Giai Toan Dien Xoay Chieu NTL

Gio vin: Nguyn Thnh Long Email: [email protected] 01694 013 498

1

(DNG CHO N THI TN C H 2011)

Gi tng: aotrangtb.com

Bm sn. 10.04.2011

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BI TON 1: TNH CC I LNG C TRNG CA DNG IN XOAY CHIU

Dng 1: Dng in xoay chiu v hiu in th xoay chiu trong cc loi on mch:

* on mch ch c in tr thun R: Ru cng pha vi i, 0u i : UIR

v 00UIR

Lu : in tr R cho dng in khng i i qua v c UIR

* on mch ch c cun thun cm L: Lu nhanh pha hn i l ,2 2u i

: L

UIZ

v 00L

UIZ

vi ZL = L l cm khng Lu : Cun thun cm L cho dng in khng i i qua hon ton (khng cn tr).

* on mch ch c t in C: Cu chm pha hn i l ,2 2u i

: C

UIZ

v 00C

UIZ

vi 1CZ C l dung khng.

Lu : T in C khng cho dng in khng i i qua (cn tr hon ton). Ch : Vi mch hoc ch cha L, hoc ch cha C, hoc cha LC khng tiu th cng sut ( 0P )

0 0

0 0

cos cos( )

cos cos( - ) u i u i i uNeu i I t thi u U t

VoiNeu u U t thi i I t

on mch nh lut m cho on mch

Quan h gia u v i Gin vecto

Ch

Ch c R .R R

UI U I RR

Ru lun ng pha i ( 0)R

RU in p hiu dng hai u in tr R

00

0 0.

R

R

U IR

U I R

Cun dy thun cmch c L

.L L LL

UI U I ZZ

*Vi cm khng: . ( )LZ L

* Ch : Nu cun khng thun cm ( c in tr thun LR )

2 2L LZ R Z day

Lu lun nhanh pha so vi i gc

2 ( ) 2L

LU in p hiu dng hai u cun thun cm L

00

0 0.

L

L

L L

U IZ

U I Z

Ch c C

.C C CC

UI U I ZZ

Vi dung khng

Lu lun chm pha so vi i gc

2 ( ) 2C

CU in p hiu dng hai u t C

00

0 0.

C

C

C C

U IZ

U I Z

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1 ( ) .C

ZC

RLC ni tip

.UI U I Z Z

Vi tng tr ca mch: 2 2( ) ( )

L CZ R Z Z

* Ch : Nu cun khng thun cm ( c in tr thun LR )

2 2( ) ( )L L CZ R R Z Z

Gi s: L C L CU U Z Z

* lch pha ca u so vi i:

iu ui

L C L C

R

U U Z ZtgU R

+ Nu 0 u sm pha hn i

L CZ Z mch c tnh cm khng +Nu 0 u cham pha hn i

L CZ Z mch c tnh dung khng +Nu 0 u cung pha vi i

L CZ Z mch c thun tr.

00

0 0.

U IZ

U I Z

Vi:0

0

2

2

I I

va U

U

+ Nu cc in tr c ghp thnh b ta c:

Ghp ni tip cc in tr Ghp song song cc in tr

1 2 ... nR R R R Ta nhn thy in tr tng ng ca mch khi ln hn in tr thnh phn. Ngha l :

Rb > R1, R2

1 2

1 1 1 1...nR R R R

Ta nhn thy in tr tng ng ca mch khi nh hn in tr thnh phn. Ngha l :

Rb < R1, R2 Ghp ni tip cc t in Ghp song song cc t in

1 2

1 1 1 1...nC C C C

Ta nhn thy in dung tng ng ca mch

1 2 ... nC C C C Ta nhn thy in dung tng ng ca mch

khi ln hn in dung ca cc t thnh

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khi nh hn in dung ca cc t thnh phn. Ngha l : Cb < C1, C2

phn. Ngha l : Cb > C1, C2

Loi 1: Xc nh gi tr cc phn t R, L, C, f c trong on mch khng phn

Phng php: Da vo cc d kin cho tnh gi tri tng tr Z ca on mch ang xt ri s dng cng thc

2 2( )L CZ R Z Z nu mch c thm r th 2 2( ) ( )L CZ R r Z Z . T suy ra: , ,L CZ Z R cn

tm. D kin cho S dng cng thc Ch

Cng hiu dng I v hiu in th

1

1

L C R r

L C

UU U U UU IZ Z Z R r Z

Cho n d kin tm c (n-1) n s

Cho lch pha ui

hoc

cho u v i th

u u ii

0 0

R 0R

tan L C L C L CZ Z U U U UR U U

hoc

0RR

0

os UURcZ U U

v sin L CZ Z

Z

vi

2 2

Nu mch c R v r th : 0R 0R

0

os rr U UU UR r cZ U U

tan L CZR r

Z

Thng tnh osRZ

c

osR rZc

Cng sut P hoc nhit lng Q

22

2 2. os ( )L C

RUP R I UIcR Z Z

nu c R v r th: 2

22 2

( )( ). os( ) ( )L C

R r UP R r I UIcR r Z Z

Thng s dng tnh

PIR

nu c R v r th

PIR r

ri p dng

nh lut Ohm tnh cc tr khng cn tm

Ch : C th s dng cng thc trc tip tnh: Cng sut ca dng in xoay chiu:

2 2 22 2 2 2

2os ( )R L CU U UP UIc U I I R R Z R R Z Z R

P PZ

Nhit lng ta ra (in nng tiu th) trong thi gian t(s) : 2 . .Q I R t H s cng sut c h os oac :

22 2os ( )

. os osR

L CP U R R Rc Z R Z Z

U I U Z c c

in p hiu dng hai u mi phn t in:

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. ; . ; .R L L C CU I R U I Z U I Z vi

.

.

.

R

LL

CC

UZ RU

U UI Z ZZ U

UZ ZU

22 2 2

22 2 2

22 2 2

( )

( )

( )

L CR

L C LL

L C CC

UR Z Z RU

UR Z Z ZU

UR Z Z ZU

Ch : - Tt c cc cng thc sau khi c bin i nh trn ta c th a v gii phng trnh bc 2 hoc - a v dng 2 2A B gii. - Hy dng cng thc trn v p dng cho mch in trong bi ton. Lp ra h phng trnh sau gii. Cnphi ngh n gin vc t v cho mch in bo m h phng trnh khng b sai. Ch thm tch

.L CLZ ZC

. Khi bi ton cho cc in p hiu dng thnh phn v hai u mch, cho cng sut tiu th

nhng cha cho dng in th hy lp phng trnh vi in p hiu dng. Khi tm ra UR s tm R

PIU

sau

tm ; ; .CR LL CUU UR Z Z

I I I

Bi tp trc nghim:

Cu 1: Hai cun dy 1 1R ; L v 2 2R ; L ni tip vo mch in xoay chiu. bit t s 12

2R R

. Khi hiu

in th hiu dng gia hai u mch bng tng cc hiu in th hiu dng ca hai cun dy th t s 12

L L

bng gi tr no sau y.

A. 12

12

L L

. B. 12

4L L

. C. 12

1L L

. D. 12

2L L

Cu 2: Mt n c ghi (110V 100W) mc ni tip vi in tr R vo mt mch in xoay chiu c u 200 2 cos(100t) (V). n sng bnh thng, R phi c gi tr bng A. 1210 . B. 10/11 . C. 121 . D. 99 .

Cu 3: Cho bit: R = 40, C F410 52,

v:

80cos100 ( )AMu t V ;7200 2 cos(100 ) ( ) 12MB

u t V

r v L c gi tr l:

A. r L H3

100, B. r L H

10 3 10,

R C L, r

MA B

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C. r L H21

50, D. r L H2

50,

Cu 4: Cho mch in xoay chiu R v L mc ni tip. Bit R = 4,5 , mch t di hiu in th c biu thc l u = 110cos100 t(V). Gi tr cc i ca cng dng in l I0 = 10A. t cm ca cun dy l A. L = 1/20 (H). B. L = 1/10 (H). C. L = 1/15 (H). D. Kt qu khc. Cu 5: Mt cun dy dn in tr khng ng k c cun di v ni vo mng in xoay chiu 127V 50Hz. Dng in cc i qua n bng 10A. t cm ca cun dy l A. 0,04H. B. 0,08H. C. 0,057H. D. 0,114H. Cu 6: Cho mch in xoay chiu gm R, L mc ni tip. Hiu in th 2 u mch c dng

uAB

= 100 2 sin100t (V) v cng dng in qua mch c dng i = 2sin(10t - 3 )A. Gi tr ca R v

L l:

A. R = 25 2 , L = 610, H. B. R = 25 2 , L =

220, H.

C. R = 25 2 , L =1 H. D. R = 50, L =

0,75 H.

Cu 7: Nu mc ni tip in tr R = 50 vi cun dy thun cm c L = 12

H th cng hiu dng

trong mch l 2 A. Nu thay R bng t in c in dung C th cng dng in tng ln 2 ln. Gi tr ca in dung C l:

A. 410 4

F B. 41021

F C. 410 1

F D. 410 41

F

Cu 8: Cho mch in nh hnh, cun dy thun cm c t cm thay i c. t vo hai u on mch hiu in th ABu U 2 sin120 t(V) , trong U l hiu in th hiu dng,

R = 30 3. Bit khi L = 3 H 4

th R3U U

2 v mch c tnh dung khng.

in dung ca t in l: A. 221F B. 0,221F C. 2,21F D. 22,1F Cu 9: Cho mch nh hnh v: B L R C A Cun dy thun cm

uAB = 220 2 cos100t(V); C = F 3

103 , V2 ch 220 3 V; V1 ch 220V.

in tr cc vn k rt ln. R v L c gi tr:

A. 20 3 v 51 H B. 10 3 v

51 H

C. 10 3 v 1 H D. Tt c u sai

Cu 10: Mch nh hnh v A R,L N R,L B uAB = 80 2 cos100 t(V), R = 160 , ZL = 60 Vn k ch UAN = 20V. Bit rng UAB = UAN + UNB in tr thun R v t cm L c gi tr:

A. R = 160 (); L = 21 H B. R = 160/3 (); L =

31 H

C. R = 160 (); L = 51 H D. R = 160/3 (); L =

51 H

R L C

A BM N

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Cu 11: Cho mt on mch xoay chiu gm cun dy c in tr r, t cm L mc ni tip vi in tr thun 5R . Bit hiu in th gia hai u on mch v cng dng in qua mch c biu thc:

100 2 cos(100 ) ( ), 2 2 cos(100 ) ( )6 2

u t V i t A . Gi tr ca r bng:

A. 20, 6 B. 36, 6 C. 15, 7 D. 25,6Cu 12: Cho on mach xoay chiu gm R, L mc ni tip, hiu in th hai u on mch c dng

100 2 sin100 ( )u t V v cng dng in qua mch c dng 2cos(100 )( ) 4

i t A . R, L

c nhng gi tr no sau y:

A. 150 ,R L H

B. 250 2 ,2

R L H

C. 150 ,2

R L H

D. 1100 ,R L H

Cu 13: Cho mch in xoay chiu gm R, C ghp ni tip, hiu in th hai u mch c

dng 50 2 cos100ABu t (V) v cng dng in qua mch 2 cos(100 3i t ) (A). R, C

c nhng gi tr no sau y?

A. 10 3

50 ;5

R C F

B. 23.10 25 ;

25R C F

C. 102

25 ;25 3

R C F

D. 35.10 50 ;R C F

Cu 14: Cho mch in xoay chiu nh hnh v bit R = 50 ;C = 2 .10-4 F ; uAM = 80cos 100t (V);

uMB = 200 2 cos(100t +2 )V . Gi tr r v L ln lt l

A. 176,8 ;0,56H B. 250 ;0,8H C. 250 ;0,56H D. 176,8 ;0,8H

Loi 2: Quan h gia cc gi tr hiu dng ca cc in p (S o ca Vn- k):

S CH CC IN K

a. Tc dng cc in k in k s dng trong mch in xoay chiu l vn k nhit v ampe k nhit o cc gi tr hiu dng ca in p v cng dng in b. S ch cc in k - Nu vn k c in tr v cng ln v ampe k c in tr khng ng k th vn k ch in p trn on mch song song vi n, ampe k ch cng dng in trong mch ni tip vi n - Nu vn k c in tr hu hn, ampe k c c in tr khc khng th ta coi chng nh nhng in tr v kho st mch bnh thng

Phng php: Cch 1: - S dng cng thc: U I .Z ; U IR R ; L LU IZ ; C CU IZ ; U = U0/ 2 .

Hoc 2 2 2( )R L CU U U U . Trong mch R, L, C ni tip lun c UR U

v ; osL C RR

U U Utg c

U U

MAC r,LR

B

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- Hoc s dng cc cng thc cho tng loi on mch:

V d:

2 2 2

2 2 2

2 2

2 2 2

(1)

(2)( ) (3)

( ) (4)

RL R L

RC R L

LC L C

R L C

U U UU U UU U UU U U U

Gii cc phng trnh trn tm ra , , .............. R L CU U U hoac so ch cua Von KeCch 2: S dng gin vec-t Fresnel - V gin vec-t Fresnel v nn v theo quy tc 3 im( V cc vec- t lin tip nhau)- p dng nh l hm s cos(hoc sin) tnh cos ( sin )hoac- Da vo h thc lng trong tam gic tnh , , , ......R L CU U U U

Bi tp trc nghim:

Cu 1: Cho mt on mch xoay chiu ni tip gm in tr R, cun dy thun cm L v t C. t vo hai u on mch hiu in th u 100 2 sin(100 t)V , lc L CZ 2Z v hiu in th hiu dng hai u in tr l U VR 60 . Hiu in th hiu dng hai u cun dy l: A. 60V B. 160V C. 120V D. 80V Cu 2: Ngi ta o c cc hiu in th UAN = UAB = 20V; UMB = 12V. Hiu in th UAM, UMN, UNB ln lt l: A. UAM = 12V; UMN = 32V; UNB =16V B. UAM = 12V; UMN = 16V; UNB =32V C. UAM = 16V; UMN = 24V; UNB =12V D. UAM = 16V; UMN = 12V; UNB =24V Cu 3: Trong on mch xoay chiu ch c t in , nu ng thi tng tn s ca in p ln 4 ln v gim in dung ca t in 2 ln ( 0U khng i ) th cng hiu dng qua mch A. tng 2 ln . B. tng 3 ln . C. gim 2 ln . D. gim 4 lnCu 4: Cho mch in xoay chiu nh hnh v. Ngi ta o c cc hiu in th UAM = 16V, UMN = 20V, UNB = 8V. Hiu in th gia hai u on mch AB l: A. 44V B. 20V C. 28V D. 16V Cu 5: on mch in xoay chiu RLC ni tip. Hiu in th hiu dng hai u on mch l U = 123V, UR = 27V; UL = 1881V. Bit rng mch c tnh dung khng. Hiu in th hiu dng hai u t in l A. 200V. B. 402V. C. 2001V. D. 201V. Cu 6: Cho mch in gm cun dy thun cm mc ni tip vi in tr thun. t vo hai u on mch mt hiu in th xoay chiu n nh. Dng vn k xoay chiu ln lt t vo hai u cun cm v in tr, s ch ln lt l 120V v 160V. Nu t vn k vo hai u on mch th s ch ca vn k l A. 140V. B. 40V. C. 200V. D. 280V. Cu 7: Mt hiu in th xoay chiu 25V, 50Hz c t vo hai u on mch gm in tr R mc ni tip vi cun dy thun cm L. Hiu in th hiu dng gia hai u R bng 20V. Hiu in th hiu dng gia hai u cun dy thun cm L l A. 5V. B. 10V. C. 15V. D. 12V. Cu 8: on mch RLC ni tip, gm in tr thun 30 , mt cun dy thun cm 191mH, mt t in 53F, c u vo mng in xoay chiu 120V, 50Hz. Hiu in th gia hai u t in l A. 60V. B. 120V. C. 240V. D. 48V. Cu 9: Cho mch in xoay chiu RLC, cun dy thun cm. Hiu in th hiu dng gia hai u mch l

200V, UL =

38 U

R = 2U

C. Hiu in th hiu dng gia hai u in tr R l:

R L C

A M N BHnh 49

R L C

A M N BHnh 50

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A. 180V. B. 120V . C. 145V. D. 100V. Cu 10: (C 2007) t hiu in th 0 sinu U t (U0 v khng i) vo hai u on mch RLC khng phn nhnh. Hiu in th hiu dng hai u in tr thun l 80V, hai u cun dy thun cm (cm thun) l 120 V v hai u t in l 60 V. Hiu in th hiu dng gia hai u on mch ny bng: A. 140 V B. 100 V C. 220 V D. 260 V Cu 11: (C 2008) Khi t hiu in th 0 sin ( )u U t V vo hai u on mch RLC khng phn nhnh th hiu in th hiu dng gia hai u in tr, hai u cun dy v hai bn t in ln lt l 30 V, 120 V v 80 V. Gi tr ca U0 bng: A. 30V B. 50 2 V C. 30 2 V D. 50 V Cu 12: (C 2008) Mt on mch gm cun dy thun cm (cm thun) mc ni tip vi in tr thun. Nu t hiu in th 15 2 sin100 ( )u t V vo hai u on mch th hiu in th hiu dng gia hai u cun dy l 5V. Khi hiu in th hiu dng gia hai u in tr bng A.10 2V B. 5 2V C.10 3V D. 5 3V Dng 2: Tnh tng tr Tnh cng dng in

+ Tnh tng tr bng cng thc thao cu to hoc cng thc nh ngha

22 L CZ R Z Z ; 0 0

UU ZI I

+ Tnh cng hay hiu in th t cng thc ca nh lut m:

UIZ

hay 00UIZ

+ Tnh cng dng in hoc in p t nh lut Ohm: 1

1

L C R

L C

UU U UU IZ Z Z R Z

+ Gia cc hiu in th, c th dng h thc lin lc sau thc hin tnh ton: i vi on mch c ba phn t RLC mc ni tip T 2 2( )L CZ R Z Z

22 2 R L CU U U U Hay

22 20 0R 0L 0CU U U U

T 2 2( ) ( )L CZ R r Z Z 2 2 2 2( ) ( )R r L CU U U U U

i vi on mch c hai trong ba phn t mc ni tip T 2 2 2 2RL L RL R LZ R Z U U U

T 2 2 2 2RC C RC R CZ R Z U U U T LC L C LC L CZ Z Z U U U + Cng c th tnh da vo gin vect quay biu din tnh cht cng ca cc hiu in th.

u = u1 + u2 0 01 02

1 2

U U U

U U U

Ch : - Nu on mch khng c ba phn t R, L, C th phn t thiu c tr khng bng khng

on mch

Tng tr 2 2

CR Z 2 2

LR Z L CZ Z

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tg CZ

R LZ

R 2

2

- Nu on mch c nhiu phn t cng loi ni tip th gi tr cc in tr trong cng thc theo cu to l tng cc in tr:

1 2

1 2

1 2 ......

...n

n

n

L L L L

C C C C

R R R RZ Z Z Z

Z Z Z Z

- Nu cun t cm c cm khng ZL v in tr hot ng R th cun t cm ny tng ng vi on mch gm cun thun cm ZL ni tip vi in tr thun R

Bi tp trc nghim:

Cu 1: (H 2008) Cho on mch gm in tr thun R ni tip vi t in c in dung C. Khi dng in xoay chiu c tn s gc chy qua th tng tr ca on mch l

A. 2

2 1R .C

B.

22 1R .

C

C. 22R C . D. 22R C .

HD:

V on mch ch c R v C mc ni tip nn 2

2 10LZ Z R C

Cu 2: (C 2010) t in p 0u U cos( t ) (V) 6

vo hai u on mch gm in tr thun R v

cun cm thun c t cm L mc ni tip th cng dng in qua on mch l

05i I sin( t ) (A) 12

. T s in tr thun R v cm khng ca cun cm l

A. 12

. B. 1. C. 3 2

. D. 3 .

HD:

0 05sin( ) cos( )12 12

i I t I t ; tan tan 14 4

LZR

Cu 3: (C 2010) t in p xoay chiu vo hai u on mch gm in tr thun 40 v t in mc

ni tip. Bit in p gia hai u on mch lch pha 3 so vi cng dng in trong on mch.

Dung khng ca t in bng

A. 40 3 B. 40 3 3

C. 40 D. 20 3

HD:

on mch ch cha R v C m theo gi thit lch pha ca u so vi I l 3 , suy ra u phi tr pha so vi i

tc l tan tan 3 40 33 3

CC

Z ZR

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Cu 4: Mt on mch gm mt in tr thun R = 50 , mt cun cm c 1L

. Hiu in th, v mt t

in c in dung 42 .10C F

, mc ni tip vo mng in xoay chiu c U = 120V, tn s

f = 50Hz. Tng tr ca on mch c th nhn gi tr no sau y? A. Z = 50 2 B. Z = 50 C. Z = 25 2 D. Z = 100Cu 5: Cho mch in xoay chiu R, L, C mc ni tip. Cun dy ch c h s t cm 0,1L

. Hiu in

th; in tr thun R = 10 v mt t in c in dung 500C F

. t vo hai u on mch mt hiu

in th xoay chiu c tn s f = 50Hz v hiu in th hiu dng U = 100V. Tng tr Z ca mch in c th nhn gi tr no trong cc gi tr sau? A. Z = 15,5 B. Z = 20 C. Z = 10 D. Z = 35,5

Cu 6: Cho mt mch in gm R, L, C mc ni tip. Bit: R = 6 ; 23 1010

L H F

3 ; C =

12; Hiu in

th hai u on mch uAB = 120sin100 t. Tng tr ca mch in c th nhn gi tr no sau y? A. Z = 8 B. Z = 12 C. Z = 15 D. Z = 12,5Cu 7: Cho on mch gm in tr thun R = 100 v t in c in dung C mc ni tip. Bit biu thc hiu in th gia 2 u on mch u = 100cos(100t)V v cng hiu dng trong mch I = 0,5 A. Tnh tng tr ca on mch v in dung ca t in?

A. Z = 100 2 ; C = F410 1

B. . Z = 200 2 ; C = F410 1

C. Z = 50 2 ; C = F410 1

D. . Z = 100 2 ; C =310 F

Cu 8: (C 2010) t in p u = U0cost vo hai u cun cm thun c t cm L. Ti thi im in p gia hai u cun cm c ln cc i th cng dng in qua cun cm bng

A. 02U

L . B. 0

2U

L . C. 0U

L . D. 0.

HD: max max 0 0 tU W W i

Cu 9: Mt on mch gm cun dy ch c t cm L = 1

Hiu in th v in tr thun R = 100mc ni tip. t vo hai u on mch mt hiu in th mt chiu U = 50V. Cng dng in trong mch nhn gi tr no trong cc gi tr sau? A. I = 0,25A B. I = 0,5A C. I = 1A D. I = 1,5A. Cu 10: t vo hai u in tr R = 50 mt hiu in th xoay chiu c biu thc: u = 100 2 sin100 t (V). Cng dng in hiu dng c th nhn gi tr no trong cc gi tr sau? A. I = 2A B. I = 2 2 A C. I = 2 A D. Mt gi tr khc. Cu 11: Mt bp in c in tr l 25 v t cm khng ng k c th s dng hiu in th xoay chiu hoc mt chiu. Ni bp in vi dng in xoay chiu c hiu in th cc i 100 2 . Dng in hiu dng qua bp c th nhn gi tr no sau y? A. I = 4A B. I = 8A C. I = 4 2 A D. Mt gi tr khc. Cu 12: t vo hai u mt on mch nh hnh v mt in p xoay chiu c gi tr hiu dng v tn s khng i c th thy in p hiu dng gia hai u cc on mch MB v AB l nh nhau. Bit cun dy ch cm khng ZL = 100. Dung khng ca t nhn gi tr no sau y? A. 50. B. 200. C. 150. D. 120.

CL, r A BR M

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Cu 13: t vo hai u on mch hiu in th xoay chiu

ABu U 2 sin120 t(V) , 14L = H;

3 24 L = H; 0

r = 30; R = 90 . Tng tr ca on mch AB l:

(L1ntL2) A. 514,8 B. 651,2 C. 760 D. 520Cu 14: (C 2007) t hiu in th 125 2 sin100 ( )u t V ln hai u on mch gm in tr thun R

= 30, cun dy thun cm (cm thun) c t cm 0, 4 L H

v ampe nhit mc ni tip. Bit ampe k

c in tr nh khng ng k. S ch ca ampe k l A. 3,5 A B. 1,8 A C. 2,5 A D. 2,0 A Dng 3: Cun cm c in tr thun

Cun cm c qun t cc dy kim loi do chng lun c in tr. Trong mt s trng hp in tr ny khng nh so vi dung khng. do ta phi kho st cun dy nh cun cm thun ni tip vi in tr a. Tnh cht cun dy khng thun cm

- Tng tr dy 2 2d LZ r Z

- in p dy 2 2d LU I r Z

- lch pha ud v id l tan LZ r

b. Mch in khi cun dy khng thun cm - Tng tr mch: ZAB = 2 2( ) ( )L CR r Z Z

- in p ton mch: UAB = 2 2( ) ( )R r L CU U U U - lch pha gia in p ton mch v cng dng in

tan = R r

CL C L

U UUU

R rZ Z

; cos =

AB

rR

AB UUU

ZR r

c. Chng minh cun cm c in tr thun Cch 1: Gi s cun cm khng c in tr thun dng lp lun ch ra iu v l Cch 2: Lp lun c kt qu vi phm tnh cht ca cun cm nh lch pha ca ud v cng dng in nh hn /2, Cch 3: Ch ra c cun dy tiu th in (to nhit) Ch : Chng minh cun dy c hoc khng c in tr thun th da vo cc du hiu quan h in p hoc gc lch pha gia dng in vi in p, gc lch pha gia cc in p vi nhau. Nn dng gin vc t d thy trong trng hp gc lch pha.

Dng 4: on mch RLC: trng hp 1 phn t in (nh R hoc L hoc C) b on mch tnh cng hiu dng I khi bit hiu in th hiu dng U (hay ngc li)?

HIN TNG ON MCH RU

LU d

U

RU

LU d

U

a. Hin tng on mch - Xt on mch AB c mt dy khng in tr ni

A B|

A B|

RU

LU

ABU

CU

rU

RU

LU

ABU

CU

rU

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13

tt nh hnh v cng dng . Ta c UAB = I.0 = 0 VA = VB trn mch tng ng vi A trng B ng thi khi cng dng in chy qua on di I = 0/Z = 0 hay ni cch khc on mch di b xa b khi mch b. Phng php bin i mch tng ng - Xc nh cc im trng nhau ( 2 u dy ni) - Xp cc im trn mt phng sao cho 2 ngun 2 u, cc im trng nhau v chung mt im- t cc linh kin vo mch mi sao cho gi nguyn 2 u nh mch c - Nu ch ra on dy song song vi on no th b on mch c. Gii mch Cc gi tr U,I c xc nh theo mch mi, s ch cc in k phi da vo mch c Phng php: Nu c mt phn t in (thuc mch RLC) b on mch th ta phi loi b phn t , ngha l trong cc cng thc ni trn ta phi cho in tr tng ng bng 0. Cc trng hp on mch thng gp: 1. Trng hp hai u phn t in b chp vi nhau:

Th d: cun L b on mch 0 LZ , lc : 2 2CR Z

UZU I

2. Trng hp 2 u ca phn t in mc song song kho in K (c RK = 0) m kho in K ng li:

Th d: K ng 0CZ , lc : 2 2

LR Z

UZU I

MCH IN THAY I DO NG NGT KHA K:

Hin tng on mch: Xt mt on mch c tng tr l XZ v mt dy ni AB c in tr khng ng k theo hnh bn. V in tr ca dy ni khng ng k nn: + in th ti A ( )AV gn bng in th ti B ( )BV : A BV V + Ton b dng in khng i qua phn t XZ m i qua dy ni AB.Hin tng trn gi l hin tng on mch Kt qu: + Khi c hin tng on mch phn t no ta ci th xem nh khng c ( khuyt) phn t trong mch.+ Ni (chp) hai im A, B hai u dy ni ri v mch li.

Ch : Khi kha K mc song song vi L hoc C, khi ng hay m th Ing = Im

a. Kha / / :K C Zm = Zng 2 2 2 20

( )2

CL C L

C L

ZR Z Z R Z

Z Z

b. Kha / / :K L Zm = Zng 2 2 2 20

( )2

LL C C

L C

ZR Z Z R Z

Z Z

Bi tp trc nghim:

Cu 1: 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 hiu in th hiu dng hai u in tr R bng A. 10V. B. 10 2 V. C. 20V. D. 30 2 V.

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Cu 2: Mt mch in RCL ni tip mc vo mch xoay chiu c hiu in th khng i, hiu in th hiu dng trn cc phn t R, L, C u bng nhau v bng 10V. Nu ni tt hai bn cc ca t in th hiu in th hiu dng trn cun t cm L s bng ?

A. 2

10 V B. 20V C. 10V D. 10 2

Cu 3: on mch nh hnh v, uAB = 100 2 cos100t (V). Khi K ng, I = 2A, khi K m dng in qua

mch lch pha 4 so vi hiu in th hai u mch. Cng dng in hiu dng qua mch khi K m l:

A. 2A B. 1A C. 2 A D. 2 2 A Cu 4: Cho mt on mch xoay chiu nh hnh v. Hiu in thhai u AB c gi tr hiu dng U 240 2 V. Bit C LZ 2Z . B qua in tr ca cc dy ni v kha K. Khi kha K ngt, dng in qua mch l:

1i 4 2 cos(100 t )A. 3

Khi kha K ngt, dng in qua mch l: 2i 4 2 cos(100 t )A. 6

Gi tr ca R l:

A. 30 2 B. 60 C. 60 2 D. mt gi tr khc Cu 5: Cho on mch xoay chiu nh hnh v. Bit

ABu U 2 cos t (V) ; R = 40; r = 20. Khi K ng hay m th dng in qua R u lch pha /3 so vi u. Cm khng ca cun dy A. 60 3 . B. 100 3 . C. 80 3 . D. 60.

Bi ton nu c 2 cun dy hoc 2 t in

+ 1 2L nt L 1 cun dy c 1 21 2 1 2L L LL L L Z Z Z L L L

+ 1 21 2 1 2

1 21 2

1 2 1 2

1 1 1 1 1 1/ / : L LLL L L L L

Z Z L LL L Z LZ Z Z Z Z L L L L L

+ 1 2

1 21 2

1 2 1 2

1 1 1: C C CC CC nt C Z Z Z C

C C C C C

+ 1 2

1 2 1 2

1 2 1 21 1 1/ / : C CC

C C C C C

Z ZC C Z C C C

Z Z Z Z Z

Bi tp trc nghim:

Cu 1: Hai cun thun cm L1 v L2 mc ni tip trong mt on mch xoay chiu c cm khng l:

A. 1 2( )LZ L L B. 1 2( )LZ L L

RB

CLA

K

L2R L1

A B R, L

K

C

R CA B

L

K

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15

C. 1 2( )LL LZ

D. 1 2( )

LL LZ

BI TON 2: LIN QUAN N LCH PHA GIA CC IN P 1u V 2u

Dng 1. lch pha ca u i vi i Khi bit lch pha ca u i vi i, ta dng hai cng thc sau tm kt qu:

tan L CZ Z R

cos R Z

Ch : Nu bit lch pha ca i i vi u ta cn lch pha ca u i vi i ri mi p dng cng thc trn v nh i u

u i

Dng 2. lch pha ca hai on mch lch pha ca hai on mch Xt on mch AM v on mch NB trn cng on mch AB - Biu thc hiu in th gia hai u A, M l u1 = U01cos 1t

- Biu tc hiu in th hai u N, B l u2 = U02cos 2t + khi 1 2 ; u1 cng pha vi u2 1 2tan tan

+ khi 1 2 2

; u1 vung pha u2 1 2 tan tan( ) 2

1 2 2 1tan cotan tan tan 1

+ khi 1 2 2

hai gc lch pha ca i so vi u hai u hai on mch l hai gc ph nhau

2 1tan tan 1

+ khi bit u hai on mch lch pha so vi nhau gc , ta c th v phc gin vct tm lch pha ca u1 hoc u2 i vi i. T tm kt qu

Phng php: - S dng cng thc lch pha gia hai in p 1 2va uu : 1 1 2

2

u u uu i i

Trong : 1

2

:

:

1

2

olech pha cua u so vi i

olech pha cua u so vi i

ui

ui

Ch : - C th dng phng php gin vec-t Fresnel gii dng ton trn. - Nu 1 2va uu lch pha nhau 2

hay1 1 2

22u u uu i i

. Ta lun c:

1 2( ).( ) 1u u

i itg tg

- Xt hai on mch bt kh X1 v X2 cng trn mt mch inu1 v u2 cng pha :

C1 L1 R1 M

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1 = 2 Z = Z1 + Z2 hoc U = U1 + U2 v 1 1 2 21 2

L C L CZ Z Z ZR R

V d: Xt on mch theo hnh bn. Bit lch pha ca

2vi laAN MBu so u . Tm h thc lin h gia , ,L CR Z Z .

Hng dn: Ta c ( ).( ) 1 1AN MB

L Cu u

i i

Z Ztg tgR R

Kt qu::(CTTN) 2 2.L CLR Z Z hay RC

Dng 3: Hai on mch AM gm R1L1C1 ni tip v on mch MB gm R2L2C2 ni tip mc ni tip vi nhau Ta c UAB = UAM + UMB uAB; uAM v uMB cng pha tanuAB = tanuAM = tanuMB

Dng 4: Hai on mch R1L1C1 v R2L2C2 cng u hoc cng i c pha lch nhau

Vi 1 111

tan L CZ Z

R

v 2 22

2

tan L CZ Z

R

(gi s 1 > 2)

C 1 2 = 1 21 2

tan tan tan1 tan tan

Trng hp c bit = /2 (vung pha nhau) th tan1.tan2 = -1.

VD: - Mch in hnh 1 c uAB v uAM lch pha nhau . y 2 on mch AB v AM c cng i v uAB chm pha hn uAM

AM AB = tan tan tan

1 tan tan

AM AB

AM AB

- Nu uAB vung pha vi uAM th tan tan =-1 1L L CAM ABZ Z ZR R

Dng 5: Mch in hnh 2: Khi C = C1 v C = C2 (gi s C1 > C2) th i1 v i2 lch pha nhau y hai on mch RLC1 v RLC2 c cng uAB Gi 1 v 2 l lch pha ca uAB so vi i1 v i2 th c 1 > 2 1 - 2 = - Nu I1 = I2 th 1 = -2 = /2

- Nu I1 I2 th tnh 1 21 2

tan tan tan1 tan tan

Dng 6: on mch RLC: cho bit cc hiu in th hiu dng UR, UL, UC. Tm Umch? lch pha ?u / i

Phng php: Cch 1: (dng cng thc): Theo nh lut m: U = IZ

2 2R

2 2 2 2R ( ) ( R) ( ) ( )L C L C L C U I Z Z I IZ IZ U U U U

V R

/( )

UU U

IRI Z Ztg L C L Cu i

u / i

R L CMA B

Hnh 1

R L CMA B

Hnh 2

C1 L1 R1 A M

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Cch 2: (dng gin vect) Hiu in th tc thi: u = uR + uL + uC RUU U UL C + V gin vect hiu in th (theo gi tr hiu dng)

+ T gin vect 2 2R ( )CU LU U U

V R

/ UUU

tg L Cu i

u / i

Ch : - Bi ton cng c ca cc in p hiu dng thnh phn : mun cng c cc in p thnh phn vi nhau th cc in p phi cng pha ngha l lch pha gia cc in p vi dng in phi nh nhau. 1 2 1 2tan tan .

- Bi ton lin quan n lch pha gia hai in p bng 2 th tan gc lch pha ny bng cotan gc

lch pha kia. Ngha l 1 1 21 2 2

L C

L C

Z Z RR Z Z

.

Bi tp gii mu:

Cu 1: (H 2008) Cho on mch in xoay chiu gm cun dy mc ni tip vi t in. lch pha

ca hiu in th gia hai u cun dy so vi cng dng in trong mch l 3 . Hiu in th hiu

dng gia hai u t in bng 3 ln hiu in th hiu dng gia hai u cun dy. lch pha ca hiu in th gia hai u cun dy so vi hiu in th gia hai u on mch trn l

A. 0. B. 2 . C.

3

. D. 23 .

HD:

2 2 2 2 23 3.3 3

32 3.3. 3

23

Lcd L L C

CC L r C L

cd

Ztg tg Z r Z Zr tgrZ rU U U Z Z r

Cu 2: (H 2008) Cho on mch in xoay chiu gm cun dy c in tr thun R, mc ni tip vi t

in. Bit hiu in th gia hai u cun dy lch pha 2 so vi hiu in th gia hai u on mch. Mi

lin h gia in tr thun R vi cm khng ZL ca cun dy v dung khng ZC ca t in l A. 2 C L CR Z Z Z . B. 2 C C LR Z Z Z . C. 2 L C LR Z Z Z . D. 2 L L CR Z Z Z .

HD: 2. . 1L L Ccd L C LZ Z Ztg tg R Z Z ZR R

Cu 3: (H 2007) t vo hai u on mch in RLC khng phn nhnh mt hiu in th xoay chiu

c tn s 50 Hz. Bit in tr thun R = 25 , cun dy thun cm (cm thun) c L =1

H. hiu in th

hai u on mch tr pha 4 so vi cng dng in th dung khng ca t in l

A. 100 . B. 150 . C. 125 . D. 75 .

IRU

CU LU

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Gii:

Hiu in th hai u on mch tr pha 4 so vi cng dng in

14

tg Z Z RR

ZZC L

CL

1

Cm khng ca cun dy

100 12 2 .50.

Z L fLL 100 25 125 CZCu 4: (H 2009) Mt on mch in xoay chiu gm in tr thun, cun cm thun v t in mc ni tip. Bit cm khng gp i dung khng. Dng vn k xoay chiu (in tr rt ln) o in p gia hai u t in v in p gia hai u in tr th s ch ca vn k l nh nhau. lch pha ca in p gia hai u on mch so vi cng dng in trong on mch l

A. 4 . B.

6 . C.

3 . D.

3

.

HD: 22 2 tan 1

4L C

R C L cZ Z R RU U Z Z R

R R

Cu 5: (H 2010) Mt on mch AB gm hai on mch AM v MB mc ni tip. on mch AM c

in tr thun 50 mc ni tip vi cun cm thun c t cm 1

H, on mch MB ch c t in vi

in dung thay i c. t in p u = U0cos100t (V) vo hai u on mch AB. iu chnh in dung

ca t in n gi tr C1 sao cho in p hai u on mch AB lch pha 2 so vi in p hai u on

mch AM. Gi tr ca C1 bng

A. 54.10 F

B.

58.10 F

C.

52.10 F

D.

510 F

HD:

C FR

Z ZRZL L C

AM AB510 8tan tan 1 . 1

Cu 6: (C 2009) t in p 0u U cos( t ) 4

vo hai u on mch ch c t in th cng

dng in trong mch l i = I0cos(t + i). Gi tr ca i bng

A. 2

. B. 34

. C. 2 . D. 3

4 .

HD: lch pha ca u so vi i

34 2 4u i i u

(v vi hai u on mch ch cha t in th 2

)

Cu 7: (C 2010) t in p 220 2 cos100u t (V) vo hai u on mch AB gm hai on mch AM v MB mc ni tip. on AM gm in tr thun R mc ni tip vi cun cm thun L, on MB ch c t in C. Bit in p gia hai u on mch AM v in p gia hai u on mch MB c gi tr hiu

dng bng nhau nhng lch pha nhau 23

. in p hiu dng gia hai u on mch AM bng

A. 220 2 V. B. 2203

V. C. 220 V. D. 110 V.

Gii: 2 1 33 6 3

LAM MB AM R L

R

UU U

U

(1)

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Mt khc 2 2 2R L CU U U (2) (do UAM = UMB).

Thay (1) vo (2) ta c 2

CL

UU (3)

22 2 R L CU U U U (4). Thay (2), (3) vo (4) ta c 220AM MB CU U U V

Cu 8: Vo cng mt thi im no , hai dng in xoay chiu cos( )1 0 1 i I t

v cos( )2 0 2 i I t u c cng gi tr tc thi l 00,5 2I nhng mt dng in ang gim, cn mt dng in ang tng. Kt lun no sau y l ng ? A. Hai dng in dao ng cng pha. B. Hai dng in dao ng ngc pha. C. Hai dng in dao ng lch pha nhau gc 1200. D. Hai dng in dao ng vung pha (lch pha nhau gc 900) Cu 9: Vo cng mt thi im no , hai dng in xoay chiu

cos( )1 0 1 i I t v cos( )2 0 2 i I t u c cng gi tr tc thi l 0,50 I nhng mt dng in ang gim, cn mt dng in ang tng. Hai dng in ny lch pha nhau mt gc bng

A. 6 . B.

32 . C.

65 . D.

34 .

Cu 10: Trong mch RLC mc ni tip ang xy ra cng hng nu tng tn s dng in v gi nguyn cc thng s khc th kt lun no sau y l sai A. H s cng sut ca on mch gim B. Cng hiu dng ca on mch gim C. in p gia hai bn t tng D. in p trn in tr thun gim Cu 11: Dung khng ca mt mch RLC mc ni tip ang c gi tr nh hn cm khng. Mun xy ra hin tng cng hng in trong mch ta phi A. gim tn s dng in xoay chiu. B. tng in dung ca t in C. tng h s t cm ca cun dy. D. gim in tr ca mch. Cu 12: on mch xoay chiu ni tip theo th t sau: Cun dy thun L = 1mH, in tr thun R, t in C = 10 F. Gi u1 l hiu in th trn hai u cun dy v in tr, u2 l hiu in th trn hai in tr v t in . u1 vung pha vi u2 th R phi c gi tr no sau y: A. R = 0,01. B. R = 0,10. C. R = 100. D. R = 10. Cu 13: Mt mch in xoay chiu gm RLC ni tip mt hiu in th xoay chiu c tn s f = 50Hz. Bit

R = 100 , cun dy c t cm L = 1

H, in tr r. T in c in dung C = -410

2 F. Bit in p gia

hai u mch tr pha 2 so vi in p gia hai u cun dy khi gi tr ca r l :

A. 100 . B. 50 C. 50 2 . D. 200 . Cu 14: Mt mch in xoay chiu (hnh v) gm RLC ni tip mt in p xoay chiu c tn s f = 50Hz.

Bit R l mt bin tr, cun dy c t cm L = 1

H, in tr r = 100. T in c in dung C = 410

2

F.

iu chnh R sao cho in p gia hai u on mch AM sm pha 2

so vi in p gia hai im MB, khi

gi tr ca R l : A. 85 . B.100 . C. 200 . D. 150 .

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Cu 15: t vo hai u on mch RLC khng phn nhnh mt in p xoay chiu c tn s 50Hz. Bit

in tr thun R = 25, cun dy thun cm c t cm L = 1

H. in p hai u mch tr pha 4 so

vi in p gia hai u in tr th in dung ca t in phi l

A. -410

F. B. 80 F

. C.

-4104

F. D. 12

F.

Cu 16: Khi ch mc vo hai u mt on mch cha in tr thun R ni tip vi t in c in dung C vo

ngun in xoay chiu 0u = U cost(V) th thy dng in i sm pha 4

so vi in p t vo mch. Khi on

mch c c in tr thun R, t in c in dung C v cun dy thun cm c t cm L v cng t vo hai

u mch in p trn th thy dng in i chm pha 4

so vi in p hai u on mch. Chn biu thc

ng?

A. ZC = 2ZL = R B. R = ZL = 12

ZC. C. ZL = ZC = R. D. R = 2ZL = 3ZC.

Cu 17: Cho mch in xoay chiu nh hnh v, cun dy thun cm .Bit UAM = 80V ; UNB = 45V v lch pha gia uAN v uMB l 900, Hiu in th gia A v B c gi tr hiu dng l : A. 69,5V. B. 35V. C. 100V. D. 60V. Cu 18: Cho on mch RLC, t vo hai u on mch mt in p xoay chiu c U = 100V th thy in p gia hai u cun cm vung pha vi in p hai u on mch, khi in p hiu dng gia hai u in tr thun l

A. UR = 100V. B. UR = 50V. C. UR = 0. D. UR = 100

3V.

Cu 19: Mch in xoay chiu gm in tr thun R = 30 mc ni tip vi cun dy. t vo hai u mch mt hiu in th xoay chiu u =U 2 sin(100 t) V. Hiu in th hiu dng hai u cun dy l Ud

= 60V. Dng in trong mch lch pha 6 so vi u v lch pha

3 so vi ud. Hiu in th hiu dng hai

u mch U c gi tr A. 60 3 V. B. 120 V. C. 90V. D. 60 2 V.

Cu 21: Mt mch in gm R,L,C mc ni tip. Cun cm c t cm L =1 H, t in c in dung C =

42.10 F. Chu k ca dng in xoay chiu trong mch l 0,02s. Cng dng in trong mch lch pha

6 so vi hiu in th hai u mch th in tr R c gi tr l

A.3

100 B. 100 3 C. 50 3 D.

350

Cu 22: t vo hai u on mch RLC khng phn nhnh mt hiu in th xoay chiu u = U0sint. K hiu UR, UL, UC tng ng l hiu in th hiu dng hai u in tr thun R, cun dy thun cm (cm thun) L v t in C. Nu UR = 0,5UL = UC th dng in qua on mch A. tr pha /2 so vi hiu in th hai u on mch. B. tr pha /4 so vi hiu in th hai u on mch. C. sm pha /2 so vi hiu in th hai u on mch. D. sm pha /4 so vi hiu in th hai u on mch.

M

L R C B A N

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21

4L C

R

U U HD : tanU

Cu 23: Mt on mch in gm t in c in dung 10-4/ F mc ni tip vi in tr 125 , mc on mch vo mng in xoay chiu c tn s f. Tn s f phi bng bao nhiu dng in lch pha /4 so vi hiu in th hai u mch. A. f = 503 Hz B. f = 40 Hz C. f = 50Hz D. f = 60Hz

122 1 40

fLfCHD : tan f Hz

R

Cu 24: Cho mch in xoay chiu nh hnh v. Bit ZL = 20 ; ZC = 125 . t vo hai u mch in mt in p xoay chiu u 200 2 cos100t (V). iu chnh R uAN v uMB vung pha, khi in tr c gi tr bng: A. 100 . B. 200 . C. 50 . D. 130 . Cu 25: Cho mch in xoay chiu nh hnh v. Bit R = 100 2 ; C = 100 / F . t vo hai u mch in mt in p xoay chiu u 200 2 cos100t (V). iu chnh L uAN v uMB lch pha nhau gc

/ 2 . t cm khi c gi tr bng:

A. .H 1

B. .H 3

C. .H 2

D. H.21

Cu 26: Cho mch in RLC mc ni tip. Bit dung khng ZC = 48 . Hiu in th hiu dng hai u mch l U, tn s f. Khi R = 36 th u lch pha so vi i gc 1 v khi R = 144 th u lch pha so vi i gc

2 . Bit 1 + 2 = 900. Cm khng ca mch l

A. 180 . B. 120 . C. 108 . D. 54 . Cu 27: Cho mch in R, L, C mc ni tip vi u 200 2 cos100 tAB (V). S ch trn hai vn k l nh

nhau nhng gi tr tc thi ca chng lch pha nhau 3

2 . Cc vn k ch gi tr no sau y?

A. 100V B. 200V C. 300V D. 400V

Cu 28: Mt t in c dung khng 30(). Chn cch ghp t in ny ni tip vi cc linh kin in t khc di y c mt on mch m dng in qua n tr pha so vi hiu th hai u mch mt gc

4

A. mt cun thun cm c cm khng bng 60 B. mt in tr thun c ln 30 C. mt in tr thun 15 v mt cun thun cm c cm khng 15 D. mt in tr thun 30 v mt cun thun cm c cm khng 60Cu 29: on mch nh hnh v, uAB = 100 2 cos100t(V). Khi K ng, I = 2(A), khi K m dng in qua

mch lch pha 4 so vi hiu in th hai u mch. Cng dng in hiu dng qua mch khi K m l:

A. 2A B. 1(A) C. 2 A D. 2 2 A

Cu 30: on mch xoay chiu nh hnh v. RV , vn k (V1) ch 80(V), vn k (V2) ch 100(V) v vn k (V) ch 60(V).

R C L

M NBA

R C L

M NBA

RB

CLA

V1 V2

RB

CLA

K

BCL

A

V1 V2

M

V

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lch pha uAM vi uAB l: A. 37 B. 53 C. 90 D. 45

Cu 31: Mt on mch in xoay chiu c dng nh hnh v.Bit hiu in th uAE

v uEB

lch pha nhau 900.Tm mi lin h gia R, r, L, C.

A B C r

R,L E

A. R = C.r.L B. r = C. R..L C. L = C.R.r D. C = L.R.r Cu 32: Cho mch in xoay chiu RLC nh hnh v u U ft VAB 2 cos 2 . Cun dy thun cm c t

cm L H35

, t din c C F24

103 .Ht u

NB v u

AB lch pha nhau 900 . Tn s f ca dng in xoay

chiu c gi tr l

A C R

L B

M

A. 120Hz B. 60Hz C. 100Hz D. 50Hz Cu 33: Cho mch in gm RLC ni tip.in p hai u mch u = 120 2 cos100 t (V). in

tr R = 50 3 , L l cun dy thun cm c L = H 1

, in dung C thay i c.Thay i C cho in p

hai u on mch nhanh pha hn hai u t mt gc 2 . Tm C .

A. C = 410 F

B. C = 410 F

C. C = 410 F

D. C = 1000 F

Cu 34: Cho mt mch in xoay chiu gm in tr thun R v t in c in dung C mc ni tip. Hiu in th t vo hai u mch l u = 100 2 cos10t (V), b qua in tr dy ni. Bit cng dng in

trong mch c gi tr hiu dng l 3 A v lch pha /3 so vi hiu in th hai u mch. Gi tr ca R v C l :

A. R = 50 3 v C = 410

F B. R = 503 v C =

410

F

C. R = 50 3 v C = 310

5

F D. R = 503 v C =

3105

F

Cu 35: on mch xoay chiu RLC mc ni tip . in tr thun R = 100 , cun dy thun cm c t

cm L, t c in dung C =

410 F. Mc vo hai u on mch hiu in th xoay hiu u =

U0sin100 t(V). hiu in th hai u on mch cng pha vi hiu in th hai u in tr R th gi tr t cm ca cun dy l

1A. L = H

B. L = 10 H

C. L = 21 H D. L =

2 H

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Cu 36: Mch in xoay chiu gm in tr thun R = 30 mc ni tip vi cun dy. t vo hai u mch mt hiu in th xoay chiu u = U 2 sin(100 t) (V). Hiu in th hiu dng hai u cun dy l

Ud = 60V. Dng in trong mch lch pha 6 so vi u v lch pha

3 so vi ud. Hiu in th hiu dng

hai u mch (U) c gi tr A. 60 3 V B. 120V C. 90 V D. 60 2 V

Cu 37: t in p u = U0cost vo hai u on mch mc ni tip gm in tr thun R, t in v cun cm thun c t cm L thay i c. Bit dung khng ca t in bng R 3 . iu chnh L in p hiu dng gia hai u cun cm t cc i, khi

A. in p gia hai u in tr lch pha 6 so vi in p gia hai u on mch.

B. in p gia hai u t in lch pha 6 so vi in p gia hai u on mch.

C. trong mch c cng hng in.

D. in p gia hai u cun cm lch pha 6 so vi in p gia hai u on mch.

Cu 38: Trong mt on mch xoay chiu gm t in c in dung C mc ni tip vi cun dy c in

tr thun R = 25 v t cm L = H 1 . Bit tn s cua dng in bng 50Hz v cng dng in qua

mch sm pha hn hiu in th gia hai u on mch mt gc 4 . Dung khng ca t in l

A. 75 B. 100 C. 125 D. 150Cu 39: Mch in xoay chiu gm in tr thun R = 10 , cun dy thun cm c t cm

L = ( )10. H

v t in c in dung C mc ni tip. t vo hai u mch mt hiu in th xoay chiu u =

U 2 cos(100 t) (V). Dng in trong mch lch pha 3 so vi u. in dung ca t in l

A. 86,5 F B. 116,5 F C. 11,65 F D. 16,5 F Cu 40: Mt on mch in xoay chiu gm mt t in c dung khng ZC = 200 v mt cun dy mc ni tip. Khi t vo hai u on mch trn mt in p xoay chiu lun c biu thc u = 120 2 cos(100t

+ 3 )V th thy in p gia hai u cun dy c gi tr hiu dng l 120 v sm pha

2 so vi in p t

vo mch. Cng sut tiu th ca cun dy l A. 72 W. B. 240W. C. 120W. D. 144W. Cu 41: Mt on mch RLC. Gi UR, UL, UC ln lt l hiu in th hiu dng hai u in tr R, cun cm L v hai bn t in C trong UR = UL = UC/2. Lc A. hiu in th hai u on mch sm pha hn dng in mt gc /4. B. hiu in th hai u on mch sm pha hn dng in mt gc /3. C. hiu in th hai u on mch tr pha hn dng in mt gc /4. D. hiu in th hai u on mch tr pha hn dng in mt gc /3. Cu 42: on mch RL c R = 100 mc ni tip vi cun thun cm L c lch pha gia u v i l /6. Cch lm no sau y u v i cng pha? A. Ni tip vi mch mt t in c ZC =100/ 3 . B. Ni tip vi mch t c ZC = 100 3 . C. Tng tn s ngun in xoay chiu. D. Khng c cch no.

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Cu 43: t mt in p xoay chiu c gi tr hiu dng U vo hai u on mch AB gm cun cm thun c t cm L, in tr thun R v t in c in dung C mc ni tip theo th t trn. Gi UL, UR v UC_ln lt l cc in p hiu dng gia hai u mi phn t. Bit in p gia hai u on mch AB lch

pha 2 so vi in p gia hai u on mch NB (on mch NB gm R v C ). H thc no di y l

ng? A. 2 2 2 2R C LU U U U . B.

2 2 2 2C R LU U U U .

C. 2 2 2 2L R CU U U U D. 2 2 2 2R C LU U U U

Cu 44: Cho mch in R, L, C mc ni tip vi u 200 2 cos100 tAB (V). S ch trn hai vn k l nh



A. 100(V) B. 200(V) C. 300(V) D. 400(V)

Cu 45: Cho hai cun dy c in tr thun (L1, r1) v (L2, r2) mc ni tip vo hai im AB. Gi U l hiu in th hiu dng gia hai u on mch, U1, U2 ln lt l hiu in th hiu dng gia hai u cun dy 1 v cun dy 2. U = U1 + U2 cn iu kin no sau y? A. L1r1 = L2r2. B. L1r2 = L2r1. C. L1L2 = r1r2. D. L1 + L2 = r1 + r2. Cu 46: Hai cun dy (R1, L1) v (R2, L2) c mc ni tip nhau v mc vo mt hiu in th xoay chiu c gi tr hiu dng U. Gi U1 v U2 l hiu in th hiu dng tng ng gia hai u cun (R1, L1) v (R2, L2). iu kin U = U1 + U2 l A. L1/ R1 = L2 / R2. B. L1/ R2 = L2 / R1 C. L1 . L2 = R1.R2 D. khng c lin h no ba trn ng.

Cu 47: mch in R = 100;410

2 C F

. Khi t vo AB

mt in p xoay chiu c tn s f = 50Hz th uAB v uAM vung pha vi nhau. Gi tr L l:

A. 2 L H

B. 3 L H

C. 3 L H

D. 1 L H

Cu 48: Cho mt mch in RLC ni tip. Bit R thay i c, L = 0,8/ H, C = 10-3/(6) F. t vo hai u on mch mt hiu in th c biu thc: u = Uocos(100t). uRL lch pha /2 so vi u th phi c A. R = 20. B. R = 40. C. R = 48. D. R = 140. Cu 49: Mt on mch in gm cun dy c in tr thun r = 10 v t cm L = (10-1/)H mc ni tip vi in tr thun R = 20 v t in C = (10-3/4)F. t vo hai u on mch mt hiu in th u = 180 2 cos(100t)V. lch pha ca ht gia hai u cun dy v hai u t in l: A. /4 B. -3/4 C. 3/4 D. /4

Cu 50: Cho mch in xoay chiu nh hnh v, trong R = 100; C = 410 F

2

; L l cun dy thun cm,

c t cm L. Nu dng in trong mch tr pha so vi hiu in th hai u on mch gc 4 th t

cm L c gi tr: A. 0,1H B. 0,95H C. 0,318H D. 30,318.10 H Cu 51: Cho mch in xoay chiu nh hnh v, in p t vo hai u mch l: AB 0u U cos100 t V .

Cun dy thun cm c t cm 1L H

. T in c in dung

RB

CLA

V1 V2

R L C

A B

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40,5.10C F

. in p tc thi uAM v uAB lch pha nhau /2. in tr thun ca on mch l:

A. 100 B. 200 C. 50 D. 75 Cu 52: Cho on mch RLC, R = 50. t vo mch c in p l u 100 2 cos t(V ) , bit in p gia hai bn t v hiu in th gia hai u mch lch pha 1 gc /6. Cng sut tiu th ca mch l A. 50 3 W B. 100 3 W C. 100W D. 50W Cu 53: Mch in xoay chiu gm in tr thun R = 30 mc ni tip vi cun dy. t vo hai u mch mt hiu in th xoay chiu u = U 2 sin(100 t) (V). Hiu in th hiu dng hai u cun dy l

Ud = 60V. Dng in trong mch lch pha 6 so vi u v lch pha

3 so vi ud. Hiu in th hiu dng

hai u mch (U) c gi tr A. 60 3 (V) B. 120 (V) C. 90 (V) D. 60 2 (V) Cu 54: Mt on mch in xoay chiu gm mt t in c dung khng ZC = 200 v mt cun dy mc ni tip. Khi t vo hai u on mch trn mt in p xoay chiu lun c biu thc u = 120 2 cos(100t

+ 3 )V th thy in p gia hai u cun dy c gi tr hiu dng l 120 v sm pha

2 so vi in p t

vo mch. Cng sut tiu th ca cun dy l A. 72 W. B. 240W. C. 120W. D. 144W Cu 55: Mt on mch gm mt t in c in dung C, in tr thun R, cun dy c in tr trong r v t cm L mc ni tip. Khi t vo hai u mch mt in p xoay chiu c biu thc u = U 2 cost (V) th cng dng in trong mch c gi tr hiu dng l I v chm pha so vi in p gia hai u mch mt gc ( 0). Cng sut tiu th trong mch ny c xc nh bng

A. 2U

R+r. B.

2U12 L-

C

. C. 2U

2(R+r). D. (R + r).I2.

Cu 56: Mch xoay chiu nh hnh v. Hiu in th hiu dng ln lt l UAB = 50V;UBC = 50V UAC = 50 3V.Ta c: A. i chm pha /4 so vi hiu in th hai u mch B. i nhanh pha /6 so vi hiu in th hai u mch C. i nhanh pha /4 so vi hiu in th hai u mch D. i chm pha /6 so vi hiu in th hai u mch

Cu 57: Mt on mch gm t c in dung ( F)12 310 C

3

ghp ni tip vi in tr R = 100 , mc

on mch vo in p xoay chiu c tn s f. dng in i lch pha 3 so vi in p u th gi tr ca f l:

A. 25 Hz B. 50 Hz C. 50 3 Hz D. 60 Hz Cu 58: Xt mch in xoay chiu RLC, hiu in th 2 u mch lch pha so vi cng dng in

qua mch 1 gc4 . Kt qu no sau y l ng?

A. ZC = 2 ZL B. Z Z RL C C. ZL = ZC D. ZL = 2ZC

Cu 59: Cho on mch xoay chiu AB nh hnh v: uAM c pha vung gc uMB th h thc lin h gia R, R0, L v C l: A. L/C = R0/R B. C/L = RR0 C. LC = RR0 D. L = CRR0 Cu 60: Mt mch xoay chiu R,L,C khng phn nhnh trong R = 50, t vo hai u mch mt hiu in th U = 120V, f 0 th i lch pha vi u mt gc 600, cng sut ca mch l

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A. 288W B. 72W C. 36W D. 144W Cu 61: Mt mch R,L,C mc ni tip trong R = 120, L = 2/(H) v C = 200/(F), hiu in th t vo mch in c tn s f thay i c. i sm pha hn u, f cn tho mn iu kin A. f >12,5Hz B. f < 25Hz C. f < 2,5Hz D. f 12,5Hz Cu 62: Mt mch R, L, C mc ni tip (cun dy thun cm) L v C khng i R thay i c. t vo hai u mch mt ngun in xoay chiu c hiu in th hiu dng v tn s khng i, ri iu chnh R n khi cng sut ca mch t cc i, lc lch pha gia u v i l A. /4 B. /6 C. /3 D. /2 Cu 63: Cho mch in LRC ni tip theo th t trn. Bit R l bin tr, cun dy thun cm c L = 4/(H), t c in dung C = 10-4/(F). t vo hai u on mch mt hiu in th xoay chiu n nh c biu thc: u = U0cos100t (V). hiu in th uRL lch pha /2 so vi uRC th R bng bao nhiu? A. R = 300. B. R = 100. C. R = 100 2 . D. R = 200.

Cu 64: Mch RLC ni tip c R =100, L = 2 3

(H). Hiu in th xoay chiu t vo on mch c

biu thc u = Uosin2ft, f thay i c. Khi f = 50Hz th i tr pha /3 so vi u. i cng pha vi u th f c gi tr l A. 100Hz B. 35Hz C. 50Hz D. 40Hz Cu 65: Cho mch gm in tr R v cun dy thun cm L ni tip, L thay i c. Hiu in th hiu dng gia hai u mch l U, tn s gc = 200rad/s. Khi L = /4H th u lch pha so vi i mt gc , khi L = 1/H th u lch pha so vi i mt gc '. Bit + ' = 90o. R c gi tr l A. 80 B. 157 C. 100 D. 50Cu 66: Cho mch in nh hnh v : A C1 R1 E L, R2 C2 B

Bit R1 = 4 , C F8

10 21

, R2 = 100 , L H1

, f = 50Hz. Thay i gi tr: C2 hiu in th UAE cng

pha vi UEB. Gi tr: C2 l:

A. C F30

12 B. C F300

12

C. C F3

10002 D. C F3

1002

Cu 67: Mch RLC: B C M L R A

R = 50 , L = 21

H, f = 50 Hz. Lc u C =

100F, sau ta gim in dung C. Gc lch pha gia UAM

v UAB lc u v lc sau c kt qu:

A. 2

rad v khng i B. 4

rad v tng dn

C. 2 rad v gim dn D.

2 rad v dn tng

Cu 68: Mch nh hnh v A L R1 M C R2 B

UAB = 120V ; L = 3 H . = 100 (rad/s) R1 = 100 , UMB = 60V v tr pha hn uAB 600. in tr

thun R2 v in dung C c gi tr:

A. R2 = 100 v C = 100 3 F

B. R2 = 200 3 v C = F 50

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C. R2 = 100 3 v C = F 4

100 D. R2 = 100 3 v C = F 50

Cu 69: Cho mch nh hnh v A R C L, r B bit uAB = 100 2 cos100 tV

K ng, dng in qua R c gi tr: hiu dng 3 A v lch pha 3 so vi uAB. K m, dng in qua R c

gi tr: ti hiu dng 1,5A v nhanh pha hn uAB l 6

. in tr thun R v t cm L c gi tr::

A. R = 3

350 () v L =

61 H B. R = 150 () v L =

31 H

C. R = 3

350 () v L =

21 H D. R = 50 2 () v L =

51 H

Cu 70: Mch nh hnh v: A C M R,L B

Bit C = F 3

10 4

; VR , uAB = 200 2 cos(100t - 6 )V.

S ch 2 vn k l bng nhau v uAM lch pha so vi uMB 3

2(rad). in tr thun R v t cm L c gi

tr:

A. R = 150 v L = 23 H B. R = 50 v L =

23 H

1C. R = 150 v L = H

D. Tt c u sai

Cu 71: Mch nh hnh v: P C N R,L M uMP = 100 2 cos100tV V2 ch 75 V ; V1 ch 125V lch pha gia uMN v uMP l:

A. 4 (rad) B.

3 (rad) C.

6 (rad) D.

2 (rad)

Cu 72: Mch nh hnh v M L R N C P Cun dy thun cm. uMP = 170cos100tV

UC = 265V ; I = 0,5A v sm pha 4 so vi uMP. in tr thun v t cm c gi tr:

A. 170 () v 1,15/H B. 170 2 () v 1 H

C. 170 () v 0,115H D. Tt c u sai Cu 73: Mch in xoay chiu nh hnh v A C R L, r B uAB = 80 2 cos100 t(V), R = 100 ,

V2 ch 30 2 V, V1 ch 50V urL sm pha hn i 1 gc 4

(rad) t cm L v in dung C c gi tr:

A. 53 H v

6103 F B.

103 H v

3103 F

C. 53 H v

33103 F D. Tt c u sai

Cu 74: Mch RLC nh hnh v: A R L M C B

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Bit uAB = 100 2 cos100 tV ; I = 0,5A uAM sm pha hn i 6 rad, uAB sm pha hn uMB

6 rad . in tr

thun R v in dng C c gi tr:

A. R = 200 v C F

125 3 B. R = 100 v C F

50 3

C. R = 100 v C F

125 3 D. R = 50 v C F

50 3

Cu 75: Mch RLC mc ni tip. i lng no sau y khng th iu chnh u v i cng pha? A. in dung. B. t cm L. C. in tr. D. Tn s dng in. Cu 76: Cho mch in R, L, C mc ni tip vi u 200 2 cos100 tAB (V). S ch trn hai vn k l nh



A. 100V B. 200V C. 300V D. 400V

Cu 77: Cho mch in nh hnh v vi UAB = 300V, UNB = 140V, dng in i tr pha so vi uAB mt gc (cos = 0,8), cun dy thun cm. Vn k V ch gi tr: A. 100V B. 200V C. 300V D. 400V

Cu 78: Mt on mch AB gm hai on mch AM v MB mc ni tip. on mch AM c in tr

thun 50 mc ni tip vi cun cm thun c t cm 1

H, on mch MB ch c t in vi in dung

thay i c. t in p u = U0 tcos100 (V) vo hai u on mch AB. iu chnh in dung ca t

in n gi tr C1 sao cho in p hai u on mch AB lch pha 2 so vi in p hai u on mch

AM. Gi tr ca C1 bng

A. F

58.10 B. F

510 C. F

54.10 D. F

52.10

Cu 79: (C 2010) t in p 220 2 cos100u t (V) vo hai u on mch AB gm hai on mch AM v MB mc ni tip. on AM gm in tr thun R mc ni tip vi cun cm thun L, on MB ch c t in C. Bit in p gia hai u on mch AM v in p gia hai u on mch MB c gi tr hiu

dng bng nhau nhng lch pha nhau 23

. in p hiu dng gia hai u on mch AM bng

A. 220 2 V. B. 220 3

V. C. 220 V. D. 110 V.

Cu 80: (C 2010) t in p 0u U cos(wt ) (V) 6

vo hai u on mch gm in tr thun R v

cun cm thun c t cm L mc ni tip th cng dng in qua on mch

l 05i I sin(wt ) (A) 12

. T s in tr thun R v cm khng ca cun cm l

A. 12

. B. 1. C. 3 2

. D. 3 .

RB

CLA

V1 V2

RB

CLA N

V

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Cu 81: (C 2010) t in p xoay chiu vo hai u on mch gm in tr thun 40 v t in

mc ni tip. Bit in p gia hai u on mch lch pha 3 so vi cng dng in trong on mch.

Dung khng ca t in bng

A. 40 3 B. 40 3 3

C. 40 D. 20 3

Cu 82: (C 2010) t in p 0u U cos wt vo hai u on mch gm in tr thun R v t in C mc ni tip. Bit in p gia hai u in tr thun v in p gia hai bn t in c gi tr hiu dng bng nhau. Pht biu no sau y l sai ?

A. Cng dng in qua mch tr pha 4 so vi in p gia hai u on mch.

B. in p gia hai u in tr thun sm pha 4 so vi in p gia hai u on mch.

C. Cng dng in qua mch sm pha 4 so vi in p gia hai u on mch.

D. in p gia hai u in tr thun tr pha 4 so vi in p gia hai u on mch.

Cu 83: (C 2007) t vo hai u on mch RLC khng phn nhnh mt hiu in th xoay chiu 0 sinu U t . K hiu UR, UL, UC tng ng l hiu in th hiu dng hai u in tr thun R, cun dy

thun cm (cm thun) L v t in C. Nu 2

LR C

UU U th dng in qua on mch l:

A. Tr pha 4 so vi hiu in th hai u on mch

B. Sm pha 2 so vi hiu in th hai u on mch

C. tr pha 2 so vi hiu in th hai u on mch

D. Sm pha 4 so vi hiu in th hai u on mch

Cu 84: (H 2008) Khi t vo hai u on mch gm cun dy thun cm (cm thun) mc ni tip vi in tr thun mt hiu in th xoay chiu th cm khng ca cun dy bng 3 ln gi tr ca in tr thun. Pha ca dng in trong on mch so vi pha hiu in th gia hai u on mch l:

A. Chm hn gc 3 B. Chm hn gc

6

C. Nhanh hn gc 3

D. Nhanh hn gc 6

Cu 85: (H 2009_ t mt in p xoay chiu c gi tr hiu dng U vo hai u on mch AB gm cun cm thun c t cm L, in tr thun R v t in c in dung C mc ni tip theo th t trn. Gi UL, UR v UC ln lt l cc in p hiu dng gia hai u mi phn t. Bit in p gia hai u on

mch AB lch pha 2 so vi in p gia hai u on mch NB (on mch NB gm R v C ). H thc no

di y l ng A. 2 2 2 2R C LU U U U B.

2 2 2 2C R LU U U U .

C. 2 2 2 2L R CU U U U D. 2 2 2 2R C LU U U U

Cu 86: (H 2007) t vo hai u on mch RLC khng phn nhnh mt hiu in th xoay chiu

0 sinu U t th dng in trong mch l i 0 sin( ) 6i I t . on mch in ny lun c:

A. ZL = ZC B. . ZL < ZC C. . ZL = R D. . ZL > ZC

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BI TON 3: LIN QUAN N HIN TNG CNG HNG

L thuyt:

T I = 22 L C

U UZ R Z Z

Do U khng i nn IMax Zmin ZL = ZC hay LC 2 = 1 hay 1LC

hay f = 12 LC

Lc ny: Zmin = R, axmin

mU UI

Z

R ,

2

ax .mUP U IR

, ax( os ) 1 0m ui

C , u cng pha vi i hay uR cng

pha vi u hay UL = UC v U = UR . C thay i ULmax. L thay i UCmax hay LU U

hoc CU U

, URmax = U Tm li: - in p hiu dng trn in tr thun t cc i : URmax = U .

- in p hiu dng trn cun cm v trn t in c gi tr bng nhau : . .L C L CU UU U Z ZR R

- Cc in p tc thi Ru u v 0L Cu u Hin tng ny gi l hin tng cng hng inCh : - Mun c cng hng in cn thay i C hoc L hoc f sao cho LC 2 = 1(vi = 2 f) - Khi thay i C n gi tr C/ c IMax th /CZ = ZL

- Khi mc C/ vi C c IMax th : ZCb = ZL 1

.b CbC

Z

+ Nu Cb < C cn mc C/ ni tip vi C, vi C/ = . b

b

C C C C

+ Nu Cb > C cn mc C/ song song vi C, vi C/ = Cb C on mch RLC: cho bit U v R tm h thc gia L, C, cng hiu dng I = max? hoc u, I cng pha? hoc h s cng sut maxcos ? Phng php:

1. Trng hp I = max: theo nh lut m: 2 2( )L CR Z Z

UZU I

Nhn xt: I = max khi Z = min 110 2

LCC

Z Z LL C

2. Trng hp u, I cng pha: lch pha 0/ u i

Vy: 0/

R

Z Ztg L Cu i 110 2

LCC

Z Z LL C

3. Trng hp cos max 1 ?

Lc cos 1ZR

( ) 12 2 2 Z rR R Z Z R Z Z LCL C L C

Kt qu chung: (hin tng cng hng in) 2 1LC I = max; u,i cng pha ( 0/ u i ); cos max 1

H qu: RU

ZU I min

max ; Do L C L CZ Z U U vi L C L C L CU U u u

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C LA NR BM

Bi tp trc nghim:

Cu 1: (C 2009) t in p xoay chiu u = U0cos2ft, c U0 khng i v f thay i c vo hai u on mch c R, L, C mc ni tip. Khi f = f0 th trong on mch c cng hng in. Gi tr ca f0 l

A. 2 LC

. B. 2LC . C. 1

LC. D. 1

2 LC.

Cu 2: Mch RLC mc ni tip ,khi tn s dng in l f th cm khng ZL = 25 v dung khng ZC = 75 Khi mch c tn s f0 th cng dng in trong mch t gi tr cc i . Kt lun no sau y l ng A. f0 = 3 f B. f = 3 f0 C. f0 = 25 3 f D. f = 25 3 f0 Cu 3: Cho on mch nh hnh v. Khi t vo hai u mch mt in p c biu thc u = 120 2 cos100 t(V) th thy in p gia hai u on NB v in p gia u on AN v c cng mt gi tr hiu dng v trong mch ang c cng hng in. in p hiu dng gia hai u in tr R l A. 30 2 V. B. 60 2 V. C. 30V. D. 60V Cu 4: Trong mch in xoay chiu R,L,C mc ni tip vi hiu in th hai 2 u mch in khng i, cng dng in t gi tr cc i khi: A. Tn s f ln nht B. Tn s f b nht C. LC4 2f2 = 1 D. LC =1 Cu 5: Cho mch in nh hnh v: R l bin tr, ng dy hot ng vi in tr khng ng k v c t cm L, t in c in dung C. Mch t di mt hiu in th n nh c biu thc u = U0cos t(V). khi R thay i m hiu in th gia hai u on mch AM khng i th ta phi c. A. LC2 = 1. B. 2L2C = 1. C. LC2 = 2. D.2LC = 1. Cu 6: Mch RLC mc ni tip c cng hng in khi A. thay i tn s f Imax. B. thay i tn s f Pmax. C. thay i tn s f URmax. D. c 3 trng hp trn u ng. Cu 7: Mch in xoay chiu RLC ni tip. Gi U l hiu in th hiu dng hai u on mch; UR; UL v UC l hiu in th hiu dng hai u R, L v C. iu no sau y khng th xy ra? A. UR > U. B. UL > U. C. UR > UC. D. U = UR = UL = UC. Cu 8: Mch in xoay chiu RLC ni tip. Hiu in th hiu dng mch in bng hiu in th hai u in tr R khi A. LC = 1. B. hiu in th cng pha dng in. C. hiu in th UL = UC = 0. D. c 3 trng hp trn u ng. Cu 9: Chn cu tr li sai. Trong mch in xoay chiu gm R, L, C mc ni tip vi cos = 1 khi v ch khi A. 1/L = C . B. P = UI. C. Z/R = 1. D. U UR. Cu 10: Mch xoay chiu RLC ni tip. Trng hp no sau y c cng hng in A. Thay i f UCmax. B. Thay i L ULmax. C. Thay i C URmax. D. Thay i R UCmax. Cu 11: Mch xoay chiu RLC mc ni tip vi R = 10 , cm khng ZL = 10 ; dung khng ZC = 5ng vi tn s f. Khi f thay i n gi tr f th trong mch c cng hng in. Ta c A. f = f. B. f > f. C. f < f. D. khng c f. Cu 12: Mt mch in c 3 phn t R, L, C mc ni tip. Mch c cng hng in. Hiu in th hiu dng gia hai u in tr R bng hiu in th hiu dng gia hai u phn t no? A. in tr R. B. T in C. C. Cun thun cm L. D. Ton mch. Cu 13: Mt on mch R, L, C mc ni tip. Cun dy thun cm. Gi U0R, U0L, U0C l hiu in th cc i hai u in tr, hai u cun dy v hai u t in. Bit U0L = 2U0R = 2U0C. Kt lun no di y v lch pha gia dng in v hiu in th gia hai u mch in l ng:

C L, rA BR . M

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C LA NR BM

A. u chm pha hn i mt gc /4 B. u sm pha hn i mt gc 3/4 C. u chm pha hn i mt gc /3 D. u sm pha i mt gc /4Cu 14: Cho on mch nh hnh v. Khi t vo hai u mch mt in p c biu thc u = 120 2 cos100 t(V) th thy in p gia hai u on NB v in p gia u on AN v c cng mt gi tr hiu dng v trong mch ang c cng hng in. in p hiu dng gia hai u in tr R l A. 30 2 V. B. 60 2 V. C. 30V. D. 60V Cu 15: Cho mch in RLC mc ni tip: cun dy thun cm c L = 0,318H v t C bin i. t vo hai u on mch mt in p xoay chiu c tn s f = 50Hz. in dung ca t phi c gi tr no sau trong mch xy ra hin tng cng hng in? A. 3,18F. B. 3,18nF. C. 38,1F. D. 31,8F. Cu 16: Mch RLC mc ni tip c cng hng in khi A. thay i tn s f Imax. B. thay i tn s f Pmax. C. thay i tn s f URmax. D. c 3 trng hp trn u ng. Cu 17: Cho mch in RLC mc ni tip, t vo hai u on mch in p xoay chiu c biu thc u U cos t0 . i lng no sau y bin i khng th lm cho mch xy ra cng hng A. in dung ca t C. B. t cm L. C. in tr thun R. D. Tn s ca dng in xoay chiu. Cu 18: Cho mch RLC mc ni tip, trong mch ang xy ra hin tng cng hng in nu ta thay i tn s ca dng in th A. I tng. B. UR tng. C. Z tng. D. UL = UC. Cu 19: Trong mch in RLC ni tip. Bit C = 10/ (F). Hiu in th gia hai u on mch khng i, c tn s f = 50Hz. t cm L ca cun dy bng bao nhiu th cng hiu dng ca dng in t cc i.(Cho R = const). A. 10/ (H). B. 5/ (H). C.1/ (H). D. 50H. Cu 20: Cho on mch RLC ni tip, gi tr ca R bit, L c nh. t mt hiu in th xoay chiu n nh vo hai u on mch, ta thy cng dng in qua mch chm pha /3 so vi hiu in th trn on RL. trong mch c cng hng th dung khng ZC ca t phi c gi tr bng A. R/ 3 . B. R. C. R 3 D. 3R. Cu 21: Cho mt on mch RLC ni tip. Bit L = 1/ H v C = 25/ F, hiu in th xoay chiu t vo hai u mch n nh v c biu thc u = U0sin100t. Ghp thm t C vo on cha t C. hiu inth hai u on mch lch pha /2 so vi hiu in th gia hai u b t th phi ghp th no v gi tr ca C bng bao nhiu? A. ghp C//C, C = 75/ F. B. ghp CntC, C = 75/ F. C. ghp C//C, C = 25 F. D. ghp CntC, C = 100 F. Cu 22: Mch RLC mc ni tip, khi tn s dng in l f th cm khng ZL = 25 v dung khng ZC = 75 . Khi mch c tn s f0 th cng dng in trong mch t gi tr cc i. Kt lun no sau y l ng A. f0 = 3 f B. f = 3 f0 C. f0 = 25 3 f D. f = 25 3 f0 Cu 23: on mch in xoay chiu RLC ni tip. Gi in p hiu dng gia u in tr, hai u cun dy thun cm v gia hai bn t ln lt l UR, UL, UC. Khi trong mch c cng hng in, biu thc no sau y l sai? A. UL- UC = 0 B. U = UR C. UL + UC = 0 D. P = U.I Cu 24: Cho mch in xoay chiu nh hnh v bn. Cun dy c

r = 10 , L= H10

1

. t vo hai u on mch mt hiu in th

dao ng iu ho c gi tr hiu dng l U = 50V v tn s f = 50Hz. Khi in dung ca t in c gi tr l C1 th s ch ca ampe k l cc i v bng 1A. Gi tr ca R v C1 l

C Rr, L

NMA

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A. R = 40 v C F

3

12.10

. B. R = 50 v C F

3

110

.

C. R = 40 v F103

1

C . D. R = 50 v C F

3

12.10

.

Cu 25: Cho mch in xoay chiu nh hnh v Bit R 80 , ABu 200 2 cos100 t (V) . Cho C thay i, khi xy ra cng hng dng in trong mch c gi tr hiu dng l 2A. Hiu in th hiu dng gia M v B l. A. 160V B. 40V C. 20V D. 0 Cu 26: Cho mch in xoay chiu nh hnh v. t vo hai u on mch AB mt hiu in th khng i. Khi t c gi tr in dung C th hiu in th hiu dng:

R L CU 30V; U 50V; U 90V . Khi t c gi tr C th mch xy ra cng hng v hiu in th gia hai u in tr l. A. 50V B. 50 2 V C. 30V D. 100V

BI TON 4: LIN QUAN TI CNG SUT

1. Cng sut, h s cng sut - Cng sut tc thi: P = UIcos + UIcos(2t + ) (1)

- Cng sut trung bnh: P = UIcos = RI2.

Trong cc bi tp ta tnh cng sut theo biu thc 2

2 2L C

RUP UIcosR (Z Z )

(2)

nu c R v r th: 2

22 2

( )( ). os( ) ( )L C

R r UP R r I UIcR r Z Z

vi ZR

cos (3)

Ta c th bin i cc dng khc 2

2 RR

UP RI U I

R (4)

2. osP ZI c hay 2

2

U RPZ

(5)

0RR

0

osUP URc

UI Z U U (6)

Nu mch c R v r th : 2P R r I v 0R 0R0

os rr U UU UR r cZ U U

Trong : I(A): cng dng in, U(V): in p; P(W): cng sut,

cos: h s cng sut, UR: in p hai u in tr R, UR = IR.

R C A B

L N M

R C

A B

L

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34

2. ngha ca h s cng sut cos

- H s cng sut c gi tr t 0 n 1. on mch khng c in tr R th khng tiu th cng sut.

- Cng sut hao ph trn ng dy ti l

Php = rI2 = 2 2

2

U cosrP

r () in tr ca ng dy ti.

- Nu h s cng sut cos nh th cng sut hao ph trn ng dy ti Php s ln, do ngi ta phi tm

cch nng cao h s cng sut. Theo quy nh ca nh nc th h s cng sut cos trong cc c s in

nng ti thiu phi bng 0,85.

- Vi cng mt in p U v dng c dng in tiu th mt cng sut P, tng h s cng sut cos gim

cng hiu dng I t gim c hao ph v ta nhit trn dy.

Dng 1: Bi ton c bn: Cho cc d kin tm cng sut P

Xc cc i lng lin quan v tnh cng sut P theo cc biu thc

P UIcos ,ZR

cos , P = RI2

Bi tp trc nghim:

Cu 1: Mt on mch gm in tr thun R = 40, t in c in dung 310

9C

F, cun dy c t

cm 3 5

L H v in tr thun khng ng k c mc ni tip vi nhau. in p t vo hai u on

mch ny c biu thc 120 2 cos100u t (V). Tnh h s cng sut v cng sut tiu th ca on mch.

A. 360W. B. 40W. C. 320 2 W. D. 230,4W.

HD: Tnh ZL= 60, ZC = 90, Z = 50, I = U/Z = 120/50 = 2,4A

Rcos 0,8 Z

, P = UIcos = 230,4W.

Cu 2: (H 2008) t vo hai u on mch in RLC khng phn nhnh mt hiu in

th u 220 2 cos t2

V th cng dng in qua on mch c biu thc l

i 2 2 cos t4

A. Cng sut tiu th ca on mch ny l

A. 440W. B. 220 2 W. C. 440 2 W. D. 220W.

HD: .cos 220 24u i

P UI W

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Cu 3: t in p u = 20cos(100t + /2)V vo hai u on mch RLC mc ni tip. Bit biu thc

cng dng in l i = 2cos(100t + /6)A. Tnh cng sut tiu th ca on mch.

A. 20W. B. 40W. C. 10 2 W. D. 10W.

HD: U = 10 2 , I = 2 A, = u i = /3 P = UIcos =1 0W.

Cu 4: t in p u = 20cos100t(V) vo hai u on mch gm in tr R = 10 mc ni tip vi mt

cun dy thun cm 0,1L

H, t in 310

2C

F. Tnh cng sut tiu th ca on mch.

A. 20W. B. 10W. C. 10 2 W. D. 20 2 W.

HD: ZL= 10, ZC = 20, Z = 10 2 , I = U/Z = 1A P = RI2 = 10W.

Cu 5: Mt on mch in xoay chiu gm mt cun dy c in tr thun R0 = 5 v t cm

L =35 .102H mc ni tip vi in tr thun R = 30. in p hai u on mch l

u = 70 2 cos100t(V). Cng sut tiu th ca on mch l

A. 35 2 W. B. 70W. C. 60W. D. 140W.

HD: ZL= 35, 2 20 LZ (R R ) Z = 35 2 , I = U/Z = 2A P = (R + r)I2 = 70W.

Dng 2: Cho cng sut P tm L hoc tm C

Cho cng sut P v cc i lng lin quan tm L hoc tm C

2RIP 2

2 2

.( )L C

RU PR Z Z

2

2L C

RUZ Z RP

Thay s xc nh ZL hoc ZC t tm L hoc C.

Bi tp trc nghim:

Cu 6: Mt on mch gm in tr thun R = 20, t in c in dung 310

4C

F, cun dy c t

cm L v in tr thun khng ng k c mc ni tip vi nhau. in p t vo hai u on mch ny

c biu thc u = 40 2 cos100t (V). Bit cng sut tiu th ca on mch l 16W. Tm L.

HD: Tnh c ZC = 40, ta c 2

2L C

RUZ Z RP

Thay s |ZL ZC| = 40 ZL= 80

0,8L

H.

Ch : Nu cho thm gi thit P, L v C tm th t 2

2L C

RUZ Z R P

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V d: (C 2009) t in p u 100 2 cost (V), c thay i c vo hai u on mch gm in

tr thun 200, cun cm thun c t cm 2536

H v t in c in dung 410

F mc ni tip. Cng sut

tiu th ca on mch l 50W. Gi tr ca l A. 150 rad/s. B. 50 rad/s. C. 100 rad/s. D. 120 rad/s.

Dng 3: on mch RLC c R thay i (L, C, hay U = const) Tm R Pmax

2RIP 2

2 2

.( )L C

RU PR Z Z

2

2L C

UP =(Z - Z )R +

R

Pmax th mu s phi nh nht, mu s l tng c hai s hng m tch ca chng l hng

s 2

2L CL C

Z ZR Z Z

R

= const. Theo h qu bt ng thc Csi ta suy ra tng ny nh nht

khi 2

L CZ ZRR

22 L CR Z Z hay R = |ZL ZC|

Khi : Z R 2 , UIR 2

, R 2cosZ 2

, 4

2

maxUP2R

, 2

maxL C

UP2 Z Z

- Khi R tng t 0 n gi tr L CR Z Z th cng sut tng t 0 n2

max 2UP

R

- Khi R tng t gi tr L CR Z Z n th cng sut gim t 2

max 2UP

R n 0.

Bi tp trc nghim:

Cu 7: (H 2008) on mch in xoay chiu gm bin tr R, cun dy thun cm c t cm L v t

in c in dung C mc ni tip. Bit hiu in th hiu dng hai u on mch l U, cm khng ZL, dung

khng ZC (vi ZC ZL) v tn s dng in trong mch khng i. Thay i R n gi tr R0 th cng sut

tiu th ca on mch t gi tr cc i Pm, khi

A. R0 = ZL + ZC. B. 2

m0

U P .R

C. 2 L

mC

Z P .Z

D. 0 L CR Z Z

HD: chn D.

Cu 8: Cho on mch xoay chiu gm bin tr R, cun thun cm L = 1 H v t in C =

4103 F mc ni

tip. t vo hai u on mch in p xoay chiu u = 120 2 cos100t(V). in tr ca bin tr phi c

gi tr bao nhiu cng sut ca mch t gi tr cc i?

A B

M N

R L C

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A. R = 120. B. R = 60. C. R = 400. D. R = 60.

HD: ZL = 100, ZC = 40 ta c R = |ZL ZC| = 60 . Chn B.

Cu 9: Cho on mch xoay chiu gm bin tr R, cun thun cm L = 1 H v t in C =

4103 F mc ni

tip. t vo hai u on mch in p xoay chiu u = 120 2 cos100t(V). iu chnh gi tr ca bin tr

cng sut ca mch t gi tr cc i. Gi tr cc i ca cng sut l bao nhiu?

A. Pmax = 60W. B. Pmax = 120W. C. Pmax = 180W. D. Pmax = 1200W.

HD: ZL= 100, ZC = 40, ta c 2

maxL C

UP2 Z Z

= 120W

Cu 10: Cho on mch xoay chiu gm bin tr R, cun thun cm L v t in C mc ni tip. t vo

hai u on mch in p xoay chiu u = 220 2 cos100t(V). iu chnh bin tr n gi tr R = 220 th

cng sut ca mch t gi tr cc i. Gi tr cc i ca cng sut l bao nhiu?

A. Pmax = 55W. B. Pmax = 110W. C. Pmax = 220W. D. Pmax = 110 2 W.

HD: Ta c 2

maxUP2R

= 110W

Cu 11: Cho on mch RLC mc ni tip: cun dy thun cm khng c t cm L, t in c in dung

C, R l mt in tr thun thay i c. t mt in p xoay chiu n nh hai u on mch AB c

biu thc: uAB = 200 2 cos100t (V). Khi R = 100 th thy mch tiu th cng sut cc i. Xc nh

cng dng in trong mch lc ny?

A. 2A. B. 2 A. C. 2 2 A. D. 2 2

A

HD: Ta c UIR 2

= 2 A.

Cu 12: Cho on mch RLC mc ni tip gm cun dy thun cm khng c t cm L, t in c in

dung C, R l mt in tr thun thay i c. t hai u mch mt in p xoay chiu n nh. iu

chnh R mch tiu th cng sut cc i. Xc nh gc lch pha ca in p v cng dng in trong

mch?

A. 2 . B.

4 . C. 0. D. 2

2

HD: Ta c 4

chn B.


C, R l mt in tr thun thay i c. t hai u mch mt in p xoay chiu n nh. iu chnh in

tr n gi tr R = 60 th mch tiu th cng sut cc i. Xc nh tng tr ca mch lc ny?

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A. 30 2 . B. 120. C. 60. D. 60 2 .

HD: Ta c Z R 2 = 60 2

Dng 4: Tm R mch c cng sut P

2RIP 2

2 2

.( )L C

RU PR Z Z

2

22L C

UR R Z Z 0P

Vi R l nghim ca phng trnh trn

y l phng trnh bc hai gii phng trnh ny ta tm c R (vi R > 0)

Bi tp trc nghim:

Cu 14: Cho don mch RLC mc ni tip: cun dy thun cm khng c t cm 1

L H, t in c

in dung -410

2C F, R l mt in tr thun thay i c. t mt hiu in th xoay chiu n nh hai

u on mch AB c biu thc uAB = 200cos100t (V). Xc nh R mch tiu th cng sut 80W.

A. 50, 200. B. 100, 400. C. 50, 200. D. 50, 200.

HD: Tnh ZL = 100, ZC = 200, ta c 2

22L C

UR R Z Z 0P

R = 50 v R = 200. Chn C.

Dng 5: Bit hai gi tr ca in tr l R1 v R2 (R1 R2) mch c cng cng sut P, tm cng sut P

- Bit hai gi tr ca in tr l R1 v R2 mch c cng cng sut P

Ta c: 2

22L C

UR R Z Z 0P

R1 v R2 l hai nghim ca phng trnh trn. Theo nh l Vit i vi phng trnh bc hai n R, ta c: 2

1 2UR RP

, 21 2 L CR R Z Z

- Vi 2 gi tr ca in tr l R1 v R2 th mch c cng cng sut P.

Khi 0R R . Tnh R0 theo R1 v R2 mch c cng sut cc i Pmax

- Vi gi tr ca in tr l R0 mch c cng sut cc i Pmax,

Ta c R = R0 = |ZL ZC|

- Vi 2 gi tr ca in tr l R1 v R2 mch c cng cng sut P th

A B

M N

R L C

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21 2 L CR R Z Z suy ra 0 1 2R R R khi 2

max1 22

UPR R

Bi tp trc nghim:

Cu 15: Cho on mch RLC mc ni tip gm cun dy thun cm khng c t cm L, t in c in

dung C, R thay i c. t mt in p xoay chiu n nh hai u on mch c U = 100V, tn s f .

iu chnh R th thy c hai gi tr 30 v 20 mch tiu th cng mt cng sut P. Xc nh P lc ny?

A. 4W. B. 100W. C. 400W. D. 200W.

HD: Ta c P = U2/(R1 + R2) = 200W.


C, R thay i c. t mt in p xoay chiu n nh hai u on mch c gi tr hiu dng U, tn s

f. iu chnh R th thy c hai gi tr 60 v 30 mch tiu th cng mt cng sut P = 40W. Xc nh U

lc ny?

A. 60V. B. 40V. C. 30V. D. 100V.

HD: Ta c 2 1 2U P R R 3600 U 60V


C, R thay i c. t mt in p xoay chiu n nh hai u on mch c gi tr hiu dng U, tn s

f. iu chnh R th thy c hai gi tr 40 v 90 mch tiu th cng mt cng sut. Xc nh R0 mch

tiu th cng sut cc i?

A. 60. B. 65. C. 130. D. 98,5.

HD: Ta c 0 1 2R R R R0 = 60.

Dng 6: Mch c RLC cun dy c in tr trong R0 (R, L, R0, C) (L, C, hay U = const)

- Tm R cng sut ton mch cc i Pmax

0 L C L C 0R R | Z Z , R Z Z | R

2

max0

UP2(R R )

, 2

maxL C

UP2 Z Z

- Tm R cng sut trn R cc i PRmax 2 2 2

0 L CR R (Z Z )

Bi tp trc nghim:


A B

M N

R 0L,R C

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L = 0,8

H, t in 410

2C

F v in tr thun R thay i c mc ni tip. in p hai u on mch

n nh. mch tiu th cng sut cc i th R phi c gi tr no sau y?

A. 100 . B. 120 . C. 60 . D. 80 .

HD: Tnh ZL = 80, ZC = 200, ta c R = |ZL ZC| R0 = 100. Chn A.


L = 0,8

H, t in 310

2C

F v in tr thun R thay i c mc ni tip. in p hai u on mch

n nh. cng sut tiu th trn R cc i th R phi c gi tr no sau y?

A. 100 . B. 120 . C. 50 . D. 80 .

HD: ZL = 80, ZC = 40, ta c R2 = R02 + (ZL ZC)2 = 2500 R = 50

Dng 7: Khi mch c cng hng

Hin tng cng hng trong mch RLC (U, R = const)

- Nu gi nguyn gi tr in p hiu dng U gia hai u on mch v thay i tn s gc (hoc thay

i f, L, C) n mt gi tr sao cho 1

L =C

(hay ZL = ZC) th c hin tng c bit xy ra trong mch: I

t gi tr cc i, gi l hin tng cng hng in.

- iu kin xy ra hin tng cng hng trong mch RLC ni tip:

ZL= ZC; 1L C

; 1LC

Lc mch c cng hng th:

Tng tr Z = Zmin = R; UR = URmax = U

Cng dng in

maxUI IR

Cng sut ca mch khi c cng hng t gi tr cc i 2

maxUP PR

Mch c cng hng th in p cng pha vi cng dng in, ngha l:

= 0; u = i ; cos = 1

in p gia hai im M, B cha L v C t cc tiu

ULCmin = 0.

Lu : L v C mc lin tip nhau

A B

M N

R L C

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Loi 1: Bi ton tnh cng sut khi mch c cng hng

Bi tp trc nghim:

Cu 20: Cho on mch RLC mc ni tip R = 20 , cun dy thun cm khng c t cm L, t in c

in dung C thay i c. t mt in p xoay chiu n nh hai u on mch AB c biu thc:

uAB = 20cos(100t)V. iu chnh C mch tiu th cng sut cc i. Tnh cng sut trong mch lc ny?

HD: Ta c P = U2/R = 100W.

Cu 21: Cho on mch RLC mc ni tip: cun dy thun cm khng c t cm L thay i c, t

in c in dung C, R = 50. t hai u mch mt in p xoay chiu n nh u = 50 2 cos(100t)V.

iu chnh L in p gia hai im M v B nh nht. Tnh cng sut tiu th ca mch lc ny?

HD: Ta c UMbmin = ULcmin = 0 khi mch c

cng hng P = Pmax = U2/R = 50W.

Loi 2: Bi ton xc nh h s cng sut khi mch c cng hng

Mch c cng hng th in p cng pha vi cng dng in, ngha l:

= 0; u = i ; cos = 1

Bi tp trc nghim:

Cu 22: (H 2008) on mch in xoay chiu khng phn nhnh gm cun dy c t cm L, in

tr thun R v t in c in dung C. Khi dng in c tn s gc 1 LC

chy qua on mch th h s

cng sut ca on mch ny

A. ph thuc in tr thun ca on mch. B. bng 0.

C. ph thuc tng tr ca on mch. D. bng 1.

HD: Pmax khi mch c cng hng th cos 1 chn D

Cu 23: Cho on mch RLC mc ni tip: R khng i, cun dy thun cm khng c t cm L, t in

c in dung C thay i c. t mt in p xoay chiu n nh hai u on mch AB c biu thc:

uAB = U0cos(t)V. iu chnh C mch tiu th cng sut cc i. Xc nh h s cng sut ca mch lc

ny?

A. 1. B. 4 . C. 0. D. 2

2

HD: Pmax khi mch c cng hng th = 0; cos = 1 chn A.

Dng 8: on mch RLC c C thay i. Tm C mch c cng sut cc i

A B

M N

R L C

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Pmax khi trong mch c cng hng th 21Z = Z C =

LL C

Bi tp trc nghim:

Cu 24: Cho on mch RLC mc ni tip: cun dy thun cm khng c t cm 0,1L

H, t in c

in dung C thay i c, R l mt in tr thun. t hai u mch mt in p xoay chiu n nh c f.

Xc nh gi tr C mch tiu th cng sut cc i.

A. 0,5/ (H). B. 0,5. (H). C. 0,5 (H). D. (H).

Dng 9: on mch RLC c L thay i. Tm L mch c cng sut cc i

Pmax khi trong mch c cng hng th 21Z = Z L =

CL C

Bi tp trc nghim:

Cu 25: Cho on mch RLC mc ni tip: cun dy thun cm khng c t cm L thay i c, t

in c in dung 310

5C

F, R l mt in tr thun. t hai u mch mt in p xoay chiu n nh c

f = 50Hz. Xc nh gi tr L mch tiu th cng sut cc i.

A. 0,5/ (H). B. 5/ (H). C. 0,5 (H). D. 5 (H).

HD: ZL = ZC = 50 L = 0,5/ (H). Chn cu A

Dng 10: Mch RLC c (hoc f) thay i

+ Tm mch c cng sut cc i Pmax:

Vi = 0 th cng sut ton mch Pmax khi trong mch c cng hng th 01

LC

+ Vi = 1 hoc = 2 th cng sut P c cng mt gi tr.

P1 = P2 I1 = I2 Z1= Z2 |ZL1 ZC1| = | ZL2 ZC2|

ZL1 ZC1 = ZC2 ZL2

ZL1 + ZL2 = ZC1 + ZC2 1.2 = 1

LC

Lin h gia 1, 2, 0.

A B

M N

R L C

A B

M NR L C

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0 1 2 1 2f f f

Bi

Giai Toan Dien Xoay Chieu NTL

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