Question #177710

At 20°C, the resistance of a silver wire is 30Ω. (Coefficient of temperature=0.0038). (a) If the wire is 1.2m- long, what is the radius of the wire? (b) What is the resistance of the same wire at 10°C? (c) What is the resistivity of the wire at 25°C?




1
Expert's answer
2021-04-02T10:34:01-0400

R=R0(1+αt)R=R_0(1+\alpha t)


(a)


R=ρlAA=ρlR=1.591081.230=6361012 (m2)R=\rho\frac{l}{A} \to A=\rho\frac{l}{R}=1.59\cdot10^{-8}\cdot\frac{1.2}{30}=636\cdot10^{-12}\ (m^2)



A=πr2r=Aπ=63610123.14=1.42105 (m)A=\pi r^2\to \frac{}{}r=\sqrt{\frac{A}{\pi}}=\sqrt{\frac{636\cdot10^{-12}}{3.14}}=1.42\cdot10^{-5}\ (m) . Answer


(b)


R=R0(1+αt)R=R_0(1+\alpha t)\to


R=R0(1+0.003820)R=R_0(1+0.0038\cdot20)


R10=R0(1+0.003810)R_{10}=R_0(1+0.0038\cdot10)\to


R/R10=(1+0.003820)/(1+0.003810)R/R_{10}=(1+0.0038\cdot20)/(1+0.0038\cdot10)


30/R10=1.0366R10=28.94 (Ω)30/R_{10}=1.0366\to R_{10}=28.94\ (\Omega) . Answer


(c)


ρ=ρ0(1+αt)\rho=\rho_0(1+\alpha t)


1.59108=ρ0(1+0.003820)1.59\cdot10^{-8}=\rho_0(1+0.0038\cdot20)


ρ25=ρ0(1+0.003825)\rho_{25}=\rho_0(1+0.0038\cdot25)


1.59108/ρ25=ρ0(1+0.003820)/ρ0(1+0.003825)1.59\cdot10^{-8}/\rho_{25}=\rho_0(1+0.0038\cdot20)/\rho_0(1+0.0038\cdot25)\to


1.59108/ρ25=0.9826ρ25=1.621081.59\cdot10^{-8}/\rho_{25}=0.9826\to \rho_{25}=1.62\cdot10^{-8} . Answer





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