\(R1=p1\dfrac{l1}{S1}\Rightarrow p1=\dfrac{R1\cdot S1}{l1}=\dfrac{12\cdot1\cdot10^{-6}}{200}=6\cdot10^{-8}\Omega m\)

Vì hai dây dẫn này cùng chất nên p1 = p2.

\(R2=p2\dfrac{l2}{S2}\Rightarrow l2=\dfrac{R2\cdot S2}{p2}=\dfrac{24\cdot2\cdot10^{-6}}{6\cdot10^{-8}}=800m\)

Chọn D

Lập tỉ lệ ta dc

\(\dfrac{R_1}{R_2}=\dfrac{\rho\dfrac{l_1}{S_1}}{\rho\dfrac{l_2}{S_2}}\Rightarrow\dfrac{12}{24}=\dfrac{\dfrac{200}{1\cdot10^{-6}}}{\dfrac{l_2}{2\cdot10^{-6}}}\Rightarrow l_2=800\left(m\right)\)

chọn D

D. 800m

\(\dfrac{S1}{S2}=\dfrac{R2}{R1}\Rightarrow S2=\dfrac{S1\cdot R1}{R2}=\dfrac{0,05\cdot10^{-6}\cdot8,5}{127,5}\approx3,\left(3\right)\cdot10^{-9}m^2\)

Bài 1: 

\(S=2mm^2=2\cdot10^{-8}m^2\) 

Điện trở của dây dẫn là:

\(R=\rho\dfrac{l}{S}=1,7\cdot10^{-8}\cdot\dfrac{100}{2\cdot10^{-8}}=85\Omega\) 

Bài 2:

\(S_1=5mm^2=5\cdot10^{-8}m^2\)

\(S_2=0,5\cdot10^{-8}m^2\)

Ta có:

\(\dfrac{S_1}{S_2}=\dfrac{R_2}{R_1}\)

\(\Rightarrow R_2=\dfrac{S_1R_1}{S_2}=\dfrac{5\cdot10^{-8}\cdot8,5}{0,5\cdot10^{-8}}=85\Omega\)

a) \(R_1=\rho\dfrac{l_1}{S_1},R_2=\rho\dfrac{l_2}{S_2}\)\(\Rightarrow\dfrac{R_1}{R_2}=\dfrac{l_1\cdot S_2}{l_2\cdot S_1}=6\)

b) \(\left\{{}\begin{matrix}\dfrac{R_1}{R_2}=6\\R_1+R_1=14\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}R_1=12\Omega\\R_2=2\Omega\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{R1}{R2}=\dfrac{\dfrac{pL1}{S1}}{\dfrac{pL2}{S2}}=\dfrac{l1.S2}{l2.S1}=\dfrac{3l2.2S1}{l2.S1}=6\\R1+R2=14\left(\Omega\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}R1=6R2\\6R2+R2=14\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}R1=12\Omega\\R2=2\Omega\end{matrix}\right.\)

Tóm tắt :

l₁ = 4m

R₁ = 2Ω

R₂ = 20Ω

l₂ = ?

                               Ta có : \(\dfrac{l_1}{l_2}=\dfrac{R_1}{R_2}\)

                                    Hay \(\dfrac{4}{l_2}=\dfrac{2}{20}\)

                                   ⇒ l₂ = \(\dfrac{4.20}{2}=40\) (m)

                                    ⇒  Chọn câu D

 Chúc bạn học tốt

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

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