Bài giải phần giải mạch P12

Chapter 12, Solution 1. (a) If Vab = 400 , then Van = 400 3 ∠ - 30° = 231∠ - 30° VVbn = 231∠ - 150° V Vcn = 231∠ - 270° V(b) For the acb sequence, Vab = Van − Vbn = Vp ∠0° − Vp ∠120° 1 3 Vab = Vp 1 + − j  = Vp 3∠ - 30°   2   2 i.e. in the acb sequence, Vab lags Van by 30°. Hence, if Vab = 400 , then Van = 400 3 ∠30° = 231∠30° VVbn = 231∠150° V Vcn =...
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Bài giải phần giải mạch P12Chapter 12, Solution 1. (a) If Vab = 400 , then 400 Van = ∠ - 30° = 231∠ - 30° V 3 Vbn = 231∠ - 150° V Vcn = 231∠ - 270° V (b) For the acb sequence, Vab = Van − Vbn = Vp ∠0° − Vp ∠120°  1 3 Vab = Vp 1 + − j  = Vp 3∠ - 30°    2 2  i.e. in the acb sequence, Vab lags Van by 30°. Hence, if Vab = 400 , then 400 Van = ∠30° = 231∠30° V 3 Vbn = 231∠150° V Vcn = 231∠ - 90° VChapter 12, Solution 2. Since phase c lags phase a by 120°, this is an acb sequence. Vbn = 160∠(30° + 120°) = 160∠150° VChapter 12, Solution 3. Since Vbn leads Vcn by 120°, this is an abc sequence. Van = 208∠(130° + 120°) = 208∠ 250° VChapter 12, Solution 4. Vbc = Vca ∠120° = 208∠140° V Vab = Vbc ∠120° = 208∠260° V Vab 208∠260° Van = = = 120∠230° V 3 ∠30° 3 ∠30° Vbn = Van ∠ - 120° = 120∠110° VChapter 12, Solution 5. This is an abc phase sequence. Vab = Van 3 ∠30° Vab 420∠0° or Van = = = 242.5∠ - 30° V 3 ∠30° 3 ∠30° Vbn = Van ∠ - 120° = 242.5∠ - 150° V Vcn = Van ∠120° = 242.5∠90° VChapter 12, Solution 6. Z Y = 10 + j5 = 11.18∠26.56° The line currents are Van 220 ∠0° Ia = = = 19.68∠ - 26.56° A Z Y 11.18∠26.56° I b = I a ∠ - 120° = 19.68∠ - 146.56° A I c = I a ∠120° = 19.68∠93.44° A The line voltages are Vab = 200 3 ∠30° = 381∠30° V Vbc = 381∠ - 90° V Vca = 381∠ - 210° V The load voltages are VAN = I a Z Y = Van = 220∠0° V VBN = Vbn = 220∠ - 120° V VCN = Vcn = 220∠120° VChapter 12, Solution 7. This is a balanced Y-Y system. + 440∠0° V ZY = 6 − j8 Ω − Using the per-phase circuit shown above, 440∠0° Ia = = 44∠53.13° A 6 − j8 I b = I a ∠ - 120° = 44∠ - 66.87° A I c = I a ∠120° = 44∠173.13° AChapter 12, Solution 8. VL = 220 V , Z Y = 16 + j9 Ω Vp VL 220 I an = = = = 6.918∠ - 29.36° ZY 3 ZY 3 (16 + j9) I L = 6.918 AChapter 12, Solution 9. Van 120 ∠0° Ia = = = 4.8∠ - 36.87° A Z L + Z Y 20 + j15 I b = I a ∠ - 120° = 4.8∠ - 156.87° A I c = I a ∠120° = 4.8∠83.13° A As a balanced system, I n = 0 AChapter 12, Solution 10. Since the neutral line is present, we can solve this problem on a per-phase basis. For phase a, Van 220 ∠0° Ia = = = 6.55∠36.53° Z A + 2 27 − j20 For phase b, Vbn 220 ∠ - 120° Ib = = = 10 ∠ - 120° ZB + 2 22 For phase c, Vcn 220 ∠120° Ic = = = 16.92 ∠97.38° ZC + 2 12 + j5 The current in the neutral line is I n = -(I a + I b + I c ) or - In = Ia + Ib + Ic - I n = (5.263 + j3.9) + (-5 − j8.66) + (-2.173 + j16.78) I n = 1.91 − j12.02 = 12.17 ∠ - 81° AChapter 12, Solution 11. Vbc VBC 220∠10° Van = = = 3 ∠ - 90° 3 ∠ - 90° 3 ∠ - 90° Van = 127 ∠100° V VAB = VBC ∠120° = 220∠130° V VAC = VBC ∠ - 120° = 220∠ - 110° V If I bB = 30 ∠60° , then I aA = 30∠180° , I cC = 30 ∠ - 60° I aA 30∠180° I AB = = ...