課件上屆第4章電路自動(dòng)化科學(xué)與電氣工程學(xué)院

第4(Circuit疊加定理(Superposition替代定理(Substitution 根定理( legen’sTheorem)互易定理(ReciprocityTheorem)對(duì)偶原理(DualPrinciple)1重重難難2 小結(jié)作3R2+_R2+_12+_12i

s=si(2 （i(2 =R

-i+R

i=i(1)+i(2)

i(1)

)= 2

R+

2R+22 = +2

u=u(1)+u(2)

1u(1)1

1u(2)=-1

i(1)=i ,u(1)=u

RS RS

RSRS 2iS 1iSu1

R+

uS

R+

i(2)=i ,u(2)=u

2uS 1uS45

+

a21x1+a22x2

+a2NxN=aN1x1+aN2

+aNNxN=

+DDD DDb

D6 u =kf1uS1+kf2uS2 +kfguS+Kf1iS1+Kf2iS2 K =kfmuSm+KfmiSm

+kf2' +

+ f g g

h+ h

S

f1

f2

fh

fm7 8ii32+

=32

3232 9 含受控源(線性)電路亦可用疊加定理，但疊加只單獨(dú)作用獨(dú)立電源，不包括受控源，受控源在電路 i 2i= 2

s+ u1

uS

R1+R2SS

I1=

66

I2=5AI3=10AI4=7.5AI5=u

uSuR2+R

u(2

R2+Ru=u(1)+u(2XXXX XX

+I(2) XX IXXX2IIX2IXX(2+1)IX

=10-2I

(1UXX2XX

XU1XUXXI(1)=XX

=2IXI=IX

XX+I(2)=2-0.6XX討論：求2W U=U(1)+U(2)=4-1.2= P=UI=2.8·1.4=

I(1)=XXXXUX=P=(U

+U(2))(I(1)+I(2) =U(1)I(1)+U(2)I(1)+U(1)I(2)+U(2)I ?U(1)I(1)+U(2) XXI+++X（1）u1=u2=5V10A，試求當(dāng)u1=u2=5V時(shí)，ix=？令u=1V單獨(dú)作用下的i為 =1V單獨(dú)作用下的i為

+3i(2)=

+i(2)=u

=5i(1)+5i(2)=

x x

u1u20V時(shí)，ixxxxu1u25Vx =i(1)

+i(3)=4.24.2(Substitution電阻為Rkuk/ik 替代前后KCL,KVL關(guān)系相同，其余支路的u、i關(guān)系用uS替代后，其余支路電壓不變(KVL)，其余支路電流也用iS替代后，其余支路電流不變(KCL)，其余支路電壓不解

(1+1)u

-1u=1 uc= =8ub=(1+1)u-1u=- 的電壓等于外電路斷開時(shí)端?處的開路電壓uoc，?電阻

R’

S S u_

S S

u=u(1)+u(2)=uoc- 電 R電 R 1RiS發(fā)?改變時(shí)，含源?端??絡(luò)的等效電路不變

等效電路可由

等效電路經(jīng)電源等效

’

uoc=0，isc=0時(shí)，等效電路為一電阻，Req值要用外加電壓法來(lái)計(jì)算。若計(jì)算結(jié)果為Req=0，則電路等效為一條短路1

a b

u2求2求1求b -

= +R3 R+ R+

R+ R+

S - SR3+ iu - - R+ R+ R+ R+ = +R3 R+ R+ i=

4(R1)uSRL+

9V

a11求Ι1=

=1AUoc=6Ι1+3Ι1=22求

U=6Ι1+6(Ι-Req=U=Ι

a9V

Ω9V

aΩaII9VIΙ1=-3

Ιsc

9A6

Ι1=

Req

UOC= 9ΩΩ9V11

{1 {

1 =-

3331 3331-1-1 +(1R

u=(u=(1+R帶入

1+R21us+R2(isc+gu3)=

·is1

u3RR =- =- +(1-R2R2R+uR(1-gR32s23R+23u3R+iu23R+s23

-

i帶入i

11求I4VUI=I4VU4V U14VU14VUUoc=200·0.01Uoc+U 2求2求

IUIUI14VUU=-180Ι+200(Ι+0.01UReq=U=-Ι

-180Ι+200Ι+4=I14V0IsΙ=-51

Ιsc=-Ι=5 IIs4V

Req=

15

4V

ΩΩ4VΙ=

PPLP=R( PPL

+R)2-2R +RP'=

+R

= 注注 或定理最方便.

40=-2i1+6i1+1+i1·4

30V

ii

u

)·

i= u=3i+

u=

30VR=u= Rii

40

=

i=9 6

30V40V=401+ix 405 （i1，i2，¨¨，ib）、（u1，u2，¨¨，ub）b條支路的電流和電壓，則對(duì)任何時(shí)間t，滿?:bbukik=

6

-i1+i2+i4=-i4+i5+i6=-i2+i3-i6= ukik

+-un1i1+(un1-un3)i2+un3i3(un1-un2)i4+un2i5+(un2-un3+un3(-i2+i3-i6)=

, , , 表示兩電路中b條支路的電流和電壓，則在任何時(shí)間t，滿足:bu?=b

?k k kb 45645623

4562345623

k?k

?bu ??kikk=1b6

-?+?+?= -?+?+?= -?+?-?= k 1 2 6u?=u?k 1 2 6k-u?+(u )?+ ?n3 n1n3

n3

n344

n2

+(un2

1k1k

?24+un2(-242+un3(-2

5353

66-66

)=

?

(uk,ik

（否 中加負(fù)號(hào)】】1A,?S2u'u' =-i=-'1i1i?i2?S2''?

11 '=

?s2

2'2'u' +u' 11 22 11 2220·4+0·?=12·(-10)+ ·2

?S2 s2 bu'

'+

'

'+Ri?=11

kkkb?'i'+ '

'+R? 11

kkkuu11' '=?'11' 'i?i1i?i2?S2純電阻網(wǎng)絡(luò)NR，當(dāng)R24W，U1=10V時(shí)，I12A，I2=1A?，F(xiàn)將R2=1W，U124V，測(cè)得I16A，求I2=？U

1

=U1=I'=-

I?

U

=R

''

'I22 '

=1A

U'

'

'

'=?'

'+ ' 10·(-6)+4I2=24·(-2)+

I2=4A 勵(lì) 11 uu= 或u?S1S22當(dāng)S2ui1i2證明: 證明:bukik b

bbb ukikk=1bbukb

=u1i1+u2i2+ukibkbb=u1i1+u2i2+Rkikik=bkb=u1i1+u2i2+ukbb=u1i1+u2i2+Rkikik=b1 2 ui?+u =u11 2 u1=uS1,u2=0, u1=0,u2=u= 或u?S1S22uS2u= 或u?S1S22uS2S1

11勵(lì) aa S2bu2=i或u1=u2當(dāng)iS2iu12d證明: 證明:1 2 ui?+u =u11 2 i1=-iS1,i2=0, i1=0,i2=-2 u1·0-2

=-u1 u2i =u1

u2 或u

=u

S2勵(lì)

a??a??S2bii2=或 =uS22當(dāng)uS2iu12cd證明: 證明:1 2 ui?+u =u11 2 i1=0,u2=2u1·0+0·?2

+uS2u1iS1=uS2

i2 i

=uS2 uS2?1

?S2 互易定理只適?于線性電阻?絡(luò)在單?電源激勵(lì)下，__+6

I

=

I=I·3 6+12·6+3·6

I3

·3=

I￠=I3-I2=1A【例4-14線性無(wú)源電阻網(wǎng)絡(luò)NR，求uab10· uab

·5

b

dWReq=10=W5

u=u=10i1=?R1無(wú)源電阻網(wǎng)絡(luò)NR，當(dāng)I12A，U1=10V，U25V。現(xiàn)將變?yōu)槿鐖D（b）所示，求I=？I2?1I2?1?R RRI=0.5I2【例4-16】無(wú)源電阻網(wǎng)絡(luò)NR，當(dāng)I12A，U1=10V，U25V?，F(xiàn)將變?yōu)槿鐖D（b）所示，求I=I2

RR

UI

?11

?1

1

I?'=

'=U2=

22

=?II

=I2=

I?'=I?=-U'

'

'

'=

'I

+ ' I=0.5

5I·(-2)+0=10I+5·(-2)

RR

?

I2I2 從口看去

U1105W I=

5V

u1+u2=i1=i2+iu2=

i1+i2=iu1=u2+ui2=i等效電阻RR1