大一各種試題電路分析第三次課

§1-1§1-2 §1-4§1-5§1-6§1-7§1-8§1-9§1-8

series(1i+

+_+_uu1uk

+ +RReq=(R1+R2+…+Rn)=ReqReq=(R1+R2+…+Rn)=+u_1+u_n +u_1+u_nuk ukuk ukRuuki u1 R u1 R1R1u - u2u2?RR 2u（4）功率關(guān)系p1=R1i ，p2=R2i ，，pn=Rnip1:p2::pn=R1:R2:: p=Reqi2=(R1+R2++Rn)i=R1i2+R2i2++Rni=p1+p2++i _ i=i1+i2++ik++設(shè)Gk=1/ （k=1，2，…， _

_ i=i1+i2++ik++in=uGeq=G1+G2++Gk++Gn=Gk=Geq=G1+G2++Gk++Gn=Gk=Geq=G1+G2++Gk++Gn=Gk=由iku/

u/ikik ikGi

i1i1 1/ i 1/R1/ R 11/1/R11/iR1ii2i2 ?1/ 1/R1/R RR 21/1/R11/iR1ip1=G1u2，p2=G2u2，，pn=Gnu2p1:p2::pn=G1:G2::Gn p=Gequ2=(G1+G2++Gn)=G1u2+G2u2++=p1+p2++RR解R=4∥(2+(3∥6))=2RRRR解R=(40∥40)+(30∥30∥30)=42422432

R5R

（c解:R12R1//R2 R34R3//R4串R12R522 R并串//R6Rab(R＋R34R7u

+sRR+sRR-32RsRR41

2

ii i i

i i i1'i1 Ri' i'3

i i ((i1,i2,i3,u12,(i1,i2,i3,u12,23,u323,u間滿足的關(guān)系間滿足的關(guān)系 1

2

i'i1 i' i'3

i iu12+u23+u31i1i2i30

u12+u23+u31=i1i2i3i1u12/R12u31/R31i2u23/R23 /R1i3u31/R31u /R

u12+u23+u31i1i2i3

T u12+u23+u31= i1i2i3

i1u1i2u23i3u31

/R12u31/R3/R23 /R1/R31u /R23

u12+u23+u31

Ti1

R3u12R2u31 Ru R

R2

i 1 32i3

R2u31R1u23RR RRRR

R2R3

R

R R2u12+u23+u31i1i2i3

T u12+u23+u31=i1i2i3

i1u1i2ui3u31

/R12u31/R3/R23 /R1/R31u /R23Π

u12+u23+u31i1i2i3 R12

R12R23 R23

R12R23

R31RR1T

Π

R1

R R12 R3

R2

R23 R1

R3

R31 R2

星形電阻R三角形中連接于ii三角形電阻

=星形中接在除i、j例題4橋T電路（bridge-T

RE

R §1-1§1-2 §1-4§1-5§1-6§1-7§1-8§1-9§1-8

R R

i +uB+uB RuR §1-9

+- +-u-

-n+u‐+u‐+‐usis+u-i +u-i+ +u-i+ -

isn nisis1is2isnisii+- +-

u u-iu iu

iiisiiss+uA Buus

RR',uR' s

Ruiis Russ

uR'iR'

R'

i

A u+u-

++ -

A uB + --(1)兩種電源模型進(jìn)行等效替代后，對(duì)外負(fù)載提供

RLuu電源B

RL

PL

54

43454

9 + + -+5+15-u- X

345

5

9s suus

45

5

9相并的電阻都可視為電源的Rs來(lái)處理。但在兩種情況下，Rs消耗的功率是不同的。i +i+iSu_+i_

例電流源短路時(shí),并聯(lián)電導(dǎo)Gi

iV76 i

912

++u-3+2

2-+u-+u-

u2i6ui1 u

13 u1