MT S PHNG PHP GII BI TON MCH CU IN TR

Trng vn thanh 0974810957

MT S PHNG PHP GII BI TON MCH CU IN TR1. khI qut v mch cu in tr, mch cu cn bngv mch cu khng cn bng.

Mch cu l mch dng ph bin trong cc php o chnh xc phng th nghim in.

Mch cu c v nh (H - 0.a) v (H - 0.b)

Cc in tr R1, R2, R3, R4 gi l cc cnh ca mch cu in tr R5 c vai tr khc bit gi l ng cho ca mch cu (ngi ta khng tnh thm ng cho ni gia A B. V nu c th ta coi ng cho mc song song vi mch cu).

Mch cu c th phn thnh hai loi( Mch cu cn bng (Dng trong php o lng in). I5 = 0 ; U5 = 0( Mch cu khng cn bng: Trong mch cu khng cn bng c phn lm 2 loi:

Loi c mt trong 5 in tr bng khng (v d mt trong 5 in tr b ni tt, hoc thay vo l mt ampe k c in tr ng khng ). Khi gp loi bi tp ny ta c th chuyn mch v dng quen thuc, ri p dng nh lut m gii.

Loi mch cn tng qut khng cn bng c c 5 in tr, th khng th gii c nu ta ch p dng nh lut m, loi bi tp ny c gii bng phng php c bit ( Trnh by mc 2.3) Vy iu kin cn bng l g ?Cho mch cu in tr nh (H1.1) ( Nu qua R5 c dng I5 = 0 v U5 = 0 th cc in tr nhnh lp thnh t l thc : = n = const

( Ngc li nu c t l thc trn th I5 = 0 v U5 = 0, ta c mch cu cn bng.

( Tm li: Cn ghi nh

( Nu mch cu in tr c dng I5 = 0 v U5 = 0 th bn in tr nhnh ca mch cu lp thnh t l thc: (n l hng s) (*) (Vi bt k gi tr no ca R5.).

Khi nu bit ba trong bn in tr nhnh ta s xc nh c in tr cn li.

( Ngc li: Nu cc in tr nhnh ca mch cu lp thnh t l thc tn, ta c mch cu cn bng v do I5 = 0 v U5 = 0

Khi mch cu cn bng th in tr tng ng ca mch lun c xc nh v khng ph thuc vo gi tr ca in tr R5 . ng thi cc i lng hiu in th v khng ph thuc vo in tr R5. Lc c th coi mch in khng c in tr R5 v bi ton c gii bnh thng theo nh lut m. Biu thc (*) chnh l iu kin mch cu cn bng.2. phng php tnh in tr tng ng ca mch cu.

Tnh in tr tng ng ca mt mch in l mt vic lm c bn v rt quan trng, cho d u bi c yu cu hay khng yu cu, th trong qu trnh gii cc bi tp in ta vn thng phi tin hnh cng vic ny.

( Vi cc mch in thng thng, th u c th tnh in tr tng ng bng mt trong hai cch sau.

( Nu bit trc cc gi tr in tr trong mch v phn tch c s mch in (thnh cc on mc ni tip, cc on mc song song) th p dng cng thc tnh in tr ca cc on mc ni tip hay cc on mc song song.

( Nu cha bit ht cc gi tr ca in tr trong mch, nhng bit c Hiu in th 2 u on mch v cng dng in qua on mch , th c th tnh in tr tng ng ca mch bng cng thc nh lut m.

Tuy nhin vi cc mch in phc tp nh mch cu, th vic phn tch on mch ny v dng cc on mch mi ni tip v song song l khng th c. iu cng c ngha l khng th tnh in tr tng ng ca mch cu bng cch p dng, cc cng thc tnh in tr ca on mch mc ni tip hay on mch mc song song. Vy ta phi tnh in tr tng ng ca mch cu bng cch no?

( Vi mch cu cn bng th ta b qua in tr R5 tnh in tr tng ng ca mch cu.

( Vi loi mch cu c mt trong 5 in tr bng 0, ta lun a c v dng mch in c cc

on mc ni tip, mc song song gii.

( Loi mch cu tng qut khng cn bng th in tr tng ng c tnh bng cc phng

php sau. Phng n chuyn mch.Thc cht l chuyn mch cu tng qut v mch in tng ng (in tr tng ng ca mch khng thay i). M vi mch in mi ny ta c th p dng cc cng thc tnh in tr ca on mch ni tip, on mch song song tnh in tr tng ng.

Mun s dng phng php ny trc ht ta phi nm c cng thc chuyn mch (chuyn t mch sao thnh mch tam gic v ngc li t mch tam gic thnh mch sao). Cng thc chuyn mch - nh l Kennli.( Cho hai s mch in, mi mch in c to thnh t ba in tr. ( H2.1a mch tam gic (() ; H2.1b - Mch sao (Y) )

( Vi cc gi tr thch hp ca in tr c th thay th mch ny bng mch kia, khi hai mch

tng ng nhau. Cng thc tnh in tr ca mch ny theo mch kia khi chng tng ng

nhau nh sau:( Bin i t mch tam gic R1, R2, R3 thnh mch sao R1, R2, R3

(1);

(2)

(3) ( y R1, R2, R3 ln lt v tr i din vi R1,R2, R3 )( Bin i t mch sao R1, R2, R3 thnh mch tam gic R1, R2, R3

(5)

(6) p dng vo bi ton tnh in tr tng ng ca mch cu ta c hai cch chuyn mch nh sau:

Cch 1:

T s mch cu tng qut ta chuyn mch tam gic R1, R3, R5 thnh mch sao :R1; R3; R5 (H2.2a) Trong cc in tr R13, R15, R35 c xc nh theo cng thc: (1); (2) v (3) t s mch in mi (H2.2a) ta c th p dng cng thc tnh in tr ca on mch mc ni tip, on mch mc song song tnh in tr tng ng ca mch AB, kt qu l:

Cch 2:T s mch cu tng qut ta chuyn mch sao R1, R2 , R5 thnh mch tam gic R1, R2 , R5 (H2.2b ). Trong cc in tr R1, R2, R3 c xc nh theo cng thc (4), (5) v(6). T s mch in mi (H2.2b) p dng cng thc tnh in tr tng ng ta cng c kt qu: Phng php dng nh lut m.

( T biu thc: suy ra

( Trong : U l hiu in th hai u on mch.

I l cng dng in qua mch chnh.

( Vy theo cng thc (*) nu mun tnh in tr tng ng (R) ca mch th trc ht ta phi tnh I theo U, ri sau thay vo cng thc (*) s c kt qu. ( C nhiu phng php tnh I theo U s c trnh by chi tit mc sau ).

( Xt v d c th: Cho mch in nh hnh H . 2.3a.

Bit R1 = R3 = R5 = 3 (, R2 = 2 (; R4 = 5 (

a. Tnh in tr tng ng ca on mch AB.

b. t vo hai u on AB mt hiu in th khng

i U = 3 (V). Hy tnh cng dng in qua

cc in tr v hiu in th hai u mi in tr.

Phng php 1: Chuyn mch.

Cch 1: Chuyn mch tam gic R1; R3 ; R5 thnh

mch sao R1 ; R3 ; R5 (H2.3b) Ta c:

Suy ra in tr tng ng ca on mch AB l : Cch 2: Chuyn mch sao R1; R2; R5 thnh mch tam gic (H2.3c). Ta c:

Suy ra:

Phng php 2: Dng cng thc nh lut m.

T cng thc:

Gi U l hiu in th hai u on mch AB ; I l cng dng in qua on mch AB

Biu din I theo U

t I1 l n s, gi s dng in trong mch c chiu nh hnh v (H2.3d)

Ta ln lt c:

U1 = R1I1 = 3 I1 (1) ; U2 = U U1 = U 3 I1 (2)

Ti nt D, ta c: I4 = I3 + I5

Thay (11) vo (7) ta c: I3 =

Suy ra cng dng in mch chnh. Thay (12) vo (*) ta c kt qu: RAB = 3 (()

b. Thay U = 3 V vo phng trnh (11) ta c:

Thay U = 3(V) v I1 = vo cc phng trnh t (1) n (9) ta c kt qu:

( c chiu t C n D)

; ( Lu

( C hai phng trnh gii trn u c th p dng tnh in tr tng ng ca bt k mch cu in tr no. Mi phng trnh gii u c nhng u im v nhc im ca n. Tu tng bi tp c th ta la chn phng php gii cho hp l.

( Nu bi ton ch yu cu tnh in tr tng ng ca mch cu (ch cu hi a) th p dng phng php chuyn mch gii, bi ton s ngn gn hn.

( Nu bi ton yu cu tnh c cc gi tr dng in v hiu in th (hi thm cu b) th p dng phung php th hai gii bi ton, bao gi cng ngn gn, d hiu v l gic hn.( Trong phng php th 2, vic biu din I theo U lin quan trc tip n vic tnh ton cc i lng cng dng in v hiu in th trong mch cu. y l mt bi ton khng h n gin m ta rt hay gp trong khi gii cc thi hc sinh gii, thi tuyn sinh. Vy c nhng phng php no gii bi ton tnh cng dng in v hiu in th trong mch cu.

3. phng php giI ton tnh cng dng in v hiu in th trong mch cu Vi mch cu cn bng hoc mch cu khng cn bng m c 1 trong 5 in tr bng 0 (hoc ln v cng) th u c th chuyn mch cu v mch in quen thuc (gm cc on mc ni tip v mc song song). Khi ta p dng nh lut m gii bi ton ny mt cch n gin.

V d: Cho cc s cc mch in nh hnh v: (H.3.1a); (H. 3.1b); (H3.1c); (H3.1d) bit cc vn k v cc am pe k l l tng.

Ta c th chuyn cc s mch in trn thnh cc s mch in tng ng, tng ng vi cc hnh H.3.1a; H.3.1b; H.3.1c; H.3.1d.

T cc s mch in mi, ta c th p dng nh lut m tm cc i lng m bi ton yu cu:

( Lu .Cc bi loi ny c nhiu ti liu trnh by, nn trong ti ny khng i su vo vic phn tch cc bi ton tuy nhin trc khi ging dy bi ton v mch cu tng qut, nn rn cho hc sinh k nng gii cc bi tp loi ny tht thnh tho.

Vi mch cu tng qut khng cn bng c c 5 in tr, ta khng th a v dng mch in gm cc on mc ni tip v mc song song. Do cc bi tp loi ny phi c phng php gii c bit - Sau y l mt s phng php gii c th:Bi ton 3:

Cho mch in h hnh v (H3.2a) Bit U = 45V

R1 = 20(, R2 = 24( ; R3 = 50( ; R4 = 45( R5 l mt bin tr1. Tnh cng dng in v hiu in th ca mi in tr

v tnh in tr tng ng ca mch khi R5 = 30(2. Khi R5 thay i trong khong t 0 n v cng, th in tr tng ng ca mch in thay i nh th no?

1. Tnh cng dng in v hiu in th ca mi in tr v tnh in tr tng ng ca

mch khi R5 = 30(Phng php 1: Lp h phng trnh c n s l dng in (Chng hn chn I1 lm n s)

Bc 1: Chn chiu dng in trn s

Bc 2: p dng nh lut m, nh lut v nt, biu din cc ilng cnl li theo n s (I1) chn (ta c cc phng trnh vi n s I1).

Bc 3: Gii h cc phng trnh va lp tm cc i lng ca u bi yu cu.

Bc 4: T cc kt qu va tm c, kim tra li chiu dng in chn bc 1

( Nu tm c I > 0, gi nguyn chiu chn.

( Nu tm c I < 0, o ngc chiu chn.

Li gii : Gi s dng in mch c chiu nh hnh v H3.2b Chn I1 lm n s ta ln lt c:

U1 = R1 . I1 = 20I1 (1) ; U2 = U U1 = 45 20I1 (2)

(9) Ti nt D cho bit: I4 = I3 + I5

(10)Suy ra I1= 1,05 (A)

Thay biu thc (10) cc biu thc t (1) n (9) ta c cc kt qu:

I1 = 1(A);I3 = 0,45 (A) ; I4 = 0,5 (A);I5 = 0,05 (A)

Vy chiu dng in chn l ng.

Hiu in th : U1 = 21(V)

U2 = 24 (V)

U3 = 22,5 (V) UBND = 22,5 (V) U5 = 1,5 (V)

( in tr tng ng

Phng php 2: Lp h phng trnh c n s l hiu in th cc bc tin hnh ging nh phng php 1. Nhng chn n s l Hiu in th. p dng (Gii c th)

Chn chiu dng in trong mch nh hnh v H3.2b Chn U1 lm n s ta ln lt c:

(1) U2 = U U1 = 45 U1

(2)

(3) (4)

(5) (6)

(7)

(8)

(9)

Ti nt D cho bit: I4 = I3 + I5

(10)Suy ra: U 1 = 21 (V)

Thay U1 = 21 (V) vo cc phng trnh t (1) n (9) ta c kt qu ging ht phng php 1Phng php 3: Chn gc in th.

Bc 1: Chn chiu dng in trong mch

Bc 2: Lp phng trnh v cng ti cc nt (Nt C v D)

Bc 3: Dng nh lut m, bin i cc phng trnh v VC, VD theo VA, VB

Bc 4: Chn VB = 0 VA = UABBc 5: Gii h phng trnh tm VC, VDtheo VA ri suy ra U1, U2, U3, U4, U5

Bc 6: Tnh cc i lng dng in ri so snh vi chiu dng in chn bc 1. p dng

Gi s dng in c chiu nh hnh v H3.2b p dng nh lut v nt C v D, ta c:

- p dng nh lut m, ta c:

Chn VD = 0 th VA = UAB = 45 (V). ( H phng trnh thnh:

Gii h 2 phng trnh (3) v (4) ta c: VC = 24(V); VD = 22,5(V)

Suy ra: U2 = VC VB = 24 (V)

U4 = VD VB = 22,5 (V)U1 = U U2 = 21 (V) U3 = U UBND = 22,5V U5 = VC VD = 1,5 (V)

T cc kt qu va tm c ta d rng tnh c cc gi tr cng dng in (nh Phng php 1).Phng php 4:

Chuyn mch sao thnh mch tam gic ( Hoc mch tam gic thnh mch sao ).

Chng h n chuyn mch tam gic R1 , R3 , R5 thnh mch sao R1 , R3 , R5 ta c s mch in tng ng H3.2c (Lc cc gi tr RAB, I1, I4, I, U2, U4,UCD vn khng i)

Cc bc tin hnh gii nh sau:

Bc 1: V s mch in mi.

Bc 2: Tnh cc gi tr in tr mi (sao R1 , R3 , R5)Bc 3: Tnh in tr tng ng ca mch

Bc 4: Tnh cng dng in mch chnh (I)

Bc 5: Tnh I2, I4 ri suy ra cc gi tr U2, U4.

Ta c: V: I4 = I I2Bc 6: Tr li mch in ban u tnh cc i lng cn li.p dng: T s mch in (H - 3.2C) ta c

in tr tng ng ca mch:

Cng dng in trong mch chnh:

Suy ra: I4 = I I2 = 1,5 1 = 0,5 (A)

U2 = I2.R2 = 24 (V) U4 = I4.R4 = 22,5 (V)

Tr li s mch in ban u (H - 3.2 b) ta c kt qu:

Hiu in th: U1 = U U2 = 21 (V) ; U3 = U U4 = = 22,5(V) ; U5 = U3 U1 = 1,5(V)

V cc gi tr dng in ; I5 = I1 I3 = 0,05 (A)Phng php 5: p dng nh lut kic sp( Do cc khi nim: Sut in ng ca ngun, in tr trong ca ngun, hay cc bi tp v mch

in c mc nhiu ngun, hc sinh lp 9 cha c hc. Nn vic ging day cho cc em hiu

y v nh lut Kic sp l khng th c. Tuy nhin ta vn c th hng dn hc sinh lp

9 p dng nh lut ny gii bi tp mch cu da vo cch pht biu sau: nh lut v nt mng.

T cng thc: I = I1+ I2+ +In(i vi mch mc song song), ta c th pht biu tng qut: mi nt, tng cc dng in i n im nt bng tng cc dng in i ra khi nt Trong mi mch vng hay mt mch.

Cng thc: U = U1+ U2+ + Un (i vi cc in tr mc ni tip) c hiu l ng khng nhng i vi cc in tr mc ni tip m c th m rng ra: Hiu in th UAB gia hai im A v B bng tng i s tt c cc hiu in th U1, U2, ca cc on k tip nhau tnh t A n B theo bt k ng i no t A n B trong mch in

( Vy c th ni: Hiu in th trong mi mch vng (mt mng) bng tng i s gim th trn mch vng

Trong gim th: UK = IK.RK ( vi K = 1, 2, 3, )

( Ch : ( Dng in IK mang du (+) nu cng chiu i trn mch

( Dng in IK mang du () nu ngc chiu i trn mch.

( Cc bc tin hnh gii.

Bc 1: Chn chiu dng in i trong mch

Bc 2: Vit tt c cc phng trnh cho cc nt mng

V tt c cc phng trnh cho cc mt mng.

Bc 3: Gii h cc phng trnh va lp tm cc i lng dng in v hiu in th trong mch.

Bc 4: Bin lun kt qu. Nu dng in tm c l:

IK > 0: ta gi nguyn chiu chn. IK < 0: ta o chiu chn.p dng: Chn chiu dng in i trong mch nh hnh v H3.2b. Ti nt C v D ta c:

Phng trnh cho cc mch vng:

( Mch vng ACBA: U = I1.R1 + I2.R2 (3) ( Mch vng ACDA: I1.R1 + I5.R5 I3.R3 = 0 (4)

( Mch vng BCDB: I4.R4 + I5.R5 I2.R2 = 0 (5) Thay cc gi tr in tr v hiu in th vo cc phng trnh trn ri rt gn, ta c h phng trnh:

Gii h 5 phng trnh trn ta tm c 5 gi tr dng in: I1 = 1,05(A); I2 = 1(A); I3 = 0,45(A); I4 = 0,5(A) v I5 = 0,05(A) Cc kt qu dng in u dng do chiu dng in chn l ng. T cc kt qu trn ta d dng tm c cc gi tr hiu in th U1, U2, U3, U4, U5 v RAB (Ging nh cc kt qu tm ra phng php 1)

2. S ph thuc ca in tr tng ng vo R5

( Khi R5 = 0, mch cu c in tr l:

( Khi R5 = (, mch cu c in tr l:

Vy khi R5 nm trong khong (0, () th in tr tng ng nm trong khong (Ro, R() Nu mch cu cn bng th vi mi gi tr R5 u c RT = R0 = R(( Nhn xt chung. Trn y l 5 phng php gii bi ton mch cu tng qut. Mi bi tp v mch cu u c th s dng mt trong 5 phng php ny gii. Tuy nhin vi hc sinh lp 9 nn s dng phng php lp h phng trnh vi n s l dng in (Hoc n s l hiu in th), th li gii bao gi cng ngn gn, d hiu v lgc hn. cho hc sinh c th hiu su sc cc tnh cht ca mch cu in tr, cng nh vic rn luyn k nng gii cc bi tp in mt chiu, th nht thit gio vin phi hng dn cc em hiu v vn dng tt c 5 phng phng php trn. Cc phng php khng ch phc v cho vic n thi hc sinh gii vt l lp 9 m c chng trnh Vt L lp 11 v n thi i hc cng gp rt nhiu bi tp phi p dng cc phng php ny m gii c.4. bi ton cu dy Mch cu dy l mch in c dng nh hnh v H4.1. Trong hai in tr R3 v R4 c gi tr thay i khi con chy C dch chuyn dc theo chiu di ca bin tr (R3 = RAC; R4 = RCB). Mch cu dy c ng dng o

in tr ca 1 vt dn. cc bi tp v mch cu dy rt a dng; phc tp v ph bin trong chng trnh Vt l nng cao lp 9 v lp 11.Vy s dng mch cu dy o in tr

nh th no? V phng php gii bi tp v mch cu dy nh th no? Phng php o in tr ca vt dn bng mch dy cu

Bi ton 4: o gi tr ca in tr Rx ngi ta dng mt in tr mu Ro,mt bin tr ACB c in tr phn b u theo chiu di, v mt in k nhy G, mc vo mch nh hnh v H4.2. Di chuyn con chy C ca bin tr n khi in k G ch s 0 o l1 ; l2 ta c kt qu: hy gii thch php o ny ? Li gii.

Trn s mch in, con chy C chia bin tr (AB) thnh hai phn.

( on AC c chiu di l1, in tr l R1 ( on CB c chiu di l2, in tr l R2 in k cho bit khi no c dng in chy qua on dy CD.

Nu in k ch s 0, th mch cu cn bng, khi in th im C bng in th im D.

Do : VA VD = VA VC Hay UAn = UAC R0I0 = R4 I1

Ta c: (1) (Vi I0, I1 ln lt l dng in qua R0 v R4) ( Tng t:

( T (1) v (2) ta c: (3)

V on dy AB l ng cht, c tit din u nn in tr tng phn c tnh theo cng thc.

Thay (4) vo (3) ta c kt qu:

( Ch .o in tr ca vt dn bng phng php trn cho kt qu c chnh xc rt cao v n gin nn c ng dng rng ri trong phng th nghim

Cc bi ton thng gp v mch dy cu.Bi ton 5Cho mch in nh hnh v H4.3. in tr ca am pe k v dy ni khng ng k, in tr ton phn ca bin tr .

a. Tm v tr uc con chy C khi bit s ch ca ampek (IA) ?b. Bit v tr con chy C, tm s ch ca ampe k ?

Phng php

Cc in tr trong mch in dc mc nh sau: (R1((RAC) nt (R2 (( RCB)

a. t x = RAC (0< x< R)

( Trng hp 1: Nu bi ton cho bit s ch ca ampe k IA = 0

Th mch cu cn bng, lc ta c iu kin cn bng.

Gii phng trnh (1) ta s tm c: RAC = x

( Trng hp 2: Am pe k ch gi tr IA ( 0

Vit phng trnh dng in cho hai nt C v D. Ri p dng nh lut m chuyn hai phng trnh v dng c n so l U1 v x.

( Nt C cho bit:

( Nt D cho bit:

(Trong cc gi tr U, Ia, R, R1, R2 u bi cho trc ) Xt chiu dng in qua ampe k (nu u bi khng cho trc), gii phng trnh (3) tm gi tr U1, ri thay vo phng trnh (2) tm x. T gi tr ca x ta tm c v tr tng ng con chy C.

b. V u bi cho bit v tr con chy C, nn ta xc nh c in tr RAC v RCB. Mch in: (R// RAC ) nt (R2 //RCB)

p dng nh lut m ta d dng tm c I1v I2. Suy ra s ch ca Ampe k: IA = (I1 - I2 ( Bi tp p dng.

Cho mch in nh hnh v H4.4. Bit U = 7V khng i.R1 = 3(, R2= 6(. Bin tr ACB l mt dy dn c in tr sut l (= 4.106 (( m), chiu di l = AB = 1,5m, tit din u: S = 1mm2a. Tnh in tr ton phn ca bin trb. Xc nh v tr con chy C s ch ca ampe k bng 0c. Con chy C v tr m AC = 2CB, hi lc ampe k ch bao nhiu?

d. Xc nh v tr con chy C ampe k ch (A)Li gii.

a. in tr ton phn ca bin tr: (()

b. Ampe k ch s 0 th mch cu cn bng, khi :

t x = RAC RCB = 6 x

. Suy ra x = 2 (()

Vi RAC = x = 2( th con chy C cch A mt on bng:

Vy khi con chy C cch A mt on bng 0,5m th ampe k ch s 0

c. Khi con chy v tr m AC = 2CB, ta d dng tnh c RAC = 4 (()

Cn RCB = 2 ((). VT RA = 0 Mch in (R1 //RAC ) nt (R2 //RCB)

in tr tng ng ca mch: (() Cng dng in trong mch chnh:

Suy ra:

V: I1 > I2, suy ra s ch ca ampe k l:

Vy khi con chy C v tr m AC = 2CB th ampe k ch 0,7 (A)

d. Tm v tr con chy C ampe k ch (A)

V: RA = 0 => mch in (R1// RAC) nt (R2 // RCB) Suy ra: Ux = U1 ( Phng trnh dng in ti nt C:

( Phng trnh dng in ti nt D:

( Trng hp 1:

Ampe k ch IA = (A) D n C

T phng trnh (2) ta tm c U1 = 3 (V)

Thay U1 = 3 (V) vo phng trnh (1) ta tm c x = 3 (() Vi RAC = x = 3( ta tm c v tr ca con chy C cch A mt on bng AC = 75 (m)

( Trng hp 2:

Ampe k ch IA = (A) chiu t C n D

T phng trnh (2) ta tm c U1

Thay U1 vo phng trnh (1) ta tm c x ( 1,16 (() Vi RAC = x = 1,16 ( , ta tm c v tr ca con chy C cch A mt on bng AC ( 29 (cm)

V ti cc v tr m con chy C cch A mt on bng 75 (cm) hoc 29 (cm) th am pe k ch .

Bi ton 6:

Cho mch in nh hnh v H4.3. Hiu in th hai u on mch l U khng i. Bin tr c in ton phn l R, vn k c in tr rt ln

a. Tm v tr con chy C, khi bit s ch ca vn kb. Bit v tr con chy C, tm s ch ca vn k

Phng php. V vn k c in tr rt ln nn mch in c dng (R1 nt R2) // RABa. Tm v tr con chy C

Vi mi v tr ca C, ta lun tm c:

Xt hai trng hp: UAC = U1 + UV v UAC = U1 - UV

Mi trng hp ta lun c:

T gi tr ca RAC ta tm c v tr tng ng ca con chy C.

b. Bit v tr con chy C, ta d dng tm c RAC v RCB v cng d dng tnh c U1 v UAC. T ch s ca vn k:

Bi tp p dng.

Cho mch in nh hnh v H4.6. Bit V = 9V khng i, R1 = 3(, R2 = 6(.

Bin tr ACB c in tr ton phn l R = 18(, vn k l l tng.

a. Xc nh v tr con chy C vn k ch s 0

b. Xc nh v tr con chy C vn k ch s 1vn

c. Khi RAC = 10( th vn k ch bao nhiu vn ?

Li gii

V vn k l l tng nn mch in c dng: (R1 nt R2) // RABa. vn k ch s 0, th mch cu phi cn bng, khi : RAC = 6 (()

b. Xc nh v tr con chy C, Uv = 1(V) Vi mi v tr ca con chy C, ta lun c:

( Trng hp 1: Vn k ch: UV = U1 UAC = 1 (V)

Suy ra: UAC = U1 UV = 3 1 = 2 (V) RAC = (()

( Trng hp 2:Vn k ch UV = UAC U1 = 1 (V)

Suy ra: UAC = U1 + UV = 3 + 1 = 4 (V) = 8 (()

Vy ti v tr m RAC = 4 (() hoc RAC = 8 (() th vn k ch 1 (V)

c. Tm s ch vn k, khi RAC = 10 (()

Khi RAC = 10(() RCB = 18 10 = 8 (() UAC = IAC . RAC = 0,5 .10 = 5 (V)

Suy ra s ch ca vn k l: UV = UAC U1 = 5 3 = 2 (V)V khi RAC = 10( th vn k ch 2(V)

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