sol2 - ECE473 HOMEWORK#2 SOLUTIONS FALL 2011 Problem...

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ECE473 HOMEWORK #2 – SOLUTIONS FALL 2011 Problem 2.1 (Problem 2.41 p. 86) S L = 3 V LL I L = 3(480)(20) = 16 . 63 kVA P L = 3 V LL I L cos ϕ = 13 . 32 kW Q L = S 2 L P 2 L = 16 . 63 2 13 . 32 2 = 9 . 977 kVAr or Q L = 3 V LL I L sin ϕ = 3(480)(20) 1 0 . 8 2 = 9 . 977 kVAr Problem 2.2 (Problem 2.43 p. 86) ¯ Z 1 = 30 + j 40 Ω ¯ Z 2 = 60 j 45 3 = 20 j 15 Ω ¯ Z L = ¯ Z 2 ¯ Z 2 = (30 + j 40)(20 j 15) 30 + j 40 + 20 j 15 = 1200 + j 350 50 + j 25 = 1250 ̸ 16 . 26 o 55 . 9 ̸ 26 . 565 o = 22 . 36 ̸ 10 . 305 o = 22 j 4 Ω ¯ Z T = 2 + j 4 + 22 j 4 = 24 + j 0 = 24 ̸ 0 o (a) ˜ I A = ˜ V AN ¯ Z T = 120 ̸ 0 o 24 ̸ 0 o = 5 ̸ 0 o A ¯ S S, 3 ph = 3 ˜ V AN ˜ I A = 3 × (120 ̸ 0 o ) × (5 ̸ 0 o ) = 1800 + j 0 VA P S, 3 ph = e { ¯ S S, 3 ph } = 1800 W Q S, 3 ph = m { ¯ S S, 3 ph } = 0 VAr (b) V L = 22 j 4 (2 + j 4) + (22 j 4) × 120 = × 500 24 × 120 = 111 . 8 V V L,LL = 3 V L = 3 × 111 . 8 = 193 . 6 V (c) I 1 ,Y = V L | 30 + j 40 | = 111 . 8 50 = 2 . 236 A I 2 , = 3 V L | 60 j 45 | = 3 × 111 . 8 75 = 2 . 582 A (d) P 1 = 3 × 30 × I 2 1 ,Y = 3 × 30 × 2 . 236 2 = 450 W Q 1 = 3 × 40 × I 2 1 ,Y = 3 × 40 × 2 . 236 2 = 600 VAr P 2 = 3 × 60 × I 2 2 , = 3 × 30 × 2 .
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