What length of German silver wire, diameter 0.050 cm, is needed to make a 28 Ωresistor, if the resistivity of German silver is 2.2 × 〖10〗^(-7) Ω m

Question

(show work)

Answer 1

Given :
Resistance R = 28Ω
A is the cross-section area and, ρ is resistivity of the wire.
To find length l
A=π(0.00025)
2
㎡=625π×10 −10
28=2.2×10 −7
So, l=28×625π×10
−3
/2.2=24.98=25m