电工学教程 多媒体课件 王蕴珊
DESCRIPTION
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 PresentationTRANSCRIPT
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:1.2. 3. 4. 1
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1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.10 1.9
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1.11 1.12
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1 1.1 (1) (2)1.
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2. : :
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: : 2.
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1. 2 ,
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E Ro R
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1.3 1.
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(2) (1) 2.
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()()(3) UI UI UI UI UI U = I R U = IR
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(a), U = IRR(b), U = IR
- 1.4 ,U = IR :1.4.1 I U U = E IR0 R0
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,U = IR :1.4.1 I U U = E IRoUI = EI IRoP = PE P : ()
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UI P = UI 0 P = UI 0UI P =UI 0 P = UI 0 1. UI 2. UI UI + UI -
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: () I < IN P < PN () () I > IN P > PN () I = IN P = PN ()
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: 1.4.2 1. I = 02. U :
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1.4.3 1. U = 02. I :
- () Uab>E I >0 Uab
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E1=4VE2=6VR1=2 R2=4R3=2VA1
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2: S A VA: (1)S (2) , b I2 = 0 VA = 0V I1 = I2 = 0 VA = 6V
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1. 5
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1abbcca 6abdaabca adbca 7 a bcd (4abd abcbcd 3
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1.5.1 (KCL)1 : = : : = 0 aI1+I2 = I3 I1+I2I3= 0 KCL
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2I =?:I = 0IA + IB + IC = 0
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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
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1 = E2 =UBE + I2R2 U = 0 I2R2 E2 + UBE = 02 U = 0 3. 1
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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
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1.6 1.6.1 :1)R =R1+R23)2)
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1.6.2 (3)(4):(1)(2)
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1R12R1221222111R12=2.68CD
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1.6.1RL=50U=220V1003Aabcde
: (1)aUL=0IL=0 (2)CRRL
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(3)dRRL (4)e
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Y-Y
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(IaIbIc)(UabUbcUca)
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Y Ra=Rb=Rc=RY Rab=Rbc=Rca= R = 3RY Y Rab=Rbc=Rca=R Ra=Rb=Rc=RY =R/3
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2 I1 abcY
- 1.9 1 : U = E IR0 R0 = 0 : U EU0=E E R0 R0
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1(2) U E(3) :(1) R0 = 0 E = 10 VRL RL= 1 U = 10 VI = 10A RL = 10 U = 10 VI = 1A
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2 U0=ISR0 OIS IS R0 : R0 = : I IS R0 >>RL I IS
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)1(2) IS (3) U :(1) R0 = IS = 10 ARL RL= 1 I = 10A U = 10 V RL = 10 I = 10A U = 100V IUISO
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3 a U = E IR0b U = ISR0 IR0
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RL= R0 R0 E R IS
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1::
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2:2:(d)
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3 1
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3: U110VIS2AR11R22R35 R1 (1) RI(2)U1IU1ISUIS(3)(1)
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(2)(a)
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(60+20)W=(36+16+8+20)W80W=80W(3)
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1.7 KCLKVL b=3 n =2 = 3 =2
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1. 2. KCL ( n1 ) 3. KVL b( n1 ) 4. b a1 I1+I2I3=012I1 R1 +I3 R3=E1I2 R2+I3 R3=E2:
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(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
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b =433312 (1) KVLKVL (2) KVL
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(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
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(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
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1.8
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(c) IS I2 = I2' + I2'' (b)E
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::I1' I1''I2' I2'' I1 = I1'+ I1''= KE1E + KS1IS I2 = I2'+ I2'' = KE2E + KS2IS
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E = 0E Is=0 Is P
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1 E =10VIS=1A R1=10 R2= R3= 5 R2 I2 IS US (b) E IS (c) IS E ( b)
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1 E =10VIS=1A R1=10 R2= R3= 5 R2 I2 IS US (b) E(c) IS(c)
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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
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1.10
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E R0 R0 a b E U0 a b
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1 E1=40VE2=20VR1=R2=4 R3=13 I3
ab
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(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
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(2) R0 1E1=40VE2=20VR1=R2=4R3=13 I3ab R1 R2 R0ab
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(3) I31E1=40VE2=20VR1=R2=4R3=13 I3
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:R0=U0/ISC(a)+E=U0
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2R1=5 R2=5 R3=10 R4=5 E=12VRG=10 IGRG
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: (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
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(3) IG
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1.11 VX 1. : (1) (2) (3)
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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
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(1) (2)
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1: S A VA: (1)S(2) , b I2 = 0 VA = 0V I1 = I2 = 0 VA = 6V
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2 (1) (2) RPAB1+12V12V+ RP I AB2 VA = IR1 +12 VB = IR2 12
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1. 2. 3. 1.12
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1. 2. (1) , (2)
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- () 1.
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2. LH, mH, H
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,
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3. CF, F, pF
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,
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u i LCR
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1 1. i u S (R)
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(b) uC
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: (1) ()(2) ()
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(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+)
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uC iL 2.
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S t = 0 t = 0 + 0iC=icuc
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1CL
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1:iC uL (2) t=0+
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1.12.3 RC1. : ()()2. ,
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(1) KVL1. uC (t 0) : , , RC1 RC
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(2) A uC RC (3) uC
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2.
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4. (2) : S(1)
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U
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t =5 uC(3)
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2 RC: RCt = 0s , u
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= + 1. uC(1) KVL RC(2)
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A
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(3) uC
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t = uC 63.2% 2. iC 4. t = 0
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3 RC1. uC : = +
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2 = + 1 = +
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t = 5 , , uC
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, uC (0 -) = UosRU+_C+_iuc
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,
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(1) (3) (2)
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C , L
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t =(0+)
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1) R0=R ; 2) R0(3) RCRL
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R0
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11.4.5 1.4.7 1.4.9 1.5.3 1.6.3 1.6.5
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5AHOME
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(4) HOME
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6A4A10A1I = HOME
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HOME
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13 5HOME
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: 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
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: U1=140V U2=90V R1=20 R2=5 R3=6 I3 2HOME
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HOME
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I= ?I'=2AI"= -1AI = I'+ I"= 1A+ HOME
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1. 2. HOME3. .4.
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HOME
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--- --- HOME UOC RO
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Uoc RoUoc=?RO=?Uoc ROHOME
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HOME
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R1= R4= 20 R3= R2= 30 U=10V
R5=10 I5=?HOME
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HOME
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UL=?Ro =50+2+5 = 57HOME
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#2HOME
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1UAB UAB = (10+8)(1//5/3) = 11.25VHOME
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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
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3I xISUS 3 3 HOME
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4UABUocRoRo=1+1//2=5/3 ,Uoc=4.5*2/3 =3VHOME,UABAB
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5I 2:.UOCUOC=64+10=34VHOMEROIsc=4+10/6 =17/3A:,I2I2=34/(6+4)=3.4AIsc.::
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1RL b) c) a2,ABUAB c-1V b) 0 c) 1V HOME
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3ABUAB -18V 18V c) - 6V a12V2A411 6A 2A c) 2A c2HOME
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5R1I2 a) b) c) c4 V b) 8.67V c) 6 VcIKVLHOME
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7 a) b) c) b8R1 R2 a) b) c) cHOME
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c9SA UA -3 V -6V c) -9V 10SAUA a)-3 V b)-6V c) -9VbHOME