電路分析基礎(chǔ)ch6習(xí)題解答(周圍修改)

習(xí)題：6-1：試確定題圖6-1所示耦合線圈的同名端。解：6-2：寫(xiě)出題圖6-2所示各耦合電感的伏安特性。解：di1Mdi2(c)uLdi1Mdi2di(a)uL1Mdi2(b)uL11dtdt11dtdt11dtdtdidididididiuM1Ldtdt2uM1Ldtdt2M1Ldtdt222u22226-3:電路如題圖6-3所示，試求電壓u。2uM1Md3(1e2t)M3(2e2tdi解：2)＝12e2tVdtdt5Ij5I100由向量模型得：1m1mj3IU1m2m(V)100,245Umj32453245ut32cost45I52451m226-4:題圖6-4所示是初始狀態(tài)為零的互感電路，電源在t＝0是施加于電路。試求電流和。i(t)i(t)12diuM1Ldtdt2di解：22diuL1M2＝0dtdi11dtdiL21di1dtitiA1Ldt11舍去直流影響產(chǎn)生的A，則itit1L21diuL1M21dtdidiuM1L22dtdi11dt22dtdi2diL12得:LuuLL21由112212dtdtdi1Luu2LL即:212dt21L2uzLLLL2it1uzdzt1201222di2LLdi2由式－式得:1L2uLu1211dt21dtdi2Luu1即LL212dt12L1LL2uzuzdzL21L1itt201216-5:耦合電感的初級(jí)端鈕為ab，次級(jí)端鈕為cd，如題圖6-5所示。穩(wěn)態(tài)電壓u10cos2000tV施加于第一線圈。當(dāng)?shù)诙€圈開(kāi)路時(shí)，穩(wěn)態(tài)ab電流1及；當(dāng)?shù)诙€圈短路時(shí)，穩(wěn)態(tài)i0.1sin2000tAu0.9cos2000tVcd電流2。試求，M及耦合系數(shù)k并標(biāo)上同名端LL，。12i0.9sin2000tA解：由題意同名端標(biāo)示如圖。1）、當(dāng)?shù)诙€圈開(kāi)路時(shí)，則i0210cos2000tLd0.1sin2000tdi1dt即uL11dtabudi10.9cos2000td0.1sin2000tdtdtcd解之得1L15.9m1.43m2）、當(dāng)?shù)诙€圈短路時(shí)，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:已知耦合電感作題圖6-6所示兩種連接試，其ab端的等效電感分別為150mH和30mH。試求該耦合電感的耦合系數(shù)k。解：150mH由圖（a）知LLMMLac11b）知LLMab1LMM22M230mHLL1由圖（LMM22ML2LL150301201ab2M12010.9515012.24L1L2MMk10.950.89512.24LLLL12126-7:試計(jì)算題圖6-7所示各互感電路的等效電感。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：如題圖6-8所示電路，已知A,,,,,M=2H。求U2I50s3rad/sR4L4HL3H12解：jIU21R545IIsRjL4j121j31151j39j39.418.43UjIj32216-9:題圖6-9所示電路中，，。試求u82cos(2t90)Vi22cos2tAssi。法一：作相量模型124Aj4Ij8j4I30it32cos2tA法二：作去耦電路圖由圖得j2Ij2IIj8sit32cos2t12AI3046-10:題圖6-10所示電路原為零狀態(tài)，開(kāi)關(guān)S于t＝0時(shí)閉合。試求：1時(shí)，1MHi(t)?2解由等效電路圖方程12i1.5ii122112i2i1i2221整理得：2i1.5i1i21212111ii2i22122由（1）式得i122i1.5i22112122i1.5idz1ti210代如（2）式得1122i1.5i222122i1.5idzti2111110兩邊導(dǎo)出得1i22i1.52i24242i32i211111i5i4i243111特征方程為S25S40S1S40S1S412則方程的通解為itBeCeS2tSt11hBetCe4t特解代如（3）式i(t)A1p4A=24A=6ititit11p1hABeCe6BetCe4t4StSt12由題意初始值為零，即i00,i0012（1）式（2）式得：t＝0時(shí)11.5ii122121－ii021212120112i011.50.52111.512將1i00,i(0)12代入（4）式得1解之B＝－4C＝－2BC6B4C12Ait64et2e4t16-11:題圖6-11所示電路中，已知，R1kR0.4k，R0.6k，12LL1,L4,0.1,U1000V，1000rad/s。試求I。212s解：MLL0.212RjLIjIUs11122jIRRjLI0L1221000j1000Ij200I10012j200I1000j4000I0121000j1000100j200j2104j2106j106j4106410610104296j5003.44149.370I21000j1000j200j2001000j4000mA6-12:題圖6-12所示電路中，耦合系數(shù)。試用互感化除法求。0.5U2j416j4解：LL0.51644Z1j4171216j416j4100164914.031716j4178.2299.4200.685.42VU100216j200j126-13:試列寫(xiě)題圖6-13所示正弦穩(wěn)態(tài)電路的網(wǎng)孔方程。解：RjLjLIjLIjMIjMIjMIjMIjMIU112122122121121132232sRjLjLIjLIjMIjMIjMIjMIjMI0223221121232131232231RjLLIjL2IU11212121323122sjLIRjLL2I＝0221323121223236-14:試用戴維南定理求解題圖6-14所示電路的次級(jí)電流i。已知2，，u1152cos314tVL3.6L0.06，M0.465，R20,R0.08,1212sR42。L解：從ab斷開(kāi)求U和ZocoV3140.465101.78914.81UjMIjoco11501150115UmAIoRjL20j3143.620j1130113089101.789s11求Z時(shí)，將用短路線代之，則次級(jí)回路對(duì)次級(jí)回路的反映阻抗為Uos22Zf2Z11RjL122RjL0.08j18.842.131040.41j0.04ZZZ113089f22222o114.81UARZ420.41j0.040.351Ioc2LoAit0.352cos314t1026-15:耦合電感電路如題圖6-15所示，已知1R7.5，L20，1，30，輸入電壓U的頻率為10kHz，假若電阻R及12L602電容C可調(diào)，問(wèn)R和C為何值時(shí)，獲得最大功率？R2121解：Z90022900Rj54000R236002ZRj60f12221L12f6280ZRjLj111111Z7.5j30j117.5j306280c6280c11ZZf1121900RR236007.5解之R602540011R23600306280C即7.5306280C111C122.562800.707uF6-16：電路如題圖6-16所示，試問(wèn)為何值時(shí)可獲得最大功率？最ZL大功率為多少？200U解：戴文南等效電壓源：ocsj210j1020245VUjZ11Z220.2j0.2k4Z10245f211等效內(nèi)阻抗：故當(dāng)2ZZjL0.2j0.2j100.2j9.8kf20時(shí)可獲最大功率：ZZ0.2j9.8kLo2022U2WP4Re[Z]40.21031ocLmax06-17：理想變壓器如題圖6-17所示，試求，II，U，U。21211IU10由圖得1150IU10IU解：11100U115I221II1102U10U2120100V31解之：0AIU31I120AU10U100I30V10321216-18:理想變壓器如題圖6-18所示,已知U200V，試求I。s解：10I641A1由原圖得：InI20A16-19:題圖6-19所示電路是含理想變壓器的單口網(wǎng)絡(luò)，試求端口輸入阻抗。ZabzUabUUZII11解：IIIUUZI2212InI21UnIZU121nUU2IInI11U1UZIn12II1nUUZI111U1UZInIZInIZUZn11nn11111U11nZII1nZ1n22nn1nU11nZI1n2nn1n1n2U1n2ZUZIZ1Iab1nn226-20：已知題圖6-20所示電路ab端的輸入阻抗為8，試求匝比n＝？解：10n2UU210n2110n210nUU1U2n2n2210n23210210230nUUIUnnI2210n22U210nI30nU2130nU210nI2U210n2Zab8I130n210n28240n即10n2240n60或n224n0.60n2424240.624578.42424.0499222取n2424.04990.02526-21：試求題圖6-21所示電路中變壓器的初級(jí)電流和次級(jí)電流i1i。2nU1U由圖知解：21n1II2U12010.25120U300V2U30010I23ARRLUUICJ50.021230＝J0.142J4.2A2JCUU1112JC3J4.25.16125.5AIII2RCI2.55.16125.512.954.5AnI12it12.92COS5t54.5A1it5.162COS5t125.5A26-22：題圖6-22所示電路中，電源功率P1kW，其一半消耗在R1003s的電阻上，已知，，U1000VR4R25，求匝比n和n。1212s由題意：P5005AR3R100P500R3I33P100010A100UPR5001001005VR3I1SU33SII5InIInI3nn232221122由圖U1US2n111U2R1Rn2RIR1RnRnI2n222222n22221111URnRnnRIS1nn12123111URIn2RI1n2n2RI.....................1S11121231I