Tóm tắt
\(R=25\Omega\\ d=0,01mm=1.10^{-5}m\\ \rho=5,5.10^{-8}\Omega.m\)
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\(l=?m\)
Giải
\(S=r^2.3,14=\left(1.10^{-5}\right)^2.3,14=3,14.10^{-10}m^2\\ R=\rho\cdot\dfrac{l}{S}\Rightarrow l=\dfrac{R.S}{\rho}=\dfrac{25.3,14.10^{-10}}{5,5.10^{-8}}\approx14,27m\)