Electric Circuit Analysis Quiz Solution Quiz2 DD08 Quiz2 Problem 1B solution Dùng nguồn tương đương, đưa mạch về dạng Quiz Solution http //www4 hcmut edu vn/~lmcuong/ECA Quiz2 DD08 html 1 of 3 04/09/2[.]
Quiz Solution of http://www4.hcmut.edu.vn/~lmcuong/ECA_Quiz2_DD08.html Electric Circuit Analysis Quiz Solution Quiz2 - DD08 Quiz2 Problem 1B solution : Dùng nguồn tương đương, đưa mạch dạng : 04/09/2013 11:28 PM Quiz Solution of http://www4.hcmut.edu.vn/~lmcuong/ECA_Quiz2_DD08.html Trong : Z = 2,4 - j0,2 ohm J = 60*6/(8 + j4)/Z = 16,712/_-21,8 A Và dòng điện theo chia dòng : KCL : Ix = (J + 5/_90)*z/(z + - j3) = 5,238/_17,35o A Problem 2A solution : Calculate the complex power : Load 1: P1 = 36 kW Q1 = 36k*tan(cos-10,82) = 25,128 kVar Load 2: P2 = 48 kW Q2 = 48k*tan(cos-10,88) = 25,9 kVar Two Loads: P = 84 kW Q = 51,028 kVar The current : I = (84000 - j51028)/240 = 409,52/_-31,3o A 04/09/2013 11:28 PM Quiz Solution of http://www4.hcmut.edu.vn/~lmcuong/ECA_Quiz2_DD08.html The phasor of the source: E = (0,1 + j0,3)*409,52/_-31,3o + 240 = 349/_13,9o Vrms Written by Le Minh Cuong 04/09/2013 11:28 PM ... 11:28 PM Quiz Solution of http://www4.hcmut.edu.vn/~lmcuong/ECA _Quiz2 _DD08.html The phasor of the source: E = (0,1 + j0,3)*409,52/_-31,3o + 240 = 349/_13,9o Vrms Written by Le Minh Cuong 04/09/2013... điện theo chia dòng : KCL : Ix = (J + 5/_90)*z/(z + - j3) = 5,238/_17,35o A Problem 2A solution : Calculate the complex power : Load 1: P1 = 36 kW Q1 = 36k*tan(cos-10,82) = 25,128 kVar Load 2: P2.. .Quiz Solution of http://www4.hcmut.edu.vn/~lmcuong/ECA _Quiz2 _DD08.html Trong : Z = 2,4 - j0,2 ohm J = 60*6/(8 + j4)/Z = 16,712/_-21,8