彭鸿才《电机原理与拖动》习题解答第一章直流电机原理3PN1410P411－3解：IAN60.87()UN230PN14P116.37(Kw)N0.855P411－4解：PUIW1NN110131430()3PN1.110N76.92%P11430pP1PN14301100330(W)UN230P421－21解：If1.53(A)Rf150IaINIf69.61.5371.13(A)EUIRVaNaa23071.130.128239.1()22pcuIaRa71.130.128647.6(W)PMEaIa239.171.1317007(W)3PN1610P118713.5(W)N0.855P421－29解：IaINIfN40.60.68339.92(A)EaUNIaRa22039.920.213211.5(V)PMEaIa211.539.928443.1(W)22pcuIaRa39.920.213339.4(W)pfIfNUN2200.683150.26(W)P1UNIN22040.68932(W)P7.5103N83.97%P189323poPMPN8443.17.510943.1(W)1彭鸿才《电机原理与拖动》习题解答PM8443.1/1000TN9550955026.88(Nm)nN3000PN7.5T2N9550955023.88(Nm)nN3000ToTNT2N26.8823.883(Nm)Po943.1/1000或者To955095503(Nm)nN3000PN27000P421－30解：IN245.5(A)UN110IaNINIfN245.55250.5(A)UNIaNRa110250.50.02CeN0.1nN1150UIR110250.50.02nNaNa1050rpmCeN0.1或者：EaNUNIaNRa110250.50.02115(V)EUIRVaNaNa110250.50.02105()EEEn1051150aNanaN1050rpmnNnEaN115P421－31解：EaNCeNnNnnNNIaIaNEaEaNUNINRa11013197(V)UEaINRa9713184(V)P421－32解：IaNINIfN2555250(A)UNIaNRa4402500.078CeN0.841nN500Ce30CTNCeN9.550.8418.032CT30PN96（1）T2N955095501833.6(Nm)nN5002彭鸿才《电机原理与拖动》习题解答（2）TNCTNIaN8.0322502008(Nm)（3）ToTNT2N20081833.6174.4(Nm)UN440（4）no523rpmCeN0.841'URNa0.078174.4(5)noo2T523521rpmCCCeNeTN0.8418.032To174.4或者：Iao21.71(A)CTN8.032'UNIaoRa44021.710.078no521rpmCeN0.841P421－33解：P1UNIN22081.717974(W)PNP1N179740.8515278(W)UN220IfN2.48(A)Rf88.8IaNINIfN81.72.4879.22(A)22pcuIaNRa79.220.1627.6(W)22pfIfNRf2.4888.8546(W)pP1PN17974152782696(W)pmpFeppcupf2696627.65461522(W)PMPNpmpFe15278152216800(W)P421－34解：PN75（1）T2N9550955