- 1. LI M U Cun Bi tp in xoay chiu c bin son bi chuyn gia Trng hc
s: Trn Duy Khoa hin ang lm vic ti Trng hc s. Ni dung ca cun sch ny
bm st chng trnh ban c bn phn in xoay chiu lp 12 ph hp vi kin thc
thi i hc hin nay. Chng in l mt chng kh v tng i chim nhiu im trong
thi i hc nhng nm gn y v bi tp in trong thi i hc tng i l kh.Nhng cc
em nu thuc l thuyt v ng dng ton tt th gii ton in xoay chiu khng phi
l tr ngi g i vi cc em.Vi quyn sch ny Khoa vit nhm gip cc bn hiu su
hn v in gip rn luyn tt k nng gii mt bi ton in tuy n vn c th cn thiu
nhng lng kin thc ny cc bn bc chn ca mnh vo in trong cc thi th v cc
thi ca b cc nm gn y. Sch gm 105 bi tp vi mc kh ngang bng nhau v mi
bi mang mt bn cht vn tng i l khc nhau to cm gic hng th khi cc em c
th lm nhng bi tp khc nhau khng b nhm chn. Mi bi tp u c mt hng dn
gii hoc nhiu hn y ch l mt hng gii quyt tng i l ti u cc em c th tm
thm nhiu phng php gii khc nhau cho cc bi ton trong quyn sch ny.
Trong qu trnh bin son d rt c gng nhng chc chn vn cn nhng ch sai st.
Mong nhn c s thng cm v xin cc bn ng gp kin ln sau ti bn c tt hn. Mi
th t thc mc xin gi v:[email protected].
2. 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.3 2 (A). 2Gii: V gin vc t nh hnh v ta c 5 MON = ( ) 3 12 12 MN =
UL + UC OM = URL = 75 2 (V) ON = URC = 50 3 (V) p dng L cosin cho
tam gic OMN:D. 3,3 (A)ULMROUR N UCR5 118 (V) 12 UR2 = ULR2 UL2 =
URC2 UC2 - UL2 UC2 = ULR2 URC2 = 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) 2 2 MN = UL + UC = U RL U RC 2.U RLU RC cos2 2 Suy ra
UL = 75 (V) UR = U RL U L 75 2 = 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 = R2 in p t
vo hai u mch; U = U R (U L U C ) 2 = 50 (V)1 3. Khi C = C2 UR = UL
U = U ' 2 (U ' L U C 2 ) 2 = 50 (V) UR = 25 2 (V). Chn p n A R1 (H)
v 4 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); Cu 3. Cho mch in xoay chiu gm 3 phn
th ni tip: in tr R; cun cm L =Gii: T biu thc ca i khi = 1 ta c 1 =
240 1 ZL1 = 240 = 60 4 Gc lch pha gia u v i : = u - i = R = ZL1
ZC1; Z1 =6 ( 12) 4 tan = 1U 45 2 45 2 I 1Z12 = R2 + (ZL ZC)2 = 2R2
R = 45 R = ZL1 ZC1 ZC1 = ZL1 R = 15 1 1 1 1 ZC1 = C= (F) 1 Z C1 240
.15 3600 1C Khi mch c cng hng 1 1 2 2 (120 ) 2 2 = 120 1 1 LC . 4
3600 Do mch cng hng nn: ZC2 = ZL2 = 2 L = 30 () U 45 2 I2 = 2 (A);
uc chm pha hn i2 tc chm pha hn u gc /2 R 45 Pha ban u ca uC2 =6 2 3
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 C 63,8 F v mt cun
dy c in tr thun r = 70, t cm L 1H . t vo hai u mt in p U=200V c tns
f = 50Hz. Gi tr ca Rx cng sut ca mch cc i v gi tr cc i ln lt l2 4.
A. 0 ;378, 4W C. 10 ;78, 4WB. 20 ;378, 4W D. 30 ;100WGii:U2 (Z Z C
) 2 R L R Vi R = Rx + r = Rx + 70 70 1 1 ZL = 2fL = 100; ZC = 50
2fC 314.63,8.10 6 3500 P = Pmax khi mu s y = R + c gi tri nh nht vi
R 70 R Xt s ph thuc ca y vo R: 3500 Ly o hm y theo R ta c y = 1 ; y
= 0 R = 50 R2 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 U 2r
378,4 W Pc = 2 r (Z L Z C ) 2 P = I2R=U 2R R 2 (Z L Z C ) 2Chn p n
ARx = 0, Pc = 378 WCu 5. Cho mch in nh hnh v L,rMRNC BAt 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 U AB U AN 3U MN 120
3(V ) . Cng dng in trong mch I 2 2 A . Gi tr ca ZL l A. 30 3B. 15
6D. 30 2C. 60V gin vc t nh hnh v:UAN NUAM MAB = UAB UAB = 120 3
(V)A AM = UAM = Ur + ULUr EUR F I 3 U B AB 5. AN = UANUAN = 120 3
(V)AE = Ur EF = MN = UMN = URUMN = UR = 120 (V)AF = Ur + UR ; EM =
FN = UL ; NB = UC NAB = MAF suy ra MAN = FAB T UAB = UMN suy ra UL2
= (UL UC)2 UC = 2UL suy ra NAF = FAB V vy MAN = ANM tam gic AMN cn
MN = AM hay UAM = UR = 120(V) Ur2 + UL2 = UAM2 = 1202 (1) (Ur +
UR)2 + (UL UC)2 = UAB2 hay (Ur + 120)2 + UL2 = 1202 (2) T (1) v (2)
ta c Ur = 60 (V); UL = 60 3 (V)U L 60 3 15 6 (), Chn p n B I 2 2 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 R 2
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 1 v lch pha gia uAM v uMB l 900. Nu t in p trn vo hai u on mch
2 LC MB th on mch ny tiu th cng sut bng: A. 85 W B. 135 W. C. 110
W. D. 170 W. Do o ZL =Gii:CR11 A trong mch c cng hng ZL = ZC LC v
cng sut tiu th ca on mch c tnh theo cng thc ZC Z U2 P= (1). Ta c:
tan1 = ; tan2 = L R1 R1 R1 R2 0 Mt khc: 2 - 1 = 90 tan1. tan2 = -1
ZC Z L Do = -1 ZL = ZC = R1 R2 (2) R1 R1 Khi t in p trn vo on mch
MB th cng sut tiu th trn on mch U 2R U 2 R2 U2 P2 = I22 R2 = 2 2 2
= 2 = P = 85W. Chn p n A R2 Z L R2 R1 R2 R1 R2Khi 2 M R2LBCu 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 sutAR M C N L, r
4B 6. tiu th ton mch bng 360W; lch pha gia uAN v uMB l 900, uAN v
uAB l 600 . 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 = EFUL UL + UCUMB EUAB FUMB = OE UMB = 120V (1) UAN = OQ UAB =
OF UAB = 120 3 (V) (2)Ur O1 UR EOQ = 900 O 0 FOQ = 60 O2 0 0 0 Suy
ra = EOF = 90 60 = 30 . U Xt tam gic OEF: EF2 = OE2 + OF2
2.OE.OFcos300 C UAN Q Thay s EF = OE = 120 (V) Suy ra UR = 120(V)
(3) 2 2 2 UAB = (UR + Ur) + (UL UC) Vi (UL UC)2 = UMB2 Ur2 ( xt tam
gic vung OO1E) UAB2 = UR2 +2UR.Ur + UMB2 . T (1); (2), (3) ta c Ur
= 60 (V) (4) Gc lch pha gia u v i trong mch: = FOO3 = 300 ( 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 BO3 UR + Urt Cu 8. t in p xoay chiu u = 100
2 cos (c thay i c trn on [100 ;200 ] ) 1 10 4 vo hai u on mch c R,
L, C mc ni tip. Cho bit R = 300 , L = (H); C = (F). in p hiu dng
gia hai u L c gi tr ln nht v nh nht tng ng l 400 100 100 V; V. v.
A.100 V; 50V. B.50 2 V; 50V. C.50V; D. 3 3 3 5 Gii:5 7. Ta c UL =
IZL; UL=UUL1 1 L 1 ( R 2 2 ) 2 L2 2 4 C C 1 Xt biu thc y = 10 8 2 X
2 7.10 4 X 2 R 2 (L 1 2 ) CUL 10 8 214 7.10 41212Vi X =12> 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 = U U 100 400 (V) 1 1 1 1 1 7 3 5 8 2 1 4 1 8 2 4 10
7.10 2 2 10 7.10 1 16 4 4 16.10 8 4 4. 2 2 UL = ULmin khi = 100. U
U 100 100 ULmin = 3 1 1 1 1 1 1 1 7 1 10 8 2 4 7.10 4 2 2 10 8 2 8
4 7.10 4 2 2 10 Chn p n D. Cu 9.. Cho mch in xoay chiu khng phn
nhnh AD gm hai on AM v MD. on mch 2 MD gm cun dy in tr thun R = 40
3 v t cm L = H. on MD l mt t 5 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 =2 R 2 Z L 80 t Y = (UAM + UMD)2. Tng (UAM + UMD) t gi tr cc i
khi Y t gi tr cc i Y = (UAM + UMD)2 = I2( ZAM2 +ZC2 + 2ZAM.ZC) =2 2
U 2 ( Z AM Z C 2Z AM Z C ) R 2 (Z L Z C ) 26 8. 2 2 U 2 (80 2 Z C
160Z C ) U 2 ( Z C 160Z C 6400) Y= 2 3.40 2 (40 Z C ) 2 Z C 80Z C
6400 2 240Z C ( Z C 160Z C 6400) Y = Ymax khi biu thc X= = 1+ 2 c
gi tr cc i 2 Z C 80Z C 6400 Z C 80Z C 6400X=240Z C = Z 80Z C 6400 2
C240 c gi tr cc i 6400 ZC 80 ZCX = 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 =120 2 (80 80) 120 2.160 U 240 2 (V) ( Z AM Z C ) = 80 Z
3.40 2 (40 80) 2UL2Ud2Chn 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 = 900
- 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 2 Do Ud = IZd = I R 2 Z L : 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=900-1 . 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
OUd1Ud2UL1Ud1 U2O 1 2 UR1UR2U1Suy ra U1U2 = Ud1Ud2 = 90 30 = 60V Do
OU1 = OU2 = U1U2/ 2UC7I 9. Suy ra U = 60/ 2 = 30 2 U0 = 60V Cu 11:
Mch in xoay chiu R, L, C mc ni tip. in p hai u on mch l u U0 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? ( 1 2 ) L12 L(1 2 ) L( 1 2 ) A. R = B. R = C. R = D. R = 2
n 1 L n2 1 n2 1 n2 1 1 1 Gii: I1 = I2 =Imax/n Z1 = Z2 1 L = - 2 L +
1C 2C 1 2 L-= m I1 = Imax/n 1C U 1U 1 2 = n2R2 = R2 +( 1 L ) = R2 +
( 1 L -2 L )2 1C nR 1 R 2 (1 L ) 1C L(1 2 ) (n2 1)R2 = ( 1 -2 )2L2
R = . Chn p n B n2 1 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 UR U1=IR = 1 2 R 2
(L ) C 1 U1 = U1max khi trong mch c s cng hng in: 12 = (1) LC U2 =
IZL =UL R 2 (L 1 2 ) CUL R 2 2 L2 1 L 2 2 C CU 2 y222L R2 2 1 1 C
L2 c gi tr cc tiu y U2 = U2max khi y2 = 2 4 2min C 2 1 1 C L t x =
2 , Ly o hm y2 theo x, cho y2 = 0 x = 2 = (2 CR 2 ) 2 C 8 10. 2 2 =
(2) L C (2 L CR 2 ) 2 2 C (2 R ) C U U U U3 = IZC = 2 y3 1 2 1 L C
R 2 (L ) C 2 ( R 2 2 L2 2 2 2 ) C C C L 1 U3 = U3max khi y3 = L24
+(R2 -2 )2 + 2 c gi tr cc tiu y3min C C 2 t y = , Ly o hm ca y3
theo y, cho y3 = 0 L 2 R2 1 R2 y = 2 = C 2 2 LC 2 L 2L 2 1 R 32 = 2
(3) LC 2 L So snh (1); (2), (3): 1 1 R2 T (1) v (3) 32 = 2 < 12
= LC 2 L LC CR 2 1 2 L (2 L CR 2 ) 2 Xt hiu 22 - 12 = = >0 LC (2
L CR 2 ) LC (2 L CR 2 ) C (2 L CR 2 ) LC (V CR2 < 2L nn 2L CR2
> 0 ) 1 2 Do 22 = > 12 = 2 LC C (2 L CR ) 2 2 1 R2 1 2 2 <
12 = < 22 = LC 2 L LC C (2 L CR 2 ) Theo th t V3, V1 , V2 ch gi
tr cc i Chn p n C Tm lai ta c 32 =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: 1 2 2,5 1 3 2 3 A. (H). B. (H). C. (H). D.
(H).Gii: Ta c ZC =100/0,5 = 200, tan Z L ZC tan 60 0 3 (ZL ZC) = R
3 RZ = U/I = 100/0,5 = 200 Z=R 2 ( Z L Z C ) 2 2 R R = 1009 11. UAM
= I.ZAM =2 U R2 ZLR (Z L Z C ) 22U R Z Z 2Z L Z C 1 R2 ZL 22 L2 CU
1400(100 Z L ) 2 100 2 Z L100 Z L = ymax c gi tr cc i 2 100 2 Z L y
= ymax khi o hm y = 0 ZL2 200ZL -100 = 0 ZL = 100(1 + 2 ) 1 2 L=
(H) Chn p n A.UAM =UAMmin khi y =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 L xoay chiu c tn s f=50Hz.
Thay i L ngi ta thy khi L=L1 v khi L=L2 = 1 th cng sut 2 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:4 3.10-4 (F) A. L1 = (H);C= 2 2 10-4 (F) C. L1 =
(H);C= 34 10-4 (F) B. L1 = (H);C= 3 1 3.10-4 (F) D. L1 = (H);C= 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 1,5ZL1 = 2ZC (1)Z L1 2Z L1 ZC Z
ZC Z Z C Z L1 Z L1 2 tan1 = L1 = v tan2 = L 2 = R R R 4R 4R 1 + 2 =
tan1. tan1 = -1 ZL12 = 16R2 ZL1 = 4R = 400 2 Z L1 4 L1 = (H) ZC =
0,75ZL1 = 300 C =1 10 4 (F) .Z C 3Chn 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 100 t (A) v i2= 12 10 12. 7 2
cos 100 t (A). nu t in p trn vo hai u on mch RLC ni tip th dng in
12 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 2 U = I1 R 2 Z L 2RI1 120
(V) 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 I 3 l I max .Cho L (H). in tr c tr
s no: 2 4 A.150. B.200. C.100. D.125. Gii: I1 = I2 Z1 = Z2 (ZL1
ZC1)2 = (ZL2 ZC2)2 ZL1 + ZL2 = ZC1 + ZC2 2 1 1 1 1 L(1 + 2) = ( ) 1
LC = ZC1 = ZL2 C 1 2 C1 2 1 2Imax =U 2 ; RI1 =U = ZU R ( Z L1 Z C1
) 22=U 2 2R 4R = 2R + 2(ZL1 ZC1) 2223 200 = 150(). Chn p n A 4 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 .R2 = (ZL1 ZL2)2 = L2 (1 - 2)2 R = L (1 - 2)
=11 13. i = - = 2 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 =
2ZLGii: Hiu pha ban u ca uL v i: UL - i =URC =2 U R' 2 Z CR' 2 ( Z
' L Z C ) 2=2 U R' 2 Z CR' 2 (2Z C Z C ) 2= U = 220V. Chn p n ACu
18: t mt in p xoay chiu u = U0cos(100t+ ) vo hai u mt on mch gm R,
L, 104 F ; R khng thay i, L thay i c. Khi 2 4 L H th biu thc ca dng
in trong mch l i I1 2cos(100t /12) A . Khi L H th biu thc ca dng in
trong mch l i I 2 2cos(100t / 4) A . in tr R c gi tr lC mc ni tip
(L l cun cm thun). Bit C A. 100 3 .B. 100.C. 200.D. 100 2 .Gii: Ta
c ZC = 100; Z ZC tan1 = L1 R Z ZC tan2 = L 2 R 2 - 1 = 4ZL1 = 200;
ZL2 = 400 100 = 1 = + R 12 300 = = 3tan1 2 = + R 4 = 12 6 1 tan(2 -
1) = tan = 6 3 tan 2 tan 1 2 tan 1 1 1 tan(2 - 1) = tan1 = 2 1 tan
2 tan 1 1 3 tan 1 3 3 100 1 = R = 100 3 () Chn p n A R 3 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. P 88 P
Thay s vo ta c: I = = = 0,5 (A); r = 2 = 352 U q cos 220.0,8 I12
14. Zqut =Uq I2 = r 2 Z L = 440Khi mc vo U = 380V: I =U U = = 2 Z
(R r) 2 Z LU 2 R 2 Rr r 2 Z L 2U 2 R2 + 2Rr + Z quat = ( ) 2 R2 +
704R +4402 = 7602 I 2 R + 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?U E = Z Z Vi E l sut in ng hiu dng gia hai
cc my pht: E = r = 0) Vi f = np n tc quay ca roto, p s cp cc t Gii:
I =2 N0 =2 2fN0 = U ( doZ = R 2 2 L2 Khi n1 = n th 1 = ; ZL1 = ZZ
Khi n3 = 3n th 3 = 3; ZL3 = 3ZZ ---->I 1 E1 Z 3 1 Z 3 1 = = I 3
E3 Z1 3 Z1 32 R 2 9Z L 2 R2 ZL2 2 6 Z L = 2R2 Z L = R2/3 ZL ==I1 1
2 2 = R2 + 9 Z L = 3R2 +3 Z L I3 3R 3- Khi n2 = 2n th 2 = 2; ZL2 =
2ZZ =2R3 Cu 21: Mt cun dy khng thun cm ni tip vi t in C trong mch
in xoay chiu c in p u U 0 .cost (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 v in p hiu dng
hai u cun dy l 90V. Bin U 0 ? 2 A. 60V . B. 30 2V C. 60 2V . D. 30V
Gii: Ud1 = 30 (V) Ud2 = 90 (V) Ud2 = 3 I2 = 3I1 Z1 = 3Z2 .Z12 =
9Z22 U d113 15. R2 + (ZL ZC1)2 = 9R2 + 9(ZL -Z C1 2 ) 2(R2 +ZL2 ) =
ZLZC1 32 2( R 2 Z L ) ZC1 = ZLU d1 Z1 U = U = Ud1 = Ud1 Z d1 Z d1
Z1R 2 ( Z L Z c1 ) 2 2 R2 ZL= Ud12 2 R 2 Z L Z C1 2 Z L Z C1 = 2 R2
ZL2 2 4( R 2 Z L ) 2 2( R 2 Z L ) 2Z L 2 2 ZL ZL 4( R 2 Z L ) 4R 2
Ud1 = Ud1 1 3 = Ud1 2 2 2 R2 ZL ZL ZL Z Z L C1 Z L ZC2 Z L Z C1 3
tan1 = ; tan1 = = R R R 2 1 1 + 2 = tan1 tan2 = -1 ( v 1 < 0) 2
2 Z Z L C1 Z L Z C1 3 = -1 (Z Z )(Z - Z C1 ) = - R2 L C1 L R R 3 2
2 2 Z C1 Z C1 2( R 2 Z L ) 4( R 2 Z L ) 2 2 2 2 2 R + ZL 4ZL + = 0
(R + ZL ) 4ZL + =0 2 3Z L 3 3 3Z L 2 R2 ZL (R + 2ZL22 2 4( R 2 Z L
) 5 4R 2 8 4( R 2 Z L ) 1 )[1- + ]=0 - =0 = 2 2 2 3 3 3 3Z L 3Z L
3Z L4R 2 4R 2 = 1 U = Ud1 1 = Ud1 2 2 2 ZL ZLDo 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 n2n2 2n 2 n 2 2 2 2 2 2 A. n0 n1.n2
B. n0 n12 n2 C. n0 2 1 2 2 D. n0 2 1 22 n1 n2 n1 n2 Gii: Sut in ng
ca ngun in: E = 2 N0 = Vi f = np n tc quay ca roto, p s cp cc t Do
P1 = P2 I12 = I22 ta c:12R 2 (1 L 1 2 ) 1C=2 2R 2 ( 2 L 1 2 ) 2C2
2fN0 = U ( do r = 0) 12 [ R 2 ( 2 L 1 2 1 2 2 ) ] = 2 [ R 2 (1 L )
] 2C 1C14 16. 2 12 2 2 L 2 2 2 2 2 2 L R L 2 2 21 = 2 R 1 2 L 2 2 2
2 C C 2 C 1 C 2 2 2 2 2 2 1 1 ( 1 )( 2 1 ) L 2 (12 2 )( R 2 2 ) = 2
( 2 12 ) = 2 2 2 2 C C 1 2 C 12 2 2 122 12 22L 1 1 - R2 )C2 = 2 2
(*) C 1 2 Dng in hiu dng qua mch U E I= Z Z (2222P = Pmac khi E /Z
c gi tr ln nht hay khi y =1 2 R ( L ) Cc gi tr ln nht21y=R 2 2 L2 1
L 2 2 C C1=1 1 C2 422R2 22L C L2 y = ymax th mu s b nht x2 L 1 t x
= 2 y = 2 ( R 2 2 ) x L2 C C Ly o hm mu s, cho bng 0 ta c kt qu x0
= T (*) v (**) ta suy ra1 1 2 2 2 2 f1 f2 f012 1hay12 2=12 0=1 2 L
C (2 R 2 ) (**) 2 C2 022n 2 n 2 1 1 2 2 2 2 n0 2 1 22 Chn p n D n1
n2 n12 n2 n0Cu 23 : t in p xoay chiu vo mch RLC ni tip c C thay i
c. Khi C= C1 =10 4F10 4 F th UC c cng gi tr. UC c gi tr cc i th C c
gi tr: 2 3.104 10 4 3.104 2.104 A. C = F. B. C = F C. C = F. D. C =
F 4 3 2 3v C= C2 =Gii: UC1 = UC2 UZ C1 R 2 ( Z L Z C1 ) 2=UZ C 2 R
2 (Z L Z C 2 ) 215 17. 2 2 ZL Z R2 ZL R2 ZL 1 1 1 1 2 -2 +1 = - 2 L
+1 (R2 + Z L )( 2 - 2 ) = 2ZL( ) 2 2 Z C1 ZC2 Z C1 ZC2 Z C1 Z C 2 Z
C1 Z C1 2Z 1 1 + = 2 L 2 (1) R ZL Z C1 Z C1UC =2 Z R2 ZL = UCmax
khi y = - 2 L +1 = ymin 2 2 2 ZC ZC R (Z L Z C )UZ C y = ymin2 R2
ZL khi ZC = ZLZ 1 = 2 L 2 (2) ZC R ZLC1 C 2 3.10 4 1 1 2 T (1) v
(2) + = C= = (F). Chn p n A 4 2 Z C1 Z C1 ZCCu 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 i1 2 6cos 100 t ( A) . Khi iu chnh in 4 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 5 5 A. i2 2 3cos 100 t B.
i2 2 2cos 100 t ( A) ( A) 12 12 C. i2 2 2cos 100 t ( A) D. i2 2
3cos 100 t ( A) 3 3 Gii: Khi C = C1 UD = UC = U Zd = ZC1 = Z1 Zd =
Z1 ZL =r 2 ( Z L Z C1 ) 2 =2 r 2 Z L ZL ZC1 = ZLZ C1 (1) 2Zd = ZC1
r2 +ZL2 = ZC!2 r2 =tan1 =Z L Z C1 r2 3Z C 1 r= 42 3Z C 1 (2) 2Z C1
Z C1 1 2 1 = 6 3 3 Z C1 2Khi C = C2 UC = UCmax khi ZC2 =2 Z2 r2 ZL
C1 2 Z C1 ZL Z C1 216 18. Khi Z2 =r 2 (Z L Z C 2 ) 2 Zc 3 2 2 Z C1
( 1 2Z C1 ) 2 3Z C1 3Z C1 4 2Z C1 2 Z C1 Z L ZC2 2 tan2 = 3 2 = r 3
3 Z C1 2 Z I 2 3 U = I1Z1 = I2Z2 I2 = I1 1 1 2 (A) Z2 3 3 Cng dng
in qua mchi2 = I2 2 cos(100t ) = 2 2 cos(100t 5 ) (A). Chn p n B 4
6 3 12 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)
U2 (50 2 ) 2 2 P=IR< = = 50W. Chn p n C: P = 25W. R 100 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.24VP = 2A; r U 20 20 P Ud = = 20V , I = d = Zd = =
10 2 Zd Zd I cos Gii: Dng in qua cun dy I =2 2 Zd = r 2 Z L ZL = Z
L r 2 = 6 U 2 I= U = IZ = I (r R) 2 Z L = 2 12 2 6 2 = 12 5 (V).
Chn p n C Z 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)17 19. Gii:
UL = IZL =UL1 2 ) C 1 2 R 2 (L ) C = y UL =ULmax khi y = min 2 R 2
(L 12 02=C L (2 -R2) (1) Vi 0 = 120 rad/s 2 CKhi f = f v f = f ta u
c U0L = UL 2 Suy ra UL = UL ' = 1 2 1 2 R 2 (L ) R 2 ( ' L ) C 'C 1
2 1 2 2 [ R 2 ( ' L ) ] = 2 [ R 2 (L ) ] ' C C '2 L 1 2 1 1 1 ( 2
-2 )( 2 -R2) = ( 2 ) = 2 ( 2 -2 )( 2 + 2 ) 2 2 C ' C C ' L 2 1 1 C2
( 2 -R ) = 2 + 2 (2) Vi = 100 rad/s C ' 2 02 1 1 2 T (1) v (2) ta c
2 = 2 + 2 2 = 2 2 2 0 0 ' = 0 2 2 2 0 =100 .120 2.100 2 2 120 2 2=
160,36 rad/s. Chn p n ACu 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: 2 R2 ZL ZL 2 2 2 2 ULUC = UR + UL UR + UL =200UL
U2 = UR2 +(UL UC)2 (100 3 )2 = UR2 + UL2 +2002 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 u U 0 .cost (V) th dng in trong
mch sm pha hn in p l 1 , in p hiu dngUC = UCmax = 200 (V) khi ZC
=18 20. 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 2 1 v in p hiu dng hai u cun dy l
90V. Bin U 0 ?Gii: Ud1 = 30 (V)Ud2 = 3 I2 = 3I1 Z1 = 3Z2 Z12 = 9Z22
U d1 Z R2 + (ZL ZC1)2 = 9R2 + 9(ZL - C1 )2 2(R2 +ZL2 ) = ZLZC1 R2 +
ZL2 = 3Ud2 = 90 (V) Z L Z C1 22 2 R 2 Z L Z C1 2 Z L Z C1 2 Z C1 =
Ud1 3 (*) 2 2 R ZL Z? Z Z L C1 Z ZC2 3 tan1 = L = R RU d1 U Z = U =
Ud1 1 = Ud1 Z d1 Z1 Z d1 tan1 =2 2Z L Z C1 ; R 1 1 + 2 = tan1 tan2
= -1 ( v 1 < 0) 2Z C1 Z L Z C1 3 = -1(Z Z )(Z - Z C1 ) = - R2 L
C1 L R R 3 2 Z Z2 Z 2 5Z L Z C1 Z Z Z Z R2 + ZL2 4ZL C1 + C1 = 0 L
C1 4ZL C1 + C1 = 0 C1 =0 3 3 3 2 6 3 3 Z 5Z L 2 Z C1 C1 = 0 ZC1 =
2,5ZL (**) U = Ud1 3 = Ud1 2 3 6 Z? ZL 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 ()U E = Z Z Vi E l sut in ng
hiu dng gia hai cc my pht: E = r = 0) Vi f = np n tc quay ca roto,
p s cp cc t 1 Z = R2 2 2 C Gii: I =2 N0 =2 2fN0 = U ( do19 21. Khi
n1 = n th 1 = ; I1 =1 E ; ZC1 = ZC = C Z1Khi n2 = 2n th 2 = 2; ZC2
= ZC1 /2 = ZC /2 I2 =I 1 E1 Z 2 1 Z 2 1 = = 2 I 2 E3 Z1 2 Z1E Z22
ZC 4 = I 1 = 1 6R2 + 1,5 Z 2 = 4R2 +4 Z 2 C C 2 6 R2 ZC I 2R2 2 2
2,5 Z C = 2R2 Z C = 2R2/2,5 = ZC =2R= 12 5 () 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 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: U Khi mc ampe k
mch RL: I1 = = 0,1 (A). Lc ny u sm pha hn i; 2 2 R ZLZL 1 R 0,2 R 2
= tan = ZL = (1) v U = I1 R 2 Z L = (V) (2) R 6 3 3 3 Khi mc vn k
mch RLC: UC = UV = 20V Z ZC 2 = - - (- ) = tan2 = L = - tan = - 3
ZC ZL = R 3 R 2 6 3 3 R 4R ZC = R 3 + = ; Z2 = R 2 ( Z L Z C ) 2 =
2R 3 3 UZ C 2U 2U 0,2 R UC = = = 20 U = = 10 3 R = 150 () Z2 3 3 3
tan1 =ZL =R 3= 50 3 2fL = 50 3 L =50 3 3 = (H) 2 .1000 40 .3 (H) ;
R = 150 () 40 . 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 105o so vi hiu in th uDB v sm pha 90o so vi hiu in th
uAB. Tnh Uo? Chn p n A: L =20 22. A. Uo = 136,6V. B. Uo = 139,3V.
C. Uo 100 2V .D. Uo = 193,2V.AGii:DHBDV gin nh hnh v. t lin tip cc
vect300 450UAD ; 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 =
750 UAB = UDB sin750 = 100 2 sin750 U0 = UAB 2 = 200sin750 =
193,18V Hay U0 = 193,2 V Chn p n DABHCu 33: Dng in i = 4cos2 t (A)
c gi tr hiu dng l bao nhiu? Gii: Ta c i = 4cos2 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 + I22 R = I2R I
=2 I12 I 2 6 (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 . in p hai u cun dy c gi tr hiu dng bng
2U v sm pha so vi 12 5 dng in l . in p hiu dng gia hai u on mch AB
ca mng in l : 12 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 5 u1 = U 2 cos(t + ). ; u2 = 2U 2
cos(t + ). 12 12 vi dng in l21 23. T gin ta tnh c 2 U AB = U2 + 4U2
- 2.2U2 cos 1200 = 7U21200UAB = U 7 . Chn p n BCu 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: UZ C Ta c UR = URmax = U v UC = UCmax = khi trong mch c cng
hng ZL = ZC R 2 R2 ZC UL = ULmax khi ZL = : (*) ZC U U ULmax = = =
2URmax = 2U 2 2 ZC R ZC ZC 1 2 1 2 ZL ZL ZL 1-ZC 1 4 = ZL = ZC (**)
4 3 ZLT (*) v (**) suy ra ZC = R 3 Do UCmax =UZ C =U 3 RU L max 2U
2 = = , Chn p n A U C max U 3 3 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 1 1 1 (1 + 2)L =
( + ) LC = (*) 1 2 C 1 2 1 Khi I = Imax; trong mch c cng hng LC = 2
(**) VyT (*) v (**) ta c = 1 2 = 120(rad/s). Chn p n D22 24. 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
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 Z L
gim hai ln cn in tr R ca ng dy khng i. Z Ta c : tand = L = tan = 3
ZL = R 3 Zd = 2R 3 R Ud = UC ZC = Zd = 2R. Z = 2R 2 3 U Do I = (*)
2R 2 3 Sau khi tng chiu di ng dy ZL =UI=R (Z ' L Z C ) 2U=2ZL R 3 =
2 2=R 3 R2 ( 2 R) 2 22U R 23 8 3(**)I' 4 2 3 = = 0,6847 I = 0,685I.
Chn p n A I 23 8 3 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 A 45,21 B 23,12
C 74,76 D 65,78 UL Gii: UL = IZL = 1 2 R 2 (L ) C1UL1 = UL2 R 2 (1
L 12 1+12 21 2 ) 1C2=L - R2 ) (*) C ULR 2 ( 2 L 1 2 ) 2C= 42C2(2UL
= ULmax khi1 2 R (L ) C 2=UL 1 2 R (L ) Cc gi tr max2223 25. 1 2 )
C = ymin 2 = 42C2(2 L - R2 ) (**) hay y = 2 C 2 2 1 1 2 1 1 T (*) v
(**) ta c 2 = 2 + 2 hay 2 = 2 + 2 1 2 f f1 f2 R 2 (L f1 f 2 2=
74,67 (Hz). Chn p n C f12 f 22 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 f=4 Gii: cos1 = 0,6 tan1 = 3 A ZL 4
4 tan1 = = ZL = (R + r) (*) 3 3 Rr 3 cos = 0,8 tan = 4 Z ZC 3 3 tan
= L = ZL ZC = (R +r) (**) 4 4 Rr ZL Z L 12 f 12 2 2 = 1 LC v 2 LC =
1 = 2 = 2 f1 = f2 ZC Z C 2 f2 * Khi ZL ZC = f1 =4 f2 7MNBZL ZCZ 3 7
16 (R +r) ZC = (R +r) L = 4 12 7 ZC= 151,2 Hz Bi ton v nghim** Khi
ZL ZC = f1 = f2CL; rRZ 3 25 16 (R +r) ZC = (R +r) L = 4 12 25 ZCZL
4 = f2. = 80Hz. Chn p n A 5 ZCCu 40: t mt in p u U 2cost (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 .24 26. Gii: PR = I2R =U 2R = ( R r ) 2 (Z L Z C ) 2U2 r 2
(Z L Z C ) 2 R 2r RPR = PRmax khi R2 = r2 + (ZL ZC)2. (1) Mt khc lc
R = 75 th PR = PRmax ng thi UC = UCmax 2 (R r) 2 Z L (R r) 2 Do ta
c: ZC = = + ZL (2) ZL ZL 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 + n2 = R2 = 752. (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
r Gii: cos = =0,8 Zd = 10 v ZL = 6, ZdP = 2 (A) r in p hiu dng 2 u
ng dy ni my pht l Cng dng in qua mch I =U=I2 ( R r ) 2 Z L = 2 12 2
6 2 = 12 5 (V) Chn p n CCu 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:
C1C 2 1 1 C// = C1 + C2; Cnt = ; = C= 2 C1 C 2 L LC 1 1 C// = 2 C1
+ C2 = 2 (*) 1 L 1 L C1C 2 1 1 1 1 1 Cnt = 2 = 2 C1C2 = 2 = 2 2 2
(**) 2 C1 C 2 2 L 2 L 2 L 1 L 1 2 L 1 1 1 T (*) v (**) C1 + 2 2 2 =
2 (***) 1 2 L C1 1 L25 27. C1 =12L(****)2L 2 1 1 1 = 2 2+ 2 2 = 2 2
2 2 2 1 2 1 L 1 2 L 1 L 4 2 2 2 2 - 2 + 1 2 = 0 (*****) 1Thay
(****) vo (***)+2 2 12 2 + 4 = 2 2 Phng trnh c hai nghim = 60 rad/s
v = 80 rad/s Chn p n DCu 43 : Mch R, L, C n i ti p . t vo 2 u ma ch
i n ap xoay chi u u = U0cost (V), vi thay i c. Thay i UCmax. Gi tr
UCmax l biu thc no sau y U U A. UCmax = C. UCmax = . 2 ZC Z2 L 1 2
1 2 ZC ZL 2U.LB. UCmax =4LC R C 2D. UCmax =22U R 4LC R 2C2Gii: UC
=UZ C R 2 (Z L _ Z C ) 2 2UC = UCmax khi 2 =UCmax =U 1 (1 ==R 4 LC
R 2 C 2 2L 242=R C R C ) L 4 L2U 1 4 L2 C 2=U R 2 (L 1 2 ) CL R2 1
C v UCmax = 2 C 2LU=1 C=U Z2 1 L 2 ZCU 2=U1 CL 1 L2 4 ( R 2 2 ) 2 2
C CU 4R 2L R4 C 4 L2 ==2 LU R 4 LC R 2 C 2U 2R R C R 4C 2 2 2 (4 LC
R C ) ) L 4 L2 4 L2 U U = = 2 L R C 2 2 2 2 (2 R ) C 1 (1 ) 2L 1 C
4 L2U 1(2L R2 )2 C L2 C 2 4 4LChn 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 qut26 28. mt bin
tr.(coi 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 sut 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/I12 198 (2) U U 220 I1 = Z1 ( R0 R1 ) 2 ( Z L Z C ) 2 268 2
( Z L Z C ) 2 Suy ra (ZL ZC )2 = (220/0,75)2 2682 ZL ZC 119 (3) Ta
c P = I2R0 (4) U U Vi I = (5) Z ( R0 R2 ) 2 ( Z L Z C ) 2U 2 R0 R0
+ R2 256 R2 58 ( R0 R2 ) 2 ( Z L Z C ) 2 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 P=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 = R2 +
ZC2 ZC2 = 3R2 ZC = R 3 UC = UR 3 = 50 3 (V) Chn p n D 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:27 29. Ta c Ur2 + UL2 = ULr2 ULr (UR +
Ur)2 + (UL UC)2 = U2 Vi U = 40 2 (V) Ur2 + UL2 = 252 (*) Ur (25+
Ur)2 + (UL 60)2 = U2 = 3200 UR 2 2 625 + 50Ur + Ur + UL -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) U UR Ur 1 Lc ny cos = = U UC 2 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 U Gii: Dng in qua cc ti l I = P I1 = I2 = 6A; I3 = 3 A R Dng in
qua dy trung tnh i = i1 + i2 + i3 I3 Dng phng php cng vc t ta cI1I
= 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:
3AI2I1I2II3 I1 I I2 I3Cu 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 V28 30. URGii: V gin vect nh hnh v. Ta
thy UC = UCmax khi = 900 tc khi uRL vung pha vi u 2 2 U C max = U2
+ U RLLKhi u = 75 6 V th uRL = 25 6 V Z = 3ZRL hay U = 3URL 2 2 2 U
C max = U2 + U RL = 10 U RL . Trong tam gic vung hai cnh gc vung U;
URL; cnh huyn UC ng cao thuc cnh huyn UR ta c: U.URL = URUC 2 3 U
RL = 10 URLUR 3URL = 10 UR = 75 10O UR URL = 25 10 (V). Do U = 75
10 (V). p n AU UCCu 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 r Gii: cos = =0,8 Zd = 10 v
ZL = 6, ZdP = 2 (A) r in p hiu dng 2 u ng dy ni my pht l Cng dng in
qua mch I =U=I2 ( R r ) 2 Z L = 2 12 2 6 2 = 12 5 (V) Chn p n CCu
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 = 0,125.10 3 (F) th tn s dao ng ring ca mch ny khi l 80 rad/s. Tn
s ca ngunin 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 1 L ZL = L; ZC = = Z L Z C = 5000 (*) C C 1 1 = = 80 L(C+
C) = (**) (80 ) 2 L(C C )UdULU UC29 31. 5000C(C+C) =1 (80 ) 2 C2
+(C)C -0,125.10 3 1 1 = 0 C2 + C=0 2 (80 ) .5000 (80 ) 2 .500010 6
.10 3 .10 3 C=0 C= F 8 8 8 2 .4 1 1 ZC = = 100 = = 80 rad/s. Chn p
n A C ZCC C2 +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/sUdGii: V gin vect ZC = ZLr = Z = 100
ZL =ZC = 50 2ULL = 5000 (2) L = 5.103C (*) C 1 1 5.103C2 + 5.103C.C
- 2 = 0 02 = L(C C ) 0ZL.ZC = 5.103C2 + 5.103 5.103C2 +
C=1,25.0,125.10 30,625. C -.C -1 80 2 2U =01 =0 6400CUC.10-4 (F); 1
1 . = = = 80 rad/s. Chn p n A 1,25 4 ZCC 100.. .10 ZC = 50 2 1,25.
-4 C = 2C C = C = 10 FCch 2 ; ZLr = Z r2 + ZL2 = r2 + (ZL ZC)2 ZC =
2ZL ZL = Khi tng thm C = C + C th ZC = ZL =ZC 230 32. =1 =
ZCC100..1 1,25= 80 rad/s. Chn p n A.104Cu 51 : t in p xoay chiu u U
2 cos(100t) V vo on mch RLC. Bit R 100 2 , 25 t in c in dung thay i
c. Khi in dung t in ln lt l C1 (F) v125 (F) th in p hiu dng trn t c
cng gi tr. in p hiu dng trn in tr R t C2 3 cc i th gi tr ca C c th
l: A. C 50B. C (F).200 (F)., C. 3C20(F).D. C 100 (F) 3Gii Ta c U C1
UZC1 R ( Z L Z C1 ) 2UC 2 2UZC 2 R ( Z L ZC 2 )2 22 2 Z C1 ZC 2 2 R
2 ( Z L Z C1 ) 2 R ( Z L Z C 2 ) 2 ZC1 = 400; ZC2 = 240 2Z L Z C1 Z
C 2 2.400.240Z L R2 + ZL2 = = = 300ZL Z C1 Z C 2 400 240 in p hiu
dng trn in tr R t cc i th trong mch c cng hng ZL = ZCUC1 = UC2 Thay
R =100 2 ; : ZC2 - 300ZC +20000 = 0 Phng trnh c hai nghim : ZC =
200 v ZC = 100 Khi ZC = 200 th C = Khi ZC = 100 th C =104 50 F F 2
104F100FChn 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: 2 2 A. 10V. B. 10 V. C. 20V. D. 20 V. Gii: Do UR = UL =
UC trong mch c cng hng , nn U = UR = 20V U Khi t b ni tt UL = UR =
= 10 2 (V). Chn p n B 231 33. 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 : f f A.f0 = 1 B. f0 =
2 C.f1.f2 = f02 D. f0 = f1 + f2 f2 f1 1 Gii: UR = Urmax khi trong
mch c cng hng in ZL = ZC f02 = (1) 4 2 LC 2 R 2 Z L2 UC = UCmax khi
ZC2 = R2 = ZL2ZC2 ZL22 (*) Z L2 2 R 2 Z C1 UL = ULmax khi ZL1 = R2
= ZL1ZC1 ZC12 (**) Z C1 T (*) v (**) suy ra ZL1ZC1 ZC12 = ZL2ZC2
ZL22 L 1 1 ZL.ZC = suy ra ZC1 = ZL2 = 2f2L f1f2 = 2 (2) 2f1C C 4 LC
T (1) v (2) ta c f1f2 = f02 Chn p n CCu 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 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 1 i = 0
i = 2 2 cos[100(t + ) -] = 0 cos(100t + -) = 0 600 6 cos100t.cos(
-) - sin100t.sin( -) = 0 cos( -) = 0 (v sin100t = 0 ) 6 6 6 = - = 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 = 2 2 220 2 cos(100t) (V) , u2 = 220 2 cos(100t + )
(V), u3 = 220 2 cos(100t ) (V), . 3 3 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 i32 34. lmchointrpha
th 1 v pha th 3 gim i mt na l: A. i = 50 2 cos(100t + ) (A) B. i =
50 2 cos(100t +) (A) 3 2 C. i = 50 2 cos(100t + ) (A) D. i = 50 2
cos(100t - ) (A) 3 3 Gii: Do cc ti tiu th l cc in tr thun nn u v i
lun cng pha I2 Khi mt cn i cc pha 220 I1 = I3 = = 100 (A) 2/3 2,2
I1 220 I2 = = 50 (A). V gin vc t : -2/3 - /3 4,4 I0 = I1 + I2 + I3
= I13 + I2 I0 I13 = I1 = I3 = 100A I0 = I13 I2 = 50 (A) I3 0 = 3 Do
biu thc cng dng in trong dy trung hoI1 3 ) (A) Chn p n D 3 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 U AM U U sin UAM = AB
UAM = AB sin sin sin UA 0 Do gc , UAB xc nh nn UAM c gi tr ln nht
khi = 90 B 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 UM U AB B I= ta thy khi
L tng th mu s tng do I gim 1 2 R 2 (L ) C Chn p n B: UAM gim, I gim
i = 50 2 cos(100t -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 in33 35. trngc rtln th cng hiu dng A. 2I B.I
2 C.I D. I/ 2 Gii: Xt thi gian mt chu k I 2R Lc ch c in tr thun R :
P = I2R = 0 2cadnginLc mc thm it, dng in qua Rch trong mt na chu ki
P = I2R = trongmchbngP I 02 R = 4 2I' 1 I = I = . Chn p n D I 2 2Cu
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 R1 ln lt l U R1 ,U C1 , cos1 . Khi bin tr c gi tr R2 th cc
gi tr tng ng ni trn ln lt l U R2 ,UC2 , cos2 bit rng s lin h:A.
1B.U R1 U R2 0, 75 v1 2U C2 U C1C. 0,49 0, 75 . Gi tr ca cos1 l:
D.3 2Gii:U R1 3 16 = UR2 = UR1 (*) U R2 4 9 UC2 3 9 = UC2 = UC1
(**) U C1 4 1616 2 2 9 2 ) U R1 + ( )2 U C1 9 16 9 16 2 2 2 - ( )2
U C1 U C1 = ( )2 U R1 16 92 2 2 2 U2 = U R1 + U C1 = U R 2 + U C 2
= ( (16 2 2 2 2 ) U R1 - U R1 = U C1 92 2 U2 = U R1 + U C1 = [(1 +
(U R1 = U16 2 2 ) ] U R1 U = 99 2 16 2 UR1 99= 0,49026 = 0,49. Chn
p n C 9 2 16 2 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 cos1 =34 36. 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 1 1 Do C = 2 = = 27,9 F. Chn p n B L (110 ) 2 .0,3 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.R 2 (Z L Z C ) 2 =Gii: Tng tr Z =215200
= 464 ()U 600 464 ZC = U UCmax = 400 2 (V) U 400 2 = 437,5 (V). Z
464 600 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 41 on mch .AB gm in tr thun R=100, cun cm thun
c t cm L= H v t in 6 10 4 c in dung C = F. Tc rto ca my c th thay i
c. Khi tc rto ca my l n 3 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 UC =Gii: Sut in ng cc i ca ngun in: E0 = N0 = 2fN0 U = E =
tr trong ca my pht khng ng k). Cng dng in qua mch I =E0 2(coi inU
ZVi f = np n tc quay ca roto, p s cp cc t Do I1 = I2 ta c:12 R 2 (1
L 1 2 ) 1C=2 2R 2 ( 2 L 1 2 ) 2C 12 [ R 2 ( 2 L 1 2 1 2 2 ) ] = 2 [
R 2 (1 L ) ] 2C 1C12 2 L 2 2 2 L 212 2 R 2 12 2 L2 2 2 2 2 2 = 2 2
C C 2 C 1 C 2 2 2 2 2 2 1 1 ( 1 )( 2 1 ) L 2 (12 2 )( R 2 2 ) = 2 (
2 12 ) = 2 2 2 2 C C 1 2 C 12 2 2 12 R 2 12 2 L2 35 37. 1121 2 2=
(24.10 3 L - R2 )C2 = (*) C 9 2 = 2f = 2np 1 1 10 1 1 1 1 1 1 10 (
2 2 )= 2 2 ( 2 + )= = (**) 2= 2 2 2 2 2 2 2 36 2 5 2 n 2 1 2 4 p n1
n2 4 p n 36 p n 9n4.10 3 9 2 10 10 = n2 = = 25 9 2 36 2 5 2 n 2 36
2 5 2 4.10 3 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 bao nhiu ? Gii: UZ C U U 1 1 UC = = = 2 2 C 1 2 C L 1
R (Z L _ Z C ) R 2 (L ) L2 4 ( R 2 2 ) 2 2 C C C L 2 R2 2 LU U 1 5U
UC = UCmax khi 2 = C 2 v UCmax = = = 2 2 C 3 2L L R 4 LC R C 4R 2 R
4 C 4 L2 36 L2 6L = 5R 4LC R 2 C 2 R2C2 4LC + 25 R 2 10 4 3,6..10 4
4.10 5 2 L 1,6 L C= = (21,6). F c 2 gi tr ca C: C1 = F v C2 = F R2
L 2 R2 2.10 4 1 2L R2 2 = C 2 = >0 C< 2 = F loi nghim C1 LC
2L2 2L R 4.10 5 10 5 2 100 2 2 R2 1 CX = C2 = F 2 = = = 2.1042 =
100 2 rad/s 2 4 2 LC 2 2L Do fX = 50 2 Hzp s CX =4.10 5F v fX = 50
2 HzCu 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? 3C0 C C C A. C = . B. C = 0 . C. C = 0 . D. C = 0 . 4 2 4 3
Gii:36 38. 2 Ta c Ud = I R 2 Z L ; Ud = Udmax khi I = Imax mch c
cng hng ZL = ZC0 (*)Udmax = 2U Zd = 2Z = 2R ( v ZL = ZC0) R2 + ZL2
= 4R2 R =ZL 3=ZC0 3(**)2 ZC0 2 ZC0 4Z C 0 R Z 3 UC = UCmax khi ZC =
= = ZC0 ZL 3 4Z C 0 3C 0 ZC = C= Chn p n A 3 4 Cu 64: t mt in p
xoay chiu u U 0 cos t (V ) 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 R r . 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 30 5 V . Gi
tr ca U0 bng: A. 120 2 V. B. 120 V. C. 60 2 V. D. 60 V. Gii: Do R =
r UR = Ur22 L2 2 Ta c :(UR + Ur)2 + U L = U AM 2 2 2 4 U R + U L =
U AMUAULM(1)2 2 U R + (UL UC)2 = U NB (2)Ur URUAM = UNB ZAM = ZNB
4R2 + ZL2 = R2 + (ZL ZC)2 3R2 + ZL2 = (ZL ZC)2 (*) uAM v uBN vung
pha tanAM.tanNB = -1RZ L Z L ZC 4R 2 = -1 (ZL ZC)2 = (**) 2 R 2R
ZLT (*) v (**) 3R2 + ZL2 = ZL4+ 3R2ZL24R 2 2 ZL 4R = 0 2ZL22U UA
BUN UC =RB22 Do UL2 = UR2 (3). T (1) v (3) 5UR2 = U AM = (30 5 )2
UR = 30 (V)UR = UL =30 (V) (4) 2 2 U R + (UL UC)2 = U NB (UL UC)2 =
(30 5 )2 302 = 4.302UAB2 = :(UR + Ur)2 + (UL UC)2 = 4UR2 + (UL UC)2
= 2.4.302 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?37
39. U RA. Na ng trn ng knhB. on thng I = kU, k l h s t l.UC. Mt na
hiperbolD. Na elip2 R2 ZLu2 i2 + 2 =1 I0 U 02UGii Ta c I =2 R ZL 2
Trn gin vc t qu tch ca u mt vc t I l mt na hiperbolU 2 R2 ZLChn 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. 50 f 50 n= = vng/s = .60 vng/min = 1000 vng/min. 3 3 p
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 : 1 1 2 2 A. 0 (12 2 ) B. 0 (1 2 ) 2 2
1 1 1 1 C. 2 = ( 2 + 2 ) D. 0 = 1 2 0 2 1 2 Gii: UL =R2 2 2 1UZ L R
2 (Z L Z C ) 2L C +1 = C2 4 1. Do UL1 = UL2 R2 22 2L C +12 R 2 (1 L
1 2 ) 1 C=2 2R 2 ( 2 L 1 2 ) 2C1 C2 4 22 L L 1 12 2 1 1 1 1 2 2 (2
- R )( 2 - 2 ) = 4 2 - 4 2 (2 - R ) = 2 C C C 2 12 2 2 1 2 C 1 C38
40. 12 1+12 2= C2 (2R2 2L - R2) (*) CL C +C2 1 L 1 + L2 c gi tr cc
tiu. 2 = (2 - R2) (**) 4 2 2 2 C C 0 1 1 1 1 T(*) v (**) suy ra: 2
= ( 2 + 2 ) . Chn p n C. Vi iu kin CR2< 2L 0 2 1 2 UL = ULmax
khiCu 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: 2 U = U 12 (U 2 U 3 ) 2 = U '1 (U ' 2 U '3 ) 2 = 100 2
(V) Suy ra : (U2 U3)2 = U2 U12 = 13600 U2 U3 = I(Z2 Z3) =100 (V)
(*) U2 U3 = I(Z2 Z3) = 13600 (V) (**)T (*) v (**) I' = IU' I'Z2 I'
13600 2 = = = U2 100 IZ 2 I13600 10013600 U2 = = 233,2 V. Chn p n A
100 Cu 69 Mc vo on mch RLC khng phn nhnh gm mt ngun in xoay chiu c
tn s thay i c. tn s f1 60 Hz , h s cng sut t cc i cos 1 . tn s f 2
120Hz , U2 =h s cng sut nhn gi tr cos 0,707 . tn s f3 90Hz , h s
cng sut ca mch bng A. 0,872B.0,486Gii: Ta c ZL1 = ZC1 1L =C. 0,6251
1 LC = 2 1C 1D. 0,781 (1)cos2 = 0,707 2 = 450 Z ZC2 tan2 = L 2 =1 R
= ZL2 - ZC2 Rtan3 =tan3 =Z L3 Z C 3 Z L3 Z C 3 = Z L2 ZC 2 R 32 1 1
3C 2 12 2 32 12 f 2 f 32 f12 = = = = 2 1 3 22 3 2 12 f 3 f 22 f12 2
L 1 2C 123 L f 2 f 32 f12 4 90 2 60 2 4 5 5 = = = (tan3)2 = 25/91 2
2 2 2 f 3 f 2 f1 3 120 60 3 12 939 41. 1 25 106 cos23 = 81/106 cos3
= 0,874. 1 2 cos 3 81 81p n ACu 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 =2 U R2 ZCR (Z L Z C ) 22= UAM = UAMmax th biu
thc y =1 R (Z L Z C ) 2 R2 ZC 221=1Z 2Z L Z C 2 R2 ZC 2 L2 Z L 2Z L
Z C = ymin o hm y = 0 2 R 2 ZC2 2 ( R 2 Z C )(-2ZL) ( Z L 2Z L Z C
)2ZC = 0 ZC2 ZLZC R2 = 0 Hay UC2 ULUC UR2 = 0 UC2 100UC 20000 = 0
UC = 200(V) (loi nghim m). Chn p n C 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: 1 1 2 = = CC
LC ( L1 L2 ) 1 2 C1 C 2 1 1 1 1 2 L1 = 2 ; 2 = L2 = 2 12 = L1C1 L2
C 2 1 C1 2 C2 1 1 1 1 1 1 C1C2 L1 + L2 = 2 + 2 = 2( + )= 2 ( v
1=2.) C2 1 C1 2 C 2 1 C1 1 C1 C2 1 12 = = 2 = 1. p n D C1C 2 ( L1
L2 ) C1 C 2 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 =
I12Rt1 = 9RT/3 = 3RT t2 = T/4: Q2 = I22Rt2 = 4RT/4 = RT40 42. t3 =
5T/12: Q3 = I32Rt3 = 12R.5T/12 = 5RT t = t1 + t2 + t3 = T l Q =
I2Rt = I2RT 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 A. 100 V. Gii: . Tm in p hiu dng hai u
mch AM khi cha thay i L? 2 B. 100 2 V. C. 100 3 V. D. 120 V.U L2 U
C 2 U L1 U C1 ; tan2 = U R1 U R2 U L1 U C1 U L 2 U C 2 1 - 2 = /2
tan1 tan2 = = -1 U R1 U R2 tan1 =2 2 2 2 2 2 (UL1 UC1)2 .(UL2 UC2)2
= U R1 U R 2 . U MB1 U MB 2 = U R1 U R 2 4 2 2 8 U MB1 = U R1 U R 2
.(*) (v UMB2 = 2 2 UMB1) 2 2 2 2 2 2 2 Mt khc U R1 + U MB1 = U R 2
+ U MB 2 (= U2) U R 2 = U R1 - 7 U MB1 (**) 4 2 2 2 2 2 T (*) v
(**): 8 U MB1 = U R1 U R 2 = U R1 ( U R1 - 7 U MB1 ) 4 2 2 4 2 2 U
R1 - 7 U MB1 U R1 - 8 U MB1 = 0 U R1 = 8 U MB12 U R1 =U U + = U2 U
+U 8 2 2 UR1 = U = 100 2 (V). Chn p n B 3 2 R12 MB122 R1Cu 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 = I0 cos(100 t ) (A). Nu ngt b t in 4
C th cng dng in qua on mch l i2 I0 cos(100t ) (A). in p hai u on 12
mch l A. u 60 2 cos(100t ) (V). B. u 60 2 cos(100t ) (V) 12 6 C. u
60 2 cos(100t ) (V). D. u 60 2 cos(100t ) (V). 12 6 Gii: Ta thy I1
= I2 (ZL ZC)2 = ZL2 ZC = 2ZL Z ZC Z Z tan1 = L = - L (*) tan1 = L
(**) 1 + 2 = 0 R R R41 43. ; 2 = u + 2u - + = 0 u = 12 12 4 4 12 Do
u 60 2 cos(100t ) , Chn p n C 12 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. 1 = u
-Gii: Khi cc in tr u sao:Id = Ip =Khi cc in tr u tam gic:Up RId =3
Ip =3U'p R=3Ud = R33U p R3UP = 3I RTng 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 u
100cos(100 t / 3)(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. ( 3 2).104 (C ) B. (1
3).104 (C ) C. ( 3 2).104 (C )D. ( 3 1).104 (C )Gii: in lng chuyn
qua in tr R bng tng ln in tch t in phng v tch in trong khong thi
gian trn t = 5.10-3 (s) = T/4 ( chu k T = 0,02s) in tch ca t in ti
thi im t: q = Cu = 2.10-4cos(100t + ) (C) 3 Khi t1 = 0 in tch ca t
in q1 = 2.10-4cos( ) (C) = 10-4C, t in phng in, t q1 n 0. 3 3 Sau t
in li tch in khi t2 = T/4 = 5.10-3(s); q2 = 2.10-4cos( + ) = -
2.10-4 (C) 2 2 3 Trong khong thi gian 5.10-3(s) k t thi im ban u,
in lng chuyn qua in tr R c ln l Q = q1 + q2 = (1 3).104 (C ) Chn p
n B Cu 77: t in p xoay chiu c gi tr hiu dng U = 30 2 V vo hai u on
mch RLC ni tip. Bit cun dy thun cm, c cm L thay i c. Khi in p hiu
dng hai u cun dy t cc i th hiu in th hiu dng hai u t in l 30V. Gi
tr hiu in th hiu dng cc i hai u cun dy l: A. 60V B. 120V C. 30 2 V
D. 60 2 V42 44. 2 2 U R 2 ZC R 2 ZC Gii: Khi L thay i ULmax khi ZL
= (1)v ULmax = R ZCTa c:U UC Z ZC30 230 2 2ZC R 2 ( Z L ZC )2 (2)
ZCR ( Z L ZC ) Th (1) vo (2) ta c: 2 4 2 R 4 Z C R 2 2Z C 0 R 2 Z C
R Z C 22UR 2 U 2 60 V. Chn p n A R Cu 78. 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 L xoay chiu c tn s f=50Hz.
Thay i L ngi ta thy khi L=L1 v khi L=L2 = 1 th cng sut 2 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:Do ULmax =4 3.10-4 (F) A. L1 = (H);C= 2 2 10-4 (F)
C. L1 = (H);C= 34 10-4 (F) B. L1 = (H);C= 3 1 3.10-4 (F) D. L1 =
(H);C= 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 1,5ZL1 = 2ZC (1)Z L1
2Z L1 ZC Z L2 Z C Z L1 Z C Z L1 Z L1 2 tan1 = = v tan2 = = R R R 4R
4R 1 + 2 = tan1. tan1 = -1 ZL12 = 16R2 ZL1 = 4R = 400 2 Z L1 4 L1 =
(H) ZC = 0,75ZL1 = 300 C =1 10 4 (F) .Z C 3Chn p n B Cu 79.:Mch in
xoay chiu gm ba in tr R, L, C mc ni tip. R v C khng i; L thun cm v
thay i c. t vo hai u on mch in p xoay chiu c biu thc u = 200 2
cos(100t) V Thay i L, khi L = L1 = 4/ (H) v khi L = L2 = 2/ (H) th
mch in c cng cng sut P = 200 W. Gi tr R bng ? ZL1 = 400; ZL2 =
200;43 45. P1 = P2 I1 = I2 (ZL1 ZC) = -((ZL2 ZC) > ZC = (ZL1 +
ZL2)/2 = 300 (200) 2 R U 2R P1 = 2 200 = 2 R2 + 1002 = 200R R = 100
2 2 R ( Z L1 Z C ) R 100 Cu 80:Mch in xoay chiu gm ba phn t, in tr
thun R, cun cm thun L v t in C mc ni tip. in tr R thay i c. t vo
hai u on mch in p xoay chiu u = 120 2 cos(100t) V iu chnh R, khi R
= R1 = 18 th cng sut trn mch l P1, khi R = R2 = 8 th cng sut P2,
bit P1 = P2 v ZC > ZL. Khi R = R3 th cng sut tiu th trn mch t cc
i. Biu thc cng dng in qua mch khi R = R3? R1 R2 Gii: P1 = P2 2 = 2
(ZL ZC)2 = 144 2 2 R1 ( Z L Z C ) R2 ( Z L Z C ) hay ZC ZL = 12 ( v
ZC > ZL) Khi R = R3 P = Pmax khi R = R3 = ZC ZL =12 Z3 =R32 ( Z
L Z C ) 2 12 2 () I3 = U/Z3 = 5 2 (A)Z L ZC = - 1 3 = R3 4 Do biu
thc i3 = 10cos(100t + ) 4 Cu 81: Cho on mch RLC mc ni tip vi C l t
in c gi tr thay i c.Gi l lch pha ca in p so vi dng in.khi iu chnh
gi tr ca C th thy Uc t gi tr cc i ng vi gc 0.khi C c gi tr C1 hoc
C2 th Uc c gi tr nh nhau ng vi gc 1 v 2.Chn p n ng: A. 1/1 + 1/2 =
2/0 B. 1 + 2 = /2 C. 1 + 2 = 20 D. 2 - 1 = /2 tan3 =Gii: Z L Z C1
ZC1 = ZL - Rtan1 R Z ZC2 tan2 = L ZC2 = ZL - Rtan2 R ZC1 + ZC2 =
2ZL R(tan1 +tan2) ZC1 ZC2 = ZL2 RZL(tan1 +tan2) + R2tan1.tan2 Z ZC0
R tan0 = L = R ZL Z ZC2 2Z 2Z 1 1 2 UC1 = UC2 + = = 2 L 2 C1 = 2 L
2 Z C1 Z C 2 Z C1 ZC2 ZC0 R ZL R ZL 2Z L R(tan 1 tan 2 ) 2Z = 2 L 2
2 2 Z L RZ L (tan 1 tan 2 ) R tan 1 . tan 2 R ZLtan1 =44 46. R ZL22
tan 0 tan 1 tan 2 2 RZ = 2 L2 = 2 = 1 - tan 1 . tan 2 R ZL 1 - tan
2 0 R 1 2 ZL tan(1+2)) = tan21+2) = 20 . Chn p n C Cu 82. Cho mch
in RLC, cun cm c in tr thun r . in p t vo hai u on mch c dng u=125
2 cos100t, thay i c. on mch AM gm R v C, on mch MB cha cun dy. Bit
uAM vung pha vi uMB v r = R. Vi hai gi tr ca tn s gc l 1= 100 v 2=
56,25 th mch c cng h s cng sut. Hy xc nh h s cng sut ca on mch. A.
0,96B. 0,85Gii: cos1 =cos2 (1+2 )L =tanAM =C. 0,91Rr Rr 1 1 = Z1 =
Z2 1L = - 2L Z1 Z2 1C 2 C1 1 1 1 ( hay ZC1 = ZL2. (1) ) LC = 1 2 C
1 2 Z C1 Z L1 ; tanMB = R ruAM vung pha vi uMB v r = R ZL1ZC1 = R2
ZL1.ZL2 = R2 L =cos1 =Rr = Z12R 4 R 2 ( Z L1 Z C1 ) 2 2Rcos1 =4 R
(1 2 ) 2D. 0,822R21 2R1 2 =2R 4 R 2 ( Z L1 Z L 2 ) 2 2=4(1 2 ) 2=2R
4 R 2 (1 2 ) 2 L2= 0,96. Chn p n A1 2Cu 83. on mch AB gm ba phn AM;
MN v NB mc ni tip nhau. on mch AM cha x cun dy thun cm L mc song
song; on mch MN cha y in tr R mc song song; on NB cha z t in C mc
song song vi 2x = z y. Mc vo on mch AN dng in mt chiu c ddin p U =
120V th cng dng in qua mch chnh IAM = 4A. Khi mc ln lt vo on mch
MB; AB ngun in xoay chiu c in p hiu dng Uhd = 100V th u thu c cng
dng in hiu dng qua mch Ihd = 2A. Khi mc on mch R, L, C ni tip vo
ngun xoay chiu ni trn th cng dng in hiu dng qua mch l Ihd = 1A. in
tr R c gi tr l: A. 50 B. 30 C. 60 D. 40 45 47. Gii: Ta c: ZAM =Z ZL
120 R ; ZMN = = = 30 () (1) v ZNB = C 4 x z yKhi UMB = UAB = 100
(V) th IMB = IAB = 2 (A) ZMB = ZAB = 50 () 2 Z MB = ((Z Z R 2 ZC 2
R 2 ) +( ) ; Z AB = ( )2 + ( L - C )2 y y z x zZ Z R 2 ZC 2 ) +( )
= 502 ( C )2 = 502 302 = 402 C = 40 () (2) y z z zZMB = ZAB (ZC 2 Z
Z Z Z Z ) = ( L - C )2 L = 2 C L = 80 () (3) z x z x z xT (1); (2);
(3) x =Z ZL R ; y= ; z= C 30 80 40Theo bi ra 2x = z y 2Z Z Z ZL R R
4R = C C - L = (ZC ZL) = . 3 80 40 30 40 40 30Khi mc mch R, L, C ni
tip: Z =R 2 ( Z L Z C ) 2 = 100V/1A = 100 () R2 + (ZL ZC)2 = 1002
R2 + (4R 2 ) = 1002 R = 60 (). Chn p n C 3Cu 84 :Mch in xoay chiu
gm ba in tr R, L, C mc ni tip. R v C khng i; L thun cm v thay i c.
t vo hai u on mch in p xoay chiu c biu thc u = 200 2 cos(100t) V
Thay i L, khi L = L1 = 4/ (H) v khi L = L2 = 2/ (H) th mch in c cng
cng sut P = 200 W. Gi tr R bng bao nhiu? Gii. ZL1 = 400; ZL2 = 200;
P1 = P2 I1 = I2 (ZL1 ZC) = -((ZL2 ZC) ZC = (ZL1 + ZL2)/2 = 300
(200) 2 R U 2R P1 = 2 200 = 2 R2 + 1002 = 200R R = 100 2 2 R ( Z L1
Z C ) R 100 Cu 85: Mt mch in gm R ni tip t in C ni tip cun dy L.
Duy tr hai u on mch mt in ap xoay chiu u = 240 2 cos(100(t)V, in tr
c th thay i c. Cho R = 80 , I = 3 A, UCL= 80 3 V, in ap uRC vung
pha vi uCL. Tnh L? A. 0,37H B. 0,58H C. 0,68H D. 0,47H Gii:46 48.
ULULUC Ta c U = 240 (V); UR = IR = 80 3 (V) U E V gin vc t nh hnh
v: UR = ULC = 80 V. Xt tam gic cn OME UC 2 L O /6 U2 = UR2 + UCL2
2URULcos = 3 M UrF UR UR = = UC /6 3 6 C N UC Xt tam gic OMN UC =
URtan = 80(V) (*) Xt tam gic OFE : EF = OE sin UL UC = Usin = 120
(V) (**) . T (*) v (**) suy ra UL = 200 (V) 6 U ZL 200 200 Do ZL =
L = L= = = 0,3677 H 0,37 H. Chn p n A 100 100 3 I 3 Cu 86. on mch
in xoay chiu nh hnh v. t vo hai u on mch mt in p xoay chiu uAB = U
2 cos(100 t ) V. R L, r C A Bit R = 80 , cun dy c r = 20 , UAN =
300V , N M UMB = 60 3 V v uAN lch pha vi uMB mt gc 900 . in p hiu
dng hai u mch c gi tr : A. 200V B. 125V C. 275V D. 180V Gii: Cch 1
R = 4r UR = 4Ur (UR + Ur)2 + UL2 = UAN2 => 25Ur2 + UL2 = 90000
(1) Ur2 + (UL UC)2 = UMB2 = 10800 (2) U UC U UL tanAM = = L ; tanMB
= L uAN lch pha vi uMB mt gc 900 UR Ur Ur 5U r tanAM tanMB =U L U L
UC 5U r 25U r2 = - 1 UL UC = (UL UC )2 = (3) 2 Ur 5U r UL UL25U r2
= 10800 Ur2 = 2700 (*) Ur = 30 3 2 90000 25U r 5U r UL2 = 90000
25Ur2 = 22500 UL = 150 (V) (**) v UC = UL + = 240 (V) (***) UL Th
(1) v (3) vo (2) ta c Ur2 +UR + Ur = 150 3 Do U2 = (UR + Ur)2 +(UL
UC)2 = 75600 U = 275 (V). Chn p n C Cch 2. V gin vc t . Do R = 4r
UR+r+ = 5Ur uAN lch pha vi uMB mt gc 900 nn hai tam gic OEF v DCO
ng dng U U L U U OE EF OF 60 3 3 = = r = C = MBr = = 5U r U AN CD
CO UL 300 5 DO47B 49. D UL =53 (UR + Ur) + UL2 = UAN2 => 25Ur2 +
UL2 = 90000 25 2 25Ur2 + Ur = 90000 Ur2 = 2700 Ur = 30 3 3 UL = 150
(V); UC = 240 (V) UR + Ur = 150 3 Do U2 = (UR + Ur)2 +(UL UC)2 =
75600 U = 275 (V). Chn p n CUAULUrULN2O UM E UrBUCUR+ CrUCUL F
UUCCu 87. t vo 2 u mt hp kn X (ch gm cc phn t mc ni tip) mt in p
xoay chiu u = 50cos(100t + /6)(V) th cng dng in qua mch i =
2cos(100t + 2/3)(A). Nu thay in p trn bng in p khc c biu thc u = 50
2 cos(200t + 2/3)(V) th cng dng in i = 2 cos(200t + /6)(A). Nhng
thng tin trn cho bit X cha A. R = 25 (), L = 2,5/(H), C = 10-4/(F).
B. L = 5/12(H), C = 1,5.1z0-4/(F). C. L = 1,5/(H), C =
1,5.10-4/(F). D. R = 25 (), L = 5/12(H). Gii: Gi s mch gm 3 phn t
thun R, thun L v t C ni tip Trong hai trng hp u v i ung pha vi nhau
nn R = 0 Z1 = ZC1 ZL1 ( ZL1 < ZC1) 2 Z 2 = u2 - i2 = Z2 = ZL2
ZC2 = 2ZL1 - C1 ( v tn s f2 = 2f1) 2 2 U U 50 25 2 Z1 = 1 = 25 ; Z2
= 2 = 50 ; I1 I2 1 2 Ta c ZC1 ZL1 = 25 ; Z 2ZL1 - C1 = 50; 2 125 5
Suy ra ZL1 = 125/3 () L = (H) 300 12 3 1,5.10 4 (F) ZC1 = 200/3 ()
C = 200.100 Chn p n B Cu 88. Cho on mch AMNB trong AM c t in C, MN
c cun dy(L,r),NB c in tr thun R. in p gia 2 u on mch l u = 50 6
cos100t (V). Thay i R n khi I =2(A) th thy 1 = u1 - i1 = -48 50.
UAM = 50 3 (V) v uAN tr pha /6 so vi uAB, uMN lch pha /2 so vi uAB.
Tnh cng sut tiu th ca cun dy ? Gii: L; r R UAM = UC = 50 3 (V) UAB
= 50 3 (V) Gc lch pha gia u v i l 3 UC UL = UAB sin = 75 (V) 3 UL =
50 3 - 75 (V) Gc lch pha gia uMN v i l Ur = UL/tan = UL 6BMNC AUM
NO E - = 2 3 6/3 UrUR3Ur = 75 37,5 3 = 10 I Cng sut tiu th ca cun
dy: Pd = I2r = 40W/6/6r=UA UABMCu 89. t in p xoay chiu u =
U0cos(120t + /3)V vo hai u mt cun cm thun c t cm 1/6(H). Ti thi im
in p gia hai u cun cm l 40 2 (V)th cng dng in qua cun cm l 1A. biu
thc cng dng in qua cun cm l ? Gii: ZL = 20 Biu thc cng dng in qua
cun cm l i = I0cos(120t + /3 -/2 ) = I0cos(120t - /6 ) U 02 i 2 Z 2
I 2i 2 i2 u2 + 2 = 1 I02 = = 2 L2 0 2 I 02 U 0 U 02 u 2 Z L I 0 u
300I02 3200 = 400 I0 = 3 (A) Do i = 3cos(120t - /6 ) (A) Cu 90. Mt
on mch gm cun dy thun cm c h s t cm L, t in c in dung C v mt in tr
thun R mc ni tip. Hai u on mch c duy tr bi in p u=Ucos(t). G s LC2=
1, lc in p hai u cun dy (UL) ln hn U khi A. Tng L dn n UL.> U B.
Gim R I tng dn n UL.> U C L C. R > D .R< UL L C Gii: Lc ny
trong mch c s cng hng UR = U. UL > U = UR th ZL > R L L L
> R R < = . Chn p n D C LCU URUC49 51. Cu 91.Mt on mch gm cun
dy thun cm c h s t cm L = 3/5(H) mc ni tip vi t in c in dung C =
10-3/14 (F). Hai u c duy tr in p u = 160cos(100t)(V). Cng sut ca on
mch l 80 W. Tm lch pha ca cng dng in so vi in p t vo hai u on mch?
Gii: ZL = 60; ZC = 140.; U = 80 2 (V) U 2R P = I2R = 2 R2 160R +
802 = 0 > R = 80 2 R (Z L Z C )P P 2 1 = 1 (A). P = UIcos cos =
= = R UI 2 2 = . Cng dng in chm pha hn in p t vo hai u mch gc 4 4
Cu 92.. Mch in gm in tr R = 100 mc ni tip vi cun dy thun cm c t cm
L = 1/ H. in p t vo hai u on mch c dng u = 400.cos2(50t) V. Tnh cng
dng in hiu dng qua mch. Gii: Ta c u = 400.cos2(50t) = 200cos(100t)
+ 200 (V) in p t vo hai u mch gm hai thnh phn: in p xoay chiu c in
p hiu dng U1 = 100 2 (V), tn s gc 100 rad/s v in p mt chiu U2 = 200
(V) Cng sut ta nhit trn din tr R: P = P1 + P2 P = I2 R ; P1 = I12R;
P2 = I22R U U 2 Vi I1 = 1 = 1(A) v ZL = 100 ; Z = R 2 Z L = 100 2
I2 = 2 = 2(A) Z R I= I=2 I 12 I 2 =5 (A)Cu 93.. t in p xoay chiu
u=220 2 cos(100t) vo 2 u on mch gm in tr R=50 , cun cm thun ZL=100
v t in ZC = 50 mc ni tip. Trong mt chu k khong thi gian in p 2 u
mch thc hin thc hin cng m l ? A) 12,5 ms B) 17,5 ms C) 15 ms D) 5
ms Gii: Chu k ca dng in T = 0,02 (s) = 20 (ms) Z = 50 2 Z ZC Gc lch
pha gia u v i: tan = L = 1 = R 4 Biu thc cng dng in qua mch i =
4,4cos(100t - ) (A) 4 Biu thc tnh cng sut tc thi: p = ui = 965 2
cos100t cos(100t - ) 4in p sinh cng m cung cp in nng cho mch khi p
< 0 hay biu thc Y = cos100t cos(100t - ) < 0 4 Xt du ca biu
thc Y = cos.cos( - ) trong mt chu k 2 4 cos > 0 khi - < <
: 2 2 Vng pha phi ng thng MM50 52. ) > 0 khi - < < 4 2 4 2
M 3 N hay khi - < < : 4 4 Vng pha trn ng thng NN Theo hnh v
du mu ng vi du ca cos du mu en ng vi du ca cos(- ) 4 ta thy vng Y
< 0 khi cos v cos( - ) tri du t N n M v t N n M 4 N T T Nh vy
trong mt chu k Y < 0 trong t = 2 = M 8 4 Do Trong mt chu k,
khong thi gian in p hai u on mch sinh cng m cung cp 20 in nng cho
mch bng: = 5 ms. Chn p n D 4 cos( -+ + ++ + +Cu 94. on mch R=100,
cun thun cm L=318,3mH v t in C=15,92F mc ni tip. Tn s dng in f =
50Hz Trong mt chu k, khong thi gian in p gia hai u on mch sinh cng
dng cung cp in nng cho mch bng: A. 12,5ms B. 15ms C. 17,5ms D. 20ms
1 Gii: Chu k ca dng in T = = 0,02 (s) = 20 (ms) f 1 ZL = 314.
318,3. 10-3 = 100; ZC = = 200; Z = 100 2 314.15,92.10 6 Z ZC Gc lch
pha gia u v i: tan = L =-1 =R 4 Gi s biu thc in p t vo hai u on mch
c biu thc: u = U0cos100t (V) ) (A) 4 Biu thc tnh cng sut tc thi: p
= ui = U0I0 cos100t cos(100t + ) (W) 4 in p sinh cng dng cung cp in
nng cho mch khi p > 0 hay biu thc Y = cos100t cos(100t + ) >
0 4 Xt du ca biu thc Y = cos.cos( + ) trong mt chu k 2 4 cos > 0
khi - < < : 2 2 Vng pha phi ng thng MM Khi biu thc cng dng in
qua mch i = I0 cos(100t +51 53. M+ ) > 0 khi - < + < N 4 2
4 2 3 hay khi 0 do cos v cos( + ) cng du . 4 T T Nh vy trong mt chu
k Y < 0 trong t = 2 = 8 4 T Suy ra Y > 0 trong khong thi gian
3 4 Do : Trong mt chu k, khong thi gian in p hai u on mch sinh cng
dng 20 cung cp in nng cho mch bng: 3. = 15 ms. Chn p n B 4 1 50 Cu
95: Cho mch in RLC (cun dy khng thun cm), L = H, C = F, R = 2r. R
mc vo hai im A, M; cun dy mc vo hai im M, N; t C mc vo hai im N, B;
Mc vo mch hiu in th uAB = U0cos(100t + ) (V), Bit UAN = 200V, hiu
in th tc thi 12 gia hai im MN lch pha so vi hiu in th tc thi gia
hai im AB l 2 a) Xc nh cc gi tr U0, R, r 200 200 100 A. 200 2 V; ;
100; B. 400V; ; ; 3 3 3 200 200 100 C. 100 2 V; ; 100; D. 200 2 V;
; ; 3 3 3 b) v vit biu thc dng in trong mch? A. i = 2 sin(100t + )
A B. i = 2sin(100t - ) A 3 3 C. i = cos(100t + ) A D. i = 2
cos(100t + ) A 3 3 Gii: L; a. Ta c: ZL = 100; ZC = 200; C R Z L ZC
r 100 tanAB = =N M A B 3r Rr cos( ++ +++ +52 54. tanMN =Z L 100 = r
r tanAB tanMN = - 1 2 100 200 100 2 Do =1r= . R = 2r = 2 3r 3 3 U
AB Z = = 1 ( V Z = ZAN = 200 ) UAB = UMN = 200V. Do U0 = 200 2 (V)
Z AN U MN Chn p n D Z ZC 100 1 b. tanAB = L == AB = - : uAB chm pha
hn i gc 3r Rr 6 6 3 U I = AB = 1 A Z Vy Biu thc dng in trong mch i
= 2 cos(100t + + ) = 2 cos(100t + ) A. Chn p n D 12 6 3 Cu 96: Cho
mch in RLC ni tip; R = 120 3 , cun dy c r = 30 3 . hiu in th hai u
on mch uAB = U0cos(100t + ) (V), R mc vo hai im A, M; cun dy mc vo
hai 12 im M, N; t C mc vo hai im N, B; UAN = 300V, UMB = 60 3 V.
Hiu in th tc thi uAN lch pha so vi uMB l . Xc nh U0, L, C? 2 10 3
10 3 1,5 1,5 A.60 42 V; H; F; B. 120V; H; F; 24 24 10 3 10 3 1,5
1,5 C. 120V; H; F; D. 60 42 V; H; F; uMN sm pha hn uAB gcZL Rr Z L
ZC tanMB = rGii: tanAN = AR ML; r NC B nn 2 Z L Z L ZC tanAN .tanMB
= -1 . = - 1 ZL(ZL ZC) = - 13500 (*) r Rr 2 2 (R r) 2 Z L U AN Z
67500 Z L 25 300 = AN = = = (**) 2 U MB Z MB 3 2700 ( Z L Z C ) r 2
( Z L Z C ) 2 60 3 T (*) v (**) ta c ZL = 150 v ZC = 240 uAN sm pha
hn uNB gc53 55. L= ZMB = Z=1,5H; C =10 3 F; 24r 2 ( Z L Z C ) 2 =
10800 = 60 3 () I = ( R r ) 2 (Z L Z C ) 2 =U0 = I0Z =U MB = 1A Z
MB75600 =30 84 ()2 .30. 84 = 6042 (V) 10 3 1,5 p s: U0 = 60 42 V; L
= H; C = F; p n A 24 Cu 97: t vo hai u on mch RLC ni tip mt in p
xoay chiu c gi tr hiu dng v tn s khng i. Ti thi im t1 cc gi tr tc
thi uL(t1) = -10 3 V, uC(t1) = 30 3 V, uR(t1) = 15V. Ti thi im t2
cc gi tr tc thi uL(t2) = 20V, uC(t2) = - 60V, uR(t2) = 0V. Tnh bin
in p t vo 2 u mch? A. 60 V. B. 50V. C. 40 V. D. 40 3 V. Gii: ) = -
U0L sint; uC = U0C cos(t - ) = U0C sint 2 2 Ti thi im t2: uR(t2) =
U0R cost2 = 0V cost2 = 0 sint2 = 1 uL(t2) = - U0L sint2 = 20V U0L =
20V (*) uC(t2) = U0C sint2 = -60V U0C = 60V (**) Ti thi imt t1:
uR(t1) = U0R cost1 = 15V. uL(t1) = - 20 sint1 = -10 3 V ; uC(t1) =
60 sint1 = 30 3 V 1 3 sint1 = cost1 = . Do U0R = 30 V (***) 2 2 2 2
2 U0 = U0R + ( U0L U0C) = 302 + 402 U0 = 50 V. Chn p n B Ta c uR =
U0R cost ; uL = U0L cos(t +Cu 98: t vo hai u on mch RLC ni tip mt
in p xoay chiu c gi tr hiu dng v tn s khng i. Ti thi im t1 cc gi tr
tc thi uL(t1) = -30 3 V, uR(t1) = 40V. Ti thi im t2 cc gi tr tc thi
uL(t2) = 60V, uC(t2) = -120V, uR(t2) = 0V. in p cc i gia hai u on
mch l: A. 50V B. 100 V C. 60 V D. 50 3 V Gii: ) = - U0L sint; uC =
U0C cos(t - ) = U0C sint 2 2 Ti thi im t2: uR(t2) = U0R cost2 = 0V
cost2 = 0 sint2 = 1 uL(t2) = - U0L sint2 = 60V U0L = 60V (*) uC(t2)
= U0C sint2 = -120V U0C = 120V (**) Ti thi imt t1: uR(t1) = U0R
cost1 = 40V. uL(t1) = - 60 sint1 = -30 3 V ; Ta c uR = U0R cost ;
uL = U0L cos(t +54 56. 1 3 cost1 = . Do U0R = 80 V (***) 2 2 2 2 2
U0 = U0R + ( U0L U0C) = 802 + 602 U0 = 100 V. Chn p n B Cu 99: Mch
in AB gm on AM v on MB . in p hai u mch n nh u = 220 2 cos100t (V).
in p hai u on AM sm pha hn cng dng in mt gc 300. on MB ch c mt t in
c in dung C thay i c. Chnh C tng in p hiu dng UAM + UMB c gi tr ln
nht. Khi in p hiu dng hai u t in l A. 440 V. B. 220 3 V. C. 220 V.
D. 220 2 V. Gii: M Z 1 tanAM = L = tan300 = B A R 3 R 2R 2 ZL = ZAM
= R 2 Z L = (*) 3 3 sint1 =t Y = (UAM + UMB)2. Tng (UAM + UMB) t gi
tr cc i khi Y t gi tr cc i Y = (UAM + UMB)2 = I2( ZAM +ZC)2 =U 2 (
Z AM Z C ) 2 U 2 ( Z AM Z C ) 2 = 2 2 2 R 2 ( Z L Z C ) 2 R Z L Z C
2Z L Z C Y = Ymax th o hm ca Y theo (ZC) Y = 0 2 2 ( R 2 Z L ZC 2Z
L ZC )2(ZAM + ZC) - (ZAM + ZC)2 2(ZC ZL) = 0. Do (ZAM + ZC) 0 nn 2
2 ( R 2 Z L ZC 2Z L ZC ) - (ZAM + ZC)(ZC ZL) = 0 (ZAM + ZL)ZC = R2
+ ZL2 + ZAMZL (**) . Thay (*) vo (**) ta c ZC = Z2 = R 2 ( Z L ZC )
2 Z =2R 3(***)2R(****) 3 Ta thy ZAM = ZMB = ZAB nn UMB = UC = UAB =
220 (V). Chn p n C Cu 100: on mch AB gm R, C v cun dy mc ni tip vo
mch c u = 120 2 cost (V); khi mc ampe k l tng G vo hai u ca cun dy
th n ch 3 A. Thay G bng vn k l tng th n ch 60V, lc in p gia hai u
cun dy lch pha 600 so vi in p gia hai u on mch AB. Tng tr ca cun dy
l: A. 20 3 B. 40 C. 40 3 D. 60 C L ,r Gii: Khi mc ampe k ta c mch
RC R A B U I1 = ZRC = 40 3 2 R2 ZC Khi mc vn k ta c mch RCLr ud =
60 2 cos(t + ) (V) 355 57. u = uRC + ud uRC = u ud V gin vect. Theo
gin ta c: 2 U RC = 1202 + 602 2.120.60 cos600 = 10800 URC = 60 3
(V) Do cng dng in qua mch U 60 3 I = RC = = 1,5 (A) Z RC 40 3 U 60
Suy ra Zd = d = = 40. Chn p n B 1,5 IUd UUR-UdCCu 101: Mch in RLC
ni tip, cun dy thun cm, cng dng in trong mch c biu thc i = Iocost.
Cc ng biu din hiu in th tc thi gia hai u cc phn t R, L, C nh hnh v.
Cc hiu in th tc thi uR, uL, uC theo th t l A. (1), (3), (2). B.
(3), (1), (2). C. (2), (1), (3). D. (3), (2), (1). Gii: Cc biu thc
ca uR; uL; uC uR = U0Rcost . Trn th (3) uL = U0Lcos(t + ). Trn th
(2) 2 uC = U0Ccos(t ). Trn th (1) 2 Chn p n D: (3); (2); (1) Cu
102: Trong my pht in xoay chiu 3 pha, c sut in ng cc i l E0 , khi
sut in ng tc thi cun 1 trit tiu th sut in ng tc thi trong cun 2 v 3
tng ng l A. E0 ; E0 . B. E0 / 2; E0 3 / 2 . C. E0 / 2; E0 / 2 . D.
E0 3 / 2; E0 3 / 2 . Gii: Ta c e1 E0cos t2 ) 3 2 e3 E0cos( t+ ) 3
e2 E0cos( t-2 2 2 E0 3 ) E0cost cos E0sint sin 3 3 3 2 E 3 2 2 2 e3
E0cos( t+ ) E0cost cos E0sint sin 0 3 3 3 2 Chn p n DKhie1 = 0 cost
= 0e2 E0cos( t-56 58. Cu 103 : 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
Ur2 + UL2 = ULr2 UL ULr (UR + Ur)2 + (UL UC)2 = U2 Vi U = 40 2 (V)
Ur2 + UL2 = 252 (*) Ur (25+ Ur)2 + (UL 60)2 = U2 = 3200 UR 2 2 625
+ 50Ur + Ur + UL -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) U UR Ur 1 Lc ny
cos = = U UC 2 P = UIcos I = 1 (A) Do r = 15 . Chn p n A Cu 104:
Mch in xoay chiu R, L, C mc ni tip. in p hai u on mch l u=Uocost.
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 2 L(1 2 ) L1 2 L(1 2 ) A. R = 1 . B. R = . C. R = . D. R =
. 2 2 2 n 1 n 1 L n 1 n2 1 Gii: U U Ta c: I1 = ; I2 = 1 2 1 2 R 2
(1 L ) R 2 ( 2 L ) 1C 2C 1 1 1 1 1 I1 = I2 1L = - (2L ) hay : (1 +
2 )L = ( + ) 1C 2C C 1 2 1 1 LC = C1 = (*) 1 2 L 2 I U U 1 2 Khi I
= Ic = I1 = I2 = c = R2 + (1L ) = n2R2 R nR 1C n 1 2 (1L ) =(n2
1)R2 (**) 1C T (*) v (**) ta c (n2 1)R2 = (1L - 2L )2 = L2 (1-
2)257 59. Do R =L(1 2 ). Chn p n B n2 1 Cu 105: Dng in i=4cos2t (A)
- Gi tr hiu dng l? - Gi tr trung bnh l? - Gi tr cc i l? Gii: Ta c i
= 4cos2t (A) = 2 (cos2t + 2) = 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) a. C gi tr hiu dng l C hai
kh nng : 1. 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) 2. Nu trong mch khng
c t th cng sut ta nhit trong mch P = P1 + P2 = I12R + I22 R = I2R I
= b. C gi tr trung bnh l2 I12 I 2 6 (A)I = 2cos2t + 2 = 0 + 2 (A)
c. C gi tr cc i l C hai kh nng : 1. Nu trong on mch c t in th thnh
phn I2 khng qua mch. Khi gi tr cc ca dng in qua mch Imax = I1max =
2 (A) 2. Nu trong mch khng c t Imax = I1max + 2 = 4 (A)58 60. NH
XUT BN TRNG HC S ***********BI TP V PHNG PHP GII BI TON IN XOAY
CHIU BIN SON: TRN DUY KHOAChu trch nhim xut bn: TRNG HC SH NI, NGY
22 THNG 1 NM 2013
