sol51 - 60 (1 . 342 10 6 ) 10 3 = 0 . 5059 / km Problem 5.4...

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ECE473 HOMEWORK #5 – SOLUTIONS FALL 2011 Problem 5.1 (Problem 4.15 p. 222) r r r r 60 o 60 o GMR’ = 49 [ r (2 r ) 6 ][ r (2 r ) 3 (2 r 3) 2 (4 r )] 6 = r 49 2 48 × 3 6 × e 7 / 4 = 2 . 1767 r Problem 5.2 (Problem 4.17 p. 222) (a) GMR’ = 9 [( re ( 1 / 4) ) × (2 r ) 2 ] 3 = 3 ( re ( 1 / 4) ) × (2 r ) 2 = 1 . 4605 r (b) GMR’ = 16 [( re ( 1 / 4) ) × (2 r ) × (4 r ) × (6 r )] 2 [( re ( 1 / 4) ) × (2 r ) 2 × (4 r )] 2 = 8 [( re ( 1 / 4) ) × (2 r ) × (4 r ) × (6 r )][( re ( 1 / 4) ) × (2 r ) 2 × (4 r )] = 2 . 1554 r (c) GMR’ = 81 [( re ( 1 / 4) ) × (2 r ) 2 × (4 r ) 2 × ( r 20) 2 × ( r 8) × ( r 32)] 4 × 81 [( re ( 1 / 4) ) × (2 r ) 3 × ( r 8) 2 × (4 r ) × ( r 20) 2 ] 4 × 81 [( re 1 / 4 ) × (2 r ) 4 × ( r 8) 4 ] = 2 . 6374 r Problem 5.3 (Problem 4.18 p. 223) GMD = 3 D · D · 2 D = D 3 2 = 8 × 3 2 = 10 . 08 m L = 2 × 10 7 ln GMD r = 2 × 10 7 ln 10 . 08 0 . 0403 × 12 × 0 . 0254 = 1 . 342 × 10 6 H/m X = ωL = 2
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Unformatted text preview: 60 (1 . 342 10 6 ) 10 3 = 0 . 5059 / km Problem 5.4 (Problem 4.20 p. 223) GMD = 3 D D 2 D = D 3 2 = 10 3 2 = 12 . 60 m GMR = 3 r d 2 = 3 (0 . 0435 12 . 0254) . 5 2 = . 1491 m L = 2 10 7 ln GMD GMR = 2 10 7 ln 12 . 60 . 1491 = . 8874 mH/km X = L = 2 60 (0 . 8874 10 3 ) = 0 . 3345 / km...
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