You are watching: Consider two wires one aluminum with resistivity

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This is college Physics Answers v Shaun Dychko. We"re said that part copper wire and some aluminum wire have the exact same resistance per length. Therefore what is the proportion of their diameters due to the fact that since castle have various resistivities, they"re going come require different diameters in stimulate to have the very same resistance every length. The aluminum is walking to have to be bigger. So resistance that the copper wire is the resistivity of copper multiplied by its length divided by its cross-sectional area. However we desire the resistivity per length. So we multiply this by one end l top top both sides. For this reason we have actually -- that"s an l there. Therefore we have actually resistance the copper per length amounts to resistivity the copper divided by the cross-sectional area and also I"ve substituted pi d squared over 4 in location of A. This is d subscript Cu to say the this is the diameter the the copper wire. Then multiply top and also bottom through four due to the fact that it"s sort of confusing looking to have actually a portion within a fraction. So we have four rho Cu over pi d Cu squared. Then because that aluminum it"s the exact same idea, we"re just substituting Al subscript everywhere. We"re told that resistivity that the copper per length equals -- I need to say the resistance, no the resistivity. So the resistance the copper per length amounts to the resistance that the aluminum per length and also so we have the right to equate this v this. That"s what we"ve excellent here. So a few things cancel here and also we"ll perform it all in one step here. We"re going to get the ratio of the diameter of the aluminum cable to the diameter the the copper cable squared, by multiply top and bottom by -- oh sorry, left and also right I have to say -- by diameter the aluminum squared and also then well, also multiply by pi. Then divide both political parties by four and also that"ll get rid of the four and the pi and also that"ll due to the fact that we"re multiplying both political parties by this fraction here. That"ll eliminate the diameter of the aluminum squared on the right and also will additionally throw in the rho Cu on the bottom there so that that cancels with that, leaving us v diameter of aluminum squared on optimal there and then the resistivity of copper will appear in the denominator ~ above the best hand side. Okay. So, climate take the square source of both sides and also we have actually the proportion of the diameters the the wires is the square root of the proportion of the resistivities the the materials. For this reason we have square root of resistivity the aluminum, 2.65 time ten to the minus eight ohm meters, separated by resistivity the copper 1.72 time ten to the minus eight ohm meters. That means that the diameter that the aluminum will need to be 1.24 times the of the copper.

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