電工電子真題講義錯誤勘誤期末考試大綱課件

Thiswatermarkdoesnotappearintheregisteredversion 第二 電路的分析方§2—1一、電阻串聯(lián)電路中流過同一電流右下圖中，有n個電阻元件依次聯(lián)接，聯(lián)接點上無分岔n R

R U+ R Uk= RK

_R U _UK R

Thiswatermarkdoesnotappearintheregisteredversion +U +U 當n=2時 U_R=R1+ U_RU1R1

+RU2R1

+Thiswatermarkdoesnotappearintheregisteredversion 二、電阻的并聯(lián)+U+U_R+U_I+U_IIR3

I I Rn n個并聯(lián)電阻也可用一個等效電阻R來代替，而且等效電阻的倒數等于各個并聯(lián)電阻的倒數之和n1=1+1+·· 1+··· = n

k=1Thiswatermarkdoesnotappearintheregisteredversion n1=1+1+·· 1+··· = n

k=1I+U_IIRI+U_IIRIRRnI=I+I+I+I=U+U+U+

=(?1

k=1 n1=? k=1 RKThiswatermarkdoesnotappearintheregisteredversion

R

G=G1+G2++GR++Gn=?GKnkn 單位：西門Thiswatermarkdoesnotappearintheregisteredversion 1 1

R

R1RR1+RRI1R1

+UR_I+UR_I1I2

R+ 特別當：R=R

R=R1= II1=I2=I

Thiswatermarkdoesnotappearintheregisteredversion _1定義及符 __理想電壓源是一個二端理想元件，在任一時刻t，元件的電壓U（即電源的電動勢E理想電壓源的輸出電流是任意的，取決于E§2— 電壓源與電流電源—向電路提供電能，如DC直流電源，AC電源電壓源電流源兩大類 激勵（源由此產生的電壓、電流 響Thiswatermarkdoesnotappearintheregisteredversion I+

+_

I= +Us_+Us_Thiswatermarkdoesnotappearintheregisteredversion

I++E_+U_實際電壓源在一定范圍內可用理想電壓源E串聯(lián)R0作為模型：（如干電池Thiswatermarkdoesnotappearintheregisteredversion +E+E_+U_（一）實際電壓源的外特性U=E-I=，U=U0=U0稱為

ERU=，I=IDER0ID稱為Thiswatermarkdoesnotappearintheregisteredversion UU0=O

R0=RoI從實際電壓源的外特性曲線來看U=E-直線斜率的絕對值為電阻 越小 特性曲線越平坦，即電壓源的端電負載能力強Thiswatermarkdoesnotappearintheregisteredversion R0<< U?Thiswatermarkdoesnotappearintheregisteredversion

如果 IS＝常數，則稱為直流電流源，它的伏安特性在U-I 平面上是一條與電壓軸平行的直線。UIOUIO+ _Thiswatermarkdoesnotappearintheregisteredversion ⑴其電流 2其輸出電壓U是任意的，取決于IS和⑶ Thiswatermarkdoesnotappearintheregisteredversion + +IS _

_Thiswatermarkdoesnotappearintheregisteredversion

I+RI+R0U_UR實際電流源在一定范圍內可用理想電流源IS并聯(lián)電阻作為模型URI=ISUR0當I=0U=0

=UI=I

=ISRThiswatermarkdoesnotappearintheregisteredversion URUR0 =00U ISU

I=III=ISI

TU=ISR0-R0RThiswatermarkdoesnotappearintheregisteredversion 當一個電流源的內阻R0RL則I=IS其外特性與理想電流源的外特性曲線相近Thiswatermarkdoesnotappearintheregisteredversion 三、實際電壓源與實際電流源的等效變換+E+U_I+IUU=E IRI=

TU=

R0'-00Thiswatermarkdoesnotappearintheregisteredversion +E+U_I+UIR0'U=E IR U=ISR0'-

E=IS R0=

Thiswatermarkdoesnotappearintheregisteredversion I+EI等效變換僅對外電路而言，對電源內部不等效 I S當電源輸出短路時，I= =SR

SThiswatermarkdoesnotappearintheregisteredversion I+EIThiswatermarkdoesnotappearintheregisteredversion E=ISThiswatermarkdoesnotappearintheregisteredversion ++- +-I+-II I

I－ +－ +3+Thiswatermarkdoesnotappearintheregisteredversion 四、電源等效變換的推廣理想電壓源與任一部分電路的并聯(lián)，對外電路II+U_EIU_++E- -Thiswatermarkdoesnotappearintheregisteredversion 理想電流源與任一部分電路的串聯(lián)，對外電路而言，可等效成為一個理想電流源，且IS 不變 E E+U

U￠SUU￠SUURL L US 1 Thiswatermarkdoesnotappearintheregisteredversion §2— a+-+-bThiswatermarkdoesnotappearintheregisteredversion a+-+-2節(jié)點數為nKCL（n-1）個獨立的KCL方程如圖所示：n=2，則可列寫一個KCL3、列寫[b-(n-1)]個獨立的KVL方 (b為電路的支路數 如圖：E1=R1I1+R3E2=R2I2+R3I3Thiswatermarkdoesnotappearintheregisteredversion 對于一般回路方程的列寫，則應保證每一個新的回路中必須包含有一條前面各回路中未曾用過的新支路4.聯(lián) n-個KCL方程與b-(n-1)個KVL方Thiswatermarkdoesnotappearintheregisteredversion R1GR2l2R1GR2l2l3B解：n=4b= 根據KCL列寫節(jié)點A：I1I2-IG

n1=3 I

+E節(jié)點B：I3IG-I4

節(jié)點CI2I4-I

Thiswatermarkdoesnotappearintheregisteredversion IAIA1I2GR2l2R3R4l3BI

+Eb (n 1)=6

網孔l1:I1R1+IGRG I3R3=0（網孔l2

I2R2 I4R4 IGR

=0（網孔l3 I3R3+I4R4= Thiswatermarkdoesnotappearintheregisteredversion IAIA1I2GR2l2R3R4l3BI

+E E（R2R3- R（R+R）（R+R）+R（R+R）+

（

+R

當R1R4 =R2R3時，IG=0這時電橋處于平衡狀態(tài)Thiswatermarkdoesnotappearintheregisteredversion 例 求各支路電3 由節(jié)點aI

+I I

IL+ IL+SEbLI1R1+I2R2=Thiswatermarkdoesnotappearintheregisteredversion I2+I3-I4=-I2+I5-I6=-I1+I4+I6=

d+-cI6b+8I3+10-2I2-8I5=8I4-8I6+2I2-10=8I5+8I6+6I1-10=

§2— —、基本思想Thiswatermarkdoesnotappearintheregisteredversion 二、節(jié)點電壓的選擇（2個節(jié)點_E_E+aI2b II+++E1_U 4例1.求上圖所示電路中，a、b節(jié)點間的電壓4解：I1

U-E1

I2=

U+E2

=UaKCL方程：I1I2IS1-IS2-I4Thiswatermarkdoesnotappearintheregisteredversion 將

代入上述方程并整理-U-E1（-U+E2）-U= -

S S_E+a_E+aI2b II+++ E1-得

+IS1-IS UU 1 1+ U

?E+??RSRRSThiswatermarkdoesnotappearintheregisteredversion 三、3個節(jié)點電壓方程的列寫方法 + +

_I _I+I5 +__ __I4

U

節(jié)點A：I1+I2-I4= 節(jié)點B：I2+I5-I3=

E-(U-UR =E1-R

I2 I3

1E3-U2

=R44R4

UI5 Thiswatermarkdoesnotappearintheregisteredversion + +

_I _I+I5 +_ _I4

U 5_3E5_3_代入節(jié)點方程12，并整理：(1+1+1)U 1

=E1+

+ ++ +U-_12

1U+(1+1+1 =-E2+ Thiswatermarkdoesnotappearintheregisteredversion AA+o--UAO

1+1+1+

=-

+o+Thiswatermarkdoesnotappearintheregisteredversion SE1+S

AA+R32UAB

=1+1+

BI1

140-U

=42 =-UAB2

=-123 =UAB3

=10§2— 疊加定一個支的電壓或電流，都可以看成是源單獨作用時，*只能用來計算線性電路（如果有內阻的話，內阻保留），電流源支路則開路Thiswatermarkdoesnotappearintheregisteredversion 例 求圖示電路中R1支路的電流I1和+U+UI+IE

I￠

U'

+1 +1E

=I'+I"=

12R1+R2 R+12Thiswatermarkdoesnotappearintheregisteredversion +U+E+U+EU'+IIS

R R U=U'+U"=

R1

R1

顯然：R1

R1+

12 R+12 = 2

（I￠+I2

+I2

Thiswatermarkdoesnotappearintheregisteredversion 例2試求圖示電路R3中的電流I+II I -

1單獨作用，IS1開路E

I

I

R3+ R

IS1單獨作用，E1R3 3

I￠=

R3+

ISR IS R

\I=I￠+I￠

R3+

R3+R4

ISThiswatermarkdoesnotappearintheregisteredversion §2—7戴維南(寧) 一、二端網絡及其等A+U-A+U-R1abb代表有源 Thiswatermarkdoesnotappearintheregisteredversion ababP+UP+U-結論：二端網絡對外電路的作用可用一個簡單代表無源結論：二端網絡對外電路的作用可用一個簡單b有源二端網絡用電壓源 有源二端網絡用電流源 Thiswatermarkdoesnotappearintheregisteredversion 任何一個線性有源二端網絡，對外電路來說，可以用一個電壓源串聯(lián)電阻支路來等效，電壓源的電壓等于原來有源二端網絡的開路電壓Uoc，而電阻等于原來有源二端網絡中所有獨立源為零時輸入電阻Ro++_babab+_a

bbabab+_a為E的理想電壓源和內阻RO串聯(lián)的電源來等效代等效電源的電動勢E就是有源線性二端網絡的開路電壓UOC等效電源的內阻RO等于有源二端網絡中所有獨立電源均除去后所得的無源網絡a,b兩端之間的等效電U+_+_U等效電已知：R1=20、R2=30WR3=30、R4=20W求：當R5=10WBBU_+A+_BASRRSA

第二步：求輸入電阻ADD U B

AB1//R2R3//1 2V

30 +_BABU_+A 2I5RI5RS52240.059A4A4+8V45EB33+_CD求求Thiswatermarkdoesnotappearintheregisteredversion 第一步：求開端電壓\C\4+8V

4441008ABO

5

B119第二步求輸入電阻 C第二步求輸入電阻W_A

4E

5

AB50AB504//457B4+8V

45E

33B

等效電S

_ U 57_ U 第三步：求解未+_UUL57′93.3例例A+U J J

>2mA時 電阻是5KW）-

A解+解求開路電壓

-Thiswatermarkdoesnotappearintheregisteredversion (111

40100

10+

-

+

+-RRAB=10//30//60=-

+U

I=26.7/(5+6.67)=2.3mA結論 繼結論 繼電器觸點閉BAA bb任何一個有源二端線性網絡都可以用一個電流為的理想電流源和內阻RO并聯(lián)的電源來等效代替流ISC等效電源的內阻RO等于有源二端網絡中所有獨立電源均除去后所得的無源網絡a,b兩端之間的等效電阻Thiswatermarkdoesnotappearintheregisteredversion 、被等效的有源二端網絡必須是線性的，而外部電路則可以是非線性負載2Thiswatermarkdoesnotappearintheregisteredversion II1例解解ISEISERS4RR0=I

（

I

R0E1R0R R R Thiswatermarkdoesnotappearintheregisteredversion 例例 RL 解Thiswatermarkdoesnotappearintheregisteredversion 解60Uaboc

1.554.3 1 50(1.56050)54.32 21.5 2

I= Ro+RL

=0.86+I Uoc

Ro 作業(yè)P.78§2— 壓和電流源的電流制，這種電源稱為受控電源。 如下學期將要學到的三極管及場效應管等電子元器件，都根據控制量（電壓或電流）及受控電源（電壓控制電壓源（voltagecontrolledvoltage（voltagecontrolled 電流控制電流源Thiswatermarkdoesnotappearintheregisteredversion 1、電壓控制電壓源+++__+++__U2= Thiswatermarkdoesnotappearintheregisteredversion 2、電流控制電壓源 _ _=

rI_ U rI

r的單位為Thiswatermarkdoesnotappearintheregisteredversion 3、電壓控制電流源++++_

I+UI2 gU

g的單位為Thiswatermarkdoesnotappearintheregisteredversion 4、電流控制電流源_ _ =

bI

I2=ccIc=bIBThiswatermarkdoesnotappearintheregisteredversion

+++_+=+=+_Thiswatermarkdoesnotappearintheregisteredversion ++++__

I1=0 受控電流源：其輸出電阻（相當于電流源的內阻）無窮大，輸出電流恒定I1 +++_受控量與控制量成正比例關系，即μ、rg、?Thiswatermarkdoesnotappearintheregisteredversion +UR= Thiswatermarkdoesnotappearintheregisteredversion

_ _bR=

=Uo SS SS

IsE_ bThiswatermarkdoesnotappearintheregisteredversion 二、含有受控源電路的分析例 應用疊加原理求圖中

2 2

10I1 +壓U和電流I

I1'=

2'

6+

=2

+

+10I1U￠=-

'=-

I1"=

6+6

′10=-4

+10I1+I2"=6+4′10=6U"=-10I1"+4I2"=

Thiswatermarkdoesnotappearintheregisteredversion U=U'+U"=I2=I2'例 應用戴維寧定理求圖中電流I10

10I1

I =-10 =20-6I￠=20+60= Thiswatermarkdoesnotappearintheregisteredversion 求短路電流I

2+10I

IS 6+10

求等效電源的內阻

=U

=6 R0R0 3求電流II

4+

=8Thiswatermarkdoesnotappearintheregisteredversion

+ 1U1

b b

+

+

1、求開路電壓I =2+

U11=2I=Uo=2U11+U11=3U11=Thiswatermarkdoesnotappearintheregisteredversion 2、求內阻①除去二端網絡內的電源，外加電壓 +b = +b U=I￠+U1