cableswire
I have a microphone cable with 2 conductors in shielded twisted pair in the center. Material is copper.
Each conductor is made of 28 strands x 0.10 mm thick.
They're covered in braided shield with 84 x 0.10 mm.
Outer diameter of the cable is 6 mm.
What is the AWG of the cable?
(28*0.10)*2+84*0.10 OD:6 mm
I might be barking up the wrong tree here, but if you've got any extra space at all at the far end of the cable, you'd probably do better to digitize the data there. You can then use a robust signaling method over a single conductor pair to deliver the data upstream. Coaxial antenna cable can be gotten in really tiny diameters, it's very noise-resistant, and can be quite strong.
Most of the major analog IC manufacturers sell multi-channel serial ADC chips that you could trivially integrate (again, assuming you have space).
I suspect it's an issue with your measurement technique. The specifications from the product link provided list the resistance, and there are only two values, one for the shield, and one for the conductors. That each conductor isn't listed separately implies they are all the same.
Further, that this is sold as a microphone further implies that the resistance is the same for each conductor. Were it not, it would not work as a balanced cable.
Measuring small resistances (certainly <1Ω) is difficult with an ordinary ohmmeter. In this range, the resistance of the test leads and the contact resistance can be significant. The precision of some meters may not be sufficient in this range. It can take some care to get an accurate measurement.
Best Answer
From your comment:
You don't. A single 0.10mm strand is AWG38.
The formula is based on AWG36 wire being exactly 0.005" (5 mils) in diameter. All other values are based on an exponential curve:
$$diam = 5\text{ mils} \cdot 92^{\frac{36 - AWG}{39}}$$
Solving for AWG as a function of diameter gives:
$$AWG = 36 - 39\cdot log_{92} \frac{diam}{5\text{ mils}}$$
28 strands of AWG38 has the same cross-sectional area as AWG24.