Given : l = 6 m, D = 0.5 mm,
r = 0.25 mm = 0.25 × 10-3 m, R = 50 Ω
To find :
i. Resistivity (ρ)
ii. Conductivity (σ)
Formulae :
Calculation : From formula (i),
= {antilog [log 50 + log 3.142 + 21og 0.25 - log 6]} × 10-6
= {antilog [ 1.6990 + 0.4972 + 2(1.3979) -0.7782]} × 10-6
= {antilog [2.1962 + 2 .7958 – 0.7782]} × 10-6
= {antilog [0.9920 – 0.7782]} × 10-6
= {antilog [0.2138]} × 10-6
= 1.636 × 10-6 Ω/m
From formula (ii),
….(Using reciprocal from log table)
= 6.157 × 105 m/Ω
