Question

VA ROG!!!!Dau coroana.

Answer 1

$\displaystyle\\\rm\\\rho_1=28,2 \ n \Omega \times m = 28,2 \times 10^{-9} \ \Omega \times m \\\\l_1 = 2,5 \ m \\\\a = 1 \cm = 10^{-2} \ m \\\\b = 5 \ cm = 5 \times 10^{-2} \ m \\\\R_1 = \frac{\rho_1 \times l_1 }{S_1} = \frac{\rho_1 \times l_1}{a \times b} = \frac{28,2 \times 10^{-9} \times 5}{2 \times 10^{-2} \times 5 \times 10^{-2}} = \frac{141}{10^6} \ ohmi$

$\displaystyle \\\rm\\\rho_2= 98 \ n \Omega \times m = 98 \times 10^{-9} \ \Omega \times m \\\\ d = 15 \ mm = 15 \times 10^{-3} \ m \\\\R_2 =R_1 = \frac{141}{10^6} \ \Omega \\ \\S_2 = 2 \pi r^2 = 2 \pi \left(\frac{d}{2}\right)^2 \\\\R_2 = \frac{\rho_2 \times l_2}{S_2} \implies l_2 = \frac{R_2 \times S_2}{\rho_2}$

$\displaystyle\\\rm\\l_2 = \frac{141 \times 2\pi \times \left(\frac{15 \times 10^{-3}}{2}\right)^2}{98 \times 10^{-9} \times 10^6} = \frac{199233}{392000} = 0,5 \ m$

Matei.