解ï¼éç¨åè·¯çµæµæ³ãå 为ææç©çéé½æ¯ç¸éï¼æ以ç¨I1ãI2.....表示ç¸éã
åè·¯ä¸ï¼j2Ãï¼I1-I3ï¼+2Ãï¼I1-I2ï¼=2â 0°ã
åè·¯äºï¼ï¼-j2ï¼Ãï¼I2-I3ï¼+2I2+2Ãï¼I2-I1ï¼=0ã
åè·¯ä¸ï¼I3=1â 0°ã
解æ¹ç¨ç»ï¼å¾ï¼I1=1.2-j0.6ï¼I2=0.8-j0.4ã
åæ¯è·¯çµæµä¸ºï¼
çµæçµæµï¼IL=I1-I3=1.2-j0.6-1=0.2-j0.6ï¼
çµå®¹çµæµä¸ºï¼Ic=I2-I3=0.8-j0.4-1=-0.2-j0.4ï¼
å³ä¾§2Ωçµé»çµæµï¼I2=0.8-j0.4ï¼æ¹ååä¸ï¼
å³è¾¹2Ωçµé»çµæµï¼I1-I2=ï¼1.2-j0.6ï¼-ï¼0.8-j0.4ï¼=0.4-j0.2ï¼
çµåæºçµæµï¼I1=1.2-j0.6ï¼æ¹ååä¸ã
图中用回路法,就用回路法吧。 (2+j2)*I1-2*I2-j2*I3=2∠0 (1), -2*I1+(2+2-j2)*I2+j2*I3=0 (2), I3=1∠0 (3). 将(3)代入(1),(2+j2)*I1-2*I2=2+j2 ,(3)代入(2), -2*I1+(4-j2)*I2=-j2
解得 I2=4/6 A, I1=4/3+j7/3 A, Ic=I2-I3=-1/3 A, IL=I1-I3=1/3+j7/3=2.457∠81.87