sol3 - ECE473 HOMEWORK #3 SOLUTIONS FALL 2011 Problem 3.1...

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ECE473 HOMEWORK #3 – SOLUTIONS FALL 2011 Problem 3.1 (Problem 4.3 p. 220) A = 19 × πd 2 4 = 19 π × (1 . 5 × 10 - 3 ) 2 4 = 0 . 33576 × 10 - 4 m 2 R = ρ × 1 . 05 l A = 1 . 72 × - 8 × 1 . 05 × 3 × 10 3 0 . 33576 × 10 - 4 = 1 . 614 Ω Problem 3.2 (Problem 4.4 p. 220) (a) A = 795 kcmil = 795 × 1000 cmil 1 kcmil × π/ 4 mil 2 1 cmil × ( 1 in 1000 mil × 2 . 54 cm 1 in × 1 m 100 cm ) 2 = 4 . 028 × 10 - 4 m 2 or A = d 2 cmil = d = A = 795 , 000 = 891 . 63 mil A = πd 2 4 = π 4 × ( 891 . 63 mil × 1 in 1000 mil × 2 . 54 cm 1 in × 1 m 100 cm ) 2 = 4 . 028 × 10 - 4 m 2 (b) R ac, 50 o C = R ac, 75 o C × 50 + T 75 + T = 0 . 0880 × 50 + 228 . 1 75 + 228 . 1 = 0 . 08074 Ω / km Problem 3.3 (Problem 4.5 p. 221) Using Table A.4, the 54/3 ACSR conductor, code name Canary, has an area of 900 kcmil and a 60-Hz resistance at 50 o C equal to R ac, 50 o C = 0
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This document was uploaded on 11/01/2011 for the course ECE 473 at Boise State.

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