\(R=\rho\dfrac{l}{S}=1,7.10^{-8}.\dfrac{100}{0,1.10^{-6}}=17\left(\Omega\right)\)

\(I=\dfrac{U}{R}=\dfrac{220}{17}\left(A\right)\)

\(R=p\dfrac{l}{S}=1,7.10^{-8}\dfrac{100}{0,1.10^{-6}}=17\left(\Omega\right)\)

\(\Rightarrow I=\dfrac{U}{R}=\dfrac{220}{17}\simeq12,9\left(A\right)\)

\(R=\rho\dfrac{l}{S}==1,10.10^{-6}.\dfrac{60}{0,15.10^{-6}}=440\left(\Omega\right)\)

\(I=\dfrac{U}{R}=\dfrac{220}{440}=0,5\left(A\right)\)

a)Điện trở dây: 

\(R=\rho\cdot\dfrac{l}{S}=0,4\cdot10^{-6}\cdot\dfrac{100}{0,5\cdot10^{-6}}=80\Omega\)

b)Cường độ dòng điện qua dây dẫn:

\(I=\dfrac{U}{R}=\dfrac{120}{80}=1,5A\)

\(R=\rho\dfrac{l}{S}=1,7.10^{-8}\dfrac{50}{0,34.10^{-6}}=2,5\left(\Omega\right)\)

\(I=\dfrac{U}{R}=\dfrac{220}{2,5}=88\left(A\right)\)

\(R=p\dfrac{l}{S}=1,7.10^{-8}\dfrac{50}{0,34.10^{-6}}=2,5\Omega\)

\(\Rightarrow I=U:R=220:2,5=88A\)

\(1.\Rightarrow\left\{{}\begin{matrix}\dfrac{U1}{U2}=\dfrac{I1}{I2}\Leftrightarrow\dfrac{48}{U2}=\dfrac{1,2}{1,6}\Rightarrow U2=64V\\\dfrac{U1}{U2}=\dfrac{I1}{I2}\Leftrightarrow\dfrac{48}{U2}=\dfrac{1,2}{2,4}\Rightarrow U2=96V\\\end{matrix}\right.\)

\(2.\Rightarrow\left\{{}\begin{matrix}V=SL\Rightarrow L=\dfrac{V}{S}=\dfrac{\dfrac{m}{D}}{S}=\dfrac{\dfrac{0,5}{8900}}{10^{-6}}=56m\\R=\dfrac{pL}{S}=\dfrac{1,7.10^{-8}.56}{10^{-6}}=0,952\Omega\\\end{matrix}\right.\)

Điện trở: \(R=p\dfrac{l}{S}=1,7.10^{-8}\dfrac{100}{2.10^{-6}}=0,85\Omega\)

Cường độ dòng điện: \(I=U:I=3,4:0,85=4A\)

\(R=\rho\cdot\dfrac{l}{S}=1,7\cdot10^{-8}\cdot\dfrac{100}{2\cdot10^{-6}}=0,85\Omega\)

Mắc vào hđt 3,4V cần một dòng điện qua dây:

  \(I=\dfrac{U}{R}=\dfrac{3,4}{0,85}=4A\)

Đổi : \(0,3mm^2=3.10^{-7}m^2\)

\(R=\rho.\dfrac{l}{S}=1,10.10^{-6}.\dfrac{30}{3.10^{-7}}=110\left(ÔM\right)\)

\(\Rightarrow I=\dfrac{U}{R}=\dfrac{220}{110}=2\left(A\right)\)