together on both sides. Therefore we have actually -- that's an l there. So we have actually resistance of copper per length equates to resistivity of copper separated by the cross-sectional area and I've substituted pi d squared over four in location of A. This is d subscript Cu come say that this is the diameter the the copper wire. Climate multiply top and bottom by four due to the fact that it's sort of messy 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 same idea, we're just substituting Al subscript everywhere. We're told that resistivity the the copper every length amounts to -- I must say the resistance, no the resistivity. So the resistance of copper per length equals the resistance the the aluminum every length and also so we deserve to equate this through this. That's what we've excellent here. Therefore a few things cancel here and we'll execute it all in one action here. We're walking to acquire the ratio of the diameter of the aluminum wire to the diameter of the copper wire squared, by multiplying top and bottom through -- five sorry, left and right I have to say -- by diameter the aluminum squared and also then well, likewise multiply by pi. Then division both sides by four and that'll get rid of the four and also the pi and also that'll because we're multiply both political parties by this portion here. That'll get rid of the diameter the the aluminum squared top top the right and also will also throw in the rho Cu on the bottom there so the that cancels through that, leaving us with diameter that aluminum squared on peak there and then the resistivity of copper will show up in the denominator top top the appropriate 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 source of the ratio of the resistivities of the materials. Therefore we have actually square root of resistivity the aluminum, 2.65 times ten to the minus eight ohm meters, separated by resistivity of copper 1.72 time ten come the minus eight ohm meters. That method that the diameter of the aluminum will need to be 1.24 times that of the copper.">


You are watching: Consider two wires one aluminum with resistivity

disclose Answer

Why is this button here? Quiz mode is a opportunity to shot solving the problem first on your own before viewing the solution. Among the complying with will more than likely happen: You acquire the answer. Congratulations! that feels good! There can still be an ext to learn, and also you could enjoy compare your problem solving method to the best practices demonstrated in the solution video. Friend don"t get the answer. This is OK! In fact it"s awesome, in spite of the an overwhelming feelings you might have around it. When you don"t obtain the answer, your mind is ready for learning. Think about how lot you really want the solution! her mind will gobble it up once it watch it. Attempting the difficulty is like trying to rally the piece of a puzzle. If girlfriend don"t gain the answer, the gaps in the puzzle are concerns that are ready and searching to it is in filled. This is an active process, wherein your psychic is turn on - discovering will happen! If you great to show the answer automatically without having actually to click "Reveal Answer", you may produce a cost-free account. Quiz mode is disabled by default, but you can check the allow Quiz setting checkbox when editing and enhancing your file to re-enable it any kind of time friend want. University Physics Answers cares a lot about academic integrity. Quiz mode is encourage to usage the solutions in a way that is most valuable for her learning.


*

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.
1PE2PE3PE4PE5PE6PE7PE8PE9PE10PE11PE12PE13PE14PE15PE16PE17PE18PE19PE20PE21PE22PE23PE24PE25PE26PE27PE28PE29PE30PE31PE32PE33PE34PE35PE36PE37PE38PE39PE40PE41PE42PE43PE44PE45PE46PE47PE48PE49PE50PE51PE52PE53PE54PE55PE56PE57PE58PE59PE60PE61PE62PE63PE64PE65PE66PE67PE68PE69PE70PE72PE73PE74PE75PE76PE77PE78PE79PE80PE81PE82PE83PE84PE85PE86PE87PE88PE89PE90PE91PE92PE93PE95PE96PE

Get the recent updates from college Physics Answers


Email deal with
Get updates


See more: The Most Distinctive Architectural Paradigm For The Hindu Temple Is

*