中国石油大学(华东)电工电子学(一)课后习题答案.docx

电工电子学课后习题答案目录第一章 电路的基本概念、定律与分析方法1第二章 正弦交流电13第三章 电路的暂态分析29第四章 常用半导体器件41第五章 基本放大电路43第六章 集成运算放大器及其应用46练习与思考1.4.2（b）-第一章 电路的基本概念、定律与分析方法-+42V6+a-1.4.2(c)a36V6Vbb52A5+40Vb-baa6A1.4.3（a）1A52A+-515Va ab bU =5+5´2=15VabR =5Wab1.4.3 (b)6+42V-baU= 6 + 6 ´ 6 = 42VabR= 6Wab1.4.3 (c)a5+40V-bU =UabR+10=6´5+10=40VR =5Wab1.4.3 (d)3+-us 6 VabuuI =s I=s1429KVL : 2I1+ u- 3I= 0ab2uu= - s V ab6R= 3´ 6 +1 = 3Wab3 + 61.4.4 (2)R1+-+9VabRR243R1+6V5V-5V-RR215V15 -V15 -VI2 =R2b I=b4R415 -V5 + VI3 =R3b I=b4R1KCL :I = I12+ I + I34V + 515 - V6 - V5 - Vb=b +RR12R b +R b43求方程中VbbI= 6 - V=2R6 - V50kV +9V +9b I=b=b126 - VV+ 9R100k1KCL:b =bV= 1V50k100kbKVL :- 6 + 50KI+100KI- 9=0100KI=15 I=1A 10kV=6 - 50k ´b110k=1V习题1.1 （a） I = 5 + 4 - 2 = 7 Ax（b） I= 0.4 - 0.7 = -0.3 Ax1I= -0.3 + 0.2 + 0.2 = 0.1Ax 2(C)Ix 4= 0.2 ´ 30 = 0.1A60I= 0.2 + 0.1 = 0.3Ax3I= 10´0.3+ 0.2´30 = 0.6Ax 21.5I= 0.3 + 0.6 = 0.9 Ax11.2I3I4I61.3= 0.01 + 0.3 = 0.31A= 9.61 - 0.31 = 9.3 A= 0.3 + 9.3 = 9.6 AP = -14 ´ 2 = -28w1发出功率P = 1´10 = 10w吸收功率2P = 4 ´ 2 = 8w吸收功率3P = -(-1´10) = 10w吸收功率4P发P吸1.6=28w=28wP =P发吸6R =50mv= 120（a) uR1.7= 6V(b)uR= 12 ´60= 4V60 +120（a） us(b)I= 1´ 4 = 4V= 2 A-5 + u- 2 ´ 2 = 0 u= 9V1.8us1sss= 2 ´ 2 +10 = 14V10I=s 210- 2 = -1AP = -2 ´14 = -28w1P21.9= -10 ´(- 1) = 10wu6.3´ 4xR = 123 = 168I = 0.45 A I= 0.3 A I= 0.45 - 0.3 = 0.15 AI30.1511.10(a) :u223= u = 16V1510 - 6.3´ 4 - 6.3R =y0.45= 174.4(b) :u2= 16 ´45 + 55= 1.6V(c) :u2= 1.6´= 0.16V45 + 55(d) :u21.11= 0.16´= 0.016V45 + 5u= uR + R2p= 8.41V211 R + R + R12pRu= u221 R+ R2 + R = 5.64V12p1.12R= 6 ´ 3 = 0.5CD6 + 3R=0.5BDR= 199 + 0.5 = 199.5ADI = 10 ´ 1 = 5mA I122= 5mAI = 5´333.6= 4.2mAI = 5´ 0.6 = 0.8mA = I - I43.623u= 10 ´199.5 = 1.995mVADu=1´ 5=5mVBDP = -10 ´1.995 = -0.01995ws1.13 (a)AR13AR22 A2WBA5A2+10VB-AB(b)9+3V-A3+15V-6-12V+BAB1.1534A+33A3I6V- 3+1A 1.5I6V-1.5 3+1.5VI6V-I = 1.5 - 6 = -1A1.5 + 31.1610V-10+10V-5I332I =10 +10= 1A310 + 5 + 3 + 2I = I - I1s13= 1-1 = 0I = -2 - 0 = -2 A0I = -1+ 2 = 1A262I = 1´=A46 + 3331I = 1´=A56 + 331.1712A3 2A62V1 a-+I 224A2b+- aIb22V+-+I8V2-I =8 - 2= 1A2 + 2 + 21.18I + I = I123KVL :- u + I R + I R = 011 13 3- I R - I R + u = 03 32 221612I = 75 = 0.213I6= 75= 0.08I322= 75 = 0.2931.19n = 2I +18 = I + I132KVL :- 140 + 20I1+ 6I = 025I - 6I = 032I = 4 A I12= 10 A I3= 12 Au = 20I11= 80Vu = 6I22= 60V电压源 P = -140 ´ 4 = -560w 发出功率电流源 P = -60 ´18 = -108w 发出功率1.20n = 2I + I12+10 = IKVL :0.8 I1-120 +116 - 0.4I = 020.4I2-116 + 4I = 02I = 8.75A I12= 9.375A I3= 28.125A120V : P = -120 ´ 9.375 = -1125w116V : P = -12016 ´ 0.75 = -1015w10 A : P = -10 ´ 4 ´ 28.125 = -1175wR : P = I 2 R = 28.1252 ´ 4 = -3078.125wL1.21I + 0.5 = I122 - 3 -V-1-V3I =2 =213VI = 22427V =21.221R3+-us2R42RRIus+ Ru6I=21R2s= 40 + 20 = 0.1A4122R3R4IsRRIIsI = 0.1+ 0.1 = 0.2 A2R204I= I ´22sR + R24= 0.3´= 0.1A40 + 201.23+-0.5I1V3 111I=1´ 1 = 0.25A 310.5 + 0.5 +1 1+10.5I3 111IsI= 2 ´1´ 1 = 0.5 A 320.5 + 0.5 +12I = 0.25 + 0.5 = 0.75 A31.24 (a)KCL : I + I + I = 0123KVL1: 2I1-130 +120 - 2I = 02KVL2 : 2I2-120 - 4I = 03I = 15A, I12= 10 A, I3= -25A(b)开关合在b 点时，求出 20V 电压源单独作用时的各支路电流：122I ' 3I ' 2+b-20VI '4I ' = -120´4= -4 A2 + 2 / /44 + 220I ' = 6 A22 + 2 / /4202I ' = -´= -2 A 32 + 2 / /44 + 2：所以开关在b 点时，各支路电流为I = 15 - 4 = 11A1I = 10 + 6 = 16 A2I = -25 - 2 = -27 A31.25+-2W（b） 等效变换3AA2.5 ABU= (3 + 2.5) ´ 2 = 11Vab（c） 等效变换4A2W6W2A6W4A1.5W2A6Wa ab bU= (4 + 2) ´1.5 = 9Vab1.26戴维宁:u = 220´ 1 = 110Va2R= 25abI = 110´1= 22 AL25 + 5015：诺顿I= 220 = 22 Aab505R= 25abI = 22 ´25= 22 A L1.28R525 + 5015R122AIRR34+-U = 10V(1) 求二端网络的开路电压：bR2a2AIRR34+-U = 10VU= 10 - 4I = 10 - 2 ´ 4 = 2V Uabab= 10 - 4I = 10 - 2 ´ 4 = 2V(2) 求二端网络的等效内阻（电压源短路、电流源开路）R= Rab2= 4WRabRU1ab(3) 得到戴维南等效电路+U I =ab= 2 A » 0.154 A1R + R13ab11.32 （a）I = I + I12350 -VI =A11050 + VI =A25I =AV32050 -VA =50 + VVA + A10V = -A520100V72.3.2(a) 取电源为参考向量··U = I R··U2= I (- jX )ctan 600 = IRIX= R =3XCcR1X=, 又 X =C3C2pfc2pfcR= 3（b）··U2= I ( jX )c··U= I RRIRtan 600 =3IXC X =LR , 又X3L=2pfL R= 32pfL第二章 正弦交流电课后习题2.3.2(a) 取电源为参考向量=··UI R··U= I (- jX )2c· IRR3I·tan 600 =60 o·U·U RIXXCc3X= RC1, 又XC1= 2pfc32U2pfcR=（b）··U = I ( jX )2c··U= I RRIR3U260o·I· tan 600 =IXC3· RU X1L=, 又X=2pfLLU ·RR32pfL =习题2.2I· = 5 + j5, I11= 5 2 Ai = 10sin(1000t + 450 ) A1I = 5 2Ð450 A1I· = 5 - j5 = 5 2Ð - 450 A2·I3I·1·I4I·22.3·U·Ui = 10sin(1000t - 450 ) A2I· = -5 + j5 = 5 2Ð1350 A3i = 10sin(1000t +1350 ) A3I· = -5 - j5 = 5 2Ð -1350 A4i = 10sin(1000t -1350 ) A4·2U130 o（1）·0·0U = 6Ð30 V U128= 6Ð30 Vj = arctan=5306U·=U·1+U·2=10Ð830Vu = 10 2 sin(wt + 830 )V·I260 o30 o·I·I1(2)I· = 10Ð - 300 A I· = 10Ð600 A12j = arctan1 =450 I· = 10 2Ð - (450 + 300 ) = 10 2Ð - 750 A i = 20sin( wt - 750 ) A2.4·I(a) 以电流为参考向量·UUjL··UR10j = arctan=450 10U = 10 2V = 14.1VU· = 10 2Ð450V·I（b）以电流I为参考向量··UUR·UCU 2 = U 2 + U 2CR1002 - 602U=C= 80Vu =2U sin(wt)i =2I11i =2Isin(wt + 900 ) A sin(wt - 450 ) A22·I(c) 以电流为参考方向L·UI·U·UCU· = (200 -100)Ð900 = 100Ð900VU = 100V·U（d） 以电压I·为参考方向R·U52 + 42·U（e） 以电压·II·I·CR为参考方向U·52 + 52I = 5 2A = 7.07A·U（f） 以电压·C·I为参考方向·UII = I- ILC I= I+ I = 18ALC2.5 （1）I= 7 2 sin(wt) ACI· = 7Ð00 A,w = 2pf = 314rad / sLX= wL = 31.4WL=jUI··wL = 7Ð00 ´ 314´100´10-3 Ð900VLC= 219.8Ð900V309.9u =sin(wt + 900 )VL310（2）309.9u =L310sin(wt + 900 )V= w L =U127I4 ALL314 ´100 ´10-3XL2.6(1)U· = 220Ð00V Ucc= 220VX = 1 =1= 796.2WCwC2 ´ 3.14´5 0´ 4´10 -6U220I =c = 0.28ACX796.2c· 0I = 0.28Ð90Ci = 0.39sin( wt + 900 ) Ac· = 0.1Ð - 600 ´Ð - 900 ´ 796.2 = 79.6Ð -150 0V(2) Uc2.9 (1)u = 220 2 sin(wt + 300 )V· 0U = 220Ð30 Vw = 2pf = 314rad / si = 10 2 sin(wt - 300 ) A·I = 10Ð - 300 A（2）· 0Z = U·I220Ð30= 22Ð600 = 11+11 3 jW 10Ð - 300X11 3 L =（3）wL = 2p´ 50 = 60.7mHS = UI = 220 ´10 = 2200V · AP = UI cosj = 220 ´10 ´ cos60 0 = 1100Wj = sin j = 2200 ´sin 600 = 1905var2.10+Ru+1uC2(1)电容两端| Z |= 2000Wf=1000HZ·I以 1 为参考向量j=-600R =| Z | cosj = 2k ´ 1 = 1KW2X = - | Z | sin j = -2k ´(-c3 ) = 1707W2X = 1c =1= 0.1uFcwcwXc（2） 电阻两端·I130 0··U= UR2·· U 1U0j=-300R =| Z | cosj=2k ´3 =1707W2X = | Z | sinjc=-2K（ ）=1000W12c = 1wc= 0.16uF2.12-+RL·UCRLCZ=R+(XL- X ) j = 10WCU =½Z½I =10´1 =10 V2.13U =½Z½I =10´1 =10V·U = 100Ð200V100 2 sin(1000t + 200 )ui =RR300=0.47sin(1000t + 200 )X= wL = 1000 ´ 0.4 = 400WL·· U100Ð200I= 0.25Ð - 700 ALjX400Ð900Li = 0.35sin(1000t - 700 ) ALX= 1 =1= 500WCwC1000 ´ 2 ´10-6·· U100Ð200I = 0.2Ð1100 Ac- jX500Ð - 900Ci = 0.28sin(1000t +1100 ) Ac1 = 1 +1+1ZRjX- jXLc=1 + (- j 1 + j 1 )30040050=0.003 - 0.0005 j = 0.003Ð - 90Z = 300Ð90·· U100Ð200 I = 0.33Ð110 A Z300Ð902.14以U· 为参考向量·I·U·IC·ILX= XCL I= ICLII ·····=+ I + I= I1CLRRI= URR= 10 A U = 10RVI = I3C= U = 10R = 10 A XXCC···I= I + I2RC= 10Ð450 A I = 10 2 A22.15 (1)Z = R - jXC=30 - j ´11000 ´ 25´10-6=30 - 40 jW = 50Ð - 530 Wu= iRSR = 300 2 sin（1000t-300）VI· = 10Ð - 300 As··U= I (- jX ) = 10Ð - 300 ´ 40Ð - 300 ´ 40Ð - 900CsC=400Ð -1200Vu= 400 2 sin(1000t -1200 )VC=UI·· Z = 10Ð - 300 ´ 50Ð - 530 = 500Ð - 830Vsu = 500 2 sin(1000t - 830 )V(2)（3）P = U I = 300 ´10 = 3000WR SQ = Q = -U I = -400´10 = -4000VarCC SS = UI = 500´10 = 5000V · ASP = S co（s -530）=3009WQ = S sin（-530）=-3993Var2.18·I·UR·UL·UC8R = UR = 2WU12I4wL =L = 3WI4L = 310= 0.3H····U = U + U + URLC=8 + (12-U ) jC82 + (12-U )2 = 102CU = 6V 或 18VUCX =C = 1.5 或4.5WCI1 =XC =1= 0.067F或1=0.022FwCC10 ´1.510 ´ 4.52.19 (a)(1)j3´ j6-18Z = 2 jW1j3 + j69 j(2) Z = 4 - 4 j -1 j + 2 j=4 - 3 j = 5Ð - 370 W·· U10Ð00I = 2Ð370 A Z5Ð - 370U· = 2Ð370 ´ 4 = 8Ð370V1U· = 2Ð370 ´ 4Ð - 900 = 8Ð - 530V2U· = 2Ð370 ´1Ð - 900 = 2Ð - 530V3U· = 2Ð370 ´ 2Ð900 = 4Ð1270V4·=· U4Ð1270I4 = 1.33Ð370 A1j33Ð900·=· U4Ð1270I4 = 0.67Ð370 A2j66Ð900(3)P = UI cos j = 10 ´ 2 ´ cos( -370) = 16W(b)(1)(- j4)(- j6)-241Z =- j3 - j6= -10 j = -2.4 jWZ = 3.4 + 4 j + 2 j - 2.4 j=4 + 3 j = 5Ð370 W(2)·· U10Ð00I = 2Ð - 370 A Z5Ð370U· = 2Ð - 370 ´3.4Ð900 = 2Ð - 370V1U· = 2Ð - 370 ´ 4 = 8Ð - 370V2U· = 2Ð - 370 ´ 2Ð900 = 4Ð530V3U· = 2Ð - 370 ´ 2.4Ð - 900 = 4.8Ð -1270V4·=· U4.8Ð -1270I4 = 1.2Ð - 370 A1-4 j4Ð - 900·=· U4.8Ð -1270I4 = 0.8Ð - 370 A2- j66Ð - 900(3)P = UI cos j = 10 ´ 2 ´ cos(370) = 16W2.20Z = 5 - 5 j = 5 2Ð - 450 W1Z = (5 - 5 j) ´10 j = 10W2（5-5 j) +10 jU100 I =2 =1| Z |102A读数为10A=10A2Z=10+10j = 10 2Ð450 WU = I11| Z |= 10´10= 100 2VV 读数为141.4V12.23X = 1 =1= 1WCwC104 ´102 ´10-621´(- j)- j11Z =Ð - 450 =-j11+ (- j)1- j222X = wL = 104 ´104 = 1WLZ = 1+ j + Z1= 1.5 - 0.5 j2.24=U··1Ð00Ic = 0.5Ð900 Ac- j22Ð - 900·=· U1Ð00Ic = 1Ð00 ARR1···I = I + IcR= 0.5Ð900 +1Ð00 = 1+ 0.5 j A = 1.1Ð26.60 A··U = I R = (1+ 0.5 j) ´ 2 = 2+j VR··U = I j2 = (1+ 0.5 j) j2 = 2 j -1 = -1+2 j VL····U = U + U + URLc= 2 + j -1+ 2 j +1 = 2 + 3 j = 3.61Ð56.30 VP = UI cos j = 3.61´1.1´cos(56.30 - 26.60 ) = 3.45W2.29P = UI cos j ，cos j = P =1.21 K= 0.511 UI11´ 220j = 600，j12= 24.50C =P (tan j - tan j）wU 211.21K (1.732- 0.456)= 1022uF2 ´ 3.14 ´ 50 ´ 2202I = P UR= 1K220= 4.54 AS = UI = 380 ´ 4.54 = 1727.3V · Aj = S sin j = 177.3´（-0.82）=-1408 Var2.30Z= R + jwL = 12 +12.56 j = 17.37Ð46.30 WRL122 +12.562I = U =1 ½Z ½220= 12.67WRL11Z = - j ´= -31.58WCjwc2 ´ p´ 50 ´100 ´10-6U220Z ½cI =½= 31.85 = 6.91AcZ =Z总RL/ /Z= ZRLZC = 25.15Ð14.40 WCZ+ ZRLcI = U = 220 =8.75AZ25.15总P = UI cosj=220 ´8.75´ cos14.40 =1864.5w Q = UI cosj=220 ´8.75´ sin14.40 =478.73w S=UI=220´8.75=1925V · Acosj = 0.96862.31380UP =3= 220VI= I = 2.2 APL| Z |= UPI220= 100W2.2Pj = arccos 0.8 = 370Z = 100Ð370 = 80 + 60 jW2.32(1)Z = 8 + 6 j = 10Ð370 WU38033U=L = 220VPUI=P =P| Z |22010= 22 AI= ILP= 22 A(2)UL= 220V UP= U = 220VLPI = U= 220 = 22 AP| Z |10I = 3ILP= 38 A(3)UL= 380V UP= U = 380VLPI = U= 380 = 38 AP| Z |10I =3ILP= 66 A(4)UN2.33= 220V ,ULUL= 380V ,Y 行= 220V，D形ARI·NNBI·BXCCI·CU = 380V ULP= 220V½Z½=10W设U· U· U·= 220Ð00 VAB= 220Ð -1200 VC= 220Ð1200 V·I· =U220Ð00A = 22Ð00 AAR10·I· =U220Ð -1200B = 22Ð - 300 ABR10Ð - 900·I· =U220Ð1200C = 22Ð300 ACR10Ð900=+····IIIINABB= 22Ð00 + 22Ð - 300 + 22Ð300 = 60.1Ð00 AP = I2.342 R = 222 ´10 = 4840WAZ = 31+ 22 j = 38Ð35.60U = 380V ,ULPU= 220V220I = I