Mathcad - 13-01 to 06 - Problems 13-1 through 13-6...

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R 2 ρ al l A 2 = c. R 3 ρ al l A 3 = R 2 0.137ohm = R 1 0.166ohm = R 3 0.638ohm = d. R 4 ρ cu l A 4 = e. R 5 ρ cu l A 5 = R 5 0.041ohm = R 4 0.062ohm = 13-3: Temperature corrections for copper and aluminum at 40 deg C are: α cu 234.5 K = α al 228.1 K = for copper at 40deg C: factor cu α cu 40 K + α cu 20 K + = factor cu 1.079 = for aluminum at 40deg C: factor al α al 40 K + α al 20 K + = factor al 1.081 = Problems 13-1 through 13-6 Solutions: 13-1: the area in CM of a conductor is the diameter (in mils) squared: a. d 1 250 mil = A 1 d 1 2 = A 1 62.5 kcmil = b. d 2 350 mil = A 2 d 2 2 = A 2 122.5 kcmil = c. From Table 8, the area of #6 AWG Aluminum is 16510 CM A 3 26240 CM = d. From Table 8, the area of #3/0 AWG Copper is 167.8 kcmil A 4 167.8 kcmil = e. By definition, the area of a 250 kcmil conductor is 250 kcmil! A 5 250 kcmil = 13-2: Resistance is given by: R ρ l A = ρ cu 10.37 ohm CM ft = ρ al 16.73 ohm
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This note was uploaded on 03/26/2012 for the course ECET 231 taught by Professor Piller during the Spring '12 term at Purdue University-West Lafayette.

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Mathcad - 13-01 to 06 - Problems 13-1 through 13-6...

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