l on both sides. Therefore we have -- that's an l there. Therefore we have resistance of copper every length equals resistivity the copper divided by the cross-sectional area and also I've substituted pi d squared over four in place of A. This is d subscript Cu come say that this is the diameter that the copper wire. Then multiply top and also bottom through four due to the fact that it's type of messy looking to have a fraction within a fraction. So we have 4 rho Cu end pi d Cu squared. Then because that aluminum it's the very same idea, we're just substituting Al subscript everywhere. We're told the resistivity the the copper every length equates to -- I should say the resistance, not the resistivity. So the resistance of copper every length equates to the resistance of the aluminum every length and also so we deserve to equate this with this. That's what we've excellent here. Therefore a few things release here and we'll execute it all in one action here. We're walking to acquire the proportion of the diameter of the aluminum cable to the diameter that the copper wire squared, by multiplying top and also bottom by -- five sorry, left and right I should say -- through diameter that aluminum squared and then well, also multiply through pi. Then division both political parties by four and also that'll eliminate the four and also the pi and that'll due to the fact that we're multiplying both political parties by this fraction here. That'll remove the diameter the the aluminum squared on the right and also will also throw in the rho Cu on the bottom over there so the that cancels v that, leaving us through diameter the aluminum squared on optimal there and also then the resistivity of copper will appear in the denominator top top the ideal hand side. Okay. So, climate take the square root of both sides and we have actually the proportion of the diameters that the wires is the square source of the proportion of the resistivities of the materials. So we have square root of resistivity that aluminum, 2.65 times ten come the minus eight ohm meters, separated by resistivity that copper 1.72 time ten come the minus eight ohm meters. That method that the diameter that the aluminum will need to be 1.24 times the of the copper.">

You are watching: Consider two wires one aluminum with resistivity This is university Physics Answers with Shaun Dychko. We"re said that part copper wire and also some aluminum wire have actually the very same resistance every length. For this reason what is the proportion of your diameters because since they have various resistivities, they"re going to require various diameters in stimulate to have the very same resistance per length. The aluminum is walk to need to be bigger. For this reason resistance the the copper wire is the resistivity the copper multiplied by that is length separated by its cross-sectional area. However we desire the resistivity every length. So we multiply this by one over l ~ above both sides. Therefore we have -- that"s an l there. For this reason we have actually resistance that copper every length amounts to resistivity the copper divided by the cross-sectional area and also I"ve substituted pi d squared over four in place of A. This is d subscript Cu come say that this is the diameter of the copper wire. Then multiply top and bottom through four due to the fact that it"s sort of confusing looking to have a fraction within a fraction. So we have 4 rho Cu over pi d Cu squared. Then because that aluminum it"s the very same idea, we"re simply substituting Al subscript everywhere. We"re told that resistivity that the copper every length equates to -- I must say the resistance, no the resistivity. So the resistance that copper per length equates to the resistance of the aluminum every length and also so we can equate this through this. That"s what we"ve excellent here. Therefore a few things release here and we"ll execute it every in one action here. We"re walking to obtain the proportion of the diameter the the aluminum wire to the diameter of the copper wire squared, by multiply top and also bottom through -- oh sorry, left and right I must say -- through diameter of aluminum squared and then well, additionally multiply through pi. Then division both sides by four and also that"ll eliminate the four and also the pi and that"ll because we"re multiplying both political parties by this portion here. That"ll remove the diameter the the aluminum squared on the right and also will also throw in the rho Cu top top the bottom over there so the that cancels through that, leaving us with diameter the aluminum squared on top there and also then the resistivity that copper will show up in the denominator ~ above the ideal hand side. Okay. So, then take the square source of both sides and we have actually the ratio of the diameters the the wires is the square root of the ratio of the resistivities of the materials. For this reason we have actually square root of resistivity that aluminum, 2.65 time ten come 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 of the aluminum will have to be 1.24 times the of the copper.
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