Analysis for Challenge 7

First, the resistance of a length of wire is given by

             R = ρl/A

where ρ is the resistivity, l is the length, and A is the cross-sectional area.

The resistivity of gold is approximately 2.2×10-8 Ω•meter.  So, for the wire in question we write

             R = (2.2×10-8 Ω•meter)(190mils)/[π(0.5mil)2]

This gives

             R = 5.3×10-6 Ω•meter/mil

There are 25.4 microns per mil, and 106 microns per meter.  Thus,

             R = (5.3×10-6 Ω•meter/mil)(1mil/25.4micron)(106micron/meter)

             R ~ 0.21 Ω

This works!  It is well within the specified window of 0.15Ω ≤ RSense ≤ 0.25Ω.