Đoạn dây được cắt thành hai đoạn bằng nhau nên \(R_1=R_2=\dfrac{R}{2}\)

Điện trở của đoạn mạch:

\(R_{tđ}=\dfrac{R_1\cdot R_2}{R_1+R_2}=\dfrac{\dfrac{R}{2}\cdot\dfrac{R}{2}}{\dfrac{R}{2}+\dfrac{R}{2}}=\dfrac{R^2}{2}:\dfrac{2R}{2}=\dfrac{R}{2}\)

Ta có  \(l=\dfrac{RS}{\rho}\Leftrightarrow R=\dfrac{l\rho}{S}\)

Cắt l thành n mảnh \(\Leftrightarrow l'=\dfrac{l}{4}\) \(\Leftrightarrow R'=\dfrac{\dfrac{l}{n}\rho}{S}=\dfrac{R}{n}=\dfrac{216}{n}\\ \Leftrightarrow R_{td}=\dfrac{R'.R'^n}{R'+R'^n}\Leftrightarrow6=\dfrac{\dfrac{216}{n}.\left(\dfrac{216}{n}\right)^n}{\dfrac{216}{n}+\left(\dfrac{216}{n}\right)^n}\Leftrightarrow n\approx0,27\)

\(a,\Rightarrow R=\dfrac{pl}{S}=\dfrac{0,4.10^{-6}.37,5}{0,5.10^{-6}}=30\Omega\)

b,\(\Rightarrow Rtd=\dfrac{x\left(30-x\right)}{x+30-x}=\dfrac{x\left(30-x\right)}{30}\)

\(\Rightarrow30Rtd=x\left(30-x\right)\Rightarrow-x^2+30x-30Rtd=0\)

\(\Rightarrow\Delta\ge0\Rightarrow30^2-4\left(-30Rtd\right).\left(-1\right)\ge0\)

\(\Rightarrow900-120Rtd\ge0\Leftrightarrow-120Rtd\ge-900\Leftrightarrow Rtd\le7,5\Omega\)

\(\Rightarrow Max\left(Rtd\right)=7,5\Leftrightarrow x=15\left(\Omega\right)\)

\(\Rightarrow\)gia tri 2 phan lan luot la \(\left\{{}\begin{matrix}R1=15\Omega\\R2=30-15=15\Omega\end{matrix}\right.\)

Câu 41.

Điện trở dây:

\(R=\rho\cdot\dfrac{l}{S}=\rho\cdot\dfrac{l}{\pi\cdot\dfrac{d^2}{4}}=2,8\cdot10^{-8}\cdot\dfrac{3,14}{\pi\cdot\left(\dfrac{2\cdot10^{-3}}{2}\right)^2}=0,028\Omega\)

Câu 39.

\(R_{tđ}=\dfrac{R_1\cdot R_2}{R_1+R_2}=\dfrac{8\cdot6}{8+6}=\dfrac{24}{7}\Omega\)

\(I=\dfrac{U}{R}=\dfrac{3,2}{\dfrac{24}{7}}=\dfrac{14}{15}A\)

Câu 40.

Điện trở qua dây: \(R_2=\dfrac{U}{I}=\dfrac{12}{1,5}=8\Omega\)

Chiều dài dây:

\(\dfrac{R_1}{R_2}=\dfrac{l_1}{l_2}\Rightarrow l_2=\dfrac{R_2\cdot l_1}{R_1}=\dfrac{8\cdot6}{2}=24m\)

Câu 36.

\(R=\rho\cdot\dfrac{l}{S}=1,7\cdot10^{-8}\cdot\dfrac{2}{2\cdot10^{-6}}=0,017\Omega\)

Hỏi thằng e họ nó bảo B:)

Chọn đáp án D.

6\(\Omega\)