You are watching: Consider two wires one aluminum with resistivity

reveal Answer

Why is this button here? Quiz mode is a chance to shot solving the problem very first on her own before viewing the solution. One of the complying with will probably happen: You get the answer. Congratulations! it feels good! There might still be more to learn, and also you could enjoy to compare your trouble solving strategy to the finest practices prove in the systems video. Girlfriend don"t gain the answer. This is OK! In reality it"s awesome, regardless of the difficult feelings you can have around it. Once you don"t obtain the answer, her mind is all set for learning. Think about how much you really desire the solution! your mind will certainly gobble it up as soon as it watch it. Attempting the difficulty is like trying to rally the piece of a puzzle. If you don"t gain the answer, the gaps in the puzzle are concerns that room ready and also searching to it is in filled. This is an active process, wherein your psychic is rotate on - learning will happen! If you wish to present the answer automatically without having actually to click "Reveal Answer", you may develop a complimentary account. Quiz mode is disabled through default, but you can inspect the enable Quiz mode checkbox when editing and enhancing your profile to re-enable it any time friend want. College Physics Answers cares a lot about academic integrity. Quiz setting is encouragement to use the remedies in a way that is most beneficial for your learning.

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.

1PE2PE3PE4PE5PE6PE7PE8PE9PE10PE11PE12PE13PE14PE15PE16PE17PE18PE19PE20PE21PE22PE23PE24PE25PE26PE27PE28PE29PE30PE31PE32PE33PE34PE35PE36PE37PE38PE39PE40PE41PE42PE43PE44PE45PE46PE47PE48PE49PE50PE51PE52PE53PE54PE55PE56PE57PE58PE59PE60PE61PE62PE63PE64PE65PE66PE67PE68PE69PE70PE72PE73PE74PE75PE76PE77PE78PE79PE80PE81PE82PE83PE84PE85PE86PE87PE88PE89PE90PE91PE92PE93PE95PE96PE

## Get the recent updates from university Physics Answers

Email resolve

Get updates

See more: *.Local, 169.254/16 - Your Browser Can'T Play This Video