I suppose 1μA x 1MΩ would result in 1V.

It's mostly about the unloaded output of the pickups. Guitar pickups are usually in the 2K to 20K range, and that's resistance of the copper wire. The string moving in the magnetic field is the generator. The coils turn that into voltage. Unloaded output was what I was trying to measure.

 

As gopher suggests, you are not going to hear the difference.
The signal from a guitar pickup might be too low for a soundcard input.
Have you tried it?

its good and low noise
i tried to put preamp(20k input imp) before soundcard, i amplify it by 1,000 still low noise, but when i tried to switch to1M input impedance shoot noise everywhere,but i hear theres a difference it got more high or im just hearing the noise
i want to know what is the consequences if 2:1 ratio of impedance

 

i want to know what is the consequences if 2:1 ratio of impedance

Nothing.
The basic function of dividing by two happens at audio frequencies without any special consequences.

 

I suppose 1μA x 1MΩ would result in 1V.

Guitars with "active pickups" are pushing 3 v (peak-to-peak) out and most amps just are not made for that. It is nice with long cables (cheap cables) but really changes the sound as the amp starts clipping way before it is expected.

 

You may find this of interest:

The Secrets of Electric Guitar Pickups

 

i tested it by sturmming many times and get the average frequency, and tested it on audiotransformer feeding sweep analyzer, theres a bump 1k in 1M input impedance
thats because pickup has a capacitance or inductance?

 

thats because pickup has a capacitance or inductance?

read the link in post #25