Bi tp vt l lp 9Bi 1 Mt m un nc bng in c 3 dy l xo, mi ci c in tr R=120 , c mc song song vi nhau. m c mc ni tip vi in tr r=50 v c mc vo ngun in. Hi thi gian cn thit un m ng y nc n khi si s thay i nh th no khi mt trong ba l xo b t? Bi1: *Lc 3 l xo mc song song: in tr tng ng ca m: R1 =R = 40 () 3U R1 + r

Dng in chy trong mch:I1 =

Thi gian t1 cn thit un m nc n khi si: Q = R1.I2.t1 t1 = Q = R1I 2 Q U R1 R +r 1 2

hay t1 =

Q( R1 + r ) 2 (1) U 2 R1

*Lc 2 l xo mc song song: (Tng t trn ta c )R = 60 () 2 U I2 = R2 + r

R2 =

Q( R2 + r ) 2 t2 = (2) U 2 + R2t1 t1 R2 ( R1 + r ) 2 60(40 + 50) 2 243 = = = 1 *Vy t1 Lp t s ta c: t2 t 2 R1 ( R2 + r ) 2 40(60 + 50) 2 242

t2

Bi 2 trang tr cho mt quy hng, ngi ta dng cc bng n 6V-9W mc ni tip vo mch in c hiu in th U=240V chng sng bnh thng. Nu c mt bng b chy, ngi ta ni tt on mch c bng li th cng sut tiu th ca mi bng tng hay gim i bao nhiu phn trm?

Bi2:

1

2 Ud = 4() in tr ca mi bng: R= Pd

S bng n cn dng chng sng bnh thng: n=

U = 40 (bng) Ud

Nu c mt bng b chy th in tr tng cng ca cc bng cn li l: R = 39R = 156 ( ) Dng in qua mi n by gi: I=U 240 = = 1,54 ( A) R 156

Cng sut tiu th mi bng by gi l: P = I2.R = 9,49 (W) Cng sut mi bng tng ln so vi trc: Pm - P = 9,49 - 9 = 0,49 (W) Ngha l tng ln so vi trcl:0,49.100 .% 5,4% 9

Bi 3:(2,5im) Cho mch in nh hnh v RV U1=180V ; R1=2000 ; R2=3000 . V a) Khi mc vn k c in tr Rv song song vi R1, vn k ch U1 = 60V.Hy xc R1 nh cng dng in qua cc in tr R1 A v R2 . b)Nu mc vn k song song vi in tr R2,von ke chi bao nhieu?

R2 B

C

+ U

Bi 4: (2,5im) Dng ngun in c hiu in th khng i U0 = 32V thp sng mt b bng n cng loi (2,5V-1,25W).Dy ni trong b n c in tr khng ng k. Dy ni t b bng n n ngun in c in tr l R=1 a) Tm cng sut ti a m b bng c th tiu th. b) Tm cch ghp bng chng sng bnh thng.Bi 4: a)Gi I l dng in qua R, cng sut ca b n l : P = U.I RI2 = 32.I I2 hay : I2 32I + P = 0 2

M

n

N

A

B

Hm s trn c cc i khi P = 256W Vy cng sut ln nht ca b n l Pmax = 256W b)Gi m l s dy n, n l s n trong mt dy: *Gii theo cng sut : Khi cc n sng bnh thng : I d = 0,5( A) v I = m . I d = 0,5m T : U0 . I = RI2 + 1,25m.n Hay 32. 0,5m = 1 (0,5)2 = 1,25m.n 64 = m + 5n ; m, n nguyn dng (1) Gii phng trnh (1) ta c 12 nghim sau : n m 1 2 3 4 5 6 7 8 9 19 10 14 11 9 12 4

59 54 49 44 39 34 29 24 *Gii theo phng trnh th :U0 =UAB + IR vi : UAB = 2,5n ; IR = 0,5m.1 = 0,5m Ta c phng trnh (1) bit 64 = 5n + m *Gii theo phng trnh dng in : nR 5n RAB = d = V I = m. I d = 0,5m m m U0 32 32 m = = 5n m + 5n Mt khc : I = R + R AB 1+ m Hay : 0,5m =32 m m + 5n

64 = 5n + m

Cu5: Cho 2 bng n 1 (12V - 9W) v 2 (6V - 3W). a. C th mc ni tip 2 bng n ny vo hiu in th U = 18V chng sng bnh thng c khng? V sao? - U + b. Mc 2 bng n ny cng vi 1 bin tr o o c con chy vo hiu in th c (U = 18V) nh hnh v th phi iu chnh bin tr c in tr l bao nhiu 2 n sng bnh thng? 2 c. By gi tho bin tr ra v thay vo 1 l 1 in tr R sao cho cng sut tiu th trn n 1 gp 3 ln cng sut tiu th trn n 2. Tnh R? (Bit hiu in th ngun vn khng i) Cu 5: (3,0 im) a. Cng dng in nh mc qua mi n: Rb Pm1 = Um1.Im1 Pdm 1 9 => Im1 = = = 0,75(A) U dm 1 1 2 Pdm 2 3 Im2 = = = 0,5(A) U dm 2 6 Ta thy Im1 Im2 nn khng th mc ni tip 3

2 n sng bnh thng. b. 2 n sng bnh thng th: U1 = Um1 = 12V; I1 = Im1 = 0,75A v U2 = Um2 = 6V; I2 = Im2 = 0,5A Do n 2 // Rb => U2 = Ub = 6V Cng dng in qua bin tr: I1 = I2 + Ib => Ib = I1 I2 = 0,75 0,5 = 0,25(A). Ub 6 Gi tr in tr ca bin tr lc bng: Rb = = 0,25 = 24 ( ) Ib c. Theo ra ta c: P1 = 3P2 I12.R1 = 3I22.R2I 1 I 2

U dm 2 .Pdm1 3R2 I1 9 3 6 2 .9 = = 3. 2 = 3. 2 = => = 2I1 = 3I2 (1) R1 I2 4 2 12 .3 U dm1 .Pdm 22

2

M I1 = I2 + IR nn (1) 2(I2 + IR) = 3I2 2I2 + 2IR = 3I2 => I2 = 2IR (2) Do n 2 // R nn U2 = UR I2.R2 = IR.R U 2 dm 2 62 Thay (2) vo ta c 2.IR.R2 = IR.R => R = 2R2 = 2. = 2. = 24 ( ) 3 Pdm 2 Cu 6: Hai in tr R1 v R2 c mc vo mt hiu in th khng i bng cch ghp song song vi nhau hoc ghp ni tip vi nhau. Gi Pss l cng sut tiu th ca on mch Pss 4. khi ghp song song, Pnt l cng sut tiu th khi ghp ni tip. Chng minh : Pnt Cho bit: R1 + R2 2 R1 .R2 Cu 6: (2,0 im) U2 P = R1 R2 . - Cng sut tiu th ca on mch khi hai in tr mc song song: ss R1 + R2 - Cng sut tiu th ca on mch khi hai in tr mc ni tip: Pnt = Pss ( R1 + R2 )2 = - Lp t s: ; Pnt R1 R2 - Do : R1 + R2 2 R1 R2 => (R1 + R2)2 4 ( R1 .R2 )2 , nn ta c:2 Pss 4( R1 R2 ) Pnt R1 R2

U2 . R1 + R2

Pss 4 Pnt

Bi 7 : Vt AB t cch thu knh hi t mt on 30cm.nh A1B1 l nh tht.Di vt n v tr khc,nh ca vt l nh o cch thu knh 20cm.Hai nh c cng ln. Tnh tiu c ca thu knh. 4

Bi 7 :

2 im B2 B' I F O A1

B A A2 , F

A

B1 * Vt v tr 1 : v nh A1B1 ca vt l nh tht ,chng t vt AB s c t ngoi khong tiu c . t : OA=d1=30cm (khong cch t vt v tr (1) n thu knh) OA1=d1 OF=OF = f (khong cch nh ca vt v tr (1) n thu knh) (tiu c) nn: nn:A1B1 OA1 d1' = = AB OA d1A1B1 F ' A1 OA1 F 'O d1' f = ' = = OI FO F 'O f

Ta c : OAB OA1B1 FOI FA1B1

(1) (2)

M OI = AB ,do t (1) & (2) ta c:

d '1 d1' f = d1 ff =

d1.d1' d1 + d1'

(a)

* Vt di n v tr 2 : v nh cho l nh o nn vt phi c di n gn thu knh v nm trong khong tiu c f. Gi s vt di i 1 on AA = a t : OA = d2 = 30-a (khong cch vt t v tr 2 n thu knh)' A2 B2 OA2 d 2 = = A' B ' OA ' d 2' A2 B2 F'A OA2 + F 'O d 2 + f = ' 2 = = OI FO F 'O f

OA2= d2 = 20cm (khong cch nh ca vt v tr 2 n thu knh) Ta c : OAB OA2B2 FOI FA2B2 nn: nn: (3) (4)

M OI = AB ,do t (3) & (4) ta c:

' d '2 d2 + f = d2 f

5

' d 2 .d 2 f = ' d2 d2

(b)

V tiu c ca thu knh khng thay i nn t biu thc (a) ,(b) Ta c :

d1.d1' d .d ' = '2 2 d1 + d1' d2 d2A1 B1 AB = 2 2 AB A' B '' d2 20 600 = 30. = cm d2 30 a 30 a

(5)

Mt khc do 2 nh c ln nh nhau ,nn : T (1) ,(2) c :d1' d' = 2 d1 d 2' d1 = d1.

Thay cc gi tr d1 , d1 ,d2 , d2 vo biu thc (5) v bin i ta c phng trnh : a2 110a + 1800 = 0 = (-110)2 4.1800 = 4900= 702

a1,2 =

( 1 1 ) 7 02 a1 = 9 0c m 0 = a2 = 2 0c m 2(loi nghim a = 90cm)

v a = AA = 90 cm > OA =d1 = 30 cm

Vy vt di i mt on a =20cm vo trong khong tiu c ca thu knh. OA = d2 = 30 a = 30 20 = 10 cm Thay d2 = 10 cm , d2 = 20 cm vo biu thc (b)' d 2 .d 2 10 .20 = 20 cm f = ' = 20 10 d2 d2

Cu 8: Cho mch in nh hnh v: U = 12V; R1 = 6 ; R2 = 6 ; R3 = 12 ; R4 = 6 a. Tnh cng dng in qua mi in tr v hiu in th gia hai u mi in tr. b. Ni M v N bng mt vn k (c in tr rt ln) th vn k ch bao nhiu? Cc dng ca vn k phi c mc vi im no? c. Ni M v N bng mt ampe k (c in tr khng ng k) th ampe k ch bao nhiu? Cu8: 6

R1

M

R3

A R2 N R4

B

+

U

-

a. Tnh c: I1 = I3 =

2 A; I2 = I4 = 1A; U1 = 4V; U3 = 8V; U2 = U4 = 6V 3

b. UAM = UAN + UNM => UNM = UAM UAN = 4 6 = -2V hay UMN = 2V Vy vn k ch 2V v cc dng ca vn k c mc vo im M. c. Lp lun v tnh c: I1 = 0,85V; I3 = 0,58A Do I1>I3 nn dng I1 n M mt phn r qua ampe k (dng Ia) mt phn qua R3 (dng I3), ta c Ia = I1 I3 = 0,85 0,58 = 0,27A Vy ampe k ch 0,27A.

Cu 9: (1,5 im). Cho hai gng phng G1 v G2 vung gc vi nhau. t mt im sng S v im sng M trc hai gng sao cho SM song song vi gng G2 (hnh v bn).

G1 S M

O

G2

a) Hy v ng i ca tia sng t S ti gng G1 phn x ti gng G2 ri qua M. Gii thch cch v. b) Nu S v hai gng c v tr c nh th im M phi c v tr th no c th v c tia sng nh cu a. Cu 9: G1 S1 S M x I O S2 K G2 M

a) V hnh ng : V S1 l nh ca S qua G1; y S1 l im i xng ca S qua mt phng gng G1. V S2 l nh ca S1 to bi G2 ; S2 l im i xng ca S1 qua mt gng G2. V G1 vung gc vi G2 nn S2 l im xuyn tm ca S qua O

7

Nhn xt: Gi s ta v c tia sng theo yu cu ca bi ton l SIKM xut pht t S, phn x trn G1 ti I n K, tia phn x IK ti I trn G1 coi nh xut pht t nh S1. Tia phn x KM ti K trn G2 c coi nh xut pht t nh S2 . T nhn xt trn ta suy ra cch v ng truyn tia sng nh sau: - Ly S1 i xng vi S qua mt G1; - Ly M i xng vi M qua mt gng G2; - Ly S2 i xng vi S1 qua mt gng G2; - Ni MS2 ct G2 ti K; - Ni S1 vi K ct G1 ti I; - Ni SIKM ta c ng i ca tia sng cn tm b) v c tia sng nh cu a th S2M phi ct G2 ti K. Mun vy M phi nm trn on Sx. Bi 10 Cho mch in nh hnh v. Bit UAB = 10V; R1 = 2 ; Ra = 0 ; RV v cng ln ; RMN = 6 . A +M

Con chy t v tr no th ampe k ch 1A. Lc ny vn k ch bao nhiu?

R1 AD

VN

B -

Bi 10 *V in tr ca ampe k Ra = 0 nn: UAC = UAD = U1 = I1R1. = 2.1 = 2 ( V ) ( Ampe k ch dng qua R1 phn MD l x th:

(0,5)*Gi in tr

*Gii ra c x = 2 . Con chy phi t v tr chia MN thnh hai phn MD c gi tr 2 v DN c gi tr 4 . Lc ny vn k ch 8 vn ( Vn k o UDN.

2 2 I x = ;I DN = I 1 + I x = 1+ x x 2 U DN = 1+ ( 6 x) x 2 U AB = U AD + U DN = 2 + 1+ ( 6 x) = 10 x

8

Cu11: (2 im) Hai in tr R= 4 v r mc ni tip vo hai u hiu in th U=24V. Khi thay i gi tr ca r th cng sut ta nhit trn r thay i v t gi tr cc i. Tnh gi tr cc i . Bai11:Gi I cng dng in qua mch. Hiu in th hai u r: Ur = U RI = 24 4I Cng sut tiu th trn r: P = Ur.I = (24 4I) I 4I2 24I + P = 0 (1) = 242 4P V phng trnh (1) lun c nghim s nn 0 => 242 4P 0 => P 36 => Pmax = 36W

Cu12: (2,5 im) Cho mch in nh hnh v: U R0 C Rb B

Trong R0 l in tr ton phn ca bin tr, Rb l in tr ca bp in. Cho R0 = Rb , in tr ca dy ni khng ng k, hiu in th U ca ngun khng i. Con chy C nm chnh gia bin tr.Tnh hiu sut ca mch in. Coi hiu sut tiu th trn bp l c ch. Bai12:in tr RCB = ( R0.R0/2 )/ (R0 + R0/2) = R0/3 Cng dng in chy trong mch chnh: I= U/(R0/2 +R0/3) = 6U/ 5R0 Cng sut tiu th ca bp l : P= U2CB/ R0 = 4U2/25R0 Hiu sut ca mch in l : H = P/UI = ( 4U2 /25R0) : (U.6U/ 5R0) = 2/15Vy H = 13,3 %

9