solex1f09

# solex1f09 - ECE473 EXAM#1 – SOLUTIONS FALL 2009...

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Unformatted text preview: ECE473 EXAM #1 – SOLUTIONS FALL 2009 Instructions : 1. Closed-book, closed-notes, open-mind exam. 2. Work each problem on the exam booklet in the space provided. 3. Write neatly and clearly for partial credit. Cross out any material you do not want graded. Name : Problem 1 : /20 Problem 2 : /30 Problem 3 : /25 Problem 4 : /25 Total : /100 1 Problem 1 (20 Points) I ~ + ~ 2 V =120 0 V o- 240-VA 0.96-pf lead Load 2.4 j61.8 Ω Ω +- V 1 ~ (a) Find the complex power ¯ S 2 = P 2 + jQ 2 (VA) absorbed by the load. ¯ S 2 = 240 6- cos- 1 . 96 = 240 6- 16 . 26 o = 230 . 4- j 67 . 2 VA (b) Find the line current ˜ I (A) in both polar and rectangular forms. ˜ I = ¯ S * 2 ˜ V * 2 = 240 6 16 . 26 o 120 6 o = 2 6 16 . 26 o = 1 . 92 + j . 56 A (c) Find the source voltage ˜ V 1 (V) in polar form. ˜ V 1 = (2 . 4 + j 61 . 8) ˜ I + ˜ V 2 = (2 . 4 + j 61 . 8)(1 . 92 + j . 56) + (120 + j 0) = 90 + j 120 = 150 6 53 . 13 o V (d) Find the complex power ¯ S 1 = P 1 + jQ 1 (VA) delivered by the source. ¯ S 1 = ˜ V 1 ˜ I * = (150 6 53 . 13 o )(2 6- 16 . 26 o ) = 300 6 36 . 87 o = 240 + j 180 VA or P 1 = 2 . 4 | ˜ I | 2 + 230 . 4 = 2 . 4 × 2 2 + 230 . 4 = 240 W Q 1 = 61 .....
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solex1f09 - ECE473 EXAM#1 – SOLUTIONS FALL 2009...

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