Help with resistivity and wires?

Question

An aluminum wire has the same resistance as a copper wire of the same length. Find the ratio of the aluminum wire's diameter to the copper wire's diameter. (The resistivities of aluminum and copper are 2.82  *10^-8 Ω-m and 1.72 * 10^-8 Ω-m, respectively.)

bonobo · Accepted Answer

R = ρ•L ⁄ A = ρ•L ⁄ (π•D² ⁄ 4) = 4•ρ•L ⁄ (π•D²)

 Ra = 4•(ρa)•L ⁄ [π•(Da)²]

 Rc = 4•(ρc)•L ⁄ [π•(Dc)²]

 Ra = 4 • [2.82 ×10(-8)] • L ⁄ [π • (Da)²]

 Rc = 4 • [1.72 ×10(-8)] • L ⁄ [π • (Dc)²]

        4 • [1.72 ×10(-8)] • L ⁄ [π • (Dc)²]   [1.72 ×10(-8)] • (Da)² 
 Rc ⁄ Ra = 1 = ─────────────────────── = ─────────────── =
        4 • [2.82 ×10(-8)] • L ⁄ [π • (Da)²]   [2.82 ×10(-8)] • (Dc)²

 
 [2.82 ×10(-8)] • (Dc)² = [1.72 ×10(-8)] • (Da)² 

 (Da)² ⁄ (Dc)² = [2.82 ×10(-8)] ⁄ [1.72 ×10(-8)]
         ________________________
 (Da) ⁄ (Dc) = √[2.82 ×10(-8)] ⁄ [1.72 ×10(-8)]

 (Da) ⁄ (Dc) = 1.28

lenore · Answer

You have got the equation wrong. It is Rho*l/a. l is length, and if you Doublee the length, resistance doubles. So meticulously calculate the resistance, as you know rho, l and a. Let this be R. Coming to next part, when length is halved, resistance is halved. When two of these are placed adjacent and ends are tied together, resistance becomes R/4 Now if you cut one of the wires ( to get say 10cms and 18 cms in one of these two wires), resistance is due to one 28cm wire as the others resistance is infinite. So resistance is back to R/2 (should you cut both? That is not asked for, but you can do it. and you will find resistance shoots up t infinity.....

Help with resistivity and wires?

2 Antworten