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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.

Respuesta :

caylus

Answer:

Hello,

9/2

Step-by-step explanation:

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

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

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