西安交大《電工技術(shù)》答案

1、西安交大電工技術(shù)答案第一章習(xí)題1.1解：)6k5k6k8k4k4k13k6k52/17k154/17kR9.1kb）10.511110.5110.50.51.5R1.5kc)3001/54900150西安交大電工技術(shù)答案300300600150900300900600150120207.7.2解：依圖所示,Ua0201.224VUb06318VUabUaUb24(18)6VaUabb1.3解:根據(jù)估算:I1mA,I10.99mA，I20.01mA.4解:a）1)U10V,I1210/23A)電流源PUI10220W發(fā)出功率電壓源PUI110(3)30W發(fā)出功率電阻RPUI10550W發(fā)出功率2

2、/54西安交大電工技術(shù)答案電流源發(fā)出功率+電壓源發(fā)出功率=電阻吸收功率I2AI1+10V-R2U2電阻RPUI428W吸收功率b）)因?yàn)榕c電流源串聯(lián)的電壓源相當(dāng)與無(wú)效,故2A,U=RI22=42）電壓源PUI10220W發(fā)出功率電流源PUI6212W吸收功率L1.7解:當(dāng)開(kāi)關(guān)S斷開(kāi)時(shí)，I8V2k2kIA8k20VI2081228m,Ua208112V當(dāng)開(kāi)關(guān)S閉合時(shí)，I18VI22k2kI1A8k20VI20228mA,Ua20824V18解:12VI8ab10-12V2)I12(12)24810R18R,Ua128I,Ub1210I當(dāng)減小時(shí),則I增大,Ua減小,Ub增大.3/54西安交大電工技

3、術(shù)答案3)當(dāng)=6時(shí),I12(12)86101A,Ua128I1284V,Ub1210I121012V19解:根據(jù)題意，+I1R0+URLE-230V22-122A,UIR1022220VIE23010RR0L內(nèi)部功率損耗PUII2R1021100W0內(nèi)阻壓降U0IR010110V同理如圖,+I230A1R0U-RL22I123010A122，UIR1022220內(nèi)部功率損耗P0UI220(23010)48400W內(nèi)阻壓降U0IR02201220V1.10解:根據(jù)題意,I1I21I32324I51+-10VI48A1利用節(jié)點(diǎn)電壓法,23322(111110)UU131311U(1)U824/54

4、解得，14V,U2V則IU/14/14A,2,IUU33，I8A,IU/17/17A西安交大電工技術(shù)答案I1043A21174211A3152.11解：I4R4I2+-US1R2R3R5I5ISR1I1III124III325III5S4I3URIRIRIS1223311RIRIRI22445512解：根據(jù)題意，I11I4R42R1IS3I3I5R2I6+R3R5R6US1-利用節(jié)點(diǎn)法IS6RRRRRRRRRRR50.15555151041111US1()UUI1212344121111U()UI12S64456111120()UU100121111U()U402S310.4U0.2U1041

5、2化簡(jiǎn)方程得，解得U18.66V,U269.58V,I86.6A，I15.6A,I6.958A,I17.4A0.2U0.55U4012I2.268A13456.13解：根據(jù)題意,化簡(jiǎn)電路圖，15/54西安交大電工技術(shù)答案I1R01+I2R02+IRL-E1-E2利用支路電流法III21代入II2I1利用節(jié)電電壓法RRRRRERIRI2300.5I5.5I1011L1ERIRI2260.3I5.5I2022L1解得I120A,I220A，I40A111EE()U1210102L010222604603R0.5REE12RR0102U220V11111022RRR3110102LEU230220E

6、UI1120AI21201,0212262200.320AIUR1L2205.540A1.14解:根據(jù)題意,利用節(jié)電電壓法1+R3US1-R22+-US2R1I4R2RRRRRRRRR1111U()UUS11212331111UU()US2I1232321.16解:46/54西安交大電工技術(shù)答案+RUISUS無(wú)源網(wǎng)絡(luò)-設(shè)未知數(shù),BUAUSBIS34A0解得A=0.75，B3.522AB當(dāng)U4V，Is=4A時(shí),U0.75US3.5BIS0.7543.5411V1.17解:根據(jù)與電壓源并聯(lián)的元件可忽略,與電流源串聯(lián)的元件可忽略，故有2A+2VI4I2R2II242-2R44代入I4I222RIRI

7、22I4I22442解得I4=1A,I21A電流源上的電壓UR4I42R141228V,4PUI8216W產(chǎn)生功率118解:根據(jù)迭加定理,當(dāng)僅有電壓源作用時(shí),+R1R2U-R3R4U4141RUU4RR42當(dāng)僅有電流源作用時(shí),R2ISR4U42R1R37/5442RRRR424UUURR西安交大電工技術(shù)答案RIRR24U2SRI4S4242RURRI24S41綜上所述,421.19解:利用迭加定理,當(dāng)僅有電壓源作用時(shí),2+4V34-I1I1442230.67A當(dāng)僅有電流源作用時(shí)，21A4I22I24421130.33A綜上所述,III1221331A20解:化簡(jiǎn)電路+33-6V4R23A2AR

8、22A8/54西安交大電工技術(shù)答案32R+10V-當(dāng)R=5時(shí),可獲得最大功率PU21004R455W1.2解：化簡(jiǎn)電路+-2I2A410V+2-10V2A2.5A+-2I10V424+-I+10V226+2R-18VIU8621A12解:化簡(jiǎn)電路-8V9/54西安交大電工技術(shù)答案4+4+2565A20V40V8-I5A10A26+44I30V8-28+30V30V-8I=0A1.23解:化簡(jiǎn)電路+6I41018V48-3A63AI44R8-24V+10/54西安交大電工技術(shù)答案4463AIR122A1AR4II=.A1.24解：化簡(jiǎn)電路1k500+1kIv-2V+-2V5002mA2mAIv+

9、2VIv1k1k-1k得到U=V,R=1k，將電壓源在外特性曲線(xiàn)與二極管在伏安特性曲線(xiàn)畫(huà)在同一坐標(biāo)系下,那么,兩條曲線(xiàn)在交點(diǎn)的坐標(biāo)即為所求二極管中的電流及端電壓.故Uv1,v=mIV/mA21.510.5UV/V0.511.521.2解：根據(jù)題意，利用節(jié)點(diǎn)電壓法，11/54西安交大電工技術(shù)答案2112U12+10V2U111U0221102-111110()UU2U11U（）U2U11111111014UU510解得：U14V4U11V011UU01第二章習(xí)題2.1解：+-S(t=0)ER1R1i1i2g0=0Ci1)當(dāng)t0時(shí)，UC(0)0V，當(dāng)t0時(shí),UC(0)UC(0)0V這時(shí)電容C兩端相

10、當(dāng)于短路ii2E100100AR11,i10A，UR1E100V,UR20V2)當(dāng)S閉合后達(dá)到穩(wěn)態(tài)時(shí)，ii1A，這時(shí)電容C相當(dāng)于開(kāi)E100i1ARR199121路，i20A,UCR992E10099VRR19912R1E1001VUR11RR19912,UR2UR199V3）當(dāng)用電感元件替換電容元件后,如圖所示，12/54西安交大電工技術(shù)答案+S(t=0)R1i1i2-ER1Li這時(shí)電感兩端相當(dāng)于開(kāi)路RR1991A當(dāng)閉合瞬間,即t=0+時(shí)i(0)i(0)0Ai(0)LL2ii11E2100，R299V，RRUR1RE1100RE9910011VU2RR1991991212當(dāng)S閉合后達(dá)到穩(wěn)態(tài)后,

11、這時(shí)電感相當(dāng)于短路,RUU99VLR2iiE211001100Ai0A1，UR1E100V,UR2U0VL2.2解:根據(jù)題意,31+U1i1iKU2iL-12V(t=0)ULt0時(shí),1當(dāng)ii(0)L12313A,UL0V當(dāng)t0時(shí),iL(0)iL(0)3AU(0)133V2,U2(0)UL(0)=0，UL(0)3V,1i(0)i(0)i(0)U12V1i(0)1234A，kL1i(0)i(0)i(0)431Ak1L.解：根據(jù)題意13/54i240UCi1+-西安交大電工技術(shù)答案80ULiL2010V40(t=0)iCCi(0)128040/40當(dāng)t0時(shí),電感相當(dāng)于短路L100.05AU(0)0V

12、C當(dāng)t0時(shí)，iL(0)iL(0)0.05A，UL(0)UC(0)0Vi(0)i(0)0.05ACL(4020)i(0)U(0)U(0)40iLLC80i40i1012iii(0)12L解得UL(0)1V.4解:根據(jù)題意:2+Si1R2i3-USR1Li2C當(dāng)t0時(shí)，L相當(dāng)于短路，C相當(dāng)于開(kāi)路i(0)LURS21001001A,UC(0)US100V當(dāng)t0時(shí)，iL(0)iL(0)1Ai2(0),UC(0)UC(0)100Vi(0)1U(0)100C0.5AR2001，又i1(0)i2(0)i3(0)00.51i(0)03，i3(0)1.5ARi(0)U(0)U(0)22LC,1001UL(0)1

13、00,UL(0)0V從而，i1(0)0.5A，i2(0)1A,i3(0)1.5A，UL(0)0V,UC(0)100V2.5解:根據(jù)題意，14/54西安交大電工技術(shù)答案U1S(t=0)2R3R1i(t)R2C當(dāng)t0時(shí)，相當(dāng)于開(kāi)路33i(0)LURR12101mA10203，120U(0)Ri(0)20VC220t0時(shí),C3，i(0)3R203當(dāng)20U(0)U(0)VCU(0)112mAi()0mA,RCR2/(R1R3)C1010100mF0.1F1133故tti(t)i()i(0)i()e00e0.1e10tA26解:根據(jù)題意，10+-S(t=0)20V102FUC當(dāng)t0時(shí),U(0)U(0)0

14、V，U()1)當(dāng)t0時(shí),UC(0)0VCCRC52106105C1010202010V則UC(t)UC()UC(0)UC()et10010et1010etUC(2)1010e28.65V，UC(2)UC(2)8.65V2)當(dāng)t=2時(shí),當(dāng)t2時(shí),1RC1021062105，UC()0VU(t)U()U(0)U()eCCCCt2108.65et10521058.65e0.55104tV2.解：根據(jù)題意,15/54西安交大電工技術(shù)答案150V68ia3Ua0.01F-30V+Ua+-683150V-30V+0.01Fi-當(dāng)t0時(shí)，Ua(0)0V,i(0)0A當(dāng)t0時(shí)，Ua(0)Ua(0)0V，RC(

15、3/68)0.010.163i()0A，U()15030360Va此時(shí)電容相當(dāng)于短路,利用疊加定理63/88336/886，i(0)115033065Ai(0)2A2i(t)i()i(0)i()e030ei(0)i(0)i(0)523At12tt0.13e10tAttU(t)U()U(0)U()e60060e60(1e10t)Vaaaa28解:根據(jù)題意，i1R1iC+-UR2SCUC當(dāng)t0時(shí)，CU(0)R230U106VR1R22030當(dāng)t0時(shí),UC(0)UC(0)6V，UC()U10VR1C20103101060.2，R1i(0)UC(0)U16/54i(0)UU(0)R20，i()i()0

16、mA西安交大電工技術(shù)答案106C0.2mA1i(0)i(0)0.2mAC11tti(t)i(t)i()i(0)i()e00.20e0.20.2e5tmAC1U(t)U()U(0)U()e10610eCCCC2.9解：根據(jù)題意，tt0.2104e5tVR1S(t=0)12R2R3CUC(t)當(dāng)t0時(shí)，U(0)CR3RR133E53V23當(dāng)t0時(shí)，U(0)U(0)3V,CCC13U()R3RR3E53V23RC（RR/R)C(32/3)0.21068.4107213tU(t)U()U(0)U()eCCCC.10解:根據(jù)題意,1(t=0)t8.4107333e36e1.19106tVR1US2V+-

17、21kUS1V+C-UCR2當(dāng)t0時(shí),U(0)CR2RR12U212S132V當(dāng)t0時(shí),U(0)U(0)2V,CCC12R2U()URR2101253S2VRC1212100031062103U(t)U()U(0)U()e1010t104CCCC.11解:根據(jù)題意，t2e2103e500tV333317/54西安交大電工技術(shù)答案10+-20ViL10S1HLUL當(dāng)t0時(shí),L10101）當(dāng)t0時(shí),iL(0)0Ai(0)i(0)0AL，那么電感L上的電壓U(0)102010ALi()105101010/5）開(kāi)關(guān)閉合很久后,R10L1，i(t)i()i(0)i()e1eA開(kāi)關(guān)閉合很久后，電感相當(dāng)于短

18、路L=0V201AL510/2tt102.1解:根據(jù)題意i(0)i(0)11933A，i1(0)i1(0)3A，i2(0)i2(0)3Ai1()4.5A,i2()0A210.015103,0.01222103i(t)i()i(0)i()e4.51.5e200tAi(t)i()i(0)i()e3e50tA6根據(jù)三要素法得：t1111t222221解:根據(jù)題意i(0)i(0)122ALL631183.8Ai()L(333612)632351L10.25R23ii()i(0)i()e3.81.8e5tAdt由三要素法得:tLLLLdiULL9e5tVLUU3i11.43.6e5tVLL18/54故西

19、安交大電工技術(shù)答案12U1211.43.6e5ti(t)0.21.2e5tA33100，i()60103A2.14解：根據(jù)題意UC(0)UC(0)45V，iL(0)iL(0)15103Ai(0)151034510315.45103AL100103R11031100106104ti(t)i()i(0)i()e60103(15.4560)103e10000t6010344.55103e10000tA2.5解：根據(jù)題意U(0)U(0)0VCC，iL(0)iL(0)ILIS+-USR1R2U(0)RIRII(RR)S1S2SS12ISIS,Us()R1IS，R1CR1CU(t)RIRIeS1S2StC

20、R1VLi(t)2e2e90tA2.解：根據(jù)題意362Ai(0)i(0)2ALLtLU(t)12e90tVR,UR(0)12V，19/540.11R990西安交大電工技術(shù)答案第三章習(xí)題.解：（1）314rad/s，f31450Hz22，1T0.02sfUm220Vm2，uU380V相量圖Um220600U，3(2）波形圖U(V)1902/3-/30/6t-380+jU（3)Um380600,U2206000/3+13.2解：(1）電流滯后于電壓450(300)750ui（)U220ej45，I10ej3000U220450,I10300Ui2100-45030045012001350Uit波形

21、圖20/54西安交大電工技術(shù)答案U450300+1U220cos0,U220，U22000I相量圖(3)若電流的方向相反,則電流的相位超前電壓60,波形圖和相量圖可雷同作出。3.3解：根據(jù)題意，I10cos14j10sin452j5210450I42sin(900)j4242900（3.4解：根據(jù)題意,U220cos60jsin60)I5cos(53.10)jsin(53.10)I5cos(126.870)jsin(126.870)2001235解：XLL314101033.14U220600I70.06300AjXj3.14L，Imax70.06299.03Ai99.03sin(314t30

22、0)，QUI22070.0615.41Kvar最大磁場(chǎng)能量即無(wú)功：Q15.41KvarU31030036解：m,jXjC11106jj132.7C31424U310300Im2.3361200AmjXj132.7C，Im2.336A3102.336QUI362.1var22最大磁場(chǎng)能量即無(wú)功：Q362.1var，i2.336sin(314t1200).7解：根據(jù)題意，IU|Z|，即0.03380162(314L)2,解得L40.34HXL12.67kL，PI2R0.0144W，PUIcos，21/54cos西安交大電工技術(shù)答案P0.01440.0013UI3800.03QPtan0.0144t

23、an11.4var.解:根據(jù)題意，abIRU1024402，uab40sin3tjIXj10U232jabL602，ubc60sin(3t900)UUUabUcacabc40602j250.99123.690uca72.11sin(3t123.690)3.9解:根據(jù)題意(1）RR1R230020320，jXLjLj3141.5j471ZRjX320471j569.4355.810LIU22000Z569.4255.8100.38655.810，I0.386AUIR0.386300115.91VU（2)R11，R1115.9155.810UI(RjX)0.38655.810(20471j)181

24、.9731.75850RL2L這兩個(gè)電壓加起來(lái)等于220V(3)PI2R0.386232047.68WUI10解：(1）PUIcos，cosP94022050.8545I|Z|U220544r|Z|cos37.6，L|Z|sin44sinL|Z|sin1000.073H(2)VI2R11.51000.115A,r和組成的|Z|V|Z|3I11.50.115100，20(11.511.5cos)2(11.5sin)20.274H|Z|UI解得，cos0.512,r|Z|cos51.2，XLL|Z|sinL|Z|sin(3)VI3R0.351000350V，r和L組成Z3503500,20.10.

25、4(0.350.1cos)2(0.1sin)2cos0.393,r|Z|cos35000.393137522/54XLL|Z|sin，L西安交大電工技術(shù)答案|Z|sin10.25H3.1解：根據(jù)題意，由R和組成的Z為Rj1C,則有tan600C111C0.219FR，Rtan6008400314tan6001U8400201|8400j|3140.2191060.354V8R6Z，23.2解:(a）A表讀數(shù)為1A(b）表讀數(shù)為3V（c)V表讀數(shù)為(d)A表讀數(shù)為56A3.13解:(1）A0表讀數(shù)為628210A4ZR（2)23時(shí)電流表讀數(shù)最大，此時(shí)為+6=14A(3)8jX6Z，234ZjXL

26、2(4)此時(shí)A2表讀數(shù)為0。L時(shí),A表讀數(shù)最小,此時(shí)讀數(shù)為2A0R10U22202P4840WIR11131解:,U22022AR101|RjX|20j102|PI2R1613.34W220220I8.98A12L,R222PPP1613.3448406453.34WR1R220.1126mF3.1解：根據(jù)題意，V表讀數(shù)220110IC2220220I2220XC220CR和L組成的Z，R10220|Z|1022220.0318H10L，L10A的讀數(shù)：5.A2的讀數(shù)：7A23/54西安交大電工技術(shù)答案II1I2A表讀數(shù)為i1和之和故表讀數(shù)為16.21A。22245011290016.2116

27、.320.16解:I110900A,U110000VI2102450R和L支路中電流1000055jII1I210245010j10A,I=10AUU1I(j10)1000010j10100100j1002450VI110900A，I2102450AII1I210A,I=1220I1j(X)RIjX10900510,解得X17U1002V31解:根據(jù)題意CCC220I2R22450，R28.5，XLR28.5I210200A,I110900A，II1I2109001000102450A3.18解:根據(jù)題意I102Au與i同相,總阻抗ZjXLR/(jXL)，RXCZR22得到R(X)jL虛部為0

28、R2XL,ZR2202，R102,XC102，XL52R100219解:根據(jù)題意,U10036.10VZ21j(2060)2140jU10036.10I2.21398.40AZ2140j，i3.13sin(314t98.40)VU2.21398.40（120j)44.32174.460VU2.21398.40（2060j)139.9616.830V12PI2R2.213221102.84W24/54若發(fā)生諧振西安交大電工技術(shù)答案i1002sin(314t36.10)6.73sin(314t36.10)A21UjXIj204.7636.1095.24126.10VUjXIj204.7636.10

29、95.2453.90VLLCC.20解：根據(jù)題意UR26000(）并聯(lián)兩端的電流為：IR2200AR302U6000(jX)IR26000(j)4020010053.130V讀數(shù)為：197AIIR1IR21106.2602001.9729.170A并聯(lián)兩端的電壓為:LCR1、串聯(lián)支路的電流為:()P1.9722512602230277.02W整個(gè)支路的阻抗ZR(R1jXL)/(R2jXC)71.3920.62jcos0.96(容性)QPtan80Var(4）總電流和總電壓同相，只需、L支路和R2、C支路并聯(lián)串聯(lián)諧振即可,所以需要2，XXRR1L3.21解:根據(jù)題意CIIS2002450A(a)

30、5555j55jIIS30005053.130A（)68j68j68j8j53.22解:根據(jù)題意IUZZ/Z123353.130A并聯(lián)支路電壓為:6000(410j)353.1303016.260I2320.610AI33016.260(j)0.123.6106.260A3016.2601036.8703.24解：根據(jù)題意，發(fā)生串聯(lián)諧振時(shí)XXC1111LL3140.1HL即：C,C1000314R15.72025/54RU.25解：根據(jù)題意,西安交大電工技術(shù)答案220I0.1當(dāng)頻率為f1時(shí)，電路是電容性的2|Z|在頻率為KHz時(shí),有:0.070728.2928.292202(XX)2LC即XC

31、XL20(）1100103L100103C(2）由(1）、(2）解得L0.1mH,C106FR200.5在f=0時(shí)，Q0L1001031030.13.26解:根據(jù)題意，LC2f1f1,2LC，C0.11FQ,QWW0047547847279.2并聯(lián)部分導(dǎo)納:RjLjC27解：根據(jù)題意Y11為最小在5KHz時(shí),并聯(lián)部分RjLY11.0429.79解得C10.72FY1jC0.07860.045j11Z1整個(gè)電路Z,CZZj111.0429.790j12125103C2最小34.6722.74420.0288AC5.8F2I1.28解:根據(jù)題意1Z34.67j在f=時(shí),Z234.67，C234.6

32、72.744j2PI2R（2）120240W，QI2XL290180VarPtanQ1802400.75,cos0.812090jjC0.005330.004jjCY126/54IUY(12090j)200(0.005330.0033j)1.335.110西安交大電工技術(shù)答案功率因數(shù)為0.85，C2.22FPUIcos15020.851.33239.8WQPtan148.61Var3.2解：根據(jù)題意IP2201102U2L，UL190.53V，400.3636AU110L，I0.3636UIXLU190.53XL524LP40cos0.5UI40220PUIcos,110PI2R,R302.5

33、并聯(lián)電容以提高功率因數(shù)：Y1302.5524jjC功率因數(shù).9,計(jì)算得到C3.3F3.30解;根據(jù)題意U220Ucos2200.6（1)電源的額定電流電源實(shí)際發(fā)出的電流S10103I45.455A1P8103I60.606A2電源輸出電流超過(guò)了它的額定電流(）對(duì)并聯(lián)電容提高功率因數(shù)，有計(jì)算公式U222022fCP8103(tantan)(1.3330.3287)0.53mF1電源輸出的電流源為：P8103I38.28AUcos2200.9545.454538.284039(3）此時(shí)除掉大負(fù)載，還可接的燈泡個(gè)數(shù)為:3.31解;根據(jù)題意,電流中有基波和次諧波220I1m基波分量：803006(41

34、2)j883.130I3m次分量:18006(124)j1.853.130i8sin(t83.130)1.8sin(3t53.130)81.8PUII2R()26()26201.72W2227/54西安交大電工技術(shù)答案333解:根據(jù)題意交流電源工作時(shí)：R3R4直流電源工作時(shí):R3R4R1R2直流電源中通過(guò)交流電流，但交流電源中不通過(guò)直流電源34解；根據(jù)題意8+6j8+6j8+6j380相電壓為:3220V220，相電流82622A,線(xiàn)電流=相電流3.5解:根據(jù)題意33003300330022073.33A相電流3，線(xiàn)電流=3相電流=123.6解:（1）種情況下,繞組應(yīng)該Y接;(）種情況下,繞組

35、應(yīng)該接；2206.11A（1)種情況下，相電流36，線(xiàn)電流=相電流28/54西安交大電工技術(shù)答案22018.33A12(2)種情況下，相電流,線(xiàn)電流=3相電流1.75A.1解：由磁路歐姆定律：FmRmINls，l1IsN4.解:（1)線(xiàn)圈電阻不變,電源電壓不變。由IUR知不變。由FmRmINlS中，只有S加倍，加倍。又BS，B不變,銅耗I2R也不變。（2）UE4.44fNm，m不變;S,Bm變?yōu)樵瓉?lái)的一半；BmmSN，I變?yōu)樵瓉?lái)的一半；銅耗2l1III2R(m)2Rmmm1/8。變?yōu)樵瓉?lái)的B（3)IUR,I不變；FmRmINlS,加倍;S，B加倍;銅耗(4)UE4.44fNm，N加倍,m減半;

36、S,Bm減不變。BmmSN，I變?yōu)樵瓉?lái)的14;銅耗2半;變?yōu)樵璴1III2R(m)2Rmmm來(lái)的1/32.，f減半，m加倍;BS，Bm加倍；（5)UE4.44fNmmmIlSN，I加倍;銅耗變?yōu)樵瓉?lái)的2倍。1mmmS，Bm不（）UE4.44fNm,U,f減半,m不變；BmmSN,I不變；銅耗不變。變；l1Immm4.3解:銅耗PI2R52250cuW；P電路功率因數(shù)cos2750.25UI2205，arccos0.2575.5;鐵耗PFePPcu2755022529/54W;UU22075.5，主磁感應(yīng)電動(dòng)勢(shì)若令I(lǐng)I050A，則EUIR22075.5502217.778.1,E218西安交大電

37、工技術(shù)答案E4.44fN4.44503002183.27103m4.4解:（1)無(wú)鐵心時(shí)電路的功率即為銅耗。Wb。PUIcos100100.0550cu2具有鐵心時(shí)PPUIcos10050.7350cuFe11WPPP35050300Fecu（)RP50cu2I522marccos0.7arccos0.720arccos0.7I5R14,X14.3U100RjXm1mm4.5解:(1）NN21UU2N60001N2203300400（2）SNUNIN，I1NS50103N15.2U33001NI2NS50103N227.3U2202NAPUIcosIcos227.30.8U2204.6解：P3

38、91032U214.5222N2N(）2,VU36NN255090211UUNN3550311為純電阻負(fù)載1222030U36U12I2PP362421I32332A，N550I1ININ19023022330.271A同極性端為1的上端,的下端和3的左端。.7解：(1)I1NS2103N5.26U3801NI2NS2103N18.18U1102NA()燈泡的額定電流（設(shè)燈泡為純電阻）30/54西安交大電工技術(shù)答案IdNPdNUdN150.136110A達(dá)到滿(mǎn)載運(yùn)行時(shí)能接的燈泡數(shù)為I0.136xI18.182N133.7dN取133Ucos1100.8P15NI0.170Nd()小電機(jī)的額定電

39、流達(dá)到滿(mǎn)載運(yùn)行時(shí)能接的燈泡數(shù)為xI2NIN18.180.17106.9取1064.8解:（1)為使負(fù)載獲得最大功率,變壓器的輸入阻抗應(yīng)等于電源內(nèi)阻，即:RK2R0LKR0RL5608.48()I1USRK2R0L108.9103A8.9mA560560UURK2R560560K8.4(3)PIU0.674.7644.86mA10SK2R5605V1L0LIIK8.98.474.76mA2115UU0.6V21222（4）將負(fù)載直接接在信號(hào)源上時(shí)，5608P(2USRR0L10)2R()282.5103W2.5mWLPP224.9解：(）4個(gè)繞組的同極性端為主繞組的上端和1，,3繞組的上端。（2

40、)能夠得到7種輸出電壓。(3）SNS1NS2NS3N319127139W1NU361N22.7AI2N138.9AU220U36S39IN0.18AU220NU2201K64.10解:（1）變壓器的變比20SS51035103INN1N（2),2N(3)變壓器的空載電流很小，忽略空載電流，則變壓器空載時(shí)輸入變壓器的有功功率就是鐵耗,即P60WFe。P60W,P鐵耗為不變損耗,當(dāng)負(fù)載后FeCu18060120W.(4）設(shè)負(fù)載后的電流與所加電壓成正比,則滿(mǎn)載時(shí)的輸出電流31/542N138.9131.2AU36P3122.56西安交大電工技術(shù)答案U34I2I2201則PU2I2cos234131.

41、20.73122.56WPPP3122.561802942.56W21P2942.562100%100%94.2%滿(mǎn)載時(shí)的效率1半載時(shí)原，副邊電流均變?yōu)樵瓉?lái)的一半,則銅耗PCu11P12030W4Cu4PPP306090WCuFe2211PP3122.561516.28W11PP1P11516.2890100%100%94.2%1516.28411解：（1）變壓器變比K6/31.13.15（）U1pN633.46kVU2pN3.15kVS3UI,IN1N1NI96.2A1pN1NSN3U1N10001033610396.2A2NII2pNS1000103N3U33.151032NI183.32

42、N105.8A33183.3A1N0.4kVK252pN0.23kVU0.4kV34.12解：(1）/Y連接時(shí)，相壓變比N2100K125N842110UU2N0.4U，2lN2NSII144.3A2pN83.3AIN10001033U30.4102NI2lN2N，(）/Y連接時(shí),相電壓變比3144.3A144.33KNN122100842532/54西安交大電工技術(shù)答案U2pNU1N1100.4kVK25,UU2lN2pN30.69kVU2NU2lN0.69kV2NS30.6910386.2AIN3U2N100103I2lNI2N86.2A,I2pNI2lN86.2A5.1解：三相異步電動(dòng)機(jī)

43、的3個(gè)繞組通入三相初相不同的對(duì)稱(chēng)電流后,在任一時(shí)刻,由電磁場(chǎng)理論知，這3個(gè)繞組中產(chǎn)生的合成磁場(chǎng)形成一對(duì)磁極,不同時(shí)刻電機(jī)定子磁場(chǎng)在空間的位置不同，因?yàn)槿嚯娏魇沁B續(xù)變化的,所以磁場(chǎng)在空間的位置的改變也是連續(xù)的,于是磁場(chǎng)就旋轉(zhuǎn)了起來(lái)，其轉(zhuǎn)向取決于三相繞組中電流的相序。p。磁極數(shù)增加,旋轉(zhuǎn)磁場(chǎng)的轉(zhuǎn)數(shù)較小，轉(zhuǎn)數(shù)與磁極之間的關(guān)系為n60f11.2解:這樣有可能會(huì)導(dǎo)致定子繞組燒毀。因?yàn)槿喈惒诫妱?dòng)機(jī)定子繞組和轉(zhuǎn)子繞組之間相當(dāng)于變壓器的原,副繞組,抽掉轉(zhuǎn)子之后，就成為電磁線(xiàn)路,定子繞組中的電流增加很多,容易燒毀定子繞組。.3解:當(dāng)電動(dòng)機(jī)軸上負(fù)載增加時(shí)，電動(dòng)機(jī)轉(zhuǎn)子的轉(zhuǎn)速就下降,而轉(zhuǎn)子的轉(zhuǎn)向和由定子產(chǎn)生的旋轉(zhuǎn)

44、磁場(chǎng)的轉(zhuǎn)向是同向的,所以轉(zhuǎn)子切割磁力線(xiàn)的相對(duì)速度就增大了，使得轉(zhuǎn)子繞組中的電流增大,于是定子繞組中的電流也就增大,輸入功率自然也增大，所以雖然異步電動(dòng)機(jī)定子繞組和轉(zhuǎn)子繞組之間沒(méi)有電的直接聯(lián)系，但由于有電磁耦合的關(guān)系,當(dāng)負(fù)載增加時(shí)，定子電流和輸入功率會(huì)自動(dòng)增加。54解：（1)n2940r/min0，n13000r/min（2）snn1n10.02()f2sf10.02501Hz（4)轉(zhuǎn)子旋轉(zhuǎn)磁場(chǎng)和定子旋轉(zhuǎn)磁場(chǎng)相等，等于n1；轉(zhuǎn)子旋轉(zhuǎn)磁場(chǎng)對(duì)轉(zhuǎn)子的轉(zhuǎn)速為n1n60r/min。(）轉(zhuǎn)子旋轉(zhuǎn)磁場(chǎng)對(duì)定子旋轉(zhuǎn)磁場(chǎng)為300r/min。(6)釘子旋轉(zhuǎn)磁場(chǎng)對(duì)轉(zhuǎn)子旋轉(zhuǎn)磁場(chǎng)的轉(zhuǎn)速為。55解:（1）由型號(hào)知，電動(dòng)機(jī)的極

45、數(shù)為8。n160f60501p4750r/minnn(1s)750(11.33%)740r/minPTn95509550()輸出的機(jī)械功率1477.574037kW2100%91%(3）效率P372100%P40.661（4)P40.661031cos0.793UI338078.2115.6解：（1)P1P2N5.56.43kW85.5%33/543Ucos33800.84p2電極四極，n1440西安交大電工技術(shù)答案1IP6.4310311.63ANNN60f6050n11500r/min1nn15001440s1N0.04n15001P5.5T95502N955036.5NmNN（2）TmT

46、N2.236.52.280.3Nm（3)IstIN7.011.637.081.41ATT2.236.52.236.52.280.3NmstN（4)采用Y/降壓起動(dòng)Ist11I81.4127.14A3st333T9550P2N9550n1460P3.5T9550P2N9550n1460P3.55.9解:（1）n735r/min，111TT80.326.8Nmstst5.7解:(1)若電源電壓為30V,電動(dòng)機(jī)Y連接2.818.3NmNNnn15001460s10.027n15001P3UIcos33806.30.8434833.5kW1NNNP2.82100%100%80%N1（)若電源電壓為20

47、V,電動(dòng)機(jī)連接2.818.3NmNNnn15001460s10.027n15001P3UIcos322010.90.843489W3.5kW1NNNP2.82100%100%80%N1可見(jiàn)，各項(xiàng)參數(shù)和（1）中的相同。n750r/minNP60f60501n75014(2）snn7507351Nn75010.0234/54EsE0.024208.4V西安交大電工技術(shù)答案E420V20220(3)f2sf1500.021Hzn9805.解:（）P55T95502N9550536NmNNT（2)當(dāng)電源電壓UUT643.2Nm600NmTTst5361.2643.2NmstNTNN時(shí)，st，電動(dòng)機(jī)可以

48、起動(dòng)。U0.8UT0.82T412Nm600NmN當(dāng)電源電壓時(shí)，stst電動(dòng)機(jī)不能起動(dòng)。,(）T0.652T272NmststTT5.1解:令mst，即R2x2Rx220，即2202KU21f12xf2x220120K122U2R11RR0.04120.04R2R220.042，解得：2RR0.040.020.02應(yīng)串入起動(dòng)電阻為25.12因?yàn)槿喈惒诫妱?dòng)機(jī)斷了一根電源線(xiàn)后,就成為單相電動(dòng)機(jī)，沒(méi)有起動(dòng)轉(zhuǎn)矩,所以無(wú)法起動(dòng),而在起動(dòng)后斷了一線(xiàn),因?yàn)槿杂修D(zhuǎn)矩，所以仍能繼續(xù)轉(zhuǎn)動(dòng)。前一種情況電流很大,時(shí)間一長(zhǎng)，電機(jī)容易被燒毀;后一種情況下,由于帶有額定負(fù)載,則勢(shì)必超過(guò)額定電流，時(shí)間長(zhǎng)了，也會(huì)使電機(jī)燒壞。

49、5.1單相異步電動(dòng)機(jī)的起動(dòng)方法主要有三種：電阻分相起動(dòng)，電容分相起動(dòng),和罩極式起動(dòng)。其產(chǎn)生轉(zhuǎn)矩的原理是將兩相繞組中通入不同的兩相電流,把工作繞組與起動(dòng)繞組中的電流“分相”。改變電動(dòng)機(jī)轉(zhuǎn)向的方法是：把主繞組或者副繞組中的任何一個(gè)繞組接電源的兩出線(xiàn)端對(duì)調(diào),改變氣隙旋轉(zhuǎn)磁場(chǎng)的轉(zhuǎn)向。61QFUSBFRKMKMKM35/54西安交大電工技術(shù)答案FSB1停止,SB啟動(dòng)。62當(dāng)電源電壓過(guò)低或者突然停電時(shí)，接觸器線(xiàn)圈失電,使得所有常開(kāi)觸點(diǎn)斷開(kāi),電動(dòng)機(jī)停轉(zhuǎn),電源電壓恢復(fù)正常后,必須重新按下起動(dòng)按鈕,電動(dòng)機(jī)才能起動(dòng),否則不能自行起動(dòng)。63S1S2KAFRSB2，SB2點(diǎn)動(dòng)，SB3連動(dòng)。SB3SB1BKK64順序起

50、動(dòng)運(yùn)行，按S2控制1M先起動(dòng)，隨后按下SB4,2起動(dòng)運(yùn)行。1M不起動(dòng)時(shí),M不能起動(dòng)。按SB3可單獨(dú)停止2M，按SB1可同時(shí)停止M,2M。該電路有短路保護(hù)、過(guò)載保護(hù),并有失壓和欠壓保護(hù)。65(a)按下SB,SB與KM短路。(）通電時(shí)電路不工作或者短路。(c）SB不起作用。（d）通電時(shí)電路不工作。6電路有7處錯(cuò)誤。改正如下圖所示：FUFRKMFKMRM3S1SBFK2KM1SM1KM236/54西安交大電工技術(shù)答案6.7SBSBST2KM2KM1FR1FR2QKM1ST4ST2T3KM4K3KKM2KM3KMKMST1M2ST3FR1FR22M33KM1MST4M34ST1KM4SB2為起動(dòng)按鈕,

51、當(dāng)KM1通電時(shí),2;撞到ST,KM1失電，KM3通電，此時(shí)開(kāi)始4；撞到S3,KM失電，KM2通電自鎖,2；撞到S1，KM2失電，KM通電自鎖,4；撞到T4，KM1通電自鎖,12，重復(fù)以上過(guò)程。68FUSSB2K2KM1FSB2K1M1M2S3SBK137/54西安交大電工技術(shù)答案RKM2KM2SB5S2正轉(zhuǎn)點(diǎn)動(dòng)按鈕，B反轉(zhuǎn)點(diǎn)動(dòng)按鈕，F(xiàn)U短路保護(hù),R過(guò)載保護(hù)。6.9S1SB2K1FRM1KT1KT1KM25SKT2KTKM34S6.1(1）不同,電源線(xiàn)有兩根換接，電動(dòng)機(jī)轉(zhuǎn)向相反,串聯(lián)限制電流。(2）按下SB2，K通電自鎖KM1主觸點(diǎn)閉合，電動(dòng)機(jī)起動(dòng)KT通電,T延時(shí)斷開(kāi)觸點(diǎn)瞬時(shí)閉合,KM1常閉觸點(diǎn)

52、斷開(kāi)。該控制電路，S2為直接起動(dòng)按鈕,B為反接制動(dòng)按鈕。（3)不正確。串入的電路，電動(dòng)機(jī)工作電壓低于額定電壓不能正常工作。（4)按B2，由于KM1、K串聯(lián),KM1的電壓大大低于額定電壓,M1線(xiàn)圈失電，欠壓保護(hù),M觸點(diǎn)不動(dòng)作，KT線(xiàn)圈通電,KT瞬時(shí)閉合,KM2通電,電動(dòng)機(jī)欠壓反轉(zhuǎn)。.1（1）B1B2KM1RKM13B4KM2M(）SB1SB1R38/54西安交大電工技術(shù)答案KM1KTKTKM2（3)SSB2K2MFKMKKTKM2M2（4)SB1SB2AKA1FRKA1KKM1KM1KM1KA1SB3K22SBM2KA1A2KA2S2控制M起動(dòng)，B1控制M停止,SB3控制M起動(dòng)，SB4控制1M停

53、止。6.12合上開(kāi)關(guān)K,HL燈亮，K1線(xiàn)圈通電,K線(xiàn)圈通電自鎖，A常閉觸點(diǎn)斷開(kāi)，HL燈滅，同時(shí)A常開(kāi)觸點(diǎn)閉合,KT2線(xiàn)圈通電，K2延時(shí)常閉觸點(diǎn)瞬時(shí)斷開(kāi)，KA常閉觸點(diǎn)閉合，H燈亮，K常開(kāi)觸點(diǎn)斷開(kāi),延時(shí)后T2閉合，KA通電，燈滅。KT瞬時(shí)打開(kāi)，A斷電，燈亮，延時(shí)后燈滅。重復(fù)此過(guò)程。613合上開(kāi)關(guān)，KM線(xiàn)圈通電，油泵電動(dòng)機(jī)起動(dòng)。K1線(xiàn)圈通電延時(shí)T1延時(shí)常開(kāi)觸點(diǎn)閉合K線(xiàn)圈通電自鎖，KA常閉觸點(diǎn)斷開(kāi)。39/54西安交大電工技術(shù)答案KT2線(xiàn)圈通電。KM線(xiàn)圈斷開(kāi),油泵電動(dòng)機(jī)停轉(zhuǎn)。延時(shí)K斷開(kāi)KA斷電，常閉觸點(diǎn)閉合M通電,油泵電動(dòng)機(jī)起動(dòng)。此過(guò)程為:系統(tǒng)運(yùn)行t1后,停止2，周而復(fù)始。SB作點(diǎn)動(dòng)運(yùn)行，不作自動(dòng)循環(huán)操

54、作。614按下S,KA1線(xiàn)圈通電自鎖,KA1常開(kāi)開(kāi)關(guān)閉合，KM線(xiàn)圈通電，水泵起動(dòng)注水。當(dāng)液面上升至E點(diǎn)時(shí),浮子帶動(dòng)NS磁鐵經(jīng)過(guò)E點(diǎn),接通，KA2線(xiàn)圈通電，KA2常閉斷開(kāi),KA1線(xiàn)圈失電，KA1常開(kāi)斷開(kāi),KM失電，水泵停轉(zhuǎn)。當(dāng)液面下降至F點(diǎn)時(shí)，S磁鐵浮在點(diǎn)J接通,KA1線(xiàn)圈通電自鎖,KA1常開(kāi)閉合，KM通電,水泵起動(dòng)注水。此過(guò)程周而復(fù)始。615按下起動(dòng)按鈕SB1,KM線(xiàn)圈通電并自鎖,1常閉打開(kāi)，KM不能通電。KA線(xiàn)圈通電并自鎖,KA常開(kāi)閉合。小車(chē)前進(jìn)一直到B，壓下STbST常閉開(kāi)關(guān)斷開(kāi)，KM1失電,KM1常閉開(kāi)關(guān)閉合，同時(shí)小車(chē)停止前進(jìn)。T1線(xiàn)圈通電延時(shí)T1閉合，M通電自鎖小車(chē)后退一直到A,壓下S

55、Ta常閉斷開(kāi),K2線(xiàn)圈失電，KM常閉閉合,小車(chē)停止后退。S常開(kāi)閉合，T2線(xiàn)圈通電延時(shí)KT2常開(kāi)閉合KM1通電自鎖，小車(chē)前進(jìn)。此過(guò)程如此反復(fù),直到按下停止按鈕0。KM1控制電動(dòng)機(jī)正轉(zhuǎn),帶動(dòng)小車(chē)前進(jìn),M2控制電動(dòng)機(jī)反轉(zhuǎn)，帶動(dòng)小車(chē)后退。1自動(dòng)往返運(yùn)動(dòng)的起動(dòng)按鈕,S0總停止按鈕。若沒(méi)有中間繼電器A,那么只要有物塊壓住Sa，即使沒(méi)有按下起動(dòng)按鈕,小車(chē)也會(huì)自動(dòng)前進(jìn),這樣有可能造成事故。另外，中間繼電器對(duì)按鈕S起鎖存作用,只要按一下SB1,當(dāng)SB1恢復(fù)常開(kāi)狀態(tài)時(shí)，K常開(kāi)會(huì)一直閉合,為接通電路作好準(zhǔn)備。只有按下B1系統(tǒng)才能起動(dòng),否則小車(chē)不會(huì)動(dòng)，KA使S1對(duì)控制電路有完全的控制作用。.1(）（b）0001000

56、500050()（）000001002002000500(e）90240/54西安交大電工技術(shù)答案500（f)（g）同。001000205007.2（a）()DIFU(13)1000009000000001001000TI0002000041200001200#000TIM000010011050060007000105001200TM00100DIFU(13)1100DIFD(14)1200KEEP(11)0501()D0001(b)LD001OR0005IL（2)ANDNO000D002OR00OT1LD0003LD0003N004IL(2)LD006LD00ANDOT007OUT1002

57、ORLDIM0041/54西安交大電工技術(shù)答案ANLD#100OROT0009LD05OUT0500L(02）L0006U1003LDTIM00OU1004ILC(03)7.4LD0000000010005000500ANDANDNOT000000NDNO00TIM000UT0500500TR0L050TR05010100OUT0#01TR0TIM00501RIM00#0005R0TIM1TIM01IM00TIM0502TIMLDEND(01)#0005ANOTOUTTIM1LAIMDAN42/54西安交大電工技術(shù)答案OU0502ND(1)7.500SLD000KEEP(11)D0012120

58、0KEE(11）001LCP1200TIM0000TIM000TIM10100TIM0CNT10TM00R#0003LD000ANDNOTTM#0100L0TIM001000TIM010CN10LDTIM#500500TI12TM00NT101200TI0CN10000(）00000000050LDANOANDTOUT(2)0000001050005000000#000050003050101LD0000TIM00501DTIM000000O0500OR50043/54西安交大電工技術(shù)答案ADNT0002ADNOT001OUT0500OUT00LD0001M0R050#TN0500LDTIM0

59、ANDNT003OUT00105000501OT0501（3)0000001000TIM000OR00050TIM00TIM00#T0501TM0M0OUT050IM000LD#TANDNDNOLDT(）0000002005000500050500050305100050IL(02)OR05000503D00ILC(03)5000000502000503LD000ANDNOT0002UT500OR0501ADO05OUT0IL(02）44/54西安交大電工技術(shù)答案00OR502AND50052NT00OR03AN00ADNOT050OUT050C(3)7.700000502000050200O

60、UT501LD051OR05005205000501050205010503LANDNOTOT053AO0503U052TIM000050TIM00502050DOR503ANDOTTIM00UT05030503002L03#TANN2IM00#T00000500HL050105020503KA1A2345/54西安交大電工技術(shù)答案7.000005D00005010500NDN100100TIM000#T0OU05TIMTIM0#TTIM010T000501#T1LD0501T1OR51TIM01#T1000油泵電動(dòng)機(jī)ANDNOTTIM01OUT0501KM7.900200051LD00205