\(\Rightarrow R=\dfrac{pl}{S}\Rightarrow\dfrac{U}{I}=\dfrac{60}{2}=30=\dfrac{0,4.10^{-6}.L}{0,5.10^{-6}}\Rightarrow L=37,5m\)

a. \(\left[{}\begin{matrix}R1=p1\dfrac{l1}{S1}=0,4.10^{-6}\dfrac{100}{0,5.10^{-6}}=80\left(\Omega\right)\\R2=p2\dfrac{l2}{S2}=0,4.10^{-6}\dfrac{100}{0,5S1}=0,4.10^{-6}\dfrac{100}{0,5.0,5.10^{-6}}=160\left(\Omega\right)\end{matrix}\right.\)

b. \(I=I1=I2=\dfrac{U}{R}=\dfrac{40}{80+160}=\dfrac{1}{6}\left(A\right)\left(R1ntR2\right)\)

\(\Rightarrow\left[{}\begin{matrix}U1=I1.R1=\dfrac{1}{6}.80=\dfrac{40}{3}\left(V\right)\\U2=I2.R2=\dfrac{1}{6}.160=\dfrac{80}{3}\left(V\right)\end{matrix}\right.\)

\(R=\rho\dfrac{l}{S}\Rightarrow S=\dfrac{\rho.l}{R}=\dfrac{0,4.10^{-6}.20}{160}=5.10^{-8}\left(m^2\right)\)

\(S=\dfrac{\pi d^2}{4}\Rightarrow d^2=\dfrac{4S}{\pi}=\dfrac{4.5.10^{-8}}{\pi}=6,4.10^{-8}\)

\(\Rightarrow d=2,5.10^{-4}\left(m\right)=0,25\left(mm\right)\)

Tiết diện của dây: \(R=p\dfrac{l}{S}\Rightarrow S=\dfrac{p.l}{R}=\dfrac{0,4.10^{-6}20}{160}=5.10^{-8}m^2\)

Đường kính của dây: \(S=\pi\left(\dfrac{d}{2}\right)^2\Rightarrow d^2=\dfrac{2^2S}{\pi}=\dfrac{4.5.10^{-8}}{\pi}\simeq6,4.10^{-8}\)

\(\Rightarrow d=2,5.10^{-4}m=0,25mm\)

S=πd24⇒d2=4Sπ=4.5.10−8π=6,4.10−8S=πd24⇒d2=4Sπ=4.5.10−8π=6,4.10−8

⇒d=2,5.10−4(m)=0,25(mm)

có điện trở R2 là:

ta có : \(\dfrac{R_1}{R_2}=\dfrac{l_1.S_2}{l_2.S_1}\Rightarrow R_2=R_1:\dfrac{l_1.S_2}{l_2.S_1}\Leftrightarrow50:\dfrac{20.0,5}{10.0,1}=5\)

Vậy.....

Điện trở của dây là:

\(R=\rho\dfrac{l}{S}=0,4.10^{-6}.\dfrac{20}{0,05.10^{-6}}=160\left(\Omega\right)\Rightarrow B\)

\(\text{Tóm tắt:}\)

\(l=20m\)

\(S=0,5mm^2=5.10^{-7}m^2\)

\(I=3A\)

\(f=1,7.10^{-8}\)

\(-----\)

\(U=?\)

\(\text{Điện trở của dây dẫn là:}\)

\(R=f.\dfrac{l}{S}\Rightarrow R=1,7.10^{-8}.\dfrac{20}{5.10^{-7}}=0,68\left(\Omega\right)\)

\(\text{Hiệu điện thế của dây dẫn là:}\)

\(R=\dfrac{U}{I}\Rightarrow U=R.I=0,68.3=2,04\left(V\right)\)

\(R=\dfrac{\delta}{s}l=\dfrac{0,4.10^{-6}}{0,5.10^{-6}}30=24\Omega\)