is 7k at the jack and you presume the value of the pot is 500k, but it could be 300k.
What is the value of the pickup?

An easy to remember formula for that is the 'product' over the 'difference'

i.e. 500*7/(500-7) for 500k
or 300*7/(300-7) for 300k

[rjb]

An easy to remember formula for that is the 'product' over the 'difference'

Yup. D'oh.

[big_teee]

i.e. 500*7/(500-7) for 500k

That looks real neat and everything, but it don't come out to my readings, with my DMM.
TBC
T

[Dave H]

That looks real neat and everything, but it don't come out to my readings, with my DMM.

It gives the same answer as the formula rjb used. I can only think there's a measurement error. A 5% change (1/20) implies the parallel resistor is 20 x the pickup resistance i.e. about 145k not 500k