电路分析基础ch6习题解答(周围修改)

习题：6-1：试确定题图6-1所示耦合线圈的同名端。解：6-2：写出题图6-2所示各耦合电感的伏安特性。解：di1Mdi2(c)uLdi1Mdi2di(a)uL1Mdi2(b)uL11dtdt11dtdt11dtdtdidididididiuM1Ldtdt2uM1Ldtdt2M1Ldtdt222u22226-3:电路如题图6-3所示，试求电压u。2uM1Md3(1e2t)M3(2e2tdi解：2)＝12e2tVdtdt5Ij5I100由向量模型得：1m1mj3IU1m2m(V)100,245Umj32453245ut32cost45I52451m226-4:题图6-4所示是初始状态为零的互感电路，电源在t＝0是施加于电路。试求电流和。i(t)i(t)12diuM1Ldtdt2di解：22diuL1M2＝0dtdi11dtdiL21di1dtitiA1Ldt11舍去直流影响产生的A，则itit1L21diuL1M21dtdidiuM1L22dtdi11dt22dtdi2diL12得:LuuLL21由112212dtdtdi1Luu2LL即:212dt21L2uzLLLL2it1uzdzt1201222di2LLdi2由式－式得:1L2uLu1211dt21dtdi2Luu1即LL212dt12L1LL2uzuzdzL21L1itt201216-5:耦合电感的初级端钮为ab，次级端钮为cd，如题图6-5所示。稳态电压u10cos2000tV施加于第一线圈。当第二线圈开路时，稳态ab电流1及；当第二线圈短路时，稳态i0.1sin2000tAu0.9cos2000tVcd电流2。试求，M及耦合系数k并标上同名端LL，。12i0.9sin2000tA解：由题意同名端标示如图。1）、当第二线圈开路时，则i0210cos2000tLd0.1sin2000tdi1dt即uL11dtabudi10.9cos2000td0.1sin2000tdtdtcd解之得1L15.9m1.43m2）、当第二线圈短路时，有1uLdi1Mdidt2dtab1didi2u0M1L2dt2dtcd由1式＋L2式得1diuLL22dtab121.4310310cos2000t(15.9103L1.432106)0.9dsin2000t即2dt(15.9L1.432103)0.92000cos2000t21.431015.9L0.920001.4321030.0045442L0.004544/15.90.286mH2M1.43103k0.6715.91030.286103LL126-6:已知耦合电感作题图6-6所示两种连接试，其ab端的等效电感分别为150mH和30mH。试求该耦合电感的耦合系数k。解：150mH由图（a）知LLMMLac11b）知LLMab1LMM22M230mHLL1由图（LMM22ML2LL150301201ab2M12010.9515012.24L1L2MMk10.950.89512.24LLLL12126-7:试计算题图6-7所示各互感电路的等效电感。解：（等1a）20.50.820.4432.19＝LLL2其中MLL0.44312（b）L＝LL26025234.8615.28mab12其中1MLL0.9602534.862LL20.80.672（c）LabLL2M10.820.670.761212其中MLL0.6712（d）MLL67.18mH12250004212.5LLLLL2M10050134.351.714mH12ab126-8：如题图6-8所示电路，已知A,,,,,M=2H。求U2I50s3rad/sR4L4HL3H12解：jIU21R545IIsRjL4j121j31151j39j39.418.43UjIj32216-9:题图6-9所示电路中，，。试求u82cos(2t90)Vi22cos2tAssi。法一：作相量模型124Aj4Ij8j4I30it32cos2tA法二：作去耦电路图由图得j2Ij2IIj8sit32cos2t12AI3046-10:题图6-10所示电路原为零状态，开关S于t＝0时闭合。试求：1时，1MHi(t)?2解由等效电路图方程12i1.5ii122112i2i1i2221整理得：2i1.5i1i21212111ii2i22122由（1）式得i122i1.5i22112122i1.5idz1ti210代如（2）式得1122i1.5i222122i1.5idzti2111110两边导出得1i22i1.52i24242i32i211111i5i4i243111特征方程为S25S40S1S40S1S412则方程的通解为itBeCeS2tSt11hBetCe4t特解代如（3）式i(t)A1p4A=24A=6ititit11p1hABeCe6BetCe4t4StSt12由题意初始值为零，即i00,i0012（1）式（2）式得：t＝0时11.5ii122121－ii021212120112i011.50.52111.512将1i00,i(0)12代入（4）式得1解之B＝－4C＝－2BC6B4C12Ait64et2e4t16-11:题图6-11所示电路中，已知，R1kR0.4k，R0.6k，12LL1,L4,0.1,U1000V，1000rad/s。试求I。212s解：MLL0.212RjLIjIUs11122jIRRjLI0L1221000j1000Ij200I10012j200I1000j4000I0121000j1000100j200j2104j2106j106j4106410610104296j5003.44149.370I21000j1000j200j2001000j4000mA6-12:题图6-12所示电路中，耦合系数。试用互感化除法求。0.5U2j416j4解：LL0.51644Z1j4171216j416j4100164914.031716j4178.2299.4200.685.42VU100216j200j126-13:试列写题图6-13所示正弦稳态电路的网孔方程。解：RjLjLIjLIjMIjMIjMIjMIjMIU112122122121121132232sRjLjLIjLIjMIjMIjMIjMIjMI0223221121232131232231RjLLIjL2IU11212121323122sjLIRjLL2I＝0221323121223236-14:试用戴维南定理求解题图6-14所示电路的次级电流i。已知2，，u1152cos314tVL3.6L0.06，M0.465，R20,R0.08,1212sR42。L解：从ab断开求U和ZocoV3140.465101.78914.81UjMIjoco11501150115UmAIoRjL20j3143.620j1130113089101.789s11求Z时，将用短路线代之，则次级回路对次级回路的反映阻抗为Uos22Zf2Z11RjL122RjL0.08j18.842.131040.41j0.04ZZZ113089f22222o114.81UARZ420.41j0.040.351Ioc2LoAit0.352cos314t1026-15:耦合电感电路如题图6-15所示，已知1R7.5，L20，1，30，输入电压U的频率为10kHz，假若电阻R及12L602电容C可调，问R和C为何值时，获得最大功率？R2121解：Z90022900Rj54000R236002ZRj60f12221L12f6280ZRjLj111111Z7.5j30j117.5j306280c6280c11ZZf1121900RR236007.5解之R602540011R23600306280C即7.5306280C111C122.562800.707uF6-16：电路如题图6-16所示，试问为何值时可获得最大功率？最ZL大功率为多少？200U解：戴文南等效电压源：ocsj210j1020245VUjZ11Z220.2j0.2k4Z10245f211等效内阻抗：故当2ZZjL0.2j0.2j100.2j9.8kf20时可获最大功率：ZZ0.2j9.8kLo2022U2WP4Re[Z]40.21031ocLmax06-17：理想变压器如题图6-17所示，试求，II，U，U。21211IU10由图得1150IU10IU解：11100U115I221II1102U10U2120100V31解之：0AIU31I120AU10U100I30V10321216-18:理想变压器如题图6-18所示,已知U200V，试求I。s解：10I641A1由原图得：InI20A16-19:题图6-19所示电路是含理想变压器的单口网络，试求端口输入阻抗。ZabzUabUUZII11解：IIIUUZI2212InI21UnIZU121nUU2IInI11U1UZIn12II1nUUZI111U1UZInIZInIZUZn11nn11111U11nZII1nZ1n22nn1nU11nZI1n2nn1n1n2U1n2ZUZIZ1Iab1nn226-20：已知题图6-20所示电路ab端的输入阻抗为8，试求匝比n＝？解：10n2UU210n2110n210nUU1U2n2n2210n23210210230nUUIUnnI2210n22U210nI30nU2130nU210nI2U210n2Zab8I130n210n28240n即10n2240n60或n224n0.60n2424240.624578.42424.0499222取n2424.04990.02526-21：试求题图6-21所示电路中变压器的初级电流和次级电流i1i。2nU1U由图知解：21n1II2U12010.25120U300V2U30010I23ARRLUUICJ50.021230＝J0.142J4.2A2JCUU1112JC3J4.25.16125.5AIII2RCI2.55.16125.512.954.5AnI12it12.92COS5t54.5A1it5.162COS5t125.5A26-22：题图6-22所示电路中，电源功率P1kW，其一半消耗在R1003s的电阻上，已知，，U1000VR4R25，求匝比n和n。1212s由题意：P5005AR3R100P500R3I33P100010A100UPR5001001005VR3I1SU33SII5InIInI3nn232221122由图U1US2n111U2R1Rn2RIR1RnRnI2n222222n22221111URnRnnRIS1nn12123111URIn2RI1n2n2RI.....................1S11121231I