Upload
vu-le
View
118
Download
4
Embed Size (px)
DESCRIPTION
nothing
Citation preview
BI TP V IN XOAY CHIU HAY V KH Cu 1. t mt in p xoay chiu vo hai u on mch L, R, C mc ni tip theo th t . in p hai u
cc on mch cha L,R v R,C ln lt c biu thc: uLR = 150sos(100t + /3) (V); uRC =
50 6 sos(100t - /12) (V). Cho R = 25 . Cng dng in trong mch c gi tr hiu dng bng: A. 3
(A). B. 3 2 (A) . C. 2
23 (A). D. 3,3 (A
Gii: V gin vc t nh hnh v ta c
MON = 12
5)
12(
3
MN = UL + UC
OM = URL = 75 2 (V)
ON = URC = 50 3 (V)
p dng L cosin cho tam gic OMN:
MN = UL + UC = 12
5cos.222
RCRLRCRL UUUU 118 (V)
UR2 = ULR
2 UL
2 = URC
2 UC
2 -----> UL
2 UC
2 = ULR
2 URC
2 = 3750
(UL + UC )(UL - UC ) = 3750-----> UL + UC = 3750/118 = 32 (V)
Ta c h phng trnh UL - UC =118 (V)
UL + UC = 32 (V)
Suy ra UL = 75 (V) -----> UR = 222 75 LRL UU = 75 (V)
Do I = UR/R = 3 (A). Chn p n A Cu 2. t mt n p xoay chiu vo hai u on mch gm in tr thun R, cun dy thun cm L v t in C c in dung thay i. Khi C = C1 in p hiu dng trn cc phn t UR = 40V, UL = 40V, UC =
70V.Khi C = C2 in p hiu dng hai u t l UC = 50 2 V. in p hiu dng gia hai u in tr l:
A. 25 2 (V). B. 25 (V). C. 25 3 (V). D. 50 (V).
Gii: Khi C = C1 UR = UL ----> ZL = R
in p t vo hai u mch; U = 22 )( CLR UUU = 50 (V)
Khi C = C2 ------> UR = UL
U = 2
2
2 )'(' CLR UUU = 50 (V)-----> UR = 25 2 (V). Chn p n A
Cu 3. Cho mch in xoay chiu gm 3 phn th ni tip: in tr R; cun cm L = 4
1(H) v t in C. Cho
bit in p tc thi hai u on mch u = 90cos(t + /6) (V). Khi = 1 th cng dng in chy qua
mch i = 2 cos(240t - /12) (A); t tnh bng giy. Cho tn s gc thay i n gi tr m trong mch c gi tr cng hng dng in, hiu in th gia hai bn t in lc l:
A. uC = 45 2 cos(100t - /3) (V); B. uC = 45 2 cos(120t - /3) (V);
C uC = 60cos(100t - /3) (V); D. uC = 60cos(120t - /3) (V);
Gii:
T biu thc ca i khi = 1 ta c 1 = 240
ZL1 = 2404
1= 60
Gc lch pha gia u v i : = u - i = 4
)12
(6
-----> tan = 1
O UR
N UCR
ULR
M
http://www.hocmaivn.com/
http://www.hocmaivn.com/
http://www.hocmaivn.com/
http://www.hocmaivn.com/
R = ZL1 ZC1; Z1 = 2451
245
I
U
Z12 = R
2 + (ZL ZC)
2 = 2R
2----> R = 45
R = ZL1 ZC1 ---> ZC1 = ZL1 R = 15
ZC1 = C1
1
----> C =
3600
1
15.240
11
11
CZ
(F)
Khi mch c cng hng
222 )120(
3600
1.
4
1
11
LC
----> 2 = 120
Do mch cng hng nn: ZC2 = ZL2 = 2 L = 30 ()
I2 = 245
245
R
U(A); uc chm pha hn i2 tc chm pha hn u gc /2
Pha ban u ca uC2 = 326
UC2 = I2,ZC2 = 30 2 (V)
Vy uC = 60cos(120t /3) (V). Chn p n D,
Cu 4 .Cho mt mch in gm bin tr Rx mc ni tip vi t in c 63,8C F v mt cun dy c in
tr thun r = 70, t cm 1
L H
. t vo hai u mt in p U=200V c tn s f = 50Hz. Gi tr ca Rx
cng sut ca mch cc i v gi tr cc i ln lt l
A. 0 ;378,4W B. 20 ;378,4W C. 10 ;78,4W D. 30 ;100W
Gii:
P = I2R=
R
ZZR
U
ZZR
RU
CLCL
2
2
22
2
)()(
Vi R = Rx + r = Rx + 70 70
ZL = 2fL = 100; ZC = 610.8,63.314
1
2
1
fC50
P = Pmax khi mu s y = R + R
3500 c gi tri nh nht vi R 70
Xt s ph thuc ca y vo R:
Ly o hm y theo R ta c y = 1 - 2
3500
R; y = 0 -----> R = 50
Khi R < 50 th nu R tng y gim. ( v y < 0)
Khi R > 50 th nu R tng th y tng
Do khi R 70 th mu s y c gi tr nh nht khi R = 70.
Cng sut ca mch c gi tr ln nht khi Rx = R r = 0
Pc = 4,378)( 22
2
CL ZZr
rUW
Chn p n A Rx = 0, Pc = 378 W
http://www.hocmaivn.com/
http://www.hocmaivn.com/
http://www.hocmaivn.com/
http://www.hocmaivn.com/
Cu 5. Cho mch in nh hnh v
t vo hai u AB mt in p xoay chiu c gi tr hiu dng v tn s khng i. lch pha ca uAN v
uAB bng lch pha ca uAM v dng in tc thi. Bit 3 120 3( )AB AN MNU U U V . Cng dng
in trong mch 2 2I A . Gi tr ca ZL l
A. 30 3 B. 15 6 C. 60 D. 30 2
V gin vc t nh hnh v:
AB = UAB UAB = 120 3 (V)
AM = UAM = Ur + UL
AN = UAN UAN = 120 3 (V)
AE = Ur
EF = MN = UMN = UR UMN = UR = 120 (V)
AF = Ur + UR ; EM = FN = UL ; NB = UC NAB = MAF suy ra MAN = FAB T UAB = UMN suy ra UL
2 = (UL UC)
2 -------> UC = 2UL suy ra NAF = FAB
V vy MAN = ANM ----> tam gic AMN cn MN = AM hay UAM = UR = 120(V) Ur
2 + UL
2 = UAM
2 = 120
2 (1)
(Ur + UR)2 + (UL UC)
2 = UAB
2
hay (Ur + 120)2 + UL
2 = 120
2 (2)
T (1) v (2) ta c Ur = 60 (V); UL = 60 3 (V)
Do o ZL = 61522
360
I
U L (), Chn p n B
Cu 6. Mt on mch AB gm hai on mch AM v BM mc ni tip. on mch AM gm in tr thun
R1 mc ni tip vi t in c in dung C, on mch MB gm in tr thun R2 mc ni tip vi cun cm
thun c t cm L. t in p xoay chiu u = U0cos t (U0 v khng i) vo hai u on mch AB th
cng sut tiu th ca on mch AB l 85 W. Khi LC
12 v lch pha gia uAM v uMB l 900. Nu
t in p trn vo hai u on mch MB th on mch ny tiu th cng sut bng:
A. 85 W B. 135 W. C. 110 W. D. 170 W.
Gii:
Khi LC
12 trong mch c cng hng ZL = ZC
v cng sut tiu th ca on mch c tnh theo cng thc
N R
A B
M C L,r
A Ur E UR F
I
UAN N
UAM
M
UAB B
L R1 C M R2 B A
http://www.hocmaivn.com/
http://www.hocmaivn.com/
http://www.hocmaivn.com/
http://www.hocmaivn.com/
P = 21
2
RR
U
(1). Ta c: tan1 =
1R
ZC ; tan2 = 1R
Z L
Mt khc: 2 - 1 = 900
------> tan1. tan2 = -1
Do 1R
ZC
1R
Z L = -1 -------> ZL = ZC = 21RR (2)
Khi t in p trn vo on mch MB th cng sut tiu th trn on mch
P2 = I22 R2 = 22
2
2
2
LZR
RU
=
212
2
2
2
RRR
RU
21
2
RR
U
= P = 85W. Chn p n A
Cu 7: Cho mch in nh hnh v. t vo hai u on
mch in p xoay chiu u=120 6 cos(100 t)(V) n nh,
th in p hiu dng hai u MB bng 120V, cng sut
tiu th ton mch bng 360W; lch pha gia uAN v uMB
l 900, uAN v uAB l 60
0 . Tm R v r
A. R=120 ; r=60 B. R=60 ; r=30 ;
C. R=60 ; r=120 D. R=30 ; r=60
Gii: V gin vc t nh hnh v
OO1 = Ur
UR = OO2 = O1O2 = EF
UMB = OE UMB = 120V (1)
UAN = OQ
UAB = OF UAB = 120 3 (V) (2)
EOQ = 900
FOQ = 600
Suy ra = EOF = 900 600 = 300.
Xt tam gic OEF: EF2 = OE
2 + OF
2 2.OE.OFcos300
Thay s ---------> EF = OE = 120 (V) Suy ra UR = 120(V) (3)
UAB2 = (UR + Ur)
2 + (UL UC)
2
Vi (UL UC)
2 = UMB
2 Ur
2 ( xt tam gic vung OO1E)
UAB2 = UR
2 +2UR.Ur + UMB
2 . T (1); (2), (3) ta c Ur = 60 (V) (4)
Gc lch pha gia u v i trong mch:
= FOO3 = 30
0 ( v theo trn tam gic OEF l tam gic cn c gc y bng 300)
T cng thc P = UIcos ----->
I = P / Ucos 360/(120 3 cos300) = 2 (A): I = 2A (5)
Do R = UR/I = 60; r = Ur /I = 30. Chn p n B
Cu 8. t in p xoay chiu u = 100 2 cost (c thay i c trn on [100 200; ] ) vo hai u
on mch c R, L, C mc ni tip. Cho bit R = 300 , L =
1(H); C =
410 (F).
UAN Q
O3
UL
UL + UC
O
UC
Ur O1 UR
O2
UAB F UMB E
UR + Ur
L,r
R A B
C N
M
http://www.hocmaivn.com/
http://www.hocmaivn.com/
http://www.hocmaivn.com/
http://www.hocmaivn.com/
in p hiu dng gia hai u L c gi tr ln nht v nh nht tng ng l
A.100 V; 50V. B.50 2 V; 50V. C.50V; 3
100v. D. .
3
100;
53
400VV
Gii:
Ta c UL = IZL; UL=
22
4
4
282
2
2
42
22 1110.71
101
)2(11
)1
(
U
LC
LR
C
UL
CLR
LU
Xt biu thc y = 2
4228 110.710
XX
Vi X = 2
1
> 0. Ly o hm y theo X ta thy y > 0:
gi tr ca y tng khi X tng, tc l lhi 2 hay gim. Vy khi tng th UL tng
Trong khong 100 200 UL = ULmax khi = 200. --->
ULmax =
22
4
4
28 1110.71
10
U
53
400
14
7
16
1
100
1
.4
110.7
10.16
110
22
4
48
28
U
(V)
UL = ULmin khi = 100. --->
ULmin =
22
4
4
28 1110.71
10
U
3
100
171
100
1110.7
10
110
22
4
48
28
U
(V)
Chn p n D.
Cu 9.. Cho mch in xoay chiu khng phn nhnh AD gm hai on AM v MD. on mch MD gm
cun dy in tr thun R = 40 3 v t cm L = 5
2 H. on MD l mt t in c in dung thay i
c, C c gi tr hu hn khc khng. t vo hai u mch in p xoay chiu uAD = 240cos100t (V).
iu chnh C tng in p (UAM + UMD) t gi tr cc i. Gi tr cc i l:
A. 240 (V). B. 240 2 (V). C. 120V. D. 120 2 (V)
Gii:
Ta c ZL = 100 .2/5 = 40-----> ZAM = 8022 LZR
t Y = (UAM + UMD)2
.
Tng (UAM + UMD) t gi tr cc i khi Y t gi tr cc i
Y = (UAM + UMD)2 = I
2( ZAM
2 +ZC
2 + 2ZAM.ZC) = 22
222
)(
)2(
CL
CAMCAM
ZZR
ZZZZU
Y = 640080
)6400160(
)40(40.3
)16080(2
22
22
222
CC
CC
C
CC
ZZ
ZZU
Z
ZZU
http://www.hocmaivn.com/
http://www.hocmaivn.com/
http://www.hocmaivn.com/
http://www.hocmaivn.com/
Y = Ymax khi biu thc X= 640080
)6400160(2
2
CC
CC
ZZ
ZZ= 1+
640080
2402 CC
C
ZZ
Z c gi tr cc i
------->X = 640080
2402 CC
C
ZZ
Z =
806400
240
C
CZ
Z
c gi tr cc i
X = Xmax khi mu s cc tiu, -----> ZC2 = 6400 -----> ZC = 80
tng in p (UAM + UMD) t gi tr cc i khi ZC = 80
(UAM + UMD)max = )( CAM ZZZ
U = 2240
80
160.2120
)8040(40.3
)8080(2120
22
(V)
Chn p n B: (UAM + UMD)max = 240 2 (V)
Cu 10. Mt cun dy khng thun cm ni tip vi t in C trong mch xoay chiu c in p u=U0cost(V) th dng in trong mch sm pha hn in p u l 1 v in p
hiu dng hai u cun dy l 30V. Nu thay C1=3C th dng in chm pha hn u gc 2 = 90
0 - 1 v in p hiu
dng hai u cun dy l 90V. Tm U0.
Gii: Cc ch s 1 ng vi trng hp t C; ch s 2 ng vi t 3C
V gin vc t nh hnh v: Ta c ZC2 = ZC1/3 = ZC/3
Do Ud = IZd = I22
LZR : Ud1 = 30V; Ud2 = 90V
Ud2 = 3Ud1 -----> I2 = 3I1
UC1 = I1ZC
UC2 = I2ZC2 = 3I1ZC/3 = I1ZC = UC1 =UC
Trn gin l cc on OUC; Ud1U1; Ud2U2 biu in UC U1 = U2 =U in p hiu dung t vo mch. Theo bi ra 2=90
0-1 .
Tam gic OU1U2 vung cn ti O
Theo hnh v ta c cc im UC; U1 v U2 thng hng. on thng UCU1 U2 song song v bng on OUd1Ud2
Suy ra U1U2 = Ud1Ud2 = 90 30 = 60V
Do OU1 = OU2 = U1U2/ 2
Suy ra U = 60/ 2 = 30 2 -----> U0 = 60V
2 UR2 I O 1 UR1
UC
U1
U2
Ud2 UL2
Ud1 UL1
http://www.hocmaivn.com/
http://www.hocmaivn.com/
http://www.hocmaivn.com/
http://www.hocmaivn.com/
Cu 11: Mch in xoay chiu R, L, C mc ni tip. in p hai u on mch l 0u U cos t . Ch
c thay i c. iu chnh thy khi gi tr ca n l 1 hoc 2 ( 2 < 1 ) th dng in hiu
dng u nh hn cng hiu dng cc i n ln (n > 1). Biu thc tnh R l
A. R = 1 22
( )
L n 1
B. R = 1 2
2
L( )
n 1
C. R = 1 2
2
L( )
n 1
D. R = 1 2
2
L
n 1
Gii: I1 = I2 =Imax/n ------> Z1 = Z2 -----> 1 L - C1
1
= - 2 L +
C2
1
-------> 2 L-=C1
1
m I1 = Imax/n
------>
)1
(1
1
2
CLR
U
= R
U
n
1--------->n
2R
2 = R
2 +( 1 L -
C1
1
)
2 = R
2 + ( 1 L -2 L )
2
------> (n2 1)R2 = ( 1 -2 )
2L
2 -------> R = 1 2
2
L( )
n 1
. Chn p n B
Cu 12. t mt in p u = U0 cos t ( U0 khng i, thay i c) vo 2 u on mch gm R, L, C mc ni tip tha mn iu kin CR2 < 2L. Gi V1,V2, V3 ln lt l cc vn k mc vo 2 u R, L, C. Khi tng dn tn s th thy trn mi vn k u c 1 gi tr cc i, th t ln lt cc vn k ch gi tr cc i khi tng dn tn s l
A. V1, V2, V3. B. V3, V2, V1. C. V3, V1, V2. D. V1, V3,V2.
Gii: Ta gi s ch ca cc vn k l U1,2,3
U1=IR =22 )
1(
CLR
UR
U1 = U1max khi trong mch c s cng hng in: ----->12 =
LC
1 (1)
U2 = IZL = 22
2
22
22222 21
)1
(y
U
C
L
CLR
UL
CLR
LU
U2 = U2max khi y2 = 2
2
2
42
211
LC
LR
C
c gi tr cc tiu y2min
t x = 2
1
, Ly o hm y2 theo x, cho y2 = 0 ----->x = 2
1
= )2(
2
2CRC
LC
)2(
2
22
2
2
RC
LC
=)2(
22CRLC
(2)
U3 = IZC = 23
22
222222 )21
()1
(y
U
C
L
CLRC
U
CLRC
U
http://www.hocmaivn.com/
http://www.hocmaivn.com/
http://www.hocmaivn.com/
http://www.hocmaivn.com/
U3 = U3max khi y3 = L24 +(R2 -2
C
L )2 +
2
1
C c gi tr cc tiu y3min
t y = 2 , Ly o hm ca y3 theo y, cho y3 = 0
y = 2 = 2
2
2
2
2
1
2
2
L
R
LCL
RC
L
32 =
2
2
2
1
L
R
LC (3)
So snh (1); (2), (3):
T (1) v (3) 32 =
2
2
2
1
L
R
LC < 1
2 =
LC
1
Xt hiu 22 - 1
2 =
)2(
22CRLC
-LC
1=
)2()2(
)2(22
2
2
2
CRLLC
CR
CRLLC
CRLL
>0
(V CR2
< 2L nn 2L CR2 > 0 )
Do 22 =
)2(
22CRLC
> 12 =
LC
1
Tm lai ta c 32 =
2
2
2
1
L
R
LC < 1
2 =
LC
1 < 2
2 =
)2(
22CRLC
Theo th t V3, V1 , V2 ch gi tr cc i Chn p n C
Cu 13 . on mch AB gm on AM ni tip vi MB. on AM goomg in tr R ni tip vi cuonj dy thun cm c t cm L thay i c. on MB ch c t in C. in p t vo hai u mch
uAB = 100 2 cos100t (V). iu chnh L = L1 th cng dng in qua mch I1 = 0,5A, UMB = 100(V),
dng in i tr pha so vi uAB mt gc 600. iu chnh L = L2 in p hiu dng UAM t cc i. Tnh
t cm L2:
A.
21(H). B.
31(H). C.
32 (H). D.
5,2(H).
Gii:
Ta c ZC =100/0,5 = 200, 360tantan0
R
ZZ CL -----> (ZL ZC) = R 3
Z = U/I = 100/0,5 = 200
Z = RZZR CL 2)(22 ------> R = 100
UAM = I.ZAM =
2212
22222
22
100
)100(4001
2)(
L
L
L
CLCLCL
L
Z
Z
U
ZR
ZZZZR
U
ZZR
ZRU
UAM =UAMmin khi y = 22100
100
L
L
Z
Z
= ymax c gi tr cc i
y = ymax khi o hm y = 0------> ZL2 200ZL -100 = 0
-------> ZL = 100(1 + 2 )
--------> L =
21(H) Chn p n A.
http://www.hocmaivn.com/
http://www.hocmaivn.com/
http://www.hocmaivn.com/
http://www.hocmaivn.com/
Cu 14. Cho mch in RLC mc ni tip theo th t R, L, C trong cun dy thun cm c t
cm L thay i c, in tr thun R=100 . t vo hai u on mch hiu in th xoay chiu c
tn s f=50Hz. Thay i L ngi ta thy khi 1L=L v khi
12
LL=L =
2th cng sut tiu th trn on
mch nh nhau nhng cng dng in tc thi vung pha nhau. Gi tr L1 v in dung C ln lt
l:
A. -4
1
4 3.10L = (H);C= (F)
2 B.
-4
1
4 10L = (H);C= (F)
3
C. -4
1
2 10L = (H);C= (F)
3 D.
-4
1
1 3.10L = (H);C= (F)
4
Gii: Do cng sut P1 = P2 -----> I1 = I2 ------> Z1 = Z2
Do (ZL1 ZC)2 = (ZL2 ZC)
2. Do ZL1 ZL2 nn ZL1 ZC = ZC ZL2 = ZC -
2
1LZ
----> 1,5ZL1 = 2ZC (1)
tan1 = R
ZZ CL 1 = R
Z L
4
1 v tan2 = R
ZZ
R
ZZ CL
CL
21
2 = R
Z L
4
1
1 + 2 = 2
------> tan1. tan1 = -1 -----> ZL1
2 = 16R
2 ----. ZL1 = 4R = 400
----> L1 =
41 LZ
(H)
ZC = 0,75ZL1 = 300 ----> C = 3
10
.
1 4
CZ(F)
Chn p n B
Cu 15: Cho 3 linh kin gm in tr thun R=60, cun cm thun L v t in C. Ln lt t in p
xoay chiu c gi tr hiu dng U vo hai u on mch ni tip RL hoc RC th biu thc cng dng
in trong mch ln lt l i1= 2 cos 10012
t
(A) v i2= 7
2 cos 10012
t
(A). nu t in p
trn vo hai u on mch RLC ni tip th dng in trong mch c biu thc
A. 2 2 cos(100t+
3 )(A) . B. 2 cos(100t+
3 )(A).
C. 2 2 cos(100t+
4 )(A) . D. 2cos(100t+
4 )(A).
Gii: Ta thy cng hiu dng trong on mch RL v RC bng nhau suy ra ZL = ZC lch pha 1 gia u v i1 v 2 gia u v i2 i nhau. tan1= - tan2
Gi s in p t vo cc on mch c dng: u = U 2 cos(100t + ) (V). Khi 1 = (- /12) = + /12 2 = 7/12
tan1 = tan( + /12) = - tan2 = - tan( 7/12) tan( + /12) + tan( 7/12) = 0 --- sin( + /12 + 7/12) = 0 Suy ra = /4 - tan1 = tan( + /12) = tan(/4 + /12) = tan /3 = ZL/R
-- ZL = R 3
U = I1 2 2
12 120LR Z RI (V)
http://www.hocmaivn.com/
http://www.hocmaivn.com/
http://www.hocmaivn.com/
http://www.hocmaivn.com/
Mch RLC c ZL = ZC trong mch c s cng hng I = U/R = 120/60 = 2 (A) v i cng pha vi u =
U 2 cos(100t + /4) .
Vy i = 2 2 cos(100t + /4) (A). Chn p n C
Cu 16. Cho mch RLC ni tip. Khi t in p xoay chiu c tn s gc ( mch ang c tnh cm khng). Cho thay i ta chn c 0 lm cho cng dng in hiu dng c gi tr ln nht
l Imax v 2 tr s 1 , 2 vi 1 2 = 200 th cng dng in hiu dng lc ny
l ax
2
mII .Cho 3
4L
(H). in tr c tr s no:
A.150. B.200. C.100. D.125.
Gii: I1 = I2 -----> Z1 = Z2 ------> (ZL1 ZC1)
2 = (ZL2 ZC2)
2 ----> ZL1 + ZL2 = ZC1 + ZC2
L(1 + 2) = 21
21
21
)11
(1
CC
-----> LC =
21
1
------> ZC1 = ZL2
Imax = R
U 2; I1 =
Z
U =
2
11
2 )( CL ZZR
U
=
R
U
2
2
-------> 4R2 = 2R
2 + 2(ZL1 ZC1)
2
R2 = (ZL1 ZL2)
2 = L
2 (1 - 2)
2 -----> R = L (1 - 2) =
200
4
3 = 150(). Chn p n A
Cu 17: Mt mch in xoay chiu gm cc linh kin l tng mc ni tip theo th t R, C v L. t vo
hai u on mch mt in p xoay chiu u = U0cos(t /6). Bit U0, C, l cc hng s. Ban u in p hiu dng hai u in tr R l UR = 220V v uL = U0Lcos(t + /3), sau tng R v L ln gp i, khi URC bng
A. 220V. B. 220 2 V. C. 110V. D. 110 2 .
Gii: Hiu pha ban u ca uL v i: UL - i = 2
---> i =
3
-
2
= -
6
Do ta c u, i cng pha, MCH C CNG HNG: nn: ZL = ZC v U = UR = 220 (V) Khi tng R v L ln gp i th R = 2R, ZL = 2ZL
URC = 22
22
)'('
'
CL
C
ZZR
ZRU
=
22
22
)2('
'
CC
C
ZZR
ZRU
= U = 220V. Chn p n A
Cu 18: t mt in p xoay chiu u = U0cos(100t+ ) vo hai u mt on mch gm R, L, C mc
ni tip (L l cun cm thun). Bit 410
C F
; R khng thay i, L thay i c. Khi 2
L H
th biu
thc ca dng in trong mch l 1 2 os(100 t /12)i I c A . Khi 4
L H
th biu thc ca dng in
trong mch l 2 2 os(100 t / 4)i I c A . in tr R c gi tr l
A. 100 3 . B. 100. C. 200. D. 100 2 .
Gii:
Ta c ZC = 100; ZL1 = 200; ZL2 = 400
tan1 = R
ZZ CL 1 = R
100 ----.>1 = +
12
http://www.hocmaivn.com/
http://www.hocmaivn.com/
http://www.hocmaivn.com/
http://www.hocmaivn.com/
tan2 = R
ZZ CL 2 =R
300 = 3tan1 ----.>2 = +
4
-------> 2 - 1 = 4
-
12
=
6
tan(2 - 1) = tan6
=
3
1
tan(2 - 1) =3
1
tan31
tan2
tantan1
tantan
1
2
1
12
12
-----> tan1 =
3
1
-----> R
100=
3
1------> R = 100 3 () Chn p n A
Cu 19. Trong gi thc hnh mt hc sinh mc ni tip mt qut in xoay chiu vi in tr R, ri mc vo hai u mch in p xoay chiu c gi tr hiu dng 380V. Bit qut c cc gi tr nh mc 220V
88W. Khi hot ng ng cng sut nh mc th lch pha gia in p hai u qut v dng in qua
n l , vi cos = 0,8. qut hot ng ng cng sut th R =?
Gii:
Gi r l in tr ca qut: P = UqIcos = I2r.
Thay s vo ta c: I = cosqU
P =
8,0.220
88= 0,5 (A); r =
2I
P= 352
Zqut = I
U q= 22 LZr = 440
Khi mc vo U = 380V: I = Z
U=
22)( LZrR
U
=
222 2 LZrRrR
U
R2 + 2Rr +
2
quatZ = 2)(
I
U------> R
2 + 704R +440
2 = 760
2
-----> R2 + 704R 384000 = 0------> R = 360,7
Cu 20. Ni hai cc ca my pht in xoay chiu mt pha vo hai u on mch AB gm R ni tip vi L thun. B qua in tr cun dy ca my pht. Khi r to quay u vi tc n vng/pht th cng
hiu dng l 1A. Khi r to quay u vi tc 3n vng/pht th cng hiu dng l 3 A..Khi r to
quay u vi tc 2n vng/pht th cm khng ca on mch AB tnh theo R l?
Gii: I = Z
U=
Z
E
Vi E l sut in ng hiu dng gia hai cc my pht: E = 2 N0 = 2 2fN0 = U ( do r = 0) Vi f = np n tc quay ca roto, p s cp cc t
Z = 222 LR
Khi n1 = n th 1 = ; ZL1 = ZZ
Khi n3 = 3n th 3 = 3; ZL3 = 3ZZ ---->
3
1
I
I=
3
1
E
E
1
3
Z
Z=
3
1
1
3
Z
Z------->
3
1
22
22 9
L
L
ZR
ZR
=
3
1
I
I=
3
1------>R
2 + 9
2
LZ = 3R2 +3
2
LZ
62
LZ = 2R2 ------>
2
LZ = R2/3-----> ZL =
3
R
http://www.hocmaivn.com/
http://www.hocmaivn.com/
http://www.hocmaivn.com/
http://www.hocmaivn.com/
- Khi n2 = 2n th 2 = 2; ZL2 = 2ZZ = 3
2R
Cu 21: Mt cun dy khng thun cm ni tip vi t in C trong mch in xoay chiu c in p
0. osu U c t (V) th dng in trong mch sm pha hn in p l 1 , in p hiu dng hai u cun dy l
30V. Bit rng nu thay t C bng t 'C 3C th dng in trong mch chm pha hn in p l 2 1
2
v in p hiu dng hai u cun dy l 90V. Bin 0 ?U
A. 60V . B. 30 2V C. 60 2V . D. 30V
Gii: Ud1 = 30 (V)
Ud2 = 90 (V) ----> 1
2
d
d
U
U= 3 ----> I2 = 3I1 -----> Z1 = 3Z2 -------.Z1
2 = 9Z2
2
------> R2 + (ZL ZC1)
2 = 9R
2 + 9(ZL -
3
1CZ )2 ----->2(R2 +ZL2 ) = ZLZC1
------> ZC1 = L
L
Z
ZR )(2 22
1
1
d
d
Z
U=
1Z
U-------> U = Ud1
1
1
dZ
Z= Ud1
22
2
1
2 )(
L
cL
ZR
ZZR
= Ud1 22
1
2
1
22 2
L
CLCL
ZR
ZZZZR
=
Ud1 22
22
2
22222 )(22
)(4
L
L
LL
L
LL
ZR
Z
ZRZ
Z
ZRZR
= Ud1 3)(4
2
22
L
L
Z
ZR = Ud1 1
42
2
LZ
R
tan1 = R
ZZ CL 1 ; tan1 = R
ZZ CL 2 = R
ZZ CL
3
1
2 12
-----> 1 + 2 =
2
-----> tan1 tan2 = -1 ( v 1 < 0)
R
ZZ CL 1
R
ZZ CL
3
1
= -1------>(ZL ZC1)(ZL - 3
1CZ ) = - R2 ------->
R2 + ZL
2 4ZL
3
1CZ + 3
2
1CZ = 0 --------> (R2 + ZL2 ) 4ZL
L
L
Z
ZR
3
)(2 22 +
2
222
3
)(4
L
L
Z
ZR = 0
----->(R2 + ZL
2 )[1-
3
8+
2
22
3
)(4
L
L
Z
ZR ] = 0 ----->
2
22
3
)(4
L
L
Z
ZR -
3
5 = 0----->
2
2
3
4
LZ
R =
3
1
---->2
24
LZ
R = 1------> U = Ud1 1
42
2
LZ
R = Ud1 2
Do U0 = U 2 = 2Ud1 = 60V. Chn p bn A
Cu 22 Ni hai cc my pht in xoay chiu mt pha vo hai u mch ngoi RLC, b qua in tr dy
ni, coi t thng cc i gi qua cun dy l khng i Khi rto quay vi tc n0 vng/pht th cng sut mch ngoi cc i.Khi rto quay vi tc n1 vng/pht v n2 vng/pht th cng sut mch ngoi
c cng gi tr Mi lin h gia n1, n2 v n0 l
http://www.hocmaivn.com/
http://www.hocmaivn.com/
http://www.hocmaivn.com/
http://www.hocmaivn.com/
A. 20 1 2.n n n B.
2 2 2
0 1 2n n n C. 22
2
1
2
2
2
12
0nn
nnn
D.
2
2
2
1
2
2
2
12
0
2
nn
nnn
Gii: Sut in ng ca ngun in: E = 2 N0 = 2 2fN0 = U ( do r = 0) Vi f = np n tc quay ca roto, p s cp cc t Do P1 = P2 -----> I1
2 = I2
2 ta c:
2
1
1
2
2
1
)1
(C
LR
=2
2
2
2
2
2
)1
(C
LR
-------> ])1
([ 2
2
2
22
1C
LR
= ])1
([ 2
1
1
22
2C
LR
---> C
L
CLR 2122
2
2
122
2
2
1
22
1 2
=
C
L
CLR 2222
1
2
222
2
2
1
22
2 2
---> )2)(( 2222
1C
LR = )(
12
2
2
1
2
1
2
2
2
C =
2
2
2
1
2
1
2
2
2
1
2
2
2
))((1
C
-----> (2C
L- R
2 )C
2 =
2
2
2
1
11
(*)
Dng in hiu dng qua mch
I = Z
E
Z
U
P = Pmac khi E2 /Z
2 c gi tr ln nht hay khi y =
22
2
)1
(C
LR
c gi tr ln nht
y =
2
22
222 21
1
C
L
CLR
=
2
2
2
42
211
1
LC
LR
C
y = ymax th mu s b nht
t x = 2
1
---> y =
22
2
2
)2( LxC
LR
C
x
Ly o hm mu s, cho bng 0 ta c kt qu x0 = 20
1
=
2
1C
2(2 )
2RC
L (**)
T (*) v (**) ta suy ra 2
2
2
1
11
=
2
0
2
2
0
2
2
2
1
211
fff hay
2
0
2
2
2
1
211
nnn ------>
2
2
2
1
2
2
2
12
0
2
nn
nnn
Chn p n D
Cu 23 : t in p xoay chiu vo mch RLC ni tip c C thay i c. Khi C= C1 = 410
F v C=
C2 = 410
2
F th UC c cng gi tr. UC c gi tr cc i th C c gi tr:
http://www.hocmaivn.com/
http://www.hocmaivn.com/
http://www.hocmaivn.com/
http://www.hocmaivn.com/
A. C = 43.10
4
F . B. C = 410
3
F C. C = 43.10
2
F. D. C = 42.10
3
F
Gii:
UC1 = UC2 ------>2
1
2
1
)( CL
C
ZZR
UZ
=
2
2
2
2
)( CL
C
ZZR
UZ
---->
2
1
22
C
L
Z
ZR - 2
1C
L
Z
Z +1 =
2
2
22
C
L
Z
ZR - 2
2C
L
Z
Z +1 ------>
(R2 + 2LZ )( 2
1
1
CZ-
2
2
1
CZ) = 2ZL(
1
1
CZ -
1
1
CZ) ------>
1
1
CZ +
1
1
CZ =
22
2
L
L
ZR
Z
(1)
UC = 22 )( CL
C
ZZR
UZ
= UCmax khi y = 2
22
C
L
Z
ZR - 2
C
L
Z
Z +1 = ymin ------>
y = ymin khi ZC = L
L
Z
ZR 22 ------>
CZ
1=
22
L
L
ZR
Z
(2)
T (1) v (2)-----> 1
1
CZ +
1
1
CZ =
CZ
2------> C =
2
21 CC = 43.10
4
(F). Chn p n A
Cu 24: Mt on mch gm cun cm c t cm L v in tr thun r mc ni tip vi t in c in dung C thay i c. t vo hai u mch mt hiu in th xoay chiu c gi tr hiu dng U v
tn s f khng i. Khi iu chnh in dung ca t in c gi tr C = C1 th in p hiu dng gia
hai u t in v hai u cun cm c cng gi tr v bng U, cng dng in trong mch khi c
biu thc 1 2 6 os 100 ( )
4i c t A
. Khi iu chnh in dung ca t in c gi tr C = C2 th in
p hiu dng gia hai bn t in t gi tr cc i. Cng dng in tc thi trong mch khi c biu thc l
A. 2
52 3 os 100 ( )
12i c t A
B.
2
52 2 os 100 ( )
12i c t A
C. 2 2 2 os 100 ( )3
i c t A
D. 2 2 3 os 100 ( )3
i c t A
Gii: Khi C = C1 UD = UC = U-------> Zd = ZC1 = Z1
Zd = Z1 -----> 2
1
2 )( CL ZZr = 22
LZr --------> ZL ZC1 = ZL
-----> ZL = 2
1CZ (1)
Zd = ZC1 -----> r2 +ZL
2 = ZC!
2 ----->r
2 =
4
3 21CZ -------> r = 2
3 21CZ (2)
tan1 = 3
1
2
3
2
1
1
1
1
C
CC
CL
Z
ZZ
r
ZZ----> 1 = -
6
http://www.hocmaivn.com/
http://www.hocmaivn.com/
http://www.hocmaivn.com/
http://www.hocmaivn.com/
Khi C = C2 UC = UCmax khi ZC2 = 11
2
1
22
2
2
C
C
C
L
L ZZ
Z
Z
Zr
Khi Z2 = 12
1
2
112
1
2
2
2 33)22
(4
3)( CCCCCL ZZZ
ZcZZZr
tan2 = 3
2
3
22
1
1
1
2
C
CC
CL
Z
ZZ
r
ZZ----> 2 = -
3
U = I1Z1 = I2Z2 -------> I2 = I1 23
32
3
1
2
1 I
Z
Z(A)
Cng dng in qua mch
i2 = I2 )364
100cos(2
t = 2 )12
5100cos(2
t (A). Chn p n B
Cu 25. t vo hai u mch in gm hai phn t R v C vi R = 100 mt ngun in tng hp c
biu thc u = 100 + 100cos(100t + /4) (V). Cng sut ta nhit trn in tr R c th l:
A. 50W. B. 200W. C. 25W, D, 150W
Gii: Ngun in tng hp gm ngun in mt chiu c U1chieu = 100V v ngun in xoay chiu c
in p hiu dng U = 50 2 (V). Do on mch cha t C nn dng in 1 chiu khng qua R. Do
cng sut ta nhit trn R < Pmax (do Z > R)
P = I2R <
R
U 2 =
100
)250( 2 = 50W. Chn p n C: P = 25W.
Cu 26: Mt mch tiu th in l cun dy c in tr thun r = 8 ,tiu th cng sut P=32W vi h s
cng sut cos = 0,8 .in nng c a t my pht in xoay chiu 1 pha nh dy dn c in tr R=
4.in p hiu dng 2 u ng dy ni my pht l
A.10 5 V B.28V C.12 5 V D.24V
Gii: Dng in qua cun dy I = r
P = 2A;
Ud = cosI
P = 20V , I =
d
d
Z
U=
dZ
20-----> Zd =
2
20 = 10
Zd = 22
LZr -----> ZL = 22 rZ L = 6
I = Z
U-----> U = IZ = I 22)( LZRr = 2
22 612 = 12 5 (V). Chn p n C
Cu 27 Cho on mch xoay chiu RLC mc ni tip.t vo 2 u mch 1 in p xoay chiu c tn s
thay i c.Khi tn s ca in p 2 u mch l f0 =60Hz th in p hiu dng 2 u cun cm thun t cc i .Khi tn s ca in p 2 u mch l f = 50Hz th in p 2 u cun cm l
uL=UL 2 cos(100t + 1 ) .Khi f = f th in p 2 u cun cm l uL =U0L cos(t+2 ) .Bit UL=U0L
/ 2 .Gi tr ca bng:
A.160(rad/s) B.130(rad/s) C.144(rad/s) D.20 30 (rad/s)
Gii: UL = IZL = 22 )
1(
CLR
LU
http://www.hocmaivn.com/
http://www.hocmaivn.com/
http://www.hocmaivn.com/
http://www.hocmaivn.com/
UL =ULmax khi y = 2
22 )1
(
C
LR
= ymin
-------> 2
0
1
=
2
2C(2
C
L-R
2) (1) Vi 0 = 120 rad/s
Khi f = f v f = f ta u c U0L = UL 2 Suy ra UL = UL ------>
22 )
1(
CLR
= 22 )
'
1'(
'
CLR
------>
2 [ 22 )'
1'(
CLR
] = 2 [ 22 )
1(
CLR
]
( 2 -2 )( 2C
L-R
2) =
2
1
C(
2
2
'
-
2
2'
) =
2
1
C( 2 -2 )(
2'
1
+
2
1
)
-----> C2 ( 2
C
L-R
2) =
2'
1
+
2
1
(2) Vi = 100 rad/s
T (1) v (2) ta c 2
0
2
=
2'
1
+
2
1
-------> 2 =
2
0
2
2
0
2
2
= 2
0
2
0
2
-------> =
2222 120100.2
120.100
= 160,36 rad/s. Chn p n A
Cu 28. t in p xoay chiu u = 100 6 cos(100t) (V); vo hai u on mch mc ni tip gm in
tr thun R, cun cm thun c t cm L v t in c in dung C thay i c. iu chnh C in p hiu dng hai u t t gi tr cc i th thy gi tr cc i bng 200 V. in p hiu dng
hai u cun cm l bao nhiu vn?
Gii:
UC = UCmax = 200 (V) khi ZC = L
L
Z
ZR 22 ----->
ULUC = UR2 + UL
2 ------.> UR
2 + UL
2 =200UL
U2 = UR
2 +(UL UC)
2 -------> (100 3 )2 = UR
2 + UL
2 +200
2 400UL
-----> 30000 = 200UL + 40000 400UL ----> UL = 50 (V)
Cu 29. Mt cun dy khng thun cm ni tip vi t in C trong mch in xoay chiu c in p
0. osu U c t (V) th dng in trong mch sm pha hn in p l 1 , in p hiu dng hai u cun dy l
30V. Bit rng nu thay t C bng t 'C 3C th dng in trong mch chm pha hn in p l 2 12
v in p hiu dng hai u cun dy l 90V. Bin 0 ?U
Gii: Ud1 = 30 (V)
Ud2 = 90 (V) ----> 1
2
d
d
U
U= 3 ----> I2 = 3I1 -----> Z1 = 3Z2 -------.Z1
2 = 9Z2
2
R2 + (ZL ZC1)
2 = 9R
2 + 9(ZL -
3
1CZ )2 ----->2(R2 +ZL2 ) = ZLZC1 -----> R2 + ZL
2 =
2
1CL ZZ
http://www.hocmaivn.com/
http://www.hocmaivn.com/
http://www.hocmaivn.com/
http://www.hocmaivn.com/
11
d
d
Z
U=
1Z
U-------> U = Ud1
1
1
dZ
Z= Ud1 22
1
2
1
22 2
L
CLCL
ZR
ZZZZR
= Ud1 3
2
?
1 Z
ZC (*)
tan1 = R
ZZ CL 1 ; tan1 = R
ZZ CL 2 = R
ZZ CL
3
1
2 12
-----> 1 + 2 =
2
-----> tan1 tan2 = -1 ( v 1 < 0)
R
ZZ CL 1
R
ZZ CL
3
1
= -1------>(ZL ZC1)(ZL - 3
1CZ ) = - R2 ------->
R2 + ZL
2 4ZL
3
1CZ + 3
2
1CZ = 0 --------> 2
1CL ZZ 4ZL3
1CZ + 3
2
1CZ = 0 ---> 3
2
1CZ - 6
5 1CL ZZ = 0
---->3
1CZ - 6
5 LZ = 0 ----> ZC1 = 2,5ZL (**) ------> U = Ud1 32
?
1 Z
ZC = Ud1 2
Do U0 = U 2 = 2Ud1 = 60V.
Cu 30. Ni hai cc ca mt my pht in xoay chiu mt pha vo hai u on mch AB gm in tr
thun R = 30 , mc ni tip vi t in C. B qua in tr cc cun dy ca my pht. Khi r to quay
vi tc n vng /pht th cng hiu dng trong on mch l 1A. . Khi r to quay vi tc 2n
vng /pht th cng hiu dng trong on mch l 6 A. Nu r to quay vi tc 3n vng /pht th
dung khng ca t in l:
A. 4 5 () B. 2 5 () C. 16 5 () D. 6 5 ()
Gii: I = Z
U=
Z
E
Vi E l sut in ng hiu dng gia hai cc my pht: E = 2 N0 = 2 2fN0 = U ( do r = 0) Vi f = np n tc quay ca roto, p s cp cc t
Z = 22
2 1
CR
Khi n1 = n th 1 = ; I1 = 1Z
E; ZC1 = ZC =
C
1
Khi n2 = 2n th 2 = 2; ZC2 = ZC1 /2 = ZC /2 ----> I2 = 2Z
E
2
1
I
I=
3
1
E
E
1
2
Z
Z=
2
1
1
2
Z
Z------->
2
1
22
2
2
4
C
C
ZR
ZR
=2
1
I
I=
6
1------> 6R
2 + 1,5 2CZ = 4R
2 +4 2CZ
2,52
CZ = 2R2 ------>
2
CZ = 2R2/2,5 = -----> ZC =
5
2R = 12 5 ()
- Khi n3 = 3n th 3 = 3; ZC3 = ZC /3 = 4 5 (). Chn p n A
Cu 31: Mch in xoay chiu, gm in tr thun R, cun dy thun cm c t cm L v t in c in dung C mc ni tip. t vo 2 u on mch mt in p xoay chiu u tn s 1000Hz. Khi mc 1
ampe k A c in tr khng ng k song song vi t C th n ch 0,1A. Dng in qua n lch pha so
http://www.hocmaivn.com/
http://www.hocmaivn.com/
http://www.hocmaivn.com/
http://www.hocmaivn.com/
vi in p hai u on mch gc /6 rad. Thay ampe k A bng vn k V c in tr rt ln th vn k
ch 20 V, in p hai u vn k chm pha hn in p hai u on mch /6 rad. t cm L v in
tr thun R c gi tr:
A. 3 /(40)(H) v 150 B. 3 /(2)v 150
C. 3 /(40) (H) v 90 D. 3 /(2)v 90
Gii:
Khi mc ampe k mch RL: I1 = 22
LZR
U
= 0,1 (A). Lc ny u sm pha hn i;
tan1 = R
Z L = tan6
=
3
1---> ZL =
3
R (1) v U = I1
22
LZR = 3
2,0 R (V) (2)
Khi mc vn k mch RLC: UC = UV = 20V
2 = -2
- (-
6
) = -
3
tan2 =
R
ZZ CL = - tan3
= - 3 ----> ZC ZL = R 3
----> ZC = R 3 + 3
R =
3
4R; Z2 =
22 )( CL ZZR = 2R
UC = 2Z
UZC = 3
2U----->
3
2U = 20 ---> U =
3
2,0 R= 10 3 ----> R = 150 ()
ZL = 3
R = 50 3 ----->2fL = 50 3 -----> L =
1000.2
350
=
.40
3
(H)
Chn p n A: L =.40
3
(H) ; R = 150 ()
Cu 32. Cho mch in nh hnh v: uAB = Uocost; in p hiu dng UDH = 100V; hiu in th tc thi
uAD sm pha 150o so vi hiu in th uDH, sm pha 105
o so vi hiu in th uDB v sm pha 90
o so vi
hiu in th uAB. Tnh Uo?
A. Uo = 136,6V. B. Uo = 139,3V. C. oU 100 2V . D. Uo = 193,2V.
Gii:
V gin nh hnh v. t lin tip cc vect
UAD ; UDH ; UHB
UAB = UAD + UDH + UHB Tam gic DHB vung cn.
UHB = UDH = 100V
UDB = 100 2 (V)
Tam gic ADB vung ti A c gc D = 75
0 ----->
UAB = UDB sin750 = 100 2 sin75
0
U0 = UAB 2 = 200sin750 = 193,18V
Hay U0 = 193,2 V
Chn p n D
Cu 33: Dng in i = 24cos t (A) c gi tr hiu dng l bao nhiu?
A D H B
300 45
0
H B
D
A
http://www.hocmaivn.com/
http://www.hocmaivn.com/
http://www.hocmaivn.com/
http://www.hocmaivn.com/
Gii: Ta c i = 24cos t = 2cos2t + 2 (A) Dng in qua mch gm hai thnh phn
- Thnh phn xoay chiu i1 = 2cos2t, c gi tr hiu dng I1 = 2 (A)
- Thnh phn dng in khng i I2 = 2 (A) C hai kh nng : a. Nu trong on mch c t in th thnh phn I2 khng qua mch. Khi gi tr hiu dng ca dng
in qua mch I = I1 = 2 (A) b. Nu trong mch khng c t th cng su ta nhit trong mch
P = P1 + P2 = I12R + I2
2 R = I
2R --------> I = 622
2
1 II (A)
Cu 34.
on mch AB gm mt ng c in mc ni tip vi mt cun dy. Khi t vo hai u AB mt in
p xoay chiu th in p hai u ng c c gi tr hiu dng bng U v sm pha so vi dng in l 12
.
in p hai u cun dy c gi tr hiu dng bng 2U v sm pha so vi dng in l 12
5. in p hiu
dng gia hai u on mch AB ca mng in l :
A. U 5 . B. U 7 . C. U 2 . D. U 3 .
Gii: Gi u1,u2 l in p gia hai u ng c v cun dy
u1 = U 2 cos(t + 12
). ; u2 = 2U 2 cos(t +
12
5).
T gin ta tnh c 2
ABU = U2
+ 4U2 - 2.2U
2 cos 120
0 = 7U
2
UAB = U 7 . Chn p n B
Cu 35: Cho mch xoay chiu R,L,C, c cun cm thun, L thay i c.iu chnh L thy ULmax= 2URmax.
Hi ULmax gp bao nhiu ln UCmax?
A 2/ 3 . B. 3 /2. C. 1/ 3 . D. 1/2
Gii:
Ta c UR = URmax = U v UC = UCmax = R
UZC khi trong mch c cng hng ZL = ZC
UL = ULmax khi ZL = C
C
Z
ZR 22 : (*)
ULmax =
122
22
L
C
L
C
Z
Z
Z
ZR
U=
L
C
Z
Z
U
1
= 2URmax = 2U
-----> 1 -L
C
Z
Z =
4
1-----> ZL =
3
4ZC (**)
T (*) v (**) suy ra ZC = R 3 Do UCmax = R
UZC = U 3
1200
http://www.hocmaivn.com/
http://www.hocmaivn.com/
http://www.hocmaivn.com/
http://www.hocmaivn.com/
Vy max
max
C
L
U
U=
3
2
U
U =
3
2, Chn p n A
Cu 36: Cho mch in xoay chiu RLC mc ni tip. in p xoay chiu t vo hai u on mch c
biu thc u = U 2 cost tn s gc bin i. Khi = 1 = 40 rad/s v khi = 2 = 360 rad/s th cng dng in hiu dng qua mch in c gi tr bng nhau. cng dng in trong mch t
gi tr ln nht th tn s gc bng
A 100(rad/s). B 110(rad/s). C 200(rad/s). D 120(rad/s).
Gii: I1 = I1 ----> Z1 = Z1 ------> (ZL1 ZC1)
2 = (ZL2 ZC2)
2
Do 1 2 nn (ZL1 ZC1) = - (ZL2 ZC2) ----> ZL1 + ZL2 = ZC1 + ZC2
(1 + 2)L = C
1 (
1
1
+
2
1
) ------> LC =
21
1
(*)
Khi I = Imax; trong mch c cng hng LC = 21
(**)
T (*) v (**) ta c = 21 = 120(rad/s). Chn p n D
Cu 37: Cho on mch xoay chiu mc ni tip gm on dy khng thun cm (L,r) ni vi t C Cun dy l mt ng dy c qun u vi chiu di ng c th thay i c.t vo 2 u mch mt HDT
xoay chiu.Khi chiu di ca ng dy l L th HDT hai u cun dy lch pha /3 so vi dng in. HDT
hiu dng 2 u t bng HDT hiu dng 2 u cun dy v cng dng in hiu dng trong mch l I..Khi tng chiu di ng dy ln 2 ln th dng in hiu dng trong mch l:
A. 0,685I B. I C. 2I/ 7 D. I/ 7
Cc thy cho e hi Khi tng chiu di ng dy ln 2 ln th L tng 2 ln th R c tng ko
Gii: Khi tng chiu di ng dy ln 2 ln (L tng 2 ln); th s vng dy ca mt n v chiu di n gim i 2 ln, t cm ca ng dy L gim 2 ln nn cm khn ZL gim hai ln cn in tr R ca ng dy
khng i.
Ta c : tand = R
Z L = tan3
= 3 -----> ZL = R 3 ----> Zd = 2R
Ud = UC -------> ZC = Zd = 2R. --------> Z = 2R 32
Do I = 322 R
U (*)
Sau khi tng chiu di ng dy ZL = 2
LZ = 2
3R
I= 22 )'( CL ZZR
U
=
22 )22
3( RR
R
U
= 3823
2
R
U (**)
I
I '=
3823
324
= 0,6847 --------> I = 0,685I. Chn p n A
Cu 38 : 1 on mch RLC . khi f1 =66 Hz hoc f2 =88 Hz th hiu in th 2 u cun cm khng i , f
= ? th ULmax
http://www.hocmaivn.com/
http://www.hocmaivn.com/
http://www.hocmaivn.com/
http://www.hocmaivn.com/
A 45,21 B 23,12 C 74,76 D 65,78
Gii: UL = IZL = 22 )
1(
CLR
LU
UL1 = UL2 -----> 2
1
1
2
1
)1
(C
LR
= 2
2
2
2
2
)1
(C
LR
2
1
1
+
2
2
1
= 42C2(2
C
L - R
2 ) (*)
UL = ULmax khi 22 )
1(
CLR
LU
=
2
22 )1
(
C
LR
UL
c gi tr max
hay y = 2
22 )1
(
C
LR = ymin ------> 2
2
= 42C2(2
C
L - R
2 ) (**)
T (*) v (**) ta c 2
2
=
2
1
1
+
2
2
1
hay
2
2
f =
2
1
1
f+
2
2
1
f
f = 2
2
2
1
21 2
ff
ff
= 74,67 (Hz). Chn p n C
Cu 39:
Cho mch in nh hnh v. in p t vo hai u on mch c gi tr hiu dng khng i nhng tn s thay i c. Khi tn s f = f1 th h s cng sut trn on AN l k1 = 0,6, H s cng sut trn ton mch l k = 0,8. Khi f = f2 = 100Hz th cng sut trn ton mch cc i. Tm f1 ?
A. 80Hz B. 50Hz C. 60Hz D. 70Hz
Gii: cos1 = 0,6 ------> tan1 = 3
4
tan1 = rR
Z L
=
3
4 -----> ZL =
3
4(R + r) (*)
cos = 0,8 ------> tan = 4
3
tan = rR
ZZ CL
=
4
3 ------> ZL ZC =
4
3(R +r) (**)
C
L
Z
Z =
2
1 LC v 2
2 LC = 1 ------> C
L
Z
Z=
2
2
2
1
=
2
2
2
1
f
f-----> f1 = f2
C
L
Z
Z
* Khi ZL ZC = 4
3(R +r) ------> ZC =
12
7(R +r) --->
C
L
Z
Z =
7
16
----> f1 = 7
4 2f = 151,2 Hz Bi ton v nghim
** Khi ZL ZC = - 4
3(R +r) ------> ZC =
12
25(R +r) -->
C
L
Z
Z =
25
16
C L; r R
N B A M
http://www.hocmaivn.com/
http://www.hocmaivn.com/
http://www.hocmaivn.com/
http://www.hocmaivn.com/
f1 = f2 C
L
Z
Z = f2.
5
4 = 80Hz. Chn p n A
Cu 40: t mt in p 2 osu U c t (U, khng i) vo on mch AB ni tip. Gia hai im AM l mt bin tr R, gia MN l cun dy c r v gia NB l t in C. Khi R = 75 th ng thi c bin
tr R tiu th cng sut cc i v thm bt k t in C no vo on NB d ni tip hay song song vi t in C vn thy UNB gim. Bit cc gi tr r, ZL, ZC, Z (tng tr) nguyn. Gi tr ca r v ZC l: A. 21 ; 120 . B. 128 ; 120 . C. 128 ; 200 . D. 21 ; 200 .
Gii: PR = I2R =
22
2
)()( CL ZZrR
RU
=
rR
ZZrR
U
CL 2)( 22
2
PR = PRmax khi R2 = r
2 + (ZL ZC)
2. (1)
Mt khc lc R = 75 th PR = PRmax ng thi UC = UCmax
Do ta c: ZC = L
L
Z
ZrR 22)( =
LZ
rR 2)( + ZL (2)
Theo bi ra cc gi tr r, ZL ZC v Z c gi tr nguyn ZC nguyn th (R+r)
2 = nZL (3) (vi n nguyn dng)
Khi ZC = n + ZL ------> ZC ZL = n (4)
Thay (4) vo (1) r2 + n
2 = R
2 = 75
2. (5)
Theo cc p n ca bi ra r c th bng 21 hoc 128. Nhng theo (5): r < 75
Do vy r c th r = 21 T (5) -----> n = 72.
Thay R, r, n vo (3) ---> ZL = 128 Thay vo (4) ----> ZC = 200
Chn p n D: r = 21 ; ZC = 200 .
Cu 41: Mt mch tiu th in l cun dy c in tr thun r= 8 m, tiu th cng sut P=32W vi h
s cng sut cos=0,8. in nng c a t my pht in xoay chiu 1 pha nh dy dn c in tr
R= 4. in p hiu dng 2 u ng dy ni my pht l
A.10 5 V B.28V C.12 5 V D.24V
Gii: cos =dZ
r=0,8 -----> Zd = 10 v ZL = 6,
Cng dng in qua mch I = r
P = 2 (A)
in p hiu dng 2 u ng dy ni my pht l
U = I 22)( LZrR = 2 22 612 = 12 5 (V) Chn p n C
Cu 42. Mch xoay chiu RLC gm cun dy c (R0, L) v hai t C1, C2. Nu mc C1//C2 ri ni tip vi cun dy th tn s cng hng l 1 = 48 (rad/s). Nu mc C1 ni tip C2 ri ni tip cun dy th tn s
cng hng l 2 = 100 (rad/s). Nu ch mc ring C1 ni tip cun dy th tn s cng hng l A = 70 rad/s B. = 50 rad/s C. = 74 rad/s D = 60 rad/s
Gii:
C// = C1 + C2; Cnt = 21
21
CC
CC
; =
LC
1-----> C =
L21
C// = L21
1
------> C1 + C2 =
L21
1
(*)
http://www.hocmaivn.com/
http://www.hocmaivn.com/
http://www.hocmaivn.com/
http://www.hocmaivn.com/
Cnt = L22
1
------->
21
21
CC
CC
=
L22
1
----> C1C2 =
L22
1
L21
1
=
22
2
2
1
1
L(**)
T (*) v (**)-------> C1 + 222
2
1
1
L 1
1
C =
L21
1
(***)
C1 = L2
1
(****)
Thay (****) vo (***) L2
1
+
22
2
2
1
2
L
L
=
L21
1
------>
2
1
+
2
2
2
1
2
=
2
1
1
-----> 22
2
1 + 4 = 22
2 -----> 4 - 222 + 22
2
1 = 0 (*****)
Phng trnh c hai nghim = 60 rad/s v = 80 rad/s Chn p n D
Cu 43 : Mch R, L, C ni tip . t vo 2 u mch in p xoay chiu u = U0cost (V), vi thay i c. Thay i UCmax. Gi tr UCmax l biu thc no sau y
A. UCmax = 2
C
2
L
U
Z1
Z
C. UCmax = 2
L
2
C
U.
Z1
Z
B. UCmax = 2 2
2U.L
4LC R C
D. UCmax = 2 2
2U
R 4LC R C
Gii:
UC = 22 )_( CL
C
ZZR
UZ
=
C
1
22 )1
(C
LR
U
= C
1
2
2242 1)2(CC
LRL
U
UC = UCmax khi 2 =
2
2
2
2
L
RC
L
v UCmax = C
1
2
42
4
4
L
RC
LR
U
= 224
2
CRLCR
LU
UCmax = 224
2CRLC
L
R
U
=
)4(4
22
2
2
CRLCL
R
U
=
)4 2
242
L
CR
L
CR
U
=
)4
1(12
242
L
CR
L
CR
U
=
22
)2
1(1L
CR
U
=
2
222
4
)2(
1L
CRC
L
U
=
22
4
22
4
)2(
1 CLL
RC
L
U
=
2241 CL
U
=
2
2
1C
L
Z
Z
U
Chn p n C.
Cu 44: Trong mt gi thc hnh mt hc sinh mun mt qut in loi 180 V - 120W hot ng bnh
thng di in p xoay chiu c gi tr hiu dng 220 V, nn mc ni tip vi qut mt bin tr.(coi
http://www.hocmaivn.com/
http://www.hocmaivn.com/
http://www.hocmaivn.com/
http://www.hocmaivn.com/
qut in tng ng vi mt on mch r-L-C ni tip) Ban u hc sinh bin tr c gi tr 70 th o thy cng dng in hiu dng trong mch l 0,75A v cng sut ca qut in t 92,8%.
Mun qut hot ng bnh thng th phi iu chnh bin tr nh th no? A. gim i 20 B. tng thm 12 C. gim i 12 D. tng thm 20
Gii : Gi R0 , ZL , ZC l in tr thun, cm khng v dung khng ca qut in. Cng su nh mc ca qut P = 120W ; dng in nh mc ca qut I. Gi R2 l gi tr ca bin tr khi
qut hot ng bnh thng khi in p U = 220V
Khi bin tr c gi tri R1 = 70 th I1 = 0,75A, P1 = 0,928P = 111,36W
P1 = I12R0 (1) ------> R0 = P1/I1
2 198 (2)
I1 = 2222
101 )(268
220
)()( CLCL ZZZZRR
U
Z
U
Suy ra
(ZL ZC )2 = (220/0,75)
2 2682 ------> ZL ZC 119 (3)
Ta c P = I2R0 (4)
Vi I = 22
20 )()( CL ZZRR
U
Z
U
(5)
P = 22
20
0
2
)()( CL ZZRR
RU
--------> R0 + R2 256 ------> R2 58
R2 < R1 ----> R = R2 R1 = - 12
Phi gim 12. Chn p n C Cu 45: t mt in p xoay chiu u vo hai u ca mt on mch gm in tr R mc ni tip vi mt t in c in dung C. in p t thi hai u in tr R c biu thc
uR = 50 2 cos(2ft + fi) (V). Vo mt thi im t no in p tc thi gia hai u on mch v hai
u in tr c gi tr u = 50 2 V v uR = -25 2 V. Xc nh in p hiu dng gia hai bn t in.
60 3 V. B. 100 V. C. 50 2 V. D. 50 3 V
Gii:
uR = 50 2 cos(2ft + ) (V). -----> UR = 50 (V)
Ti thi im t: u = 50 2 ;(V) uR = -25 2 (V) u = 2uR-----> Z = 2R
Z2 = R
2 + ZC
2 ------> ZC
2 = 3R
2 -----> ZC = R 3 ----->
UC = UR 3 = 50 3 (V) Chn p n D
http://www.hocmaivn.com/
http://www.hocmaivn.com/
http://www.hocmaivn.com/
http://www.hocmaivn.com/
Cu 46 : t mt in p u = 80cos(t) (V) vo hai u on mch ni tip gm in tr R, t in C v cun dy khng thun cm th thy cng sut tiu th ca mch l 40W, in p hiu dng UR = ULr = 25V; UC = 60V. in tr thun r ca cun dy bng bao nhiu? A. 15 B. 25 C. 20 D. 40
Gii: Ta c Ur
2 + UL
2 = ULr
2
(UR + Ur)2 + (UL UC)
2 = U
2
Vi U = 40 2 (V) Ur
2 + UL
2 = 25
2 (*)
(25+ Ur)2 + (UL 60)
2 = U
2 = 3200
625 + 50Ur + Ur2 + UL
2 -120UL + 3600 = 3200
12UL 5Ur = 165 (**) Gii h phng trnh (*) v (**) ta c
* UL1 = 3,43 (V) ----> Ur1 = 24,76 (V)
nghim ny loi v lc ny U > 40 2 * UL = 20 (V) ----> Ur = 15 (V)
Lc ny cos = U
UU rR = 2
1
P = UIcos -----> I = 1 (A)
Do r = 15 . Chn p n A
Cu 46: Mng in 3 pha c hiu in th pha l 120 V c ti tiu th mc hnh sao, cc ti c in tr l
R1 = R2 = 20 ; R3 = 40 . Tnh cng dng in trong dy trung ho.
A. 6 A B. 3 A C. 0 A D. 2 3 A
Gii: Dng in qua cc ti l I = R
U P I1 = I2 = 6A; I3 = 3 A
Dng in qua dy trung tnh i = i1 + i2 + i3 Dng phng php cng vc t ta c
I = I1 + I2 + I3
Gc gia i1, i2., i3 l 2 /3
t lin tip cc vc t cng dng in nh hnh v,
ta c tam gic u Theo hnh v ta c I = I3 = 3 A
Chn p n B: 3A
Cu 47: Cho mch in RLC, t in c in dung C thay i.
iu chnh in dung sao cho in p hiu dng ca t t gi tr cc i, khi in p hiu dng trn R l 75 V. Khi in
p tc thi hai u mch l 75 6 V th in p tc thi ca
on mch RL l 25 6 V in p hiu dng ca on mch l
A. 75 10 V. B. 75 3 V C. 150 V. D. 150 2 V
Gii: V gin vect nh hnh v. Ta thy
UC = UCmax khi = 900 tc khi uRL vung pha vi u
2
maxCU = U2 +
2
RLU
ULr
U
UC
UL
Ur UR
I3
I2 I1
I3
I
I1
I2
I
I3
I2
I1
O
C
UR
UR
L
http://www.hocmaivn.com/
http://www.hocmaivn.com/
http://www.hocmaivn.com/
http://www.hocmaivn.com/
Khi u = 75 6 V th uRL = 25 6 V ----> Z = 3ZRL hay U = 3URL
---> 2maxCU = U
2 + 2
RLU = 102
RLU .
Trong tam gic vung hai cnh gc vung U; URL; cnh huyn UC ng cao thuc cnh huyn UR ta c: U.URL = URUC
3 2RLU = 10 URLUR ----> 3URL = 10 UR = 75 10
----> URL = 25 10 (V). Do U = 75 10 (V). p n A
Cu 48: Mt mch tiu th in l cun dy c in tr thun r= 8, tiu th cng sut P=32W vi h s
cng sut cos=0,8. in nng c a t my pht in xoay chiu 1 pha nh dy dn c in tr R=
4. in p hiu dng 2 u ng dy ni my pht l
A.10 5 V B.28V C.12 5 V D.24V
Gii: cos =dZ
r=0,8 -----> Zd = 10 v ZL = 6,
Cng dng in qua mch I = r
P = 2 (A)
in p hiu dng 2 u ng dy ni my pht l
U = I 22)( LZrR = 2 22 612 = 12 5 (V) Chn p n C
Cu 49: Mt cun dy khng thun cm ni tip vi t in C thay i c trong mch in xoay chiu
c in p u = U0 cost (V). Ban u dung khng ZC, tng tr cun dy Zd v tng tr Z ton mch bng
nhau v u bng 100. Tng in dung thm mt lng C =
310.125,0 (F) th tn s dao ng ring
ca mch ny khi l 80 rad/s. Tn s ca ngun in xoay chiu bng:
A. 80 rad/s. B. 100 rad/s. C. 40 rad/s. . D.50 rad/s.
Gii: Do ZC = Zd = Z.--------> UC = Ud = U. = 100I
V gin vc t nh hnh v. ta suy bra UL = Ud/2 = 50I
---> 2ZL = -----ZL = 50
Vi I l cng dng in qua mch
ZL = L; ZC = C
1----->
C
L= CLZZ = 5000 (*)
= )(
1
CCL = 80-------> L(C+ C) =
2)80(
1
(**)
5000C(C+C) = 2)80(
1
---->
C2
+(C)C - 5000.)80(
12
= 0----> C2
+
310.125,0 C -
5000.)80(
12
= 0--->
C2
+ 8
10. 3C -
4.8
102
6
= 0 ------> C = 8
10. 3F
ZC = C
1 = 100 -----> =
CZC
1 = 80 rad/s. Chn p n A
UC
Ud
U
UL
http://www.hocmaivn.com/
http://www.hocmaivn.com/
http://www.hocmaivn.com/
http://www.hocmaivn.com/
Cu 50 Mt cun dy khng thun cm ni tip vi t in c in dung C thay i c trong mch
in xoay chiu c in p u = U0cost (V). Ban u dung khng ZC v tng tr ZLr ca cun dy v Z ca
ton mch u bng 100. Tng in dung thm mt lng C = 0,125.10-3/ (F) th tn s dao ng ring
ca mch ny khi l 80 rad/s. Tn s ca ngun in xoay chiu bng
A. 80rad/s B. 100rad/s C. 40rad/s D. 50rad/s
Gii: V gin vect
ZC = ZLr = Z = 100-----. ZL = 2
CZ = 50
ZL.ZC =C
L = 5000 (2) -----> L = 5.103C (*)
2
0 = )(
1
CCL -----> 5.10
3C
2 + 5.10
3C.C - 2
0
1
= 0
---> 5.103C
2 + 5.10
3
310.125,0 .C -
2280
1
= 0
----> 5.103C
2 +
.625,0C -
6400
1 = 0
-----> C =
.25,1.10
-4 (F);
------>. = CZC
1 =
410.25,1
..100
1
= 80 rad/s. Chn p n A
Cch 2 ; ZLr = Z ------> r2 + ZL
2 = r
2 + (ZL ZC)
2 -------> ZC = 2ZL -----> ZL =
2
CZ = 50
Khi tng thm C = C + C th ZC = ZL = 2
CZ -------> C = 2C ----> C = C =
.25,110
-4F
------>. = CZC
1 =
410.25,1
..100
1
= 80 rad/s. Chn p n A
Cu 51 : t in p xoay chiu u U 2 cos(100 t)V vo on mch RLC. Bit R 100 2 , t in c
in dung thay i c. Khi in dung t in ln lt l
251 C (F) v
3
1252 C (F) th in p hiu
dng trn t c cng gi tr. in p hiu dng trn in tr R t cc i th gi tr ca C c th l:
A.
50C (F). B.
3
200C (F)., C.
20C (F). D.
3
100C (F)
Gii
Ta c 112 2
1( )
CC
L C
UZU
R Z Z
22
2 2
2( )
CC
L C
UZU
R Z Z
UC1 = UC2 --------->> 2 2
1 2
2 2 2 2
1 2( ) ( )
C C
L C L C
Z Z
R Z Z R Z Z
UL
Ud
UC
U
C
http://www.hocmaivn.com/
http://www.hocmaivn.com/
http://www.hocmaivn.com/
http://www.hocmaivn.com/
ZC1 = 400; ZC2 = 240
-----> R2 + ZL
2 =
21
212
CC
CCL
ZZ
ZZZ
=
240400
240.400.2
LZ = 300ZL
in p hiu dng trn in tr R t cc i th trong mch c cng hng ZL = ZC
Thay R =100 2 ; :
- ZC2
- 300ZC +20000 = 0
Phng trnh c hai nghim : ZC = 200 v ZC = 100
Khi ZC = 200 th C = 410 50
2F F
Khi ZC = 100 th C = 410 100
F F
Chn p n A
Cu 52: t vo hai u mch in RLC ni tip mt hiu in th xoay chiu c gi tr hiu dng khng i th hiu in th hiu dng trn cc phn t R, L v C u bng nhau v bng 20V. Khi t b ni tt th
in p dng hai u in tr R bng:
A. 10V. B. 10 2 V. C. 20V. D. 20 2 V.
Gii: Do UR = UL = UC trong mch c cng hng , nn U = UR = 20V
Khi t b ni tt UL = UR = 2
U = 10 2 (V). Chn p n B
Cu 53: Cho mch in xoay chiu khng phn nhnh RLC c tn s thay i c.Gi f0 ;f1 ;f2 ln lt
cc gi tr tn s lm cho hiu in th hiu dung hai u in tr cc i,hiu in th hiu dung hai u cun cm cc i,hiu in th hiu dung hai u t in cc i.Ta c :
A.f0 = 2
1
f
f B. f0 =
1
2
f
f C.f1.f2 = f0
2 D. f0 = f1 + f2
Gii: UR = Urmax khi trong mch c cng hng in ZL = ZC -----> f02 =
LC24
1
(1)
UC = UCmax khi ZC2 = 2
2
2
2
L
L
Z
ZR -----> R
2 = ZL2ZC2 ZL2
2 (*)
UL = ULmax khi ZL1 = 1
2
1
2
C
C
Z
ZR -----> R
2 = ZL1ZC1 ZC1
2 (**)
T (*) v (**) suy ra ZL1ZC1 ZC12 = ZL2ZC2 ZL2
2
ZL.ZC = C
L suy ra ZC1 = ZL2 ----->
Cf12
1
= 2f2L -----> f1f2 =
LC24
1
(2)
T (1) v (2) ta c f1f2 = f02 Chn p n C
Cu 54 : Mt mch in xoay chiu gm AM ni tip MB. Bit AM gm in tr thun R1, t in C1,
cun dy thun cm L1 mc ni tip. on MB c hp X, bit trong hp X cng c cc phn t l in tr thun, cun cm, t in mc ni tip nhau. t in p xoay chiu vo hai u mch AB c tn s 50Hz
v gi tr hiu dng l 200V th thy dng in trong mch c gi tr hiu dng 2A. Bit R1 = 20 v nu
thi im t (s), uAB = 200 2 V th thi im ( t+1/600)s dng in iAB = 0(A ) v ang gim. Cng sut ca on mch MB l:
A. 266,4W B. 120W C. 320W D. 400W
http://www.hocmaivn.com/
http://www.hocmaivn.com/
http://www.hocmaivn.com/
http://www.hocmaivn.com/
Gii:
Gi s in p t vo hai u mch c biu thc u = U 2 cost = 200 2 cos100t (V). Khi cng
dng in qua mch c biu thc i = 2 2 cos(100t -) vi gc lch pha gia u v i
Ti thi im t (s) u = 200 2 (V) -----> cost = 1. Do cng dng in ti thi im ( t+1/600)s
i = 0 ------> i = 2 2 cos[100(t + 600
1) -] = 0------> cos(100t +
6
-) = 0
----> cos100t.cos(6
-) - sin100t.sin(
6
-) = 0 -----> cos(
6
-) = 0 (v sin100t = 0 )--->
= 6
-
2
= -
3
----->
Cng sut ca on mch MB l: PMB = UIcos - I2R1 = 200.2.0,5 4. 20 = 120W. Chn p n B
Cu 55: Trong li in dn dng ba pha mc hnh sao, in p mi pha l u1 = 220 2 cos(100t) (V) ,
u2 = 220 2 cos(100t + 3
2) (V), u3 = 220 2 cos(100t -
3
2) (V), . Bnh thng vic s dng in
ca cc pha l i xng v in tr mi pha c gi tr R1=R2=R3 = 4,4. Biu thc cng dng in trong dy trung ho tnh trng s dng in mt cn i lm cho in tr pha th 1 v pha th 3 gim i mt na l:
A. i = 50 2 cos(100t +3
) (A) B. i = 50 2 cos(100t +) (A)
C. i = 50 2 cos(100t +3
2) (A) D. i = 50 2 cos(100t -
3
) (A)
Gii: Do cc ti tiu th l cc in tr thun nn u v i lun cng pha
Khi mt cn i cc pha
I1 = I3 = 2,2
220 = 100 (A)
I2 = 4,4
220 = 50 (A). V gin vc t :
I0 = I1 + I2 + I3 = I13 + I2
I13 = I1 = I3 = 100A
I0 = I13 I2 = 50 (A)
0 = - 3
Do biu thc cng dng in trong dy trung ho
i = 50 2 cos(100t -3
) (A) Chn p n D
Cu 56: on mch AB gm cun dy thun cm c t cm L c th thay i mc gia A v M, in
tr thun mc gia M v N, t in mc gia N v B mc ni tip. t vo hai u A , B ca mch in mt in p xoay chiu c tn s f, in p hiu dng U n nh. iu chnh L c uMB vung pha vi
uAB, sau tng gi tr ca L th trong mch s c A. UAM tng, I gim. B. UAM gim, I gim. C. UAM gim, I tng. D. UAM tng, I tng.
Gii: V gin vect nh hnh v. Theo L hm sin
I1
-2/3
2/3
- /3
I0
I3
I2
I1
3
http://www.hocmaivn.com/
http://www.hocmaivn.com/
http://www.hocmaivn.com/
http://www.hocmaivn.com/
sinAMU =
sinABU ------> UAM =
sin
sinABU
Do gc , UAB xc nh nn UAM c gi tr ln nht khi = 900
Tc l khi uMB vung pha vi uAB th UAM c gi tr ln nht.
Do vy khi tng L th UAM gim Cng dng in qua mch
I = 22 )
1(
CLR
U AB
ta thy khi L tng th mu s tng do I gim
Chn p n B: UAM gim, I gim
Cu 57.t mt in p xoay chiu u = U0cos100t (V) vo hai u ca mt in tr thun R th trong
mch c dng in vi cng hiu dng I. Nu t in p vo hai u on mch gm in tr
thun R mc ni tip vi mt it bn dn c in tr thun bng khng v in trngc rtln th cng hiu dng ca dng in trong mch bng
A. 2I B.I 2 C.I D. I/ 2
Gii: Xt thi gian mt chu k
Lc ch c in tr thun R : P = I2R = 2
2
0 RI
Lc mc thm it, dng in qua Rch trong mt na chu ki P = I2R = 2
P=
4
2
0 RI
------> I
I ' =
2
1 -----> I =
2
I. Chn p n D
Cu 58: t mt in p xoay chiu c gi tr hiu dng U v tn s f khng i vo hai u on mch gm bin tr R mc ni tip vi t in c in dung C. Gi in p hiu dng gia hai u bin tr, gia
hai u t in v h s cng sut ca on mch khi bin tr c gi tr 1R ln lt l 1 1 1, , osR CU U c . Khi
bin tr c gi tr 2R th cc gi tr tng ng ni trn ln lt l 2 2 2, , osR CU U c bit rng s lin h:
1
2
0,75R
R
U
U v 2
1
0,75C
C
U
U . Gi tr ca 1osc l: A. 1 B.
1
2 C. 0,49 D.
3
2
Gii:
2
1
R
R
U
U =
4
3------> UR2 =
9
16UR1 (*)
1
2
C
C
U
U =
4
3------> UC2 =
16
9UC1 (**)
U2 =
2
1RU + 2
1CU = 2
2RU + 2
2CU = (9
16)2 2
1RU + (16
9)
2 21CU -------->
(9
16)2 2
1RU - 2
1RU = 2
1CU - (16
9)2 2
1CU -------->2
1CU = (9
16)
2 21RU ------>
UAM
UMB
UA
B
http://www.hocmaivn.com/
http://www.hocmaivn.com/
http://www.hocmaivn.com/
http://www.hocmaivn.com/
U2 = 2
1RU + 2
1CU = [(1 + (9
16)2] 2
1RU --------> U = 9
169 22 UR1
cos1 = U
U R1 = 22 169
9
= 0,49026 = 0,49. Chn p n C
Cu 59: t mt in p u = U 2 cos(110t /3) (V) vo hai u on mch mc ni tip gm in tr R (khng i), cun dy thun cm c L = 0,3 H v mt t in c in dung C thay i c. Cn phi iu chnh in dung ca t n gi tr no in tch trn bn t in dao ng vi bin ln nht?
A. 26,9 F. B. 27,9 F. C. 33,77 F. D. 23,5 F
Gii: Gi s in tch gia hai bn cc t in bin thin theo phg trnh q = Q0 cos(t + )
Khi dng in qua mch c biu thc: i = q = -Q0sin(t + ) = I0cos((t + + 2
)
Vi I0 = Q0-----> Q0 c gi tr ln nht khi I0 c gi tr ln nht
---> I = Ic tc l khi trong mch c s cng hng ---- > ZC = ZL
Do C = L2
1
=
3,0.)110(
12
= 27,9 F. Chn p n B
Cu 60: Cho on mch xoay chiu RLC mc ni tip. Cho cc gi tr R = 60 m; ZC =600 m; ZL=140
m.t vo hai u on mch mt in p xoay chiu c tn s f = 50Hz. Bit in p gii hn (in p nh thng) ca t in l 400V. in p hiu dng ti a c th t vo hai u on mch t in khng b nh thng l
A. 400 2 V. B. 471,4 V. C. 666,67 V. D. 942,8 V.
Gii: Tng tr Z = 22 )( CL ZZR = 215200 = 464 ()
UC = Z
UZC =
464
600U UCmax = 400 2 (V) ------> U
600
464400 2 = 437,5 (V).
Chn p n A
Cu 61. Ni hai cc ca mt my pht in xoay chiu mt pha c 5 cp cc t vo hai u on mch
.AB gm in tr thun R=100, cun cm thun c t cm L=6
41H v t in c in dung C =
3
10 4F. Tc rto ca my c th thay i c. Khi tc rto ca my l n hoc 3n th cng
dng in hiu dng trong mch c cng gi tr I. Gi tr ca n bng
A. 10vng/s B. 15 vng/s C. 20 vng/s D. 5vng/s
Gii: Sut in ng cc i ca ngun in: E0 = N0 = 2fN0 => U = E = 2
0E (coi in tr trong
ca my pht khng ng k). Cng dng in qua mch I = Z
U
Vi f = np n tc quay ca roto, p s cp cc t Do I1 = I2 ta c:
2
1
1
2
2
1
)1
(C
LR
=2
2
2
2
2
2
)1
(C
LR
-------> ])1
([ 2
2
2
22
1C
LR
= ])1
([ 2
1
1
22
2C
LR
http://www.hocmaivn.com/
http://www.hocmaivn.com/
http://www.hocmaivn.com/
http://www.hocmaivn.com/
---> C
L
CLR 2122
2
2
122
2
2
1
22
1 2
=
C
L
CLR 2222
1
2
222
2
2
1
22
2 2
---> )2)(( 2222
1C
LR = )(
12
2
2
1
2
1
2
2
2
C =
2
2
2
1
2
1
2
2
2
1
2
2
2
))((1
C
-----> 2
2
2
1
11
= (2
C
L- R
2 )C
2 =
2
3
9
10.4
(*)
= 2f = 2np
2
2
2
1
11
=
224
1
p(
2
2
2
1
11
nn ) =
224
1
p(
2
1
n +
29
1
n) =
22236
10
np =
222 536
10
n (**)
-------> 222 536
10
n =
2
3
9
10.4
------> n2 =
22 536
10
3
2
10.4
9
= 25----->
n = 5 vng /s. Chn p n D
Cu 62: Cho on mch R,L,C ni tip, in p gia hai u on mch
u = 220 2 cos2ft (V); R =100; L l cun cm thun, L = 1/(H); T in c in dung C v tn s f thay i c. iu chnh C= CX, sau iu chnh tn s, khi f = fX th in p hiu dng gia hai bn t
C t cc i; gi tr ln nht ny gp 5/3 ln in p hiu dng gia hai u on mch. Gi tr CX, v tn s fX bng
Gii:
UC = 22 )_( CL
C
ZZR
UZ
=
C
1
22 )1
(C
LR
U
= C
1
2
2242 1)2(CC
LRL
U
UC = UCmax khi 2 =
2
2
2
2
L
RC
L
v UCmax = C
1
2
42
4
4
L
RC
LR
U
= 224
2
CRLCR
LU
=
3
5U
----> 6L = 5R 224 CRLC ------> R2C2 4LC +
2
2
25
36
R
L
-----> C = 2
6,12
R
LL = (21,6).
410F ------> c 2 gi tr ca C: C1 =
410..6,3 F v C2 =
510.4 F
2 = 2
2
2
2
L
RC
L
= LC
1 -
2
2
2L
R > 0 -----> C <
2
2
R
L =
410.2 F.-----> loi nghim C1
CX = C2 =
510.4 F ----> 2 =
2
1
LC -
2
2
2L
R =
4
10 25 -
2
100 22= 2.10
42 -----> = 100 2 rad/s
Do fX = 50 2 Hz p s CX =
510.4 F v fX = 50 2 Hz
Cu 63: Cho mch in gm cun dy c in tr hot ng R ni tip t C. t vo hai u mch in
mt in p xoay chiu n nh u = U 2 cost. Khi C = C0 th in p hiu dng gia hai u cun dy ln nht bng 2U. Vi gi tr no ca C th UC t cc i?
A. C = 03C
4. B. C = 0
C
2. C. C = 0
C
4. D. C = 0
C
3.
http://www.hocmaivn.com/
http://www.hocmaivn.com/
http://www.hocmaivn.com/
http://www.hocmaivn.com/
Gii:
Ta c Ud = I22
LZR ; Ud = Udmax khi I = Imax mch c cng hng ZL = ZC0 (*)
Udmax = 2U----> Zd = 2Z = 2R ( v ZL = ZC0)-----> R2 + ZL
2 = 4R
2 ----> R =
3
LZ = 3
0CZ (**)
UC = UCmax khi ZC = L
L
Z
ZR 22 =
0
2
0
2
0
3
C
CC
Z
ZZ
= 3
4 0CZ -
---> ZC = 3
4 0CZ -----> C = 4
3 0C Chn p n A
Cu 64: t mt in p xoay chiu )(cos0 VtUu vo hai u mch in AB mc ni tip theo th t
gm in tr R, cun dy khng thun cm (L, r) v t in C vi rR . Gi N l im nm gia in tr
R v cun dy, M l im nm gia cun dy v t in. in p tc thi uAM v uNB vung pha vi nhau
v c cng mt gi tr hiu dng l V530 . Gi tr ca U0 bng:
A. 2120 V. B. 120V. C. 260 V. D. 60 V. Gii: Do R = r ---> UR = Ur
Ta c :(UR + Ur)2 + 2LU =
2
AMU
----> 4 2RU + 2
LU = 2
AMU (1)
2RU + (UL UC)2 = 2NBU (2)
UAM = UNB -----> ZAM = ZNB ------>
4R2 + ZL
2 = R
2 + (ZL ZC)
2
3R2 + ZL
2 = (ZL ZC)
2 (*)
uAM v uBN vung pha ----> tanAM.tanNB = -1
R
Z L
2 R
ZZ CL = -1---->(ZL ZC)2 = 2
24
LZ
R (**)
T (*) v (**) 3R2 + ZL2 =
2
24
LZ
R
------> ZL4 + 3R
2ZL
2 4R2 = 0 -----> ZL
2 = R
2
Do UL2 = UR
2 (3). T (1) v (3)----> 5UR
2 = 2AMU = (30 5 )
2 -----> UR = 30 (V)
UR = UL =30 (V) (4)
2
RU + (UL UC)2 = 2NBU ------>(UL UC)
2 = (30 5 )2 302 = 4.302
UAB2 = :(UR + Ur)
2 + (UL UC)
2 = 4UR
2 + (UL UC)
2 = 2.4.30
2
---> UAB = 60 2 (V)-------> U0 = UAB 2 = 120 (V). Chn p n B
Cu 65: Cho mch in RL ni tip, cun dy thun cm, L bin thin t 0 in p hiu dng t vo hai u on mch l U. Hi trn gin vc t qu tch ca u mt vc t I l ng g?
A. Na ng trn ng knh R
U B. on thng I = kU, k l h s t l.
C. Mt na hiperbol 22
LZR
U
D. Na elip
2
0
2
U
u +
2
0
2
I
i=1
UC
UA
M
UL
UAB
Ur
UR 2U
R
UN
B
http://www.hocmaivn.com/
http://www.hocmaivn.com/
http://www.hocmaivn.com/
http://www.hocmaivn.com/
Gii Ta c I = 22
LZR
U
------> Trn gin vc t qu tch ca u mt vc t I l mt na hiperbol 22
LZR
U
Chn p n C
Cu 66. Stato ca mt ng c khng ng b ba pha gm 9 cun dy, cho dng in xoay chiu ba pha tn s 50Hz vo ng c. Rto lng sc ca ng c c th quay vi tc no sau y? A. 1000vng/min. B. 900vng/min. C. 3000vng/min. D. 1500vng/min.
Gii: p dng cng thc f = np. vi p l s cp cc t. ng c khng ng b 3 pha mi cp cc t ng vi 3 cun dy stato. Do p = 3. n l tc quay ca t trng.
--------> n = p
f=
3
50 vng/s =
3
50.60 vng/min = 1000 vng/min.
Tc quay ca roto ng c n < n nn c th l n = 900 vng /min. Chn p n B
Cu 67: t in p xoay chiu u=U0cost (U0 khng i v thay i c) vo hai u on mch
gm in tr thun R,cun cm thun c t cm L v t in c in dung C mc ni tip,vi CR2<
2L.Khi = 1 hoc = 2 th in p hiu dng gia hai u cun cm c cng mt gi tr.Khi = 0
th in p hiu dng gia hai u cun cm c gi tr cc i.H thc lin h gia 1,2 v 0 l :
A. )(2
1 22
2
1
2
0 B. )(2
1210 B. 2
0
1
=
2
1 (
2
1
1
+
2
2
1
) C. 0 = 21
Gii: UL = 22 )( CL
L
ZZR
UZ
. Do UL1 = UL2 ----->
2
1
1
2
2
1
)1
(C
LR
= 2
2
2
2
2
2
)1
(C
LR
-----> 2
1
2 2
C
LR
+ 24
1
1
C=
2
2
2 2
C
LR
+ 24
2
1
C
------> (2C
L- R
2)(
2
2
1
-
2
1
1
) =
24
2
1
C-
24
1
1
C -----> (2
C
L- R
2) =
2
1
C 22
2
1
2
2
2
1
-----> 2
1
1
+
2
2
1
= C
2 (2
C
L- R
2) (*)
UL = ULmax khi 2
2 2
C
LR
+ 24
1
C + L
2 c gi tr cc tiu.----->
2
0
1
=
2
2C(2
C
L- R
2) (**)
T(*) v (**) suy ra:2
0
1
=
2
1 (
2
1
1
+
2
2
1
) . Chn p n C. Vi iu kin CR2< 2L
Cu 68: Cho mch in AB c hiu in th khng i gm c bin tr R, cun dy thun cm L v t in C mc ni tip. Gi U1, U2 , U3 ln lt l hiu in th hiu dng trn R, L v C. Bit khi U1 =
100V, U2 = 200V, U3 = 100 V. iu chnh R U1 = 80V, lc y U2 c gi tr
A. 233,2V. B. 100 2 V. C. 50 2 V. D. 50V.
Gii:
http://www.hocmaivn.com/
http://www.hocmaivn.com/
http://www.hocmaivn.com/
http://www.hocmaivn.com/
U = 2322
1 )( UUU = 2
32
2
1 )''(' UUU = 100 2 (V)
Suy ra : (U2 U3)2 = U
2 U1
2 = 13600
U2 U3 = I(Z2 Z3) =100 (V) (*)
U2 U3 = I(Z2 Z3) = 13600 (V) (**)
T (*) v (**) ------> I
I '=
100
13600------>
2
2'
U
U =
2
2'
IZ
ZI=
I
I '=
100
13600---->
U2 = 100
13600U2 = = 233,2 V. Chn p n A
Cu 69 Mc vo on mch RLC khng phn nhnh gm mt ngun in xoay chiu c tn s thay i
c. tn s 1 60f Hz , h s cng sut t cc i cos 1 . tn s 2 120f Hz , h s cng sut
nhn gi tr cos 0,707 . tn s 3 90f Hz , h s cng sut ca mch bng
A. 0,872 B.0,486 C. 0,625 D. 0,781
Gii: Ta c ZL1 = ZC1 -----> 1L = C1
1
-----> LC =
2
1
1
(1)
cos2 = 0,707 -----> 2 = 450 ---->
tan2 = R
ZZ CL 22 =1 -----> R = ZL2 - ZC2
tan3 = R
ZZ CL 33 = 22
33
CL
CL
ZZ
ZZ
=
CL
CL
2
2
3
3
1
1
= 3
2
1
1
2
1
2
2
2
1
2
3
= 3
2
2
1
2
2
2
1
2
3
=
3
2
f
f2
1
2
2
2
1
2
3
ff
ff
tan3 = 3
2
f
f2
1
2
2
2
1
2
3
ff
ff
=
3
422
22
60120
6090
=
3
4
12
5 =
9
5 -----> (tan3)
2 = 25/91---->
81
106
81
251
cos
1
3
2
-------> cos
23 = 81/106 ------> cos3 = 0,874. p n A
Cu 70: Mt on mch AB gm hai on mch AM v MB mc ni tip. on mch AM gm in tr
thun R mc ni tip vi t in C c in dung thay i c, on mch MB l cun dy thun cm c t cm L. Thay i C in p hiu dng ca on mch AM t cc i th thy cc in p hiu
dng gia hai u in tr v cun dy ln lt l UR = 100 2 V, UL = 100V. Khi in p hiu dng
gia hai u t in l:
A. UC = 100 3 V B. UC = 100 2 V C. UC = 200 V D. UC = 100V
Gii:
Ta c UAM = 22
22
)( CL
C
ZZR
ZRU
=
22
22 )(
1
C
CL
ZR
ZZR
=
22
2 21
1
C
CLL
ZR
ZZZ
UAM = UAMmax th biu thc y = 22
2 2
C
CLL
ZR
ZZZ
= ymin ------> o hm y = 0
---> (22
CZR )(-2ZL) ( CLL ZZZ 22 )2ZC = 0 ZC
2 ZLZC R2 = 0
Hay UC2 ULUC UR
2 = 0 UC
2 100UC 20000 = 0
UC = 200(V) (loi nghim m). Chn p n C
http://www.hocmaivn.com/
http://www.hocmaivn.com/
http://www.hocmaivn.com/
http://www.hocmaivn.com/
Cu 71: Mch in R1L1C1 c tn s cng hng 1 v mch R2L2C2 c tn s cng hng 2 , bit
1=2. Mc ni tip hai mch vi nhau th tn s cng hng ca mch s l . lin h vi 1v
2theo cng thc no? Chn p n ng:
A. =21. B. = 31. C. = 0. D. = 1.
Gii:
2 = LC
1 =
21
2121 )(
1
CC
CCLL
2
1 = 11
1
CL ---> L1 =
1
2
1
1
C ; 22 =
22
1
CL----->L2 =
2
2
2
1
C
L1 + L2 = 1
2
1
1
C +
2
2
2
1
C =
2
1
1
(
1
1
C +
2
1
C) =
2
1
1
21
21
CC
CC
( v 1=2.)
----> 21 =
21
2121 )(
1
CC
CCLL
= 2 --------> = 1. p n D
Cu 72. Dng in xoay chiu c chu k T, nu tnh gi tr hiu dng ca dng in trong th gian T/3 l 3(A), trong T/4 tip theo gi tr hiu dng l 2(A) v trong 5T/12 tip theo na gi tr hiu dng l
2 3 (A). Tm gi tr hiu dng ca dng in:
A. 4 (A). B. 3 2 (A). C. 3 (A). D. 5(A).
Gii: Nhit lng ta ra trn in tr R ca mch trong thi gian: t1 = T/3: Q1 = I1
2Rt1 = 9RT/3 = 3RT
t2 = T/4: Q2 = I22Rt2 = 4RT/4 = RT
t3 = 5T/12: Q3 = I32Rt3 = 12R.5T/12 = 5RT
t = t1 + t2 + t3 = T l Q = I2Rt = I
2RT
M Q = Q1 + Q2 + Q3 = 9RT-------> I2 = 9 -----> I = 3 (A). Chn p n C
Cu 73 : t in p xoay chiu c gi tr hiu dng khng i 150 V vo on mch AMB gm on AM ch cha in tr R, on mch MB cha t in c in dung C mc ni tip vi mt cun cm thun c t cm L thay i c. Bit sau khi thay i t cm L th in p hiu dng hai u mch
MB tng 2 2 ln v dng in trong mch trc v sau khi thay i lch pha nhau mt gc 2
. Tm in
p hiu dng hai u mch AM khi cha thay i L?
A. 100 V. B. 100 2 V. C. 100 3 V. D. 120 V.
Gii:
tan1 = 1
11
R
CL
U
UU ; tan2 =
2
22
R
CL
U
UU
1 - 2 = /2 -------> tan1 tan2 = 1
11
R
CL
U
UU
2
22
R
CL
U
UU = -1
(UL1 UC1)2 .(UL2 UC2)
2 =
2
1RU2
2RU .-------> 2
1MBU2
2MBU = 2
1RU2
2RU .------>
84
1MBU =2
1RU2
2RU .(*) (v UMB2 = 2 2 UMB1)
Mt khc 21RU + 2
1MBU = 2
2RU + 2
2MBU (= U2) ----->
2
2RU = 2
1RU - 72
1MBU (**)
T (*) v (**): 8 4 1MBU =2
1RU2
2RU = 2
1RU (2
1RU - 72
1MBU )
-----> 4
1RU - 72
1MBU2
1RU - 84
1MBU = 0 ------> 2
1RU = 82
1MBU
http://www.hocmaivn.com/
http://www.hocmaivn.com/
http://www.hocmaivn.com/
http://www.hocmaivn.com/
21RU + 2
1MBU = U2 ------> 2
1RU + 8
2
1RU = U2
----> UR1 = 3
22U = 100 2 (V). Chn p n B
Cu 74: t in p xoay chiu c gi tr hiu dng 60 V vo hai u on mch R, L, C mc ni tip th
cng dng in qua on mch l i1 = 0I cos(100 t )4
(A). Nu ngt b t in C th cng
dng in qua on mch l 2 0i I cos(100 t )12
(A). in p hai u on mch l
A. u 60 2 cos(100 t )12
(V). B. u 60 2 cos(100 t )
6
(V)
C. u 60 2 cos(100 t )12
(V). D. u 60 2 cos(100 t )
6
(V).
Gii: Ta thy I1 = I2 ----> (ZL ZC)
2 = ZL
2 ----. ZC = 2ZL
tan1 = R
ZZ CL = -R
Z L (*) tan1 = R
Z L (**) ----> 1 + 2 = 0
1 = u - 4
; 2 = u +
12
-------> 2u -
4
+
12
= 0 ---> u =
12
Do u 60 2 cos(100 t )12
, Chn p n C
Cu 75: Ba in tr ging nhau u hnh sao v ni vo ngun n nh cng u hnh sao nh cc ng dy dn. Nu i cch u ba in tr thnh tam gic (ngun vn u hnh sao) th cng dng in hiu dng qua mi ng dy dn:
A. tng 3 ln. B. tng 3 ln. C. gim 3 ln. D. gim 3 ln.
Gii:
Khi cc in tr u sao: Id = Ip = R
U p
Khi cc in tr u tam gic: Id = 3 Ip = 3R
U p'= 3
R
U d = 3R
U
R
UPp 3
3 = 3I
Tng ln gp 3 ln. Chn p n A
Cu 76 : Cho on mch xoay chiu ni tip RLC, in dung C = 2F. t vo hai u on mch mt
in p xoay chiu th in p gia hai bn t in c biu thc 100cos(100 / 3)( )u t V . Trong
khong thi gian 5.10-3(s) k t thi im ban u, in lng chuyn qua in tr R c ln l
A. 4( 3 2).10 ( )C B. 4(1 3).10 ( )C
C. 4(