Tóm tắt :
\(l_1=l_2=l\)
\(S_1=S_2=S\)
\(\rho_1=2,82.10^{-8}\Omega m\)
\(\rho_2=1,72.10^{-8}\Omega m\)
\(R_2=25\Omega\)
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R1 = ?
GIẢI :
Ta có : \(\dfrac{R_1}{R_2}=\dfrac{\rho_1.\dfrac{l}{S}}{\rho_2.\dfrac{l}{S}}=\dfrac{\rho_1}{\rho_2}\)
=> \(\dfrac{R_1}{25}=\dfrac{2,82.10^{-8}}{1,72.10^{-8}}\)
\(\rightarrow R_1=\dfrac{2,82.10^{-8}.25}{1,72.10^{-8}}\approx40,99\left(\Omega\right)\)
Vậy dây nhôm có điện trở là 40,99\(\Omega\)