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祝同学们 新学期愉快 学习进步!. 电工学简明教程(第二版) 秦曾煌 主编. 主讲:机电学院 薛亚茹. 第 0 章 绪论 —— 课程介绍. 课程的目的. 目的是使学生通过本课程的学习,获得电工技术必要的基本理论,基本知识和基本技能,了解电工技术的应用和电工事业的发展概况。. 课程主要内容 ELECTROTECHNICS AND ELECTRONIC. 电路分析. 电机拖动. 模拟电路. 数字电路. I. a. +. R 1. R 2. U ab. U. _. 2 . 2 . b. 电路分析 (1,2 章). - PowerPoint PPT Presentation

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  • 0

  • (1,2

  • 3,4

  • 9 ,10,11

  • (13,14

  • 10A+5VBCRY

  • :3070

  • 1.1 1.2.

  • 1.2

  • 1.3 ()

  • ABBA

  • ,

    1.()2.

  • :ababV, mV

  • u_+1.Uab 3.2. uu>0abu
  • ababab

  • abIUabIU

  • U= IRU IU ICI= 2AR=3U= (2)3=6V

  • EIU1.4 1.4.1 1. R0RABCDU=RIU=E R0I

  • 1.5 1.5.1 1. U=RIU=E R0I 2. UI=EI R0I2 [](W)(KW)=+EIUR0RABC

  • U , I :P = UI

  • P 0 P 0

  • UAB=3VP=UI = (2)3= 6WI = 2ANPN,

  • U=220V I = 5A E1 E2R01=R02=0.6 1.2

  • 3. S1US2+IPI1I2

  • EIU01.4.2 R0RABCDI=0U=U0=E P=0U0

  • EUIR11.4.3 R2R0U=0I=IS=E/ R0P = 0

  • 1.3U012VIs=30A,

  • R3abE1 cd acb ab adb abca adba adbca 1.5 E2 abR2R1:

  • 1.6.1 KCL1

    3+ -3. KCL2I1+I3+I4-I20I1+I3+I4I2

  • I1= 9A I2= 2A I4=8A I3KCLI1 I2 + I3 + I4=01.4

  • BC1.5

  • IAIBIABIBCICAKCL I = 0ICABCKCLIA= IABICAIB= IBCIABIC= ICAIBCIA + IB + IC =0

  • ,

    = -3 + 4 -2 = -1A

  • I=?

  • 131 2 + -1.6.2 KVL 2U1+U2 U3 U4 + U5 =0

  • + -1=0++-2+--=0i1R1i3R3KVL1.6

  • KVLE+ IR UAB=0UAB= E+ IRKVLUBUAUABUAB= UA UB uab a b

  • 1.6 1.6.1 R = R1+R2

  • 1.6.2 []S

  • ab1.7

  • KCLKVL 1.7

  • 2 KCL3 KVL b (n1)4 A n (n1)KCL1 b ,E1 I1 R1 +E2 I2 R2 0(2)

  • bacd1.8

  • bacd

  • B=44

  • 1.4.5 1.4.9 1.5.3 1.6.3 1.7.1

  • =1.8 =

  • 5UP+ 10A51520V+ U20V24 120V

  • + 10A51520V+ U24210A = 221.25W

  • 1.

  • 4. I3R3

  • 1.9

  • 1.9.1 1 2

  • Us =EIs = U = E R0 IR0

  • )1 IS2IS1.9.2

  • ISabUabIUab / R0= IS I R0=,I ISU0 = IS R0Uab = IS R0 I R0

  • Uab Uab I I I ----- I Uab -----Uab

  • R0R0

  • I=2A U=4V

  • 1.9.3 E= Is R0EISEUIRLR0++--

  • IS = US / R0R0 = R0

  • US Is (2)

  • 5A

  • 5AI55V5

  • 5AI

  • 1.10

  • ()()RLAB

  • NE R0R0E

  • a1UU=30VR0= 6+6V6b6A2A15U+(1)U0CU0C = 66+ 6 =42V(2)R0(3)UR0

  • E1R3R4R1+R2E2ISIR52I+(1)U0cU0C = I3 R3 E2 + IS R2

  • E1R3R4R1+R2E2ISIR5+ABR3R1R2IS(2)R0R0 =(R1// R3)+ R5+ R2(3)I

    R0R5U0C = I3 R3 E2 + IS R2

  • +3ABUAB1051059V3A100.5AAB

  • +5105+9V3A10AB1 UOC+ABI1I215 I1+9 30=015 I2+9=0 I2= 0.6A I1=1.4AUOC = UAB=1.45+10 0.6=13V

  • 105105+3A10AB2. R0UAB=13+ 0.520/3 =16.33V3. UABR0 = RAB =2(10//5) =20/3

  • R1abcdR2E1E2R3Va = E1Vc = E2Vb = I3 R3I3d:+E1 E21.11 ++

  • SaSUa= 0.408 mA= 10.3V

  • SsUa= 1.2V

  • .ABCDA210V+5V+3BCDAII==3AVC=33=9VVD= 32= 6VBVD= 5VVC=10VVCD= VCVD= 15VVB= 10+9= 1V

  • uiuC 11.12

  • 2 InductoruiiL

  • ?

  • LC

  • RCRL (1) ;(2)

  • : 1.12.2

  • : t=0

  • : K 1t=021 3.1.2 ui t=0+

  • , t=0,,

  • t=0 +

  • (1) t=0- uC(0-) iL(0-)(2) : uC(0+) = uC(0-) , iL(0+) = iL(0-) (3) t=0+ (4) t=0+

  • :1.12.3 RC:

  • RCRCt=0K uc(t) i(t) t0Rt=0baiSuC,+-U0UC(0-)= U0+-iR + uC =0p = 1/RC A A= uC(0+)= U0uC = uC(0+) e t /RC = i(0+) e t /RC

  • U0uC = uC(0+) e t /RC t0tuc i t0

  • = RC0.338 U02 > 1t=5 uC uC uC = uC(0+) e t /RC

  • RCuR +uC =UuC=K uC=Ae pt= Ae t /RC K=U uC=UuC =U+ Ae t /RC uC(0+)=0A= UuC =U Ue t /RC =U(1et /)

  • tuUuC:uCuCuC =U(1et /RC) =U Ue t /RC RCuC 00

  • RC : uC(0+) Rt=0baS+-uC = U0+e -t/ + U(1e -t/) = + f (t) = [ f(0+) f()] e -t/ + f ()

  • 1 :

  • 2

  • 3 RL RC

  • : t 0 ucic u1

  • :

  • 1.12.4 RLRt=0baUiLSuLuRiL(0-)= I0=U/RiL(0+)= I0,KVL uL + uR=0+-iL uL (t 0) Lp + R=0 p = R/L A= I0 =L / R

  • 300 kWR=0.189L=0.398HU=35V50V RV=5 kt=0SS12 i(0+)3 i uv t0 4 uv(0+)