solution chapter 2 Department of Electrical （第二章电气部门的解决方案）.pdfVIP

Chapter 2: Transformers 2-1. The secondary winding of a transformer has a terminal voltage of vs (t) = 282.8 sin 377t V . The turns ratio of the o transformer is 100:200 (a = 0.50). If the secondary current of the transformer is (7.07 sin) (377 t −36.87 ) A , what is the primary current of this transformer? What are its voltage regulation and efficiency? The impedances of this transformer referred to the primary side are Req = 0.20 Ω RC = 300 Ω X eq = 0.750 Ω XM = 80 Ω o ( IP 11.1=∠−41.0 A , VR = 6.2% and η 93.7%) 2-2. A 20-kVA 8000/480-V distribution transformer has the following resistances and reactances: RP = 32 Ω RS = 0.05 Ω XP = 45 Ω XS = 0.06 Ω RC = 250 k Ω XM = 30 k Ω The excitation branch impedances are given referred to the high-voltage side of the transformer. (a) Find the equivalent circuit of this transformer referred to the high-voltage side. (b) Find the per-unit equivalent circuit of this transformer. (c) Assume that this transformer is supplying rated load at 480 V and 0.8 PF lagging. What is this transformer’s input voltage? What is its voltage regulation? (d) What is the transformer’s efficiency under the conditions of part (c)? ((a) the secondary impedances referred to the primary side are RS 13.9 Ω, X S 16.7 =Ω The resulting equivalent circuit is (b ) the resulting per-unit equivalent circuit is as shown below: (c) VP 8185=∠0.38o V and VR = 2.31 % , (d) η 96.6%) 2-