电路分析基础第9章 答案.docx
第9章答案一、填空题9-1.滞后 9-2.超前 9-3.选出来,衰减掉 94谐振 9-5.1/VZc , l/2VZc 9-6. 100, 109-7,电压,导纳，1 9-8.选择性,越窄9-9.相等,180。9-10.相等，180。二、选择题9-1. B 9-2. C 9-3. B 9-4. D 9-5. B, A, D, E96 B, A, D, E 9-7C A, D, B9-8.A , B, D三、计算题9L (a) a)。=* (b) g=" LL92 /)= -=/1=796KHz2HLe 2x 3.14x J160x IO* * 25。x IO"_ coL _ 2兀hL 一下- R= 62.62 x 3.14x798000 *250x10-6209.3.=/0 lOOxlO-3= 10Q2x3.14x106x1000= 159 pF1 1?C" (2x3.14x106)2x159x10-121 1?C" (2x3.14x106)2x159x10-12= 159”gL _ 2兀/J _ 2 x 3.14x106x159xW6 C-Z R R10= 1009-4.9-4.(1) Q =(2) R =A=800 = 100境 810CQ (2 7ix800xl03)xl.99xl0-10Q«IOQxlOO1a)lCH«2001(2兀 x800x1()3)2 x1.99x101°9-5. L =H = 0.02Hco-C 25002x8x109-6.9-8.9-10._ 2500x0.02i=50C- 10002xl0-6=1H , 1=66.67mH49 助 2c 49 可 21-红=3.18M"z, Z0=M = 500Z。，Q = = A£ = io（）L0 RCR RR=g Q=*号="41g 二刀V3LC9-11.1 犬+（ 疗= 1092kHz, Z=- =' °）=117.7kO9-12.9-13.Us=Ur= 20V ,Us 209-14.25002C 25002 x10x10-6= 0.016HU2 52R = J = = 0.1670 始150c 1 oL 2500 x0.016（J = Z4U 0CR R 0.1679-15. R = = 10QI。 0.2L =巴丛=-=5.007 x 10-5H/（）%0.2x271x1.59 xlO6就 L (2tix1.59x106)2 x 5.007 x IO-52.003 xlO-loF2.003 xlO-loF2tix1.59x10650= 199704rad/s9-16. (1)(2福(2x3.14x540xl03)2 X300 X10-6= 289.849 pFUs 2八 cooL 2x3.14x540 x103 x300 x10-6 1一 口(J -= 127.17R8(2)(2)Q心2 L (2 x 3.14 x 600xl03)2 x300 xl0-6= 234.778 pF(3)(3)UC =QUS =127.17x1.5 = 190.755 mV%UxM= 6.979liA|Z|214.925U =22c 2 x 3.14x 600x 103 x 289.849x 10-12%UxM= 6.979liA|Z|214.925U =22c 2 x 3.14x 600x 103 x 289.849x 10-12【26.979 xlO-6=6.39mV9-17. t/ = /s7? = 25Z60°V,。=八=5/601,Ul =0.033Z 一 30°A , Ic = jcoCU = -O.O33Z - 30°A jcoL9-18. f。吟2兀5.763 x 106= 917.675 kHz2x3.14Z(jg) =Z(jg) =L 0.35 xlO-3y(jg) CR 86x1()72x25=162.79 IkQ9-19.要使电路发生谐振，那么需处-一- = 0,那么谐振频率/0=也=1=4 coC2ti 7iVZc9-20. HdB= 201g3.2 + 201g|#|-201gl + - -201g 1 +jf105ff(P = 90 arctan() arctan(-10IO5921.上下限频率分别为/=105 Hz和/尸0/，中频增益=100,转化为分贝数:20叫"| = 20炮100 = 20x 2 = 402瓦4加为实数,故匕，匕相位差为0。
