Du 14: Cho mgt may bieh the co hi^u suat 80% Cugn so cap CO

Một phần của tài liệu Bí quyết ôn luyện thi đại học đạt điểm tối đa vật lý tập 1 lê văn vinh part 2 (Trang 89)

vong, cugn t h u ca'p c6 300 vong. Hai dau cugn t h u ca'p noi v o i mgt cugn

day CO d i ^ n tra hoat dgng lOOQ, dg t y cam 1/n H . so cong suat mach

so cap bang 1. Hai dau cugn so cap dugc dat a h i ^ u d i ^ n the xoay chieu

CO U i = lOOV, tan so 50Hz. Tinh cong suat mach t h u cap va cuong dg h i f u d y n g mach so cap?

A. lOOW v a l , 5 A B. 150W va 2,5A

C. 200W va 2,5* D. 250W va 2,0A

'Phdn tich vd hu&ng ddn gidi

H l= ^ = > U 2= 2 0 0 V R2 = 100Q, Zi.2 = 100Q => Z2 = 100V2Q => h = ^ = V2 A P2 = R2I2 = 200W „ = | ^ P , = | . U , , , ^ I , = J ^ = ^ = 2 , 5 < A ) H U i 0,8.100 C h q n d a p a n C

Vi du 1 5 : M a y bie'n ap c6 so vong day 6 cuon so cap la N i = 400 vong, t h u cap la N 2 = 100 vong. Dien tro ciia cuon so cap r i = 4 Q , di^n tro 6 cugn t h i i cap r2 = 1 Q . Dien tro mac vao cupn t h u cap R = 10 . Xem mach t u la khep k i n va bo qua hao p h i . Dat vao hai cuon so cap dien ap xoay chieu

CO gia t r i hieu d u n g U i = 360V. Xac dinh di^n ap hi^u d u n g U 2 tai hai dau cugn t h i i cap va hi^u suat cua may bien the.

A. 84V; 93,3% B. 80V; 80% C. lOOV; 93,3% D . 84V; 80%

'Phdn tich vd hicorng ddn gidi

400 <=> • N 2 E 2 l2(R + r2) 100 12(10 + 1) ^ , . , N , I2 400 I2 , „ T a Uii c6: — i - = <=> = - i - => L = 41 N 2 I i 100 I i 2 3 6 0 - L . 4 ^ 3 6 0 - 4 L , . ' <=>4 = — — — ^ ( 1 ) I I L (2) T u (1) va (2) ta c6: 44I2 = 360 - 41^ o 44.4Ij = 360 - 41^ U 2 = U R = R I 2 = 1 0 . 8 , 4 = 8 4 V 8 4 . 8 , 4 I ^ = 2 , 1 A I2 = 8 , 4 A H = H ^ = 93,3% 360.2,1 C h p n d a p a n A m B A I T A P VAN DUNG:

C a u 1: Cupn so cap va t h u cap cua mot may bie'n ap c6 so vong Ian l u p t la 600 v o n g va 120 vong. Dat vao 2 dau cupn so cap m o t dien ap xoay chieu

CO gia trj hieu d u n g 380V. N o i 2 dau cupn t h u cap v o i dien tro c6 R = 100 Q. •

T i n h cuong dp dong di^n chay trong cupn so ca'p( bo qua m p i hao p h i 6 may bie'n ap)

A.0,152A B.0,76A C. 2A D . 6A

i'hdn tich vd huang ddn gidi

Theo bai ra ta c6 hinh ve n h u sau: Vi bo qua hao phi 6 may bie'n ap

N

Nen ta c6: U 2 = U j — "2 _ = 76 V Cuong dp dong dien chay trong cupn t h u cap: I2 = — = 0,76 A

R

C u o n g dp dong dien chay trong cupn so cap: I j = ^ ^ = 0,152 A

Chpn d a p a n A ' '

Cau 2: Cuon so cap cua mot may bien ap dupe noi voi mang dien xoay chieu

CO dien ap 380V. Cupn t h u cap c6 dong dien 1,5A chay qua va c6 dien ap giua hai dau day la 120V. Bie't so vong day ciia cupn t h u cap la 30. T i m so vong day cua cupn so cap va cuong dp dong d i f n chay qua no. Bo qua hao phi dien nang trong may.

A. N i =30 Ij = 4 , 7 5 A B. Ij = 4 , 7 5 A B. =95 = 4 A C. N j =95 D. N j =60 I i =4,75 A = 4 , 7 5 A

^hdn tich vd hu&ng ddn gidi

V i bo qua su hao p h i nang l u p n g nen hi?u sua't la 100%, may bien ap la l i tuong. Ta co: — - - — - - —

U , N 2 I2

Thay cac dai lupne da bie't: = — = — 120 30 1,5

N , =95 I i = 4 , 7 5 A I i = 4 , 7 5 A

C h p n dap a n C

-au 3: M o t may bie'n the'co so'vong cupn so cap gap 10 Ian cupn t h u cap. Hai dau cupn so cap mac vao nguon xoay chieu c6 di^n ap hieu dung U i = 220V.

Dien tro ciia cupn so ca'p la n w 0 va cupn t h u cap r2 w 2Q. Mach t u khep kin; bo qua hao phi do dong Fuco va buc xa. Khi hai dau cupn t h u cap mac voi dien

tro R = 20Q thi di?n ap hi^u dung giCia hai dau cuon t h u cap bang bao nhieu?

A. 18V B. 22V C. 20V D . 24V.

'Phdn tich vd hu&ng ddn gidi

Dien ap hieu d u n g giira hai dau cupn t h u cap de ha: U , = = 22V = E,

2 10 2

C u o n g dp dong di?n qua cupn t h u cap k h i noi v o i d i f n t r o R:

2 ~ R + r 20 + 2

K h i do dien ap hi?u d u n g giira hai dau cuon t h u cap c6 tai: U ' , = R l 2 =20.1 = 20V

C h g n dap an C

C a u 4 : H i ^ u dien the'duoc dua vao cuon so cap cua m p t may bien ap c6 gia trj hi^u d u n g la 220V. So v o n g day ciia cupn so cap va t h u cap t u o n g u n g la 1100 v o n g va SOvong. Mach t h u cap g o m m o t dien tro thuan 8 Q , mot cuon cam CO cam khang 2 Q va m o t tu di|n mac noi tie'p. Bie't dong di^n chay qua cuon so cap bang 0,032A, bo qua hao p h i cua may bien ap, dp l^ch pha giiia hieu di^n t h e v a cuong do d o n g di^n trong mach t h u cap la

A. — B. — hoac — C. — hoac — D . —

2 4 • 4 6 • 6 3

^hdn tich v>d huong ddn gidi

A p d u n g : ^ = i^ ^ U , = H l : ^ = ^ = 10V

l y , ^ 220.0,032 ^ 7 2 ^

10 2 U , 10

I .

Tong tro mach t h u cap: Zz = — - = —j^ = loVI Q 2

D g l^ch pha giiia hieu dien theva cuong do dong di^n trong mach t h u cap la

R + r 10 x/2 71

cos (p = = 7=^ = =>(p = ± —

10V2 2 ^ 4

C h p n dap an B

C a u 5 : Co hai may bien ap If t u o n g (bo qua m o i hao phi) cupn so cap c6 cung so vong day n h u n g cuon t h u cap c6 so v o n g day khac nhau. K h i dat di?n ap xoay chieu c6 gia trj hieu d u n g U khong d o i vao hai dau cupn t h u cap ctia may t h u nhat thi ti so giiia d i | n ap hieu d u n g 6 hai dau cupn t h u cap de ho va cupn so cap cua may do la 1,5. K h i dat d i ^ n ap xoay chieu noi tren vao hai dau cupn so cap cua may t h u hai t h i t i so'do la 2. K h i cung thay doi so vong day cua cupn t h u cap cua moi may 50 v o n g day roi lap lai thi nghiem thi ti so'dien ap noi tren cua hai may la bang nhau. So'vong day cua cupn sO cap cua m o i may la

A. 200 v o n g B. 100 v o n g C. 150 vong D . 250 v o n g

4 0 8

^hdn tich vd huong ddn gidi

Gpi so vong day cupn so cap la N , cupn t h u cap la N i va N2

N N i = 1,5N N2 = 2 N Theo bai ra ta c6 : = ^ = 1,5 N U N

De hai ti so tren bang nhau ta phai tang N i va giam N2

^ N , + 5 0 N , - 5 0 Do do —* - ^ N N , N N r i > l , 5 N + 50 = 2 N - 5 0 C h p n dap an A =5 N1+5O = N2 N = 200 vong. •50 N

C a u 6 : M o t may ha ap, cupn day so cap c6 N i = 440 vong va di^n tro n = 7,20,

• I cupn t h u ca'p c6 N2 = 254 vong va di#n tro r2 = 2,4Q. Mac vao cupn t h u ca'p mot dien tro R = 20Q, coi mach la khep k i n va hao p h i do dong Fu-c6 khong dang ke. Dat vao hai dau cupn so cap mot dien ap xoay chieu c6 gia trj hi^u d u n g Ui = 220V. Xac d j n h dien ap hieu d u n g 6 hai dau tai R. T i n h hi^u sua't cua may bien ap do.

A. 102,4V; 80,64% B. 80V; 82% C. 100V;94,2% D . 84V; 86%

'Phdn tich m huong ddn gidi

Cong thiic may bien the: — - ^ - H i N

i-ij 440 2 2 0 - I , .7,2

2 E 2 l 2(R + r2) 254 12(20 + 2,4) Mach khep kin va hao phi do dong Fu-c6 khong dang ke nen Pj =

(2) ^ ^ 440 I2 ^ ^ 440 I2 " E, N ~ 254 ' I , T u (1) va (2) ta c6 h^: 985612 +1828,8I1 =254.220 254I2 -4401, = 0 • I , =2,9563 I2 =5,121 U R = R l 2 = 1 0 2 , 4 2 V H = 2 2. = 80,64% U i l , C h p n dap an A

C S u 7 : M o t may bien theco ti so vong ^ = 5, h i f u sua't 96% nhan cong suat

hi

N2

]0(kW) 6 cupn so cap va hieu t h e o hai dau so cap la l ( k V ) , he so cong sua't cua mach t h u cap la 0,8, thi cuong dp dong dien chay trong cupn t h u cap la: A. 30(A) B. 40(A) C. 50(A) D . 60(A)

^hdn tick v>d huang dan gidi

Ta CO H = ^ = 0,96 => P2 = 0,96P. = 0,96.10 = 9,6(KW) = 9600(W) Pi

Theo cong thuc: ^ = Hi Suy ra: U , = ^ . U , = = = 2OOV

T u do: P2 = U2I2 cos (p =^ I , = — = = 60A ^ Ujcoscp 200.0,8 C h p n dap an D

C a u 8 : Dat vao hai dau cuon so cap ciia mot may bien ap l i t u p n g mot dien ap xoay chieu c6 gia tri khong doi thl hieu dien the hieu d u n g giiia hai dau mach t h u cap k h i de ho la lOOV. 6 cuon so cap, khi ta giam bat d i n vong day thi hieu di^n the hieu dung giOa hai dau mach t h u cap khi de ho la U ; neu tang n vong day 6 cuon so cap thi hieu dien the'hi^u d u n g giiia hai dau mach t h u cap khi de ho la U/2. Gia trj ciia U la:

A. 150V. B.200V C. lOOV D. 50V

^hdn tich v>d huang dan gidi

Gpi dien ap hieu d u n g dat vao cupn so cap la U i , so vong day cupn so cap va t h u cap la N i va N2 Theo bai ra ta c6: i i . i i . ,1, p, 2 U , ^ I V ^ 100 N 2 U N 2 U N 2 Lay (4): (5) - = ^ ' ^ ^ ^ i ^ N j + n = 2 N i - 2n = 3n 1 N j - n T u ( 4 ) => U = 1 0 0 . - ^ = 1 0 0 . ^ ^ = 150V N j - n 3 n - n C h p n dap an A

C a u 9 : Cupn so cap ciia mpt may bien the c6 Ni = 1000 vong, cupn t h u cap co

N2 = 2000 vong. H i e u dien the hieu dung ciia cupn so cap la Ui = l l O V va ciia cupn t h u cap khi de ho la U 2 = 216 V. Ti so'giira cam khang ciia cupn sO cap va di?n tro thuan ciia cupn nay la:

A. 0,19 B. 5,2 C. 0,1 D. 4,2 410 410

C t y TNHH MTV DWH Khang Viff

'Ph.dn tich huang dan gidi

Khi cupn t h u cap de ho thi: E2 = U2 = 216V . Ap d u n g cong thiic may bien the:

E . N . _ N , _ U,

^ 2 =I:ilE^ =-io8V N

^1 '^1 " ' 2 ^ Ei CO vai tro la di^n ap hai dau cuon cam E, = U L

Vithe'taco: =U^^ + U f — TT2 _ T T2 i T 2 _ i i n 2

Ket qua can t i m la : = =

r, U = U | - U f =110^ -108^ 108 = 20,9 V = 5,2. ., 20,88 Chon dap an B

C a u 1 0 : M o t may bien ap li tucVng c6 hieu sua't bang 1 dupe noi vao nguon dien xoay chieu c6 dien ap hieu dung 5V. Biet so vong day ciia cupn so cap va t h u cap la 100 vong va 150 vong. Do cupn so cap c6 10 vong bj quan

ngupc nen dien ap thu dupe a cupn t h u cap la:

A. 7,5V. B. 9,375 V. C. 8,33V. D. 7,783V.

0idn tich v>d huang dan gidi

A p d u n g cong thiic tinh so vong ngupc doi voi cupn so cap (n, = n = 10; = 0 ) , t a c 6 : U , _ N , - 2 n N - N 2 U 1 N , - 2 n 150.5 100-2.10 = 9,375 V 2 '^2 Chpn dap an B

C a u 1 1 : (Trich de t h i tuyen sinh vao dai hpc nam 2013): Dat vao hai dau

cupn so cap ciia may bien ap M i mpt dien ap xoay chieu co gia trj hieu d u n g 200V. Khi noi hai dau cupn so cap ciia may bien ap M2 vao hai dau cupn t h u cap ciia M i thi dien ap hieu dung 6 hai dau cupn t h u cap ciia M2 de ho b3ng 12,5 V. K h i noi hai dau cupn thii cap ciia M2 voi hai dau cupn t h i i cap ciia M I thi dien ap hieu d u n g 6 hai dau cupn so cap ciia M2 de ho b^ng 50 V. Bo qua mpi hao phi. M i co ti so giOa so vong day cupn so cap va so vong day cupn t h i i cap b5ng

A. 6. B. 15. C. 8.

<Phdn tich pd huang dan gidi

Theo bai ra:

D. 4.

May bien ap M 2 dau Ian 1:

N ' , U \5 May bien ap M2 dau Ian 2: i ^ = -H2__H2

50

(1)

Bi quyei on luyftt thi d^i hQC dat diem tS'i da Vgt li, t^p 1 - Le Van Vinh Tu(l) v a ( 2 ) t a c 6 : 1 = U 12,5.50 . U2 = 25V = U j N . U j 200 - M a y b i e h a p M i : 77T- = T 7 r " N ' l U ' l 25; = 8 Chgn dap an C

DANG 3: BAI TOAN TRUYEN TAI OIEN NANG DI XA PHl/QNG PHAP

B

1) Cong suat may phat: Pphat = Uph.itI.cos(p

2) Cong suat hao phi trong qua trinh truyen d i f n nang: AP= / R = A U . I

U ^ c o s ^

Vol. P la cong suat truyen di a noi cung cap

U la dien ap 6 noi cung cap cos(p la h^ so cong suat ciia day tai di?n I la cuong dp dong di^n chay tren duong day

R = p—' la di^n tro tong cpng cua day tai di?n {luu y: dan di?n hang2 day)

s

AU = IR la dp giam di^n ap tren duang day tai di^n

6 hinh ve tren ta c6: AU = IR = - Ug

3) Bi^n phap giam hao phi: c6 2 each Giam R: Cach nay rat ton kem chi phi

Tang U : BSng each dung may bien the^ each nay c6 h i | u qua Ta c6: Hi?u suat truyen tai di^n nang: H = ^ ^^^.100%

m v i DU MAU:

Vi du 1: Ns^uoi ta can truyen mpt cong suat 5MW t u nha may di?n de'n

noi tieu thu each nhau 5km. Hieu di^n the hi#u dung cupn t h u cap ciia may tang ap la U =100kV. Muon dp giam the'tren duong day khong qua 1%U th'i tie't di^n cua duong day dan phai thoa dieu k i f n nao? Biet digj}

Cty TNHH MTV DWH Khang Vi?t

tro suat cua day tai dien la 1,7.10 ^Qm.

A. 5,8(mm2)< S C. 8,5(mm2)<S C. 8,5(mm2)<S

B. 5,8(mm2)< S ^ 8,5 (mm^) D. 8,5(mm2) > S • ' D. 8,5(mm2) > S • '

i'hdn tick v>d huang ddn gidi

Chung ta phai hieu rang: truyen tai phai can hai day vi the chieu dai day dan: 1 = 2.5km = 10000m

•.lilt:

1000 "\:.> Theo bai thi: AU = IR < 1%U = I k V =1000V R <

I M a P = U I ^ I = : ^ = ^ : l i - = 50A

U 100.10^

50 S 20

1 7 1 0 " 10000

Thay so: S > ' ' = 8,5.10-*(m2) = 8,5(mm2). Hay S > 8,5(mm2) 20 Chgn dap an C

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