电工学教程 多媒体课件 王蕴珊

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电工学教程 多媒体课件 王蕴珊. 第 1 章电路及其分析方法. 本章要求 : 1. 理解电压与电流参考方向的意义; 2. 理解电路的基本定律并能正确应用; 3. 了解电路的有载工作、开路与短路状态, 理解电功率和额定值的意义; 4. 会计算电路中各点的电位。. 第 一 章 电路及其分析方法. 1.1 电路的作用与组成部分. 1.2 电路模型. 1.3 电压和电流的参考方向. 1.4 电源有载工作、开路与短路. 1.5 基尔霍夫定律. 1.6 电阻的串并联. 1.7 支路电流法. 1.8 叠加原理. 1.9 电压源与电流源等效变换. - PowerPoint PPT Presentation

Transcript of 电工学教程 多媒体课件 王蕴珊

  • :1.2. 3. 4. 1

  • 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.10 1.9

  • 1.11 1.12

  • 1 1.1 (1) (2)1.

  • 2. : :

  • : : 2.

  • 1. 2 ,

  • E Ro R

  • 1.3 1.

  • (2) (1) 2.

  • ()()(3) UI UI UI UI UI U = I R U = IR

  • (a), U = IRR(b), U = IR

  • 1.4 ,U = IR :1.4.1 I U U = E IR0 R0
  • ,U = IR :1.4.1 I U U = E IRoUI = EI IRoP = PE P : ()

  • UI P = UI 0 P = UI 0UI P =UI 0 P = UI 0 1. UI 2. UI UI + UI -

  • : () I < IN P < PN () () I > IN P > PN () I = IN P = PN ()

  • : 1.4.2 1. I = 02. U :

  • 1.4.3 1. U = 02. I :

  • () Uab>E I >0 Uab
  • E1=4VE2=6VR1=2 R2=4R3=2VA1

  • 2: S A VA: (1)S (2) , b I2 = 0 VA = 0V I1 = I2 = 0 VA = 6V

  • 1. 5

  • 1abbcca 6abdaabca adbca 7 a bcd (4abd abcbcd 3

  • 1.5.1 (KCL)1 : = : : = 0 aI1+I2 = I3 I1+I2I3= 0 KCL

  • 2I =?:I = 0IA + IB + IC = 0

  • 1.5.2 KVL)1 U = 0 12 E1 = I1 R1 +I3 R3I2 R2+I3 R3=E2 I1 R1 +I3 R3 E1 = 0 I2 R2+I3 R3 E2 = 0 KVL

  • 1 = E2 =UBE + I2R2 U = 0 I2R2 E2 + UBE = 02 U = 0 3. 1

  • abdaacbabcdbR6I6 R6 I3 R3 +I1 R1 = 0I2 R2 I4 R4 I6 R6 = 0I4 R4 + I3 R3 E = 0 adbca I1 R1 + I3 R3 + I4 R4 I2 R2 = 0 U = 0 cadc I2 R2 I1 R1 + E = 0

  • 1.6 1.6.1 :1)R =R1+R23)2)

  • 1.6.2 (3)(4):(1)(2)

  • 1R12R1221222111R12=2.68CD

  • 1.6.1RL=50U=220V1003Aabcde

    : (1)aUL=0IL=0 (2)CRRL

  • (3)dRRL (4)e

  • Y-Y

  • (IaIbIc)(UabUbcUca)

  • Y Ra=Rb=Rc=RY Rab=Rbc=Rca= R = 3RY Y Rab=Rbc=Rca=R Ra=Rb=Rc=RY =R/3

  • 2 I1 abcY

  • 1.9 1 : U = E IR0 R0 = 0 : U EU0=E E R0 R0
  • 1(2) U E(3) :(1) R0 = 0 E = 10 VRL RL= 1 U = 10 VI = 10A RL = 10 U = 10 VI = 1A

  • 2 U0=ISR0 OIS IS R0 : R0 = : I IS R0 >>RL I IS

  • )1(2) IS (3) U :(1) R0 = IS = 10 ARL RL= 1 I = 10A U = 10 V RL = 10 I = 10A U = 100V IUISO

  • 3 a U = E IR0b U = ISR0 IR0

  • RL= R0 R0 E R IS

  • 1::

  • 2:2:(d)

  • 3 1

  • 3: U110VIS2AR11R22R35 R1 (1) RI(2)U1IU1ISUIS(3)(1)

  • (2)(a)

  • (60+20)W=(36+16+8+20)W80W=80W(3)

  • 1.7 KCLKVL b=3 n =2 = 3 =2

  • 1. 2. KCL ( n1 ) 3. KVL b( n1 ) 4. b a1 I1+I2I3=012I1 R1 +I3 R3=E1I2 R2+I3 R3=E2:

  • (1) KCL(n-1) b=66(2) KVL(3) IG 2 a I1 I2 IG = 0abdaIG RG I3 R3 +I1 R1 = 0 b I3 I4 +IG = 0 c I2 + I4 I = 0acbaI2 R2 I4 R4 IG RG = 0bcdbI4 R4 + I3 R3 = E IGRG

  • b =433312 (1) KVLKVL (2) KVL

  • (1) KCL b =433(2) KVL(3) I1= 2A I2= 3A I3=6A 3 a I1 + I2 I3 = 7112I1 6I2 = 4226I2 + 3I3 = 0 acbd(ac)( bd)12 32KVL

  • (1) KCL b =4(2) KVL(3) I1= 2A I2= 3A I3=6A 3 a I1 + I2 I3 = 7112I1 6I2 = 4226I2 + UX = 012 33KVL3+UX3UX + 3I3 = 0

  • 1.8

  • (c) IS I2 = I2' + I2'' (b)E

  • ::I1' I1''I2' I2'' I1 = I1'+ I1''= KE1E + KS1IS I2 = I2'+ I2'' = KE2E + KS2IS

  • E = 0E Is=0 Is P

  • 1 E =10VIS=1A R1=10 R2= R3= 5 R2 I2 IS US (b) E IS (c) IS E ( b)

  • 1 E =10VIS=1A R1=10 R2= R3= 5 R2 I2 IS US (b) E(c) IS(c)

  • 2US =1VIS=1A Uo=0VUS =10 VIS=0A Uo=1VUS = 0 VIS=10A Uo=? Uo = K1US + K2 IS US =10 VIS=0A US = 1VIS=1A 0 = K1 1 + K2 1 1 = K1 10+K2 0 K1 = 0.1K2 = 0.1 Uo = K1US + K2 IS = 0.1 0 +( 0.1 ) 10 = 1V

  • 1.10

  • E R0 R0 a b E U0 a b

  • 1 E1=40VE2=20VR1=R2=4 R3=13 I3

    ab

  • (1) E1E1=40VE2=20VR1=R2=4 R3=13 I3E E = U0= E2 + I R2 = 20V +2.5 4 V= 30VE = U0 = E1 I R1 = 40V 2.5 4 V = 30V

  • (2) R0 1E1=40VE2=20VR1=R2=4R3=13 I3ab R1 R2 R0ab

  • (3) I31E1=40VE2=20VR1=R2=4R3=13 I3

  • :R0=U0/ISC(a)+E=U0

  • 2R1=5 R2=5 R3=10 R4=5 E=12VRG=10 IGRG

  • : (1) U0EE' = Uo = I1 R2 I2 R4 = 1.2 5V0.8 5 V = 2VE' = Uo = I2 R3 I1R1 = 0.8 10V1.2 5 V = 2V(2) R0abR1 R2 R3 R4

  • (3) IG

  • 1.11 VX 1. : (1) (2) (3)

  • 2. :VaVbVcVd a Va=0VVb=Uba= 106= 60VVc=Uca = 420 = 80 VVd =Uda= 65 = 30 V bVb=0VVa = Uab=106 = 60 VVc = Ucb = E1 = 140 VVd = Udb =E2 = 90 VbaUab = 106 = 60 VUcb = E1 = 140 VUdb = E2 = 90 VUab = 106 = 60 VUcb = E1 = 140 VUdb = E2 = 90 V

  • (1) (2)

  • 1: S A VA: (1)S(2) , b I2 = 0 VA = 0V I1 = I2 = 0 VA = 6V

  • 2 (1) (2) RPAB1+12V12V+ RP I AB2 VA = IR1 +12 VB = IR2 12

  • 1. 2. 3. 1.12

  • 1. 2. (1) , (2)

  • - () 1.

  • 2. LH, mH, H

  • ,

  • 3. CF, F, pF

  • ,

  • u i LCR

  • 1 1. i u S (R)

  • (b) uC

  • : (1) ()(2) ()

  • (2) ui t =0+ (1) uC( 0+)iL ( 0+) 1) t =0- uC ( 0 ) iL ( 0 ) 2) uC( 0+)iL ( 0+) 1) t =0+2) t =0+ uC = uC( 0+) t =0+ iL = iL ( 0+)

  • uC iL 2.

  • S t = 0 t = 0 + 0iC=icuc

  • 1CL

  • 1:iC uL (2) t=0+

  • 1.12.3 RC1. : ()()2. ,

  • (1) KVL1. uC (t 0) : , , RC1 RC

  • (2) A uC RC (3) uC

  • 2.

  • 4. (2) : S(1)

  • U

  • t =5 uC(3)

  • 2 RC: RCt = 0s , u

  • = + 1. uC(1) KVL RC(2)

  • A

  • (3) uC

  • t = uC 63.2% 2. iC 4. t = 0

  • 3 RC1. uC : = +

  • 2 = + 1 = +

  • t = 5 , , uC

  • , uC (0 -) = UosRU+_C+_iuc

  • ,

  • (1) (3) (2)

  • C , L

  • t =(0+)

  • 1) R0=R ; 2) R0(3) RCRL

  • R0

  • 11.4.5 1.4.7 1.4.9 1.5.3 1.6.3 1.6.5

  • 5AHOME

  • (4) HOME

  • 6A4A10A1I = HOME

  • HOME

  • 13 5HOME

  • : U1=140V U2=90V R1=20 R2=5 R3=6 1A I1-I2-I3=01 I1 R1 +I3 R3 -U1 =02 I2R2 -I3 R3 +U2 =0I1 - I2 - I3=020 I1 +6 I3 =1405 I2 - 6 I3 = -90I1 = 4AI2 = - 6AI3= 10AHOME

  • : U1=140V U2=90V R1=20 R2=5 R3=6 I3 2HOME

  • HOME

  • I= ?I'=2AI"= -1AI = I'+ I"= 1A+ HOME

  • 1. 2. HOME3. .4.

  • HOME

  • --- --- HOME UOC RO

  • Uoc RoUoc=?RO=?Uoc ROHOME

  • HOME

  • R1= R4= 20 R3= R2= 30 U=10V

    R5=10 I5=?HOME

  • HOME

  • UL=?Ro =50+2+5 = 57HOME

  • #2HOME

  • 1UAB UAB = (10+8)(1//5/3) = 11.25VHOME

  • 2 11KCL2KCLKVL I1 = 4A I2 =10A I3 = 12A2 2 I1 -I2 - I3 +18= 0 120 I1 + 6 I2 = 140 2-6 I2 +5 I3 = 0 3 1~3#HOME

  • 3I xISUS 3 3 HOME

  • 4UABUocRoRo=1+1//2=5/3 ,Uoc=4.5*2/3 =3VHOME,UABAB

  • 5I 2:.UOCUOC=64+10=34VHOMEROIsc=4+10/6 =17/3A:,I2I2=34/(6+4)=3.4AIsc.::

  • 1RL b) c) a2,ABUAB c-1V b) 0 c) 1V HOME

  • 3ABUAB -18V 18V c) - 6V a12V2A411 6A 2A c) 2A c2HOME

  • 5R1I2 a) b) c) c4 V b) 8.67V c) 6 VcIKVLHOME

  • 7 a) b) c) b8R1 R2 a) b) c) cHOME

  • c9SA UA -3 V -6V c) -9V 10SAUA a)-3 V b)-6V c) -9VbHOME