Câu 1.

Điện trở dây: \(R=\rho\cdot\dfrac{l}{S}=5,5\cdot10^{-8}\cdot\dfrac{100}{1\cdot10^{-6}}=5,5\Omega\)

Cường độ dòng điện qua dây: \(I=\dfrac{U}{R}=\dfrac{22}{5,5}=4A\)

Câu 2.

Công suất toả nhiệt của bếp: 

\(P=U\cdot I=RI^2=40\cdot1,2^2=57,6\Omega\)

Điện trở dây dẫn: \(R=\rho\cdot\dfrac{l}{S}\)

\(\Rightarrow\)Chiều dài dây: \(l=\dfrac{R\cdot S}{\rho}\)

Tiết diện dây: \(S=\pi\cdot R^2=\pi\cdot\left(0,01\cdot10^{-3}\right)^2=3,14\cdot10^{-10}m^2\)

\(l=\dfrac{25\cdot3,14\cdot10^{-10}}{5,5\cdot10^{-8}}=0,1427m=14,27cm\)

\(R=\rho\cdot\dfrac{l}{S}=1\cdot10^{-8}\cdot\dfrac{100}{2\cdot10^{-6}}=0,5\Omega\)

\(R=p\dfrac{l}{S}=1\cdot10^{-8}\dfrac{100}{2\cdot10^{-6}}=0,5\Omega\)

\(S=\pi\dfrac{d^2}{4}=\pi\dfrac{0,04^2}{4}=1,256\cdot10^{-3}mm^2=1,256\cdot10^{-3}\cdot10^{-6}m^2\)

\(R=p\dfrac{l}{S}\Rightarrow l=\dfrac{R\cdot S}{p}=\dfrac{40\cdot1,256\cdot10^{-3}\cdot10^{-6}}{5,5\cdot10^{-8}}\approx1\left(m\right)\)

\(\Rightarrow\left\{{}\begin{matrix}1.\Rightarrow L=\dfrac{RS}{p}=\dfrac{50.\left(\dfrac{2.10^{-5}}{2}\right)^2.\pi}{5,5.10^{-8}}=0,28m\\2.\Rightarrow R=\dfrac{pL}{S}=\dfrac{1,6.10^{-8}.50}{0,0005^2.\pi}\approx1,01\Omega\\3.\Rightarrow S=\dfrac{pL}{R}=\dfrac{0,4.10^{-6}.60}{\dfrac{U}{I}}=2.10^{-7}m\\\end{matrix}\right.\)

\(R=p\dfrac{l}{S}=5,5\cdot10^{-8}\dfrac{12:100}{0,000165\cdot10^{-6}}=40\Omega\)

\(\Rightarrow I=\dfrac{U}{R}=\dfrac{220}{40}=\dfrac{11}{2}A\)

\(R=\rho\dfrac{l}{S}=2,8.10^{-8}.\dfrac{100}{2.10^{-6}}=1,4\left(\Omega\right)\)