Question

Considerem-se dois fios condutores do mesmo material: o primeiro com diâmetro igual a 0,6 mm, comprimento 18 m e resistência 16 Ω e o segundo com diâmetro igual a 0,4 mm, comprimento igual a 4 m e resistência igual a x Ω.Com base nessas informações, conclui-se que x é igual a:


a) 6
b) 8
c) 10
d) 12
e) 18​

Answer 1

$\boxed{\begin{array}{l}\underline{\rm 2^a~Lei~de~Ohm}\\\huge\boxed{\boxed{\boxed{\boxed{\sf R=\rho\dfrac{\ell}{A}}}}} \\\sf d_1=0,6mm =0,6\cdot 10^{-3}m=6\cdot10^{-4}m\implies r_1=3\cdot10^{-4}m\\\sf d_2=0,4mm=0,4\cdot10^{-3}m=4\cdot10^{-4}m\implies r_2=2\cdot10^{-4}m\\\sf A_1=\pi r_1^2\\\sf A_1=\pi\cdot(3\cdot10^{-4})^2\\\sf A_1=9\pi\cdot10^{-8} m^2\\\sf R=\rho\dfrac{\ell}{A}\\\\\sf 16=\rho\cdot\dfrac{18}{9\pi\cdot10^{-8}}\\\\\sf 16=\dfrac{2\rho}{\pi}10^{8}\\\\\sf 2\rho\cdot 10^8=16\pi\\\\\sf \rho=\dfrac{16\pi}{2\cdot10^8}\\\\\sf \rho=8\pi\cdot 10^{-8}\end{array}}$

$\boxed{\begin{array}{l}\sf A_2=\pi\cdot r_2^2\\\sf A_2=\pi\cdot(2\cdot10^{-4})^2\\\sf A_2=4\pi\cdot 10^{-8}~m^2\\\sf R=\rho\cdot\dfrac{\ell_2}{A_2}\\\\\sf x=\diagup\!\!\!8 \diagdown\!\!\!\!\pi\cdot\diagup\!\!\!\!\!10^{-8}\cdot\dfrac{4}{\diagup\!\!\!4\diagdown\!\!\!\!\pi\cdot\diagup\!\!\!\!\!10^{-8}}\\\sf x=8\\\huge\boxed{\boxed{\boxed{\boxed{\sf\maltese~alternativa~b}}}} \end{array}}$