Question

The length of a constantan wire P is twice of the constantan wire Q and the ratio of the diameter of wire P to

wire Q is 1:3. Given that the resistance of the wire varies directly as the length and inversely as the square of the
diameter, find the resistance ratio of wire P to wire Q.

Answer 1

Hello,

9/2

Explanation:

Using "la loi de Pouillet", i don't know the name in USA

`R=\rho*(l)/(S) \\\\\rho: resistivity\\\\l: length\\\\S:section\ of \ the\ wire\\\\R_1=\rho*(l_1)/(S_1) \\S_1=\pi*d_1^2*(1)/(4) \\\\\\l_2=2*l_1\\d_2=3*d_1\\\\S_2=\pi*d_2^2*(1)/(4) =\pi*(3*d_1)^2*(1)/(4)=9*S_1\\\\\\(R_1)/(R_2) =(\rho*(l_1)/(S_1))/(\rho*(l_2)/(S_2)) \\\\=(\rho*(l_1)/(S_1))/(\rho*(2*l_1)/(9*S_1)) \\\\\\\boxed{(R_1)/(R_2) =(9)/(2)}`