Tham khảo tài liệu 'electric machinery fundamentals (solutions manual) part 7', kỹ thuật - công nghệ, cơ khí - chế tạo máy phục vụ nhu cầu học tập, nghiên cứu và làm việc hiệu quả | With a 20 decrease EA2 10 584 V and sin 1 E-JL sin sin -1323 sin D 2 E A2 2 10 584 V Therefore the new armature current is 146 A2 ũ ì 10. 584 70 44 0 - 2 ------- _ -------- 780-----1470 A E V _ I jXs With a 25 decrease. EA2 9. 923 V . and sin 1 A- sin sin 1323ft _V sin 2 E A 2 2 V D Therefore. the new armature current is A2 ũ ì 9. 923 70 44 0 -------1 ------ 62------ TW A E V I jXs f A MATLAB program to plot the magnitude of the armature current I A as a function of E A is shown below. M-file prob5 M-file to calculate and plot the armature current supplied to an infinite bus as Ea is varied. Define values for this generator Ea 13230 Ea Vp 7044 Phase voltage d1 pi 180 torque angle at full Ea Xs Xs ohms Calculate delta for each Ea d asin 13230 . Ea . sin d1 Calculate Ia for each flux Ea Ea . cos d j. sin d Ia Ea - Vp . j Xs Plot the armature current versus Ea figure 1 plot abs Ea 1000 abs Ia b- LineWidth title bfArmature current versus itE A rm xlabel bf itE_ A rm bf kV ylabel bf itI_ A rm bf A grid on hold off .
