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Get Full Access to Numerical Analysis - 10 Edition - Chapter 7.6 - Problem 12
Get Full Access to Numerical Analysis - 10 Edition - Chapter 7.6 - Problem 12

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# A coaxial cable is made up of a 0.1 -inch-square inner conductor and 0.5-inch-square

ISBN: 9781305253667 457

## Solution for problem 12 Chapter 7.6

Numerical Analysis | 10th Edition

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Problem 12

A coaxial cable is made up of a 0.1 -inch-square inner conductor and 0.5-inch-square outer conductor. The potential at a point in the cross section of the cable is described by Laplace's equation. Suppose the inner conductor is kept at 0 volts and the outer conductor is kept at 110 volts. Approximating the potential between the two conductors requires solving the following linearsystem. (See Exercise 5 of Section 12.1.) 4 -1 0 0 -1 0 0 0 0 0 0 0 " 1 4 -1 0 0 0 0 0 0 0 0 0 VV| ' 220 " 0 -1 4 -1 0 0 0 0 0 0 0 0 W2 110 0 0 -1 4 0 -1 0 0 0 0 0 0 VV'j 110 H'4 220 1 0 0 0 4 0 -1 0 0 0 0 0 IV'g 110 0 0 0 -1 0 4 0 -1 0 0 0 0 WY, 110 0 0 0 0 -1 0 4 0 -1 0 0 0 W-] no 0 0 0 0 0 -1 0 4 0 0 0 -1 Wg 110 VI'y 220 0 0 0 0 0 0 -1 0 4 -1 0 0 H'10 110 0 0 0 0 0 0 0 0 -1 4 -1 0 Wu 110 0 0 0 0 0 0 0 0 0 -1 4 -1 . w12 . . 220 . 0 0 0 0 0 -1 0 0 0 0 -1 4 Solve the linear system using the conjugate gradient method with TOL = 10 2 and C 1 = D 1

Step-by-Step Solution:
Step 1 of 3

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Step 2 of 3

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##### ISBN: 9781305253667

Numerical Analysis was written by and is associated to the ISBN: 9781305253667. The full step-by-step solution to problem: 12 from chapter: 7.6 was answered by , our top Math solution expert on 03/16/18, 03:24PM. This full solution covers the following key subjects: . This expansive textbook survival guide covers 76 chapters, and 1204 solutions. The answer to “A coaxial cable is made up of a 0.1 -inch-square inner conductor and 0.5-inch-square outer conductor. The potential at a point in the cross section of the cable is described by Laplace's equation. Suppose the inner conductor is kept at 0 volts and the outer conductor is kept at 110 volts. Approximating the potential between the two conductors requires solving the following linearsystem. (See Exercise 5 of Section 12.1.) 4 -1 0 0 -1 0 0 0 0 0 0 0 " 1 4 -1 0 0 0 0 0 0 0 0 0 VV| ' 220 " 0 -1 4 -1 0 0 0 0 0 0 0 0 W2 110 0 0 -1 4 0 -1 0 0 0 0 0 0 VV'j 110 H'4 220 1 0 0 0 4 0 -1 0 0 0 0 0 IV'g 110 0 0 0 -1 0 4 0 -1 0 0 0 0 WY, 110 0 0 0 0 -1 0 4 0 -1 0 0 0 W-] no 0 0 0 0 0 -1 0 4 0 0 0 -1 Wg 110 VI'y 220 0 0 0 0 0 0 -1 0 4 -1 0 0 H'10 110 0 0 0 0 0 0 0 0 -1 4 -1 0 Wu 110 0 0 0 0 0 0 0 0 0 -1 4 -1 . w12 . . 220 . 0 0 0 0 0 -1 0 0 0 0 -1 4 Solve the linear system using the conjugate gradient method with TOL = 10 2 and C 1 = D 1” is broken down into a number of easy to follow steps, and 264 words. This textbook survival guide was created for the textbook: Numerical Analysis, edition: 10. Since the solution to 12 from 7.6 chapter was answered, more than 228 students have viewed the full step-by-step answer.

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