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

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Unformatted text preview: 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 − . 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 ̸ o Ω (a) ˜ I A = ˜ V AN ¯ Z T = 120 ̸ o 24 ̸ o = 5 ̸ o A ¯ S S, 3 ph = 3 ˜ V AN ˜ I ∗ A = 3 × (120 ̸ o ) × (5 ̸ o ) ∗ = 1800 + j 0 VA P S, 3 ph = ℜ e { ¯ S S, 3 ph } = 1800 W Q S, 3 ph = ℑ m { ¯ S S, 3 ph } = 0 VAr (b) V L = 22 − j 4 (2 + j 4) + (22 − j 4) × 120 = × √ 500 24 × 120 ∼ = 111 .....
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This document was uploaded on 11/01/2011 for the course ECE 473 at Boise State.

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

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