电路与模拟电子技术-张绪光+刘在娥-部分习题答案

1、部分习题答案1-11. B2.B3.D4.A5.D1-21.X2.V3.X4.X5.X1-3(a)U=5V(b)R=10Q(c)I =-5A(d)U=-0.8V1-40.1A,50V1-60.5A,0.45A,108V1-7(1)I=20A, Url=110V, Ue=114V, PRL =2200W ,PE = 2400W ,2=2280W ;I =1.8A ； (d) U =26V(2) U=E=120V, Ur=0V；(3) U =0V , I =400A 1-9 (1) In=4A, R=12C; (2) U =0V ; (3) I =100A 1-10 E =8V,RO =15 1-

2、13 Uab =22V, Ir =0A1-14 (a) R=4C; (b); U =40V ; (c) 1-15 Us =12V , R =3kQ , R3 =6kG1-16 Ui =14V , U2 =18VUbc 7V ,U ac = -24Vac1-17 (1) I1 =-6A , I2 =-8A ,I2 =-0.8A ； (2) Lb =16V ,1-18 R =2.5Q, Ua =-5V2-11 . D2. D2-21.X2.X6. D6. X 7. X3.C4.A5.B3.X4.V5.X2-3 I =1A2-4 U =7.5V2-5 Ua=1.71V, Ub=3.80V2-6I1

3、= 1A, I2=4A2-7 I1 =0.57A , I2 =0.32A ,2-8Ii = 1A, I2=4A2-9 I1 =0.57A , I2 =0.32A ,I3 =0.25A , I4 =0.09A ,I3 =0.25A , I4 =0.09A ,2-10I =3A , RS1 =10W , PS2 =50W ,PUS1 =25WI5 =0.23A ,I5 =0.23A ,I6 =0.25AI6 =0.25A2-11 Ui = 6V , U 2=0V , U 3=6V2-13 I1 = -0.5A, I2=-0.5A, I3=1A2-14 I =1.4A2-15 Ii=3A , I2

4、=1A , I3=5A2-16 Uab=16V2-17 I =8.4A2-18 190mA2-19Uoc =42V , Req =1 复,I =8.4A2-20 (a) Uoc =2V , % =1.5G(b)U oc =10V , Req 02-21 U0c=10V, Req=2Q2-22U0c =2V ,ocReq=4.8C, -0.204A , -0.135A, -0.081A2-23 (a)U =16V , % =4C , J4A , I =1A (b) U = 267V , % =2Q ,噎=1.33A ,I - 9.67A2-24 Uoc =4V , Rl =Rq =6建，Pmax

5、 =0.67W*2-25 I = 3A*2-26 U = 24V3-1 1. C 2. B 3. A 4. D 5. B 6. C3-2 1. V 2.X 3. X 4. V 5. V 6. V 7. X 8. X 9.，10. , 11. ,12. X3-3Uc(0Q=1.667V , % (0 Q =0.417mA , u® (0+)=4.17V3-4(a) i(0»=0A, i(g)=1.5A, (b) i(06 =2A , i(g)=1.5A3-5iL(04=3A, iL(g)=6A3-6iL(04=2mA, uR(0/=10V3-7 E=8s, Uc(©

6、=3.68Vt3-8uc (t) =(-6 16et)Vt-3-9uc (t) =(8-2e0. 1)V-t_ -t3-10 uc(t) =(4+4e )V , ic(t)=-2e A3-11 h(t) =(10 -10e-500t )A , iL(3ms) =7.768A3-12 iL(t) =(3.331.33e-300t)A , uab (t) =(33.3 26.7e-300t )V8t8t3-13 iL(t)=2eA, ul (t) = 16e A1 t010t3-14 h(t)=(22e )A, u(t) =60e V4-1 1. D 2. B 3. B 4. A 5. B 6.

7、B 7. B 8. B 9. C10. B4-2 1.X 2. V 3. V 4. X 5. X 6. V 7. V 8. X 9. X10. X4-3 (1) U=220V,中=30*, f =50Hz , T =0.02s; (3) 155.5V , 177.1V4-4 i =20、2sin(314t 90 )A4-5 (1)i1 =52sin(314t+30A , i2 =1012sin(314t+30，A ; ( 2)5=04-7 u =10-2sin( t 23 )V4-10 (1) u =1.256/sin314tV ； I =20/-60 夬4-11 (1) i =0.89cos

8、314tA ; (2) U =314 M10-6/-120 ”4-12 (1)电阻；(2)电感；(3)电容；(4)电阻与电感串联4-13R=20C, L=1.19mH4-14 50Hz 时，/= 44/36.87 *,容性;100Hz 时，= 25.88/-61.93 ，感性4-15 3A4-16R=10C, L =55mH4-17 11V4-18 U =40V , I =10 J5A4-19 (1) 250rad/s； (2) 500rad/s ； (3) 1000rad/s4-20 (1) I1 =11 / 60。, I； =5.5/-60 叭,=16.5/-60 八;(2) P=1815

9、W, Q =3143Var ;(3) 0.5,感性4-21R =1kC, C =0.1 心 F,滞后4-22 cos平=0.682 , R =15C , L =51.3mH4-23 P =580.8W, Q =474.4Var , S =968V *A4-24 (1) cos平=0.5(2) C=90.3aF4-25 (1) R=250C, Rl=43.75C, L =1.48H ; (2) Pr =40W , P =47W , cos邛= 0.53； (3) C=3.46aF4-26 Z=105j45C, I=1.75A, I1=I2 =1.237A4-27 I =1.43A , I1 =0

10、.37A , I2 =1.29A , P =84.83W4-28 C=0.1aF, ur =sin5000tV , ul =40sin(5000t+90V , Uc = 40sin(5000t 901V4-29 能4-30 (1) 1062 心 F(2) 531 心1, 25A4-31 L1 =1H , C =1 心 F1 .14-32 L1 =22， L2 = -2(- 2 Tf12)C ；C5-11 . C2. D3. C4. D5. B6. C5-21 .，2.，3. V4. X5. X6. V7. X 8. V5-3(1) Il=Ip=4.4A, P=348.5W；(2) IP=4.4

11、A, IL =7.6A , P =348.5W 5-4 I L =36.5A , Z =(50.4 +j37.8)C5-5(1) I1 =I2 =I3 =0.15A ；I 23(2) Ua =Ub =190V ,不能正常发光。5-6 U =380V , I =6.6A5-7(1)略；(2) IL =I P =45.45A ；(3)略5-8 39.3A6-1 1 . C 2. A3. B 4. A5. A 6. C 7. D 8. C6-2 1.X 2.X3. V 4.，5. X6-3n =166 个,I1N=1A/2N=45.5A6-4 I1 =2.1A , U2 =38V6-5106-6 K

12、=43.5, I1N=4A, I2N=173.9A, 9 % =4.3%6-7当钳入一根零线、三根火线及四根线全部时,电流表的读数为0;当钳入两根火线时,电流表的读数为5A。6-8 I =93.9% , n =94.4%26-9 两种情况下U2 =5.5V2. B 3. A4. B5. B 6. A 7. B8. B 9. A10. B7-2 1.X11. B 12. B2. V 3. X 4. V5. X6. X7. V 8. X 9. X10. X 11. x 12. X 7-3 (1) 0V; (2) 12V7-4 VD 1处于截止状态，VD 2导通导通状态；U o=8V7-6 (1)

13、PZM = 0.2W ; (2)稳压管将因功耗过大而被烧毁；(3)稳压管将因功耗过大而被烧毁7-7 435 Q<Rl< 1.25k 7-8 467 Q < R< 1.4k Q7-10当开关合在a位置时，三极管处于放大状态；当开关合在b位置时，三极管将因功耗过大而被烧毁；当开关合在 c位置时，三极管处于饱和状态7-13 (1) N沟道耗尽型 MOS管7-14 恒流状态8-11. A2. C3. C4. C5. A6. A C7,B8. B8-21.X2.X3. X4. V5. X6. X7.X8.，9.，10. X8-3 图略8-4 (1) Vcc=10V, RC=2.5

14、kQ, RB=250k Q; (2) Au = 50 , Uo=Sin314tV8-5 (1)静态工作点：Ibq=50 NA, Icq =2.5 mA,Uceq =VCc -IcqRc =12-2.5 10” 3 103=4.5 V(2)当RL =3kQ时，电压放大倍数为：Au之一90 ;当RL时，电压放大倍数为：Au定-181(3) R ：. rbe = 830IRq = Rc= 3 k(4) us 18 mA8-6 (1) Icq =2 mA, I BQ =0.04 mA, Rb=300 kQ ；(5) Ibq =0.08 mA, Rb=150 kQ8-8 (1)Ubq=3.4V,Icq

15、-Ieq=1mA, Ibq -20kA,Uceq =4.3V；(6) rbe 电 1.5kC,Au =83.3,Ri =1.36 kQ ,Ro=RC=5kG；(7) Au 之一0.9, R %13 kG, Ro=RC=5kG8-9 (1) Ibq =22 RA, Icq 定Ieq=0.9 mA, Uceq=5.14 V；(2) Au0.99 , R =71.8 kQ , R=39C8-10 Au1 = -727.38-11 (1) Ibq =20 NA, Icq =1 mA, Uceq 定 2 V；(3) Au1 之 _0.98 右一1,入2z0.99定1, uo1 =ui =20sin314

16、tmV ,uo2 =U = 20sin314t mV8-12 (1) Ubq 定 3V, Icq WEQ=1.15mA, Ibq 定 11.4 NA, Uceq 也 7 V；(4) Au 电8, R 定3.5 kC, R=RC=5kC8-13Au1-12,入2%1，A =&Au2%12，Ri=R=10MC,R。= 39Q8-14 Au -388-15 Au -38.58-16 Au ： -26.79-1 1. B 2. A3. D 4. A 5. C 6. C 7. A 8. C9-2 12. X 3, V 4. V 5. V 6. X9-4 (1)乙类工作状态；(2) 12V; (3

17、) 2V9-5 (1)乙类工作状态；(2) 9W, 11.5W9-69-79-10(1) 9.9V,(2) 6.3V;(2) 18W24.5W; (2)14V(3)4W10-1B 2.3. B4.A 5.10-23. X4.5.6.7. V 8.10-4UoRfRR2R3R2(Ui1-ui2)10-7UoRiRfR1Ui2R111-12.3.4.H 5.6.7. E 8.11-211-32.11-41.X 5.X7.X 8x 9.X10.11-5A =66.7,詈=6.67%11-6图（a）中的Re引回的是交直流反馈。反馈组态为：电压并联负反馈。图（b）中的Rb引回的是交直流反馈。反馈组态为：

18、电压串联负反馈。图（c）中的Rs引回的是直流反馈。反馈极性为负反馈。11-7图（a）中的反馈支路是 R2和C所在的支路。所引入的反馈是交流反馈。 反馈形式为: 电压并联正反馈。图（b）中Ri和R2引入的是直流负反馈。图（c）中的反馈支路是 R2,引回的是交直流反馈。反馈组态为：电压串联负反馈。图（d）中的R3和C所在的支路是反馈支路。引回的是交流反馈。反馈形式为：电压 并联正反馈。11-8由瞬时极性法可判断出，Rf引入的反馈极性为负反馈。反馈组态为电流串联负反馈。11-9由瞬时极性法可判断出，R3引入的反馈极性为正反馈。反馈形式为电压串联负反馈。11-10 C =0.01 NF, Au =1 + R2 +RW , Rw应至少调到1k电路才能起振。R111-11图略。反馈电压取自线圈L2两端，为正反馈，符合相位平衡条件。11-12图略。反馈电压取自线圈 C2两端，为正反馈，符合相位平衡条件。只要电路参数合 适，满足振荡的幅值条件，电路就有