L=0.5 mA=3.14×10−6 m2R=2.53×10−3 ΩR=ρLAL = 0.5 \;m \\ A = 3.14 \times 10^{-6} \;m^2 \\ R = 2.53 \times 10^{-3} \; \Omega \\ R = \frac{ρL}{A}L=0.5mA=3.14×10−6m2R=2.53×10−3ΩR=AρL
ρ = resistivity of material
ρ=RAL=2.53×10−3×3.14×10−60.5=15.8884×10−9 Ω−mρ = \frac{RA}{L} \\ = \frac{2.53 \times 10^{-3} \times 3.14 \times 10^{-6}}{0.5} \\ = 15.8884 \times 10^{-9} \; \Omega-mρ=LRA=0.52.53×10−3×3.14×10−6=15.8884×10−9Ω−m
According to the resistivity chart resistivity of silver is 15.9×10−9 Ω−m15.9 \times 10^{-9} \; \Omega-m15.9×10−9Ω−m
So, silver was used for formation of wire.
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments
Leave a comment