中国石油大学(华东)电工电子学(一)课后习题答案.docx
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1、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+52=15VabR =5Wab1.4.3 (b)6+42V-baU= 6 + 6 6 = 42VabR= 6Wab1.4.3 (c)a5+40V-bU =UabR+10=65+10=40VR =5
2、Wab1.4.3 (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+ 9R100k1
3、KCL: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= 100.3+ 0.230 = 0.6Ax 21.5I= 0.3 + 0.6 = 0.9 Ax11.2I3I4I61.3= 0.01 + 0.3 = 0.31A= 9.6
4、1 - 0.31 = 9.3 A= 0.3 + 9.3 = 9.6 AP = -14 2 = -28w1发出功率P = 110 = 10w吸收功率2P = 4 2 = 8w吸收功率3P = -(-110) = 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 (
5、- 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 +
6、 3R=0.5BDR= 199 + 0.5 = 199.5ADI = 10 1 = 5mA I122= 5mAI = 5333.6= 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+10
7、V-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 =
8、 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、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+-0.5I
10、1V3 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 = -1204= -4 A2 + 2 / /44 + 220
11、I = 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 = 1101= 22 AL25 + 5015:诺顿I= 220 = 22 Aab50
12、5R= 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
13、 =50 + VVA + A10V = -A520100V72.3.2(a) 取电源为参考向量U = I RU2= I (- jX )ctan 600 = IRIX= R =3XCcR1X=, 又 X =C3C2pfc2pfcR= 3(b)U2= I ( jX )cU= 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 RU= I (- jX )2c IRR3Itan 600 =60 oUU RIXXCc3X= RC1, 又XC1= 2pfc32U2pfcR=(b)U = I ( jX
14、 )2cU= I RRIR3U260oI tan 600 =IXC3 RU X1L=, 又X=2pfLLU RR32pfL =习题2.2I = 5 + j5, I11= 5 2 Ai = 10sin(1000t + 450 ) A1I = 5 2450 A1I = 5 - j5 = 5 2 - 450 A2I3I1I4I22.3UUi = 10sin(1000t - 450 ) A2I = -5 + j5 = 5 21350 A3i = 10sin(1000t +1350 ) A3I = -5 - j5 = 5 2 -1350 A4i = 10sin(1000t -1350 ) A42U130
15、 o(1)00U = 630 V U128= 630 Vj = arctan=5306U=U1+U2=10830Vu = 10 2 sin(wt + 830 )VI260 o30 oII1(2)I = 10 - 300 A I = 10600 A12j = arctan1 =450 I = 10 2 - (450 + 300 ) = 10 2 - 750 A i = 20sin( wt - 750 ) A2.4I(a) 以电流为参考向量UUjLUR10j = arctan=450 10U = 10 2V = 14.1VU = 10 2450VI(b)以电流I为参考向量UURUCU 2 = U
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