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A lead (= 5 x 10 S/m) bar of square cross section has a hole bored along its length of 4 m so that its cross section becomes that of Figure 5.5. Find the resistance between the square ends. Solution: R= aS s = d² − xr² = 3² − x( ¹)² = 9 − −— cm². 4 R= 974 µ 5 × 10 (9 x/4) × 104 3 cr I cm 3 cm

A lead (= 5 x 10 S/m) bar of square cross section has a hole bored along its length of 4 m so that its cross section becomes that of Figure 5.5. Find the resistance between the square ends. Solution: R= aS s = d² − xr² = 3² − x( ¹)² = 9 − −— cm². 4 R= 974 µ 5 × 10 (9 x/4) × 104 3 cr I cm 3 cm

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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hi can you explane the calculation of S

Example: A lead (= 5 x 10° S/m) bar of square cross section has a hole bored along its length of 4 m so that its cross section becomes that of Figure 5.5. Find the resistance between the square ends. Solution: R aS S - - - =9 cm². 4 R= = 974 µ 5x 10°(9-x/4) × 10* 3 cm I cm 3 cm
Transcribed Image Text:Example: A lead (= 5 x 10° S/m) bar of square cross section has a hole bored along its length of 4 m so that its cross section becomes that of Figure 5.5. Find the resistance between the square ends. Solution: R aS S - - - =9 cm². 4 R= = 974 µ 5x 10°(9-x/4) × 10* 3 cm I cm 3 cm
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