Example 27.2 The Resistance of a ConductorProblem Calculate the resistance of an aluminum cylinder that has a length of 10.0 cm and a cross-sectionalarea of 2.00 x 10-4 m2. Repeat the calculation for a cylinder of the same dimensions and made of glasshaving a resistivity of 3.0 x 10102.m.SolutionFrom Equation 27.11 and Table 27.1, we can calculate the resistance of the aluminum cylinder as follows.0.100 m-=(2.82 x 10-8 N. m)R =2.00 x 10-4 m?Enter a number. differs significantly from the correctanswer. Rework your solution from the beginning andcheck each step carefully.Similarly, for glass we find the following.0.100 m(3.0 × 1010 2 · m) GR =p- =2.00 x 10-4 m2As you might guess from the large difference in resistivities, the resistances of identically shaped cylinders ofaluminum and glass differ widely. The resistance of the glass cylinder is 18 orders of magnitude greater thanthat of the aluminum cylinder. Hints: Getting Started | I'm StuckExercise 27.2What if the same amount of aluminum is used to make a cylinder with 4.0 timesthe original length. What will be the resistance of the resulting cylinder (in units ofun)?

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Answered: Problem Calculate the resistance of an…

Physics for Scientists and Engineers, Technology Update (No access codes included)

9th Edition

ISBN:9781305116399

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter27: Current And Resistance

Section: Chapter Questions

Problem 27.4OQ

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Problem Calculate the resistance of an aluminum cylinder that has a length of 10.0 cm and a cross-sectional area of 2.00 x 10-4 m². Repeat the calculation for a cylinder of the same dimensions and made of glass having a resistivity of 3.0 x 1010 N•m.