Răspuns:
Explicație:
[tex]3. R = \rho \cdot\frac{l}{S} = 6.8\cdot10^{-8}\Omega m\cdot\frac{500\cdot10^{-3}m}{3.4\cdot10^{-6}m^2} = 0.01 \Omega\\\\b)l_2 = \frac{l}{4} = 125 mm\\ S_2 = 8\cdot S = 3.4 mm^2 \cdot 8 = 27.2 mm^2\\R_2 = \rho \cdot\frac{l_2}{S_2} = 6.8\cdot10^{-8}\Omega m\cdot\frac{125\cdot10^{-3}m}{27.2\cdot10^{-6}m^2} = 0.0003125 \Omega\\\\4. L = q\cdot U = 60 V \cdot 2000\cdot 10^{-3} C = 120 J[/tex]
