1 z R^+V
BAI TAP MAU
B a l 1. C h o m a c h d i e n A B g o m R, L , C m a c n o i t i e p UAB = 40>/2 coslOOitt ( V )
R = 20 Q; C t h a y d d i ; L = — ( H ) . D i e u c h i n h C de d i e n a p h i e u d u n g h a i
71
d a u t u d i e n d a t g i a t r i cifc d a i . T i m d i e n d u n g t u d i e n v a d i e n a p h i e u d u n g h a i d a u t u d i e n l u c n a y .
Tom tat
• R - 20 Q
• U A B = 40N/2 coslOOTTt ( V ,
. L = M ( H )
T i m C v a Uc„,ax
Do Z L = Leo :
T h e v a o (*) ^ Zc
HU&ng dan gidi
A P-
R L B
D o d i e u c h i n h C de d i e n a p h i e u d u n g h a i d a u t u d i e n d a t g i d t r i ciTc d a i t h i
0,2 .lOOTt = 20 Q R ^ + Z ?
^ Zn = 2 0 ' + 2 0 '
20 = 40 O
C(o Z(,.a) 4 0 . 1 0 0 7 1 471
* Di^n dp hi§u dung hai dau ciia tu di§n:
= I.Zr Vdi
Vay:
1= 40
ZAB - Z C ) ' + (20 - 4 0 ) ' 4 0
20N/2 = V2 (A) Up =40V2(V)
Bai 2. Cho mach dien nhu' hinh:
A R L
Cdc von ke c6 dien trd rat IcJn 10 ' R = 40 Q; C =
271 (F), L thay doi UAB = 80V2cos(1007it) (V)
Tim he so tif cam ciia cupn day di:
a) Vi chi gia tri ciTc dai.
b) V2 chi gia tri cUc dai.
B
Tom tat
• R = 40 Q 10'^
. C = — (F)
271
UAB = 8OV2 cosdOOTtt) (V) a) Vi chi gia t r i cUc dai.
b) V2 chi gia t r i cifc dai.
Hitdng dan gidi Ceo = 20 Q
a) Do L thay doi de so chi Vj ciic dai nen trong mach xay ra hien tifgng cong hirorng. Luc nay ta CO
Z L = Zc -> Z L = 20 ma Z L = Leo -> L = 20
IOO71 b) Do L thay doi de cue dai
L = M ( H )
71
R' + 40' + 20'
20 = 100 Q Ma Z L = Lo) -> L = ^
(0
L = - ( H )
7t
1 B A I T A P T R A C N G H I E M
Q^u 55. Cho mach dien gom R, L, C m^c noi tiep. Dien dp ddt vdo hai dau doan mach c6 dang: u = UoCoslOOTtt (V). C thay doi R, L khong doi. Biet R =
0 2 -
40n , L = —^ (H). Dilu chinh C de dien dp hieu dung hai dau tu dien cue dai. Gid t r i ciia di^n dung tu dien trong mach la
C . H I F D . 1 2 1 F
71 27t 71 271
C&u 56. Cho mach dien gom R, L, C m^c no'i tiep. Dien dp dat vao hai dau doan mach c6 dang: u - UoCosl20 7it (V). C thay doi. R, L jchong doi. Biet L = — ( H ) . Dieu chinh C den gid t r i C =^^— (F) thi di§n dp hieu dung hai
7t 727t
dau tu di^n dat gia t r i cUc dai. Gid t r i ciia dien trd R trong mach Id A. 12Q B. 24Q C. 36Q D. 48Q
CSu 57. Cho mach dien gom R, L, C mdc noi tiep c6 C thay doi, R, L khong doi. Biet dien dp dat vao hai dau mach c6 dang: u = 200cosl00 7it (V), R = 50( Q ), L = — (H). Dieu chinh C den gid t r i sao cho di?n dp hi^u dung hai
71
dau tu dien cuon cam cure dai. Gid t r i cifc dai ciia di#n dp hi|u dung hai dau tu dien Id
A. 316,2 V B. 416,5 V C. 600,5 V D. 700,15 V Van de 9: MAY BIEN A P -- SI/ T R U Y E N TAI DIEN NANG PHl/CfNG P H A P
I. Cau tao
"7
J -Ni N2 Cuon sa Cuon thiJ II. Cac cong thu-c cap cS'p II. Cac cong thu-c cap
1 • C6ng thQc m^y b\6n
* Chu y:
U , > U , (N, > N , ): Mdy tdng dp.
^ U , < U , (N, < N j ): Mdy ha dp.
H i ^ u sua't m a y b i e n the'
V6i: P 2 = U ^ . I z C o s c p ^
Pi = U i - I , cos(|)i
L i f u y : Hieu suat may bien ap la 100% (Bo qua hao phi mdy bien ap) H = 100% = 1
U , N , I , 2. S a truySn t^i dien nSng di xa
-4 P . = A P + P„
S = Tt.r' A P =
U ' c o s > -R
Trong do:
* I: Tong chieu d a i day (m)
* p : Dien t r d suat ciia kirn loai (Q.m )
* P , U : cong suat va dien ap truyen d i Ptt: cong suat tieu t h u
p 2
Thi^dng xet: coscp = 1 k h i do A P = j^R
Truyen t a i d i xa: H ' =
B A I T A P M A U
B a i 1 . M o t mAy bien ap c6 cupn so cap gom 4000 vong, cuon thiJ cap gom 2000 vong. Biet hai dau cuon scf cap dMc noi vdri mang dien xoay chieu ma gia tri dien ap hieu dung la 200 V. Bo qua hao phi ciia may bien ap. T i m :
a) Dien ap hieu dung d hai dau cuon thuT cap.
b) N o i 2 dau cupn thu" cap v d i t d i R = 20 Q. T i m ciTcfng dp dong d i ^ n qua cupn scf cap va thu' cap.
Tom tat
• N i = 4000 vong
• N2 = 2000 vong
• U , = 200 V T i m :
a) T i m U2 = ? U i
Hitdng ddn gidi 200 4000 U2 " U2 ~ 2000
<^ U , = 100V
[a) U2 = ? b) I i = ? I 2 = ?
b) Cudng do d6ng dien qua cu^n thu" ca'p
I 2 =
R 100
20 I2 = 5 A Cifofng dp dong dien qua cupn scf cap
U , L U , I,
200 5
—— = — <->
100 I , I , = 2,5(A)
B a i 2. M o t may bien ap dung l a m may tSng ap gom cupn day 2000 v6ng v ^ cupn day 1000 vong. Bo qua hao p h i cua may bien ap. K h i n o i h a i dau cupn so cap v d i dien ap u = Uocostot (V) t h i dien dp hieu dung d hai dau cupn thiif cap la 400 V. T i m dien ap ciic dai d hai dau cupn so cap.
Tom tat
• N , = 1000 vong
• N2 = 2000 vong
• U2 = 400 V T i m Uo, = ?
U , N i
Hit&ng ddn gidi Do mdy tang ap nen
U2 > N i N2 > N i
Do do N i = 1000 vong, N2 = 2000 vong
• Dien ap hieu dung hai dau cuon so cap
ma
U, N2 400 2000
U i = ^ Uoi = U2 V2 ^ |Uo, =200V2(V)
B a i 3. M o t may phat dien truyen d i mot cong suat P = 200 k W tren day dan CO dien t r d R = 10 Q . Hieu dien the tit t r a m phdt dien chuyen d i la U = 2 kV. T i n h :
a) Cirdng dp dong dien t r e n day do may cung cap. Biet dong di$n cung pha vdi dien ap.
b) Cong suat toa nhiet tren day.
c) Cong suat noi tieu thu.
d) Hieu suat t a i dien.
Tom tat P = 200 k W
= 200000 W
• R = 10 Q
• U = 2 k V =r 2000 V T m h :
a) I = ? b) A P = ?
0 Ptt = ? d) H = ? .
d) Hieu suat t a i dien
Hiidng ddn gidi a) Cuorng dp dong dien tren day
P = UIcoscp
P 2.10"
-ằ I =
U.coscp 2.10'.l b) Cong suat t6a nhiet tren day
AP = RI=^ = 10.(100)^
c) Cong suat nofi tieu t h u
I = 100(A)
AP = 10'(W) Ptt = P - A P = 2.10=^ - 1 0 ^ ^ Ptt = 1 0 '( W )
H = 10'
2.10' = 0,5 -> H = 50%
B a i 4. K h i truyen t d i dien nSng di xa ngirdfi ta sijf dung bien dp, tSng dien aj, len den 110 k V va chuyen di mot cong suat P = 5500 k W t r e n mot n(,j each nguon mot doan 5 k m . Biet dp giam dien dp tren dudng day khon.r qua 1% dien ap noi truyen di. T f n h dien trot dudng day va tiet dien cua day dan. Biet dien t r o suat ciia day dan la p = 1,7.10"** Qm. Biet dong die,,
cung pha vdi dien ap. . Tom tat
• U = 110 k V
= 110.10^ (V) . P = 5500 k W
= 5500.10^ W
• AU < 1%U T i n h R = ? S = ?
Cho p = 1,7.lO"** Qm
Hit&ng dan gidi Cifdng dp d6ng dien t r e n day
P 5500.10' I =
U.coscp M a : A U < 1%U
1
110.10' = 50 A
RI <
100 R < 22Q
.110.10' .e^R< 1 110.10' 100 50 (*)
M a t khac: R = p - S N e n (*) p - < 22
* S
•o S > -— (Z: tong chieu dai day (m)) pi 2 2
<^s> 1,7.10''.10.10'
22 S > 7 , 7 2 . 1 0 ' ( m ' )
BAI TAPTRAC N G H I E M
C a u 58. M o t may bien ap c6 cupn sP cap gom 1000 vong va dien ap hi$u dung la 200(V). H o i cupn thuf cap c6 500 vong t h i dien ap hieu dung la bao nhien'^
Bo qua hao p h i may bien ap nay
A. 100 V B. 150 V C. 200 V D. 300 V
C a u 59. M o t may bien ap c6 cupn so cap gom 2000 vong. Cupn thiir cap c6 so vpng 1^ 1000 vong va dien ap hieu dung 200V. Bo qua hao phi may bien ap Dien ap hieu dung hai dau cupn so cap la
A. 200 V B. 300 V C. 400 V D. 500 V
C a u 60. M o t may bien ap c6 t i so vong day giCfa cuon sp cap v^ thuf cap N , = 10 . Bo qua hao p h i may bien ap. T i so giuTa d i ^ n ap h i ^ u dung gi^'''^
cupn thiJ cap va sp cap la
A. 10 B. 0,1 C. 0,2 D. 0,3
C a u 6 1 . Mot may bien ap c6 t i so giCTa dien ap hieu dung giOra cupn thuf cap so cap la 20. T i m t i so giCfa dong dien cupn so cap va thu" cap. Bo qua hao phi may bien ap
A. 20 B. 1_
20 C.
30 D.
40
62. M o t may bien dp c6 cupn so cSp gom 1000 v6ng dupe m^c vao m a n g dien CO d i e n dp hieu dung l a 220 V, cupn thuT cap c6 55 vong. Bo qua hao p h i inay bien dp. N o i hai dau cupn thu: cap vdi t a i tieu thu t h i c6 I = 0,5 A chay qua. Dong dien chay qua cupn so cap l a
A. 0,0275 A B. 36 A C. 40 A D. 50 A
(^aô P^^* "^^^^ chieu 3 pha c6 cong suat P = 660 (kW). Dong dien tir may phdt dupc truyen di xa bkng day c6 dien t r d tong cong 10 Q . Dien
^p d dau duOng day t a i la 5 (kV). Biet ciJOng dp dong dien cung pha dien ap.
f 63.1- CuOng dp dong dien t r e n day t a i l a
A. 130 A B. 132 A C. 120 A D. 140 A 63.2. Cong suat hao p h i t r e n dudng day la
A. 174240 W B. 15000 W C. 1320 W D. 20000 W 63.3. Cong suat noi tieu t h u l a
A. 40000 W B. 485760 W C. 50000 W D.6000 W 63.4. Dp giam dien ap t r e n dudng day va dien ap noi tieu t h u 1^
A. 1320 V; 368 V B. 1300 V; 2700 V C. 1320 V; 3680 V D. 1500 V; 3500 V
Cau 64. Vdi cung mot cong suat truyen t a i , neu tdng dien ap hieu dung noi truyen di l e n 10 Ian t h i cong suat hao phi se
A. Tdng 100 Ian. B. Giam 100 Ian.
C. Tang 200 Ian. D. Giam 200 Ian.
Cau 65. Cong suat hao p h i t r e n dudng day t a i se bien doi n h i i the nao k h i t&ng U l e n gap doi va giam P di mot nuTa.
A. Giam 4 B. Giam 16 C. K h o n g doi. D. Giam 8