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Bài giải phần giải mạch P12

Chapter 12, Solution 1. (a) If Vab = 400 , then Van = 400 3 ∠ - 30° = 231∠ - 30° V 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 Van = 400 3 ∠30° = 231∠30° V Vbn = 231∠150° V Vcn =. | Chapter 12 Solution 1. a If Vab 400 then 400 V -30 231Z-30 V an I Vbn 231Z -150 V Vcn 231Z -270 V b For the acb sequence Vab Van - Vbn Vp Z0 - Vp Z120 1 V3 r Vab Vp 1 --J 1 Vp 3Z-30 2 2 7 i.e. in the acb sequence Vab lags Van by 30 . Hence if Vab 400 then Van 400 - Z30 231Z30 V V3 ---------- Vbn 231Z150 V Vcn 231Z -90 V Chapter 12 Solution 2. Since phase c lags phase a by 120 this is an acb sequence. Vbn 160Z 30 120 160Z150 V Chapter 12 Solution 3. Since Vbn leads Vcn by 120 this is an abc sequence. Van 208Z 130 120 208Z250 V Chapter 12 Solution 4. V. V Z120 208Z140 V UC Id Vdb Vbc Z120 208Z260 V V d_ 73 Z30 208Z260 V3 Z30 120Z230 V V. V Z -120 120Z110 V un an Chapter 12 Solution 5. This is an abc phase sequence. V.b V.J3 30 Vab 420 Z0 or V -ab - ------------ 242.5Z - 30 V V3 Z30 V3 Z30 --------------- V. V Z -120 242.5Z -150 V bn an Vcn Van Z120 242.5Z90 V Chapter 12 Solution 6. Z Y 10 j5 11.18Z26.56 The line currents are V 220Z0 I -------------- 19.68Z - 26.56 A a Z Y 11.18Z26.56 ------------------ Ib I Z -120 19.68Z -146.56 A b a Ic Ia Z120 19.68Z93.44 A The line voltages are Vab 20073 Z30 381Z30 V Vbc 381Z - 90 V V 381Z -210 V ca The load voltages are Van I a Z y Van 220Z0 V VBN Vbn 220Z -120 V VCN Vcn 220Z120 V Chapter 12 Solution 7. This is a balanced Y-Y system. 440Z0 V Zy 6 - j8 Q Using the per-phase circuit shown above I a I b 440Z0 44Z53.13 A 6 - j8 ------------- T Z -120 44Z -66.87 A a Ic Ia Z120 44Z173.13 A Chapter 12 Solution 8. VL 220 V Z y 16 j9 Q I Vp Vl an z y 73 z y 220 73 16 j9 6.918Z-29.36 IL 6.918