Vat Ly - Chuyen de Hay Va Kho dong dien xoay chieu -

8/14/2019 Vat Ly - Chuyen de Hay Va Kho dong dien xoay chieu - http://hocmaivn.com

1/35

Gio n vt l, th vin vt l, thi vt l, thi hsg vt l, tin tc vt l - http://hocmaivn.com

1

I. CNG SUT:

Cng sut ca dng in xoay chiu: P = UIcosN = I2R =2

2

Z

RU.

- H s cng sut: cosN =

Z

R= R

U

U

- ngha ca h s cng sut cosN+ Trng hp cosN = 1 tc l N = 0: mch ch c R, hoc mch RLC c cng hng in (ZL

= ZC) th

P = Pmax = UI =R

U2

= I2R

+ Trng hp cosN = 0 tc l N = s2

T: Mch ch c L, hoc ch c C, hoc c c L v C m

khng c R th P = Pmin = 0.- R tiu th nng lng di dng to nhit, ZL v ZC khng tiu th nng lng ca ngun in

xoay chiu.* nng cao h s cng sut ca mch bng cch mc thm vo mch cun cm hoc t inthch hp sao cho cm khng v dung khng ca mch xp x bng nhau cos N} 1.

i vi cc ng c in, t lnh, nng cao h s cng sut cosN gim cng dngin.

II. C S L THUYT GII BI TON HP EN1. Cc cng thc.

+ Nu gi s: i = I0cos[t

th hiu in th hai u mch in UAB = Uocos([t + N)+ Cm khng: ZL = [L

+ Dung khng: ZC =C

1

[

+ Tng tr Z = 2CL

2 )ZZ(R

+ nh lut m: I =Z

UI

Z

U0

0!

+ lch pha gia u v i: tgN =R

ZZCL

+ Cng sut to nhit: P = UIcosN = I2R

H s cng sut: K = cosN =Z

R

UI

P!

2. Gin vc t


2/35


2

* C s:

+ V dng in lan truyn vi vn tc c 3.108m/s nn trn mt on mch in khng

phn nhnh ti mi thi im ta coi ln v pha ca cng dng in l nh nhau ti mi

im.

+ Hiu in th tc thi hai u on mch uAB = uR+ uL + uC

* Cch v gin vc tV i khng i nn ta chn trc

cng dng in lm trc gc, gc ti

im O, chiu dng l chiu quay lng

gic.

3. Cch v gin vc t trt

Bc 1: Chn trc nm ngang ltrc dng in, im u mch lm gc

( l im A).

Bc 2: Biu din ln lt hiu

in th qua mi phn bng cc vc t

NB;MN;AM ni ui nhau theo nguyn tc: R - i ngang; L - i ln; C - i xung.

Bc 3: Ni A vi B th vc t AB chnh l biu din uABNhn xt:

+ Cc hiu in th trn cc phn t c biu din bi cc vc t m ln ca cc vc t

t l vi hiu in th hiu dng ca n.

+ lch pha gia cc hiu in th l gc hp bi gia cc vc t tng ng biu din

chng.

+ lch pha gia hiu in th v cng dng in l gc hp bi vc t biu din n

vi trc i+ Vic gii bi ton l nhm xc nh ln cc cnh v gc ca tam gic da vo cc

nh l hm s sin, hm s cosin v cc cng thc ton hc.

UAB

i

+

UAN

UL UC

URA M

B

N

UL

UR

UAB

O

U +L UC

UC

i

+


3/35


3

DNG 1: Tnh cng sut tiu th bi on mch in xoay chiu

A

BC

b

a

c

Trong ton hc mt tam gic s

gii c nu bit trc ba (hai cnh 1

gc, hai gc mt cnh, ba cnh) trong su

yu t (3 gc v 3 cnh).

lm c iu ta s dng nh l hm s sin hoc Cosin.

+SinC

a

SinB

b

Sin

a!!

+ a2 = b2 + c2 - 2bccosA

b2 = a2 + c2 - 2accosB

c2 = a2 + b2 - 2abcosC

Cch gii:- p dng cc cng thc:

+ Cng thc tng qut tnh cng sut: cosP UI N! + Vi on mch RLC khng phn nhnh, c th tnh cng sut bi: P UI! cosN

+ H s cng sut (on mch khng phn nhnh): cosP R

UI ZN ! !

z Bi tpT LUN:Bi 1: Mc ni tip vi cun cm c ri mc vo ngun xoay chiu. Dng vnk crt ln o hai u cun cm, in tr v c on mch ta c cc gi tr tng ng l 100V,100V, 173,2V. Suy ra h s cng sut ca cun cm

Bi gii

Theo bi ra :

Ta c:

H s cng sut ca cun cm: 0

0 0

0 50cos 0,5100

R

LR LR

UR

Z UN ! ! ! !


4/35


4

Bi 2: t mt hiu in th xoay chiu c tn s gc vo hai u cun dy c R, L th cngsut tiu th ca on mch l P1. Nu ni tip vi cun dy mt t in C vi

22 1LC[ ! v tvo hiu in th trn th cng sut tiu th l P2. Tnh gi tr ca P2

Bi gii

Cng dng in trc khi mc t in C: 1 2 2L

UI

R Z!

Cng dng in sau khi mc thm t in C l: 2 2 2( )L C

UI

R Z Z !

Do 22 1 2L C

LC Z Z [ ! ! Suy ra 2 2 2( )L

UI

R Z!

Suy ra I2=I1> P2=P1

Bi 3 : Cho mt on mch in gm mt bin tr R mc ni tip vi mt t in c in

dung . t vo hai u on mch mt hiu in th xoay chiu vi tn s gc

. Thay i R ta thy vi hai gi tr ca th cng sut ca on mch u bngnhau. Tch bng:

Bi gii

Khi

Khi

V v

Vi:

Bi 4: Cho on mch in xoay chiu RLC mc ni tip. t vo hai u mch mt hiuin th n nh u = Uo cos(2Tft). V th biu din s bin i ca cng sut tiu th P caon mch in khi cho in tr R ca on mch thay i t 0

Bi gii:


5/35


5

RO

P

Pmax

R = b

+ Cng sut tiu th:bR

aR)ZZ(R

RURIP 22

CL2

22

!

!!

+ Ly o hm ca P theo R: 22 )bR()Rb(a

'P

!

P' = 0 R = bs + Lp bng bin thin:

+ th ca P theo R

TRC NGHIM:Bi 1: Chn cu ng. Hiu in th gia hai u mt on mch xoay chiu l:u = 100 2 cos(100Tt - T/6)(V) v cng dng in qua mch l i = 4 2 cos(100Tt - T/2)(A).Cng sut tiu th ca on mch l:

A. 200W. B. 600W. C. 400W. D. 800W.

CHNA

Bi 2: Cho mch in xoay chiu RLC mc ni tip, c R l bin tr. t vo hai u on mchhiu in th xoay chiu c biu thc 120 2 cos(120 )u tT! V. Bit rng ng vi hai gi tr ca

bin tr :R1=18 ; ,R2=32 ; th cng sut tiu th P trn on mach nh nhau. Cng sut ca onmch c th nhn gi tr no sau y: A.144W B.288W C.576W

D.282W

Bi gii

p dng cng thc: 21 2 ( )L C R R Z Z ! 1 2 24L C Z Z R R ! ! ;

Vy1

2 2

1 22 2 2 22

288( ) ( )

L C L C

U UP R R W

R Z Z R Z Z ! ! !

CHN

B

R

P'

P

0 b g

0+

Pmax0 0


6/35


6

Bi 3: Khi t mt hiu in th u = 120cos200t (V) vo hai u on mch gm cun dy c L =

200

R. Khi h s cng sut ca mch l:

A.2

2 B.

4

2C.

2

3D.

3

3

Bi 4:t mt hiu in th u = 250cos(100 tT )V vo hai u on mch gm cun cm c L =0.75

HT

v in tr thun R mc ni tip. cng sut ca mch c gi tr P =125W th R c gi

trA. 25 ; B. 50 ;

C. 75 ; D. 100 ;

CHNABi 5: Mt mch xoay chiu R,L,C khng phn nhnh trong R= 50;, t vo hai u mch mt hiuin th U=120V, f{0 th i lch pha vi u mt gc 600, cng sut ca mch lA. 288W B. 72W C. 36W D. 144W

CHN BBi 6: Mt cun cm mc ni tip vi mt t in, t vo hai u on mch mt hiu in thxoay chiu c U=100(V) th hiu in th hai u cun dy l U1=100(V), hai u t l

U2= 2.100 (V). H s cng sut ca on mch bng:

A). .23 B). 0. C). 2

2 . D). 0,5.

CHN C

CHNA


7/35

7

Bi 7: Cho on mch RLC, R = 50W. t vo mch u = 100 2 sint(V), bit hiu in th giahai bn t v hiu in th gia hai u mch lch pha 1 gcT /6. Cng sut tiu th ca mch l

A.100W B.100 3 W C.50W D.50 3 W CHN C

Dng 2: nh iu kin R,L,C cng sut t cc tr

Cch gii:- Da vo cc cng thc c lin quan, lp biu thc ca i lng cn tm cc tr di dng hmca 1 bin thch hp- Tm cc tr bng cc phng php vn dng

+ Hin tng cng hng ca mch ni tip+ Tnh cht ca phn thc i s+ Tnh cht ca hm lng gic+ Bt ng thc Cauchy+ Tnh cht o hm ca hm s

CC GI TR CC I

Cng sut cc i:

22

2 2L C

UP = RI = R

R + (Z - Z )

R i:2 2

222

I( - )( - )

!

2

Pmax khi L C R Z Z ! 2

max 2L C

UP

Z Z !

L i:2

2 2C

UP R

R + ( - Z )L=

Z

Pmax khi C- ZLZ =0 LZ = CZ Pmax=2

U

R

C i:2

2 2

L

UP R

R + (Z - )C

=

Z

Pmax khi C- ZLZ =0 CZ = LZ


8/35

8

K

W

VR

A

Dng bi tp R i:

T LUN:

Bi 1: Cho mch in xoay chiu gm cun dy c4

r 50 ; L H10

! ; !T

, v t in c in

dung

410

! T F v in tr thun R thay i c. Tt c c mc ni tip vi nhau, ri t vo

hai u on mch c hiu in th xoay chiu u 100 2 cos100 t(V)! T . ng sut tiu th trn intr R t gi tr cc i khi R c gi tr bng bao nhiu ?

Bi gii

L CZ 40 ; Z 100! ; ! ;

2 2 2

2 22 22 2L C L CL C

U R U UP

(Z Z ) (Z Z )(R r) r(R r) (Z Z )R 2r

R R R R

! ! !

p dng BT csi:

2 22 2L C

L C

r (Z Z )R 2 r (Z Z )R

u

Du = xy ra khi 2 2 2 2L CR r (Z Z ) 50 60 78.1! ! ! ;

Bi 2:Cho mch in RLC ni tip, trong cun L thun cm, R l bin tr .Hiu in thhiu dng U=200V, f=50Hz, bit ZL = 2ZC,iu chnh R cng sut ca h t gi tr ln nht thdng in trong mch c gi tr l I= . Tnh gi tr ca C, L

Bi gii

P max khi v ch khi:L C

R Z Z ! hay ( 2 )C L C

R Z doZ Z ! !

Khi , tng tr ca mch l 100 2( )U

ZI

! ! ; Hay 2 2( ) 100 2L C

R Z Z !

1 1

10010C

C

Z C mF Z [ T

! ; ! ! 2

2 200 LL C

Z Z Z L H

[ T! ! ; ! !

Bi 3: Cho mch in nh hnh v bn, cc dng c o khng nh hng g n mch in.1. K m: R=R1. Vn k ch 100V, Wat k ch 100W, ampe k ch 1,4= 2 A.

a.Tnh R1 v cm khng cun dy.b.Cho R bin thin. Cng sut tiu th mch cc i khi R bng

bao nhiu? Tnh h s cng sut ca mch lc .Bi gii

1.K m: a) U=100(V), P=PR=100W, I= 2 A.P=I2R1 100=( 2 )

2R1 R1=50()

Z=I

U= 221 LZR =50 2 ZL=50 .


9/35

9

b) P=I2R RZ

U 2)(! = 222

LZR

RU

=

R

ZR

U

L

2

2

PMax (R

ZR L

2

)min . Thy R.R

ZL2

=ZL2=hng s.

Nn ( RZ

R

L

2

)min R= RZL

2

R=ZL=50().

Cos=Z

R=

250

500,7

1. K ng: Zc=C[

1=100().

a) V gin vec t quay Frecnel. t =(OLO II R ).

Ta c: sin =OC

OL

OL

OC

U

U

I

I! (

ROOC UU ! ).

22

2. OLOCOC

OL

C

L

OL

OC

UUU

U

Z

Z

U

U

!

! (*).Mt khc: 22

L

2OOOC UUU ! , T (*) thay vo ta c: UL=U=100(V).

Theo trn: sin = 4/2

2TE !!

OC

OL

U

U

Nn: IR=IC=Uc/100= 2 UL/100= 2 (A).V IAIIII

LCL!!!! )(2422

R

2 b) Watt k ch : P=IR

2.R=200W.


10/35

10

BI TP PDNGBi 1: Cho mch in xoay chiu nh hnh v 1,

200cos100 ( )AB

u t VT! , t c in dung )(.2

10 4FC

T

! ,

cun dy thun cm c t cm )(10

8HL

T! , R bin i

c t 0 n 200; .1. Tm cng thc tnh R cng sut tiu th P ca mch cc i. Tnh cng sut cc i .

2. Tnh R cng sut tiu th P =Max

P5

3. Vit biu thc cng dng in khi .

S:1) L C maxR Z Z 120 , P 83.3W! ! ; !

2) R 40 ,i 1.58cos(100 t 1.25)(A)! ; ! T

Bi 2:Cho mch in nh hnh v , cun dy thuncm. t vo hai u on mch mt hiu in th c gitr hiu dng khng i, c dng:

u U 2 cos100 t(V)! T .1. Khi bin tr R = 30 ; th hiu in th hiu dng

UAN = 75V; UMB = 100V. Bit cc hiu in th uAN v uMB lch pha nhau gc 900. Tnh cc

gi tr L v C.2. Khi bin tr R = R1 th cng sut tiu th ca mch in l cc i. Xc nh R1 v gi tr cc

ca cng sut. Vit biu thc ca cng dng in khi .S: 1)L}0,127H, C}141,5 FQ

2)R1 = 17,5 ;,PMax=138WBi 3: Cho mch in nh hnh v. Cc vn k c

in tr v cng ln. t vo hai u AB mt hiu in th

xoay chiu: ABu 240 2 cos100 t(V)! T .1. Cho R = R1 = 80 ; , dng in hiu dng ca mch I

= 3 A, Vn k V2 ch 80 3 V, hiu in th giahai u cc vn k lch pha nhau gc T/2. Tnh L, C.

2. Gi L, C, UAB khng i. Thay i R n gi tr R2 cng sut trn on AN t cc i. Tm R2 v gi trcc i ca cng sut. Tm s ch ca vn k V1 khi .

S: 1)L } 0,37H, C} = 69 FQ ;

Bi 4: Cho mch in RLC ni tip, cun dy thun cm c t cm1

L H!T

, t c in dung

C=15,9 FQ v in tr R thay i c. t vo hai u A,B mt hiu in th ABu 200cos100 t(V)! T .

1. Chn R = 100 3 ; . Vit biu thc dng in qua mch.2. Cho cng sut ca mch l P = 80W. Tnh R? Mun cng sut ca mch ny t cc i th

phi chn R l bao nhiu? Tnh PMax khi .3. Tnh R cho uANv uMB lch pha nhau mt gc T/2.

S:1) i 1cos(100 t )A6

T! T ;

CL

B

M

C R L

N

Hnh 1

BRA

BA

V1

C

R

L,r

V2


11/35

11

2) 1 2 MAXR 200 , R 50 , R 100 P 100W! ; ! ; ! ; ! 3) R 100 2! ;

TRC NGHIM:

Bi 1: on mch xoay chiu mc ni tip gm t in410

C

!T

F , cun dy thun cm L=T2

1H v

in tr thun c R thay i. t vo hai u on mch hiu in th xoay chiu c gi tr hiu

dng U = 80V v tn s f = 50 Hz. Khi thay i R th cng sut tiu th trn mch t gi tr cc il:A. Pmax = 64W B. Pmax=100W C. Pmax=128W D. Pmax=150W

=> CHNABi 2: Cho mch in RLC ni tip, trong cun L thun cm, R l bin tr .Hiu in th hiudng U=200V, f=50Hz, bit ZL = 2ZC,iu chnh R cng sut ca h t gi tr ln nht th dngin trong mch c gi tr l I= . Gi tr ca C, L l:

A.1

10m

TF v

2H

TC.

3

10TmF v

4H

T

B.1

10TF v

2mH

TD.

1

10TmF v

4H

T

Bi gii: P UI! hay2 2

2 2( )L C

U UP

Z R Z Z

! !

Vy P max khi v ch khi:L C

R Z Z ! hay ( 2 )C L C

R Z d oZ Z! !

Khi , tng tr ca mch l 100 2( )U

Z I! ! ;

Hay 2 2( ) 100 2L C

R Z Z ! 1 1

10010C C

Z C mF Z [ T

! ; ! !

22 200 L

L C

Z Z Z L H

[ T! ! ; ! ! CHN

A

Bi 3: Mt on mch gm bin tr R mc ni tip vi mt t in C. hiu in th gia 2 u on

mch c biu thc 0 cos ( )u U t

[! . Hi phi cn iu chnh R n gi tr no cng sut to nhittrn bin tr t cc i ? Tnh cng sut cc i .

A) 2max;2

CUPC

R [[

!! B) 2max 2;1

CUPC

R [[

!!

C) 2max 5,0;2CUP

CR [

[!! D.) 2max

1; 0,5R P CU

C[

[! !

Bi 4: Cho mch in nh hnh v:

V

A R L,r C B

=>CHND


12/35

12

Von k c in tr v cng ln. ABu = 200 2cos100t (V) .

L = 1/2T (H), r = 20 ( ; ), C = 31,8.10-6 (F) . cng sut ca mch cc i th R bngA. 30 ( ; ); B. 40 ( ; ); C. 50 ( ; ); D. 60 ( ; ).

CHNA

Bi 5:Cho mch in xoay chiu nh hnh v.C = 318QF ; R l bin tr ;ly1

0.318}T . Hiu inth

Hai u on mch AB :uAB = 100 2 cos 100 Tt (V)a. Xc nh gi tr R0 ca bin tr cng sut cc i. Tnh Pmax

b. Gi R1, R2 l 2 gi tr khc nhau ca bin tr sao cho cng sut ca mch l nh nhau. Tm milin h gia hai i lng ny.

A. R0 = 10 ; ; Pmax = 500 W; R1 . R2 = R20 .

B. R0

= 100 ; ; Pmax

= 50 W; R1

. R2

= R20

.

C. R0 = 100 ; ; Pmax = 50 W; R1 . R2 = R20 .

D. R0 = 10; ; Pmax = 500 W; R1 . R2 = 2R20 .

CHNA

Bi 6: Mt mch R, L, C mc ni tip (cun dy thun cm) L v C khng i R thay i c. tvo hai u mch mt ngun in xoay chiu c hiu in th hiu dng v tn s khng i, riiu chnh R n khi cng sut ca mch t cc i, lc lch pha gia u v i lA. T/4 B. T/6 C. T/3 D. T/2

CHNA

Bi 7: Mt cun dy c in tr thun r = 15;, t cm L =T51 H v mt bin tr thun c

mc nh hnh v, 100 2 cos100 ( )AB

u t

T! A R L,r BKhi dch chuyn con chy ca bin tr. Cngsut to nhit trn bin tr c th t gi tr cc i l.

A. 130 W. B. 125 W. C. 132 W. D. 150 W

CHNB

Bi 8: Mt on mch gm bin tr R mc ni tip vi cun dy c t cm L = 0,08H v in

tr thun r = 32;. t vo hai u on mch mt hiu in th dao ng iu ho n nh c tns gc 300 rad/s. cng sut to nhit trn bin tr t gi tr ln nht th in tr ca bin tr

phi c gi tr bng bao nhiu?A. 56;. B. 24;. C. 32;. D. 40;.

CHND

Dng bi tp L,C i:T LUN:

A B

CR


13/35

13

Bi 1:Cho on mch xoay chiu sau:

R 100! ; (in tr thun)

C 31.8! QF410

}T

F

L: t cm thay i c ca mt cun thun cm

Hiu in th gia hai u AB ca on mch c biu thc:u 200cos314t( ) 200cos100 t( )! } T a)Tnh L h s cng sut ca on mch t cc i.Tnh cng sut tiu th ca on mch lc.

b)Tnh L cng sut tiu th ca on mch cc i. pht ha dng th ca cng sut tiuth P ca on mch theo L.

Bi gii:a)Tnh L trong trng hp 1:

-H s cng sut ca on mch l:2 2

L C

R Rcos

Z R (Z Z )N ! !

Khi L bin thin, cos N s c gi tr ln nht nu c: 2L CZ Z 0 LC 1 ! [ ! Do :

422

1 1 1L 0.318

10C(100 )

! ! ! }

[ TT

T

Z R ! Cng sut tiu th bi on mch l:2

2 22

200

U U 2P I R R 200W

Z R 100

! ! ! ! !

b)Tnh L trong trng hp 2:

- Cng sut tiu th bi on mch c biu thc:2 22

2 2L C

U RUP I R R

Z R (Z Z ) ! ! !

Khi L bin thin, P ln nht nu c:2

L CZ Z 0 LC 1 ! [ !

2

1L 0.318

C ! !

[

2

max

UP 200W

R ! !

- S bin thin ca P theo L:

y2

L 0 2 2

C

RUL 0 Z L 0 P 100W

R Z

! ! [ ! ! !

yL

L Z P 0g

p g p g !

L C maxL 0.318 Z Z 0 P 200W! ! !

Bi 2: Cho mch in RLC mc ni tip, vi L thay i c. Hiu in th hai u mch

l 120 2 cos(100 )u tT! (V), 30R ! ; ,410

( )C FT

! . Hy tnh L :

1. Cng sut tiu th ca mch l

L BRA C


14/35

14

2. Cng sut tiu th ca mch l cc i. Tnh 3. l cc i v tnh

Bi gii

1.

Mt khc

suy ra (c hai gi tr ca )

2. (1)

khi (c cng hng in).

Suy ra

Tnh . T (1) suy ra

3. (2)

Bin i y ta c

(3)Mun cc i th y phi cc tiu . T (3) ta thy :


15/35

15

Thay vo (2) :

Khi

Suy ra


16/35


17/35

17

suy ra = =

Thay (3) vo (2) ta c = =


18/35

18

Dng 3: Bi ton hp en

Phng php gii

gii mt bi ton v hp kn ta thng s dng hai phng php sau:

a. Phng php i s

B1: Cn c u vo ca bai ton t ra cc gi thit c th xy ra.

B2: Cn c u raca bi ton loi b cc gi thit khng ph hp.

B3: Gi thit c chn l gi thit ph hp vi tt c cc d kin u vo v u ra ca bi

ton.

b. Phng php s dng gin vc t trt.

B1: V gin vc t (trt) cho phn bit ca on mch.B2: Cn c vo d kin bi ton v phn cn li ca gin .

B3: Da vo gin vc t tnh cc i lng cha bit, t lm sng to hp kn.

* Trong mt s ti liu c vit v cc bi ton hp kn thng s dng phng php i s,

nhng theo xu hng chung th phng php gin vc t (trt) cho li gii ngn gn hn, logic

hn, d hiu hn.


19/35

19

UC0

U R0

U M N

UA M

N

ABUA B

M

i

1. Bi ton trong mch in c cha mt hp kn.

V d 1: Cho mch in nh hnh v:

UAB = 200cos100Tt(V)

ZC = 100; ; ZL = 200;

I = 2 )A( ; cosN = 1; X l on mch gm hai trong ba phn t (R0, L0 (thun), C0) mcni tip. Hi X cha nhng linh kin g ? Xc nh gi tr ca cc linh kin .

Gii

Cch 1: Dng phng php gin vc t trt.

Hng dn Li gii

B1: V gin vc t cho on mch

bit

+ Chn trc cng dng in lm

trc gc, A l im gc.

+ Biu din cc hiu in th uAB;

uAM; uMN bng cc vc t tng ng.

* Theo bi ra cosN = 1 uAB v i cng pha.

UAM = UC = 200 2 (V)

UMN = UL = 400 2 (V)

UAB = 100 2 (V)

Gin vc t trt

V UAB cng pha so vi i nn trn NB (hp X) phi cha

in tr Ro v t in Co.

B2: Cn c vo d kin ca bi ton

NB

U xin gc v tr pha so vi

i nn X phi cha Ro v Co

B3: Da vo gin URo v UCo t

+ URo = UABm IRo = 100 2

p Ro = )(5022

2100;!

+ UCo = UL - UC

A

C

BNMX


20/35

20

tnh Ro; Co p I . ZCo = 200 2

p ZCo = )(10022

2200;!

Co = )F(10

100.100

1 4

T

!T

Cch 2: Dng phng php i s

Hng dn Li gii

B1: Cn c u vo ca bi ton

t cc gi thit c th xy ra.

p Trong X c cha Ro&Lo hoc Ro

v Co

B2: Cn c u ra loi b cc

gi thit khng ph hp v ZL > ZC

nn X phi cha Co.

B3: Ta thy X cha Ro v Co ph hp

vi gi thit t ra.

* Theo bi ZAB = )(5022

2100;!

1Z

Rcos !!N

V trn AN ch c C v L nn NB (trong X) phi cha Ro,

mt khc: Ro=Zp ZL(tng) = ZC(tng) nn ZL = ZC+ZCo

Vy X c cha Ro v Co

;!!!

;!!

)(100100200ZZZ

)(50ZR

CLC

AB0

o

Co = )(10 4

T

Nhnxt: Trn y l mt bi tp cn kh n gin v hp kn, trong bi ny cho bit N

v I, chnh v vy m gii theo phng php i s c phn d dng. i vi nhng bi ton v hp

kn cha bit N v I th gii theo phng php i s s gp kh khn, nu gii theo phng php

gin vc t trt s thun li hn rt nhiu. V d 2 sau y l mt bi ton in hnh.

V d 2: Cho mch in nh hnh v

UAB = 120(V); ZC = )(310 ;

R = 10(;); uAN = 60 6 cos100 ( )t vT

UAB = 60(v)

a. Vit biu thc uAB(t)

b. Xc nh X. Bit X l on mch gm hai trong ba phn t (Ro, Lo (thun), Co) mc ni

tip

A

C

BNMX

R


21/35

21

Gii:

a. V gin vc t cho on mch bit A

Phn cn li cha bit hp kn cha g v vy ta gi s n l mt vc t bt k tin theo chiu

dng in sao cho: NB = 60V, AB = 120V, AN = 60 V3

+ Xt tham gic ANB, ta nhn thy

AB2 = AN2 + NB2, vy l tam gic vung

ti N

tgE =3

1

360

60

AN

NB!!

6

T!E UAB sm pha so vi UAN 1 gc

6

T

p Biu thc uAB(t): uAB= 120 2cos 100 6t

TT

(V)

b. Xc nh X

T gin ta nhn thy NB cho ln m trong X ch cha 2 trong 3 phn t nn X phi cha

Ro v Lo. Do ta v thm c00 LR

UvU nh hnh v.

+ Xt tam gic vung AMN:63

1

Z

R

U

Utg

CC

RT

!F!!!F

+ Xt tam gic vung NDB

)V(302

1.60sinUU

)V(3302

3.60cosUU

NBL

NBR

O

O

!!F!

!!F!

Mt khc: UR= UANsinF = 60 )(3302

1.3 !

T!

T!;!!!

;!!!

!!

)H(3

1,0

3100

10L)(

3

10

33

30

I

UZ

)(1033

330

I

UR

)A(33

10

330

I

O

L

L

R

O

O

O

O

UAB

UC

UR

A

M N

B

i

UA

N

UNB

UR0

Ul 0

D


22/35

22

A

C

BNMX

R

*Nhnxt: y l bi ton cha bit trc pha v cng dng in nn gii theo phng

php i s s gp rt nhiu kh khn (phi xt nhiu trng hp, s lng phng trnh ln p gii

rt phc tp). Nhng khi s dng gin vc t trt s cho kt qu nhanh chng, ngn gn, ... Tuy

nhin ci kh ca hc sinh l ch rt kh nhn bit c tnh cht 2NB

2

AN

2

ABUUU ! . c s

nhn bit tt, hc sinh phi rn luyn nhiu bi tp c k nng gii.V d 3: Cho mch in nh hnh v:

UAB = cost; uAN = 180 2 cos 100 ( )2

t VT

T

ZC = 90(;); R = 90(;); uAB = 60 2 cos100 ( )t VT

a. Vit biu thc uAB(t)

b. Xc nh X. Bit X l on mch gm hai trong ba phn t (RO, Lo (thun), CO) mc nitip.

Phn tch bi ton: Trong v d 3 ny ta cha bit cng dng in cng nh lch pha

ca cc hiu in th so vi cng dng in nn gii theo phng php i s s gp nhiu kh

khn. V d 3 ny cng khc v d 2 ch cha bit trc UAB c ngha l tnh cht c bit trong

v d 2 khng s dng c. Tuy nhin ta li bit lch pha gia uAN v uNB, c th ni y l mu

cht gii ton.

Gii

a. V gin vc t cho on mch bit AN. Phn cn li cha bit hp kn cha g, v

vy ta gi s n l mt vc t bt k tin theo chiu dng in sao cho uNB sm pha2

Tso vi uAN

+ Xt tam gic vung ANB

* tgE =3

1

180

60

U

U

AN

NB

AN

NB !!!

E} 800 = 0,1T(rad)

uAB sm pha so vi uAN mt gc 0,1T

* 2NB

2

AN

2

ABUUU ! = 1802 + 602} 1900 UAb = 190(V)

A

C

BNMX

R

AB

C

UR

A

M N

B

i

UAN

UNB

UR0

Uc 0

D


23/35

23

p biu thc uAB(t): uAB = 190 2 cos 100 0,12

tT

T T

= 190 2 cos 100 0,4 ( )t T T

b. T gin ta nhn thy NB cho ln m trong X ch cha hai trong 3 phn t trn X phi

cha RO v LO. Do ta v thm c OO LR UvU nh hnh v.

+ Xt tam gic vung AMN: 190

90

Z

R

U

Utg

CC

R !!!!F

F = 450

UC = UAN.cosF = 180. )A(290

290

Z

UI290

2

2

C

C !!!!

+ Xt tam gic vung NDB

)(302230R)V(230

22.60cosUU

0NBRO;!!!!F!

F = 450 ULo = URo= 30 2 (V) p ZLo = 30(;)

)(3,0

100

30L

OT

!T

!

Nhnxt: Qua ba th d trn ta hiu c phn no v phng php gii bi ton hp kn

bng gin vc t trt, cng nh nhn ra c u th ca phng php ny. Cc bi tp tip theo

ti s cp n bi ton c cha 2 hoc 3 hp kn, ta s thy r hn na u th vt tri caphng php ny.

2. Bi ton trong mch in c cha hai hp kn

V d 1: Mt mch in xoay chiu c s nh hnh v.

Trong hp X v Y ch c mt linh kin

hoc in tr, hoc cun cm, hoc l t in.

Ampe k nhit (a) ch 1A; UAM = UMB = 10V

UAB = 10 V3 . Cng sut tiu th ca on mch AB l P = 5 6W. Hy xc nh linh kin

trong X v Y v ln ca cc i lng c trng cho cc linh kin . Cho bit tn s dng in

xoay chiu l f = 50Hz.

*Phn tch bi ton: Trong bi ton ny ta c th bit c gc lch N (Bit U, I, P pN)

nhng on mch ch cha hai hp kn. Do nu ta gii theo phng php i s th phi xt rt

nhiu trng hp, mt trng hp phi gii vi s lng rt nhiu cc phng trnh, ni chung l

A BMYa X


24/35

24

vic gii gp kh khn. Nhng nu gii theo phng php gin vc t trt s trnh c nhng

kh khn . Bi ton ny mt ln na li s dng tnh cht c bit ca tam gic l: U = UMB

UAB = 10 AMU3V3 ! p tam gic AMB l ( cn c 1 gc bng 300.

Gii:

H s cng sut:UI

cos !N

42

2

310.1

65cos

Ts!N!!N

* Trng hp 1: uAB sm pha4

Tso vi i

gin vc t

V:

!!

AMAB

MBAM

U3U

UU

(AMB l ( cn v UAB = 2UAMcosE cosE =10.2

310

U2

U

AM

AB !

cosE = 0302

3!E

a. uAB sm pha hn uAM mt gc 300

UAM sm pha hn so vi i 1 gc NX = 45

0 - 300 = 150

X phi l 1 cun cm c tng tr ZX gm in tr thun RX v t cm LX

Ta c: )(101

10

I

UZ AM

X;!!!

Xt tam gic AHM:

+ 0XX

0

XR15cosZR15cosUU

X!!

RX = 10.cos150 = 9,66(;)

+ )(59,215sin1015sinZZ15sinUU 00XL

0

XL XX;!!!!

)mH(24,8100

59,2L

X !T

!

Xt tam gic vung MKB: MBK = 150 (v i xng)

i

M

URX

ULXK

UAB

UY

URY

ULY

A H

B

450

300

150 U


25/35

25

UMB sm pha so vi i mt gc NY = 900 - 150 = 750

Y l mt cun cm c in tr RY v t cm LY

+ RY =XL

Z (v UAM = UMB) RY = 2,59(;)

+XL

RZY

! = 9,66(;) LY = 30,7m(H)

b. uAB tr pha hn uAM mt gc 300

Tng t ta c:

+ X l cun cm c tng tr

ZX = )(101

10

I

UAM ;!!

Cun cm X c in tr thun RX v t cm LX vi RX = 2,59(;); RY=9,66(;)

* Trnghp2: uAB tr pha4T so vi i, khi

uAM v uMB cng tr pha hn i (gc 150 v 750).

Nh vy mi hp phi cha t in c tng tr ZX,

ZX gm in tr thun RX, RY v dung khng CX,

CY. Trng hp ny khng th tho mn v t in

khng c in tr

.

Nhnxt: n bi ton ny hc sinh bt u cm thy kh khn v n i hi hc sinh

phi c c phn on tt, c kin thc tng hp v mch in xoay chiu kh su sc. khc phc

kh khn, hc sinh phi n tp l thuyt tht k v c k nng tt trong b mn hnh hc.

Khi mc hai im A, B vo hai cc ca mt ngun in xoay chiu tn s 50Hz th Ia = 1(A), Uv1 =

60v; UV2 = 80V,UAM lch pha so vi UMB mt gc 1200, xc nh X, Y v cc gi tr ca chng.

V d 2: Cho hai hp kn X, Y ch cha 2

trong ba phn t: R, L (thun), C mc ni tip. Khimc hai im A, M vo hai cc ca mt ngun in

mt chiu th Ia = 2(A), UV1 = 60(V).

i

BK

MH

A

UAB

URY

UX

ULY

URX

ULX

300

450

UY

450

300

A

M

MB

i

A B

M

Ya

X

1 v2


26/35

26

*Phn tch bi ton: y l mt bi ton c s dng n tnh cht ca dng in 1 chiu i

vi cun cm v t in. Khi gii phi lu n vi dng in 1 chiu th [ = 0 ZL = 0 v

g![

!C

1Z

C. Cng ging nh phn tch trong v d 1 bi ton ny phi gii theo phng php

gin vc t (trt).

Gii

* V X cho dng in mt chiu i qua nn X khng cha t in. Theo bi th X cha 2

trong ba phn t nn X phi cha in tr thun (RX) v cun dy thun cm (LX). Cun dy thun

cm khng c tc dng vi dng in mt chiu nn: RX = )(302

60

I

U1V

;!!

* Khi mc A, B vo ngun in xoay chiu ZAM =2

L

2

X

V

X

1 ZR)(601

60

I

U!;!!

)(330Z30.33060ZXX L

222

L;!!! tgNAM=

0

AM

X

L603

R

ZX !N!

* V gin vc t cho on AM. on mch

MB tuy cha bit nhng chc chn trn gin n

l mt vct tin theo chiu dng in, c di =

2VU = 80V v hp vi vc t AB

uuurmt gc 1200

ta v c gin vc t cho ton mch.

T gin vc t ta thy MB buc phi

cho xung th mi tin theo chiu dng in, do

Y phi cha in tr thun (RY) v t in CY.

+ Xt tam gic vung MDB

)V(4021.8030sinUU 0MBRY !!!

)(401

40

I

UR Y

R

Y;!!!

i

UAM Ul x

Ur xA

M

AM

600 iA

Ur y

UAB

Ur x

Uc yUAM

M D

UMBUl x

300

B

300

300

1200


27/35

27

)H(34,0

100

340L

)(340Z)V(3402

3.8030cosUU

Y

L

0

MBL YY

T!

T!

;!!!!

3. Bi ton ny trong mch in c cha ba hp kn

V d: Cho mch in cha ba linh kin

ghp ni tip:

R, L (thun) v C. Mi linh kin cha trong mt hp kn X, Y, Z t vo hai u A, B ca

mch in mt hiu in th xoay chiu 8 2 cos2 ( )u ft V T!

Khi f = 50Hz, dng mt vn k o ln lt c UAM = UMN = 5VUNB = 4V; UMB = 3V. Dng ot k o cng sut mch c P = 1,6W

Khi f { 50Hz th s ch ca ampe k gim. Bit RA} O; RV}g

a. Mi hp kn X, Y, Z cha linh kin g ?b. Tm gi tr ca cc linh kin.

*Phn tch bi ton: Bi ton ny s dng ti ba hp kn, cha bit I v N nn khng th

gii theo phng php i s, phng php gin vc t trt l ti u cho bi ny. Bn cnh hc sinh phi pht hin ra khi f = 50Hz c hin tng cng hng in v mt ln na bi ton li

s dng n tnh cht a2 = b2 + c2 trong mt tam gic vung.

Gii

Theo u bi: )V(82

28U

AB

!!

Khi f = 50HzUAM = UMN = 5V; UNB = 4V; UMB = 3V

Nhn thy:+ UAB = UAM + UMB (8 = 5 + 3) ba im A, M v B thng hng

+ 2MB

2

NB

2

MNUUU ! (52 = 42 + 32) Ba im M, N, B to thnh tam gic vung ti B.

Gin vc t ca on mch c dng nh hnh v.Trong on mch in khng phn

nhnh RLC ta c CRC UvUU B

mun pha hnR

U AM

U biu din

A B

MYa X Z*

N*

UMN

UM N

UMBUAMA M B

N

MN


28/35

28

hiu in th hai u in tr R (X cha R) vNB

U biu din hiu in th hai u t in (Z cha

C). Mt khcMN

U sm pha so viAM

U mt gc NMN

Documents

Vat Ly - Chuyen de Hay Va Kho dong dien xoay chieu -