Giai Nhanh Vat Li 12 Bang FX570ES

8/4/2019 Giai Nhanh Vat Li 12 Bang FX570ES

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ng- Email: [email protected] ; [email protected] Trang 1

HY THDNG MY TNH CASIO fx570ES & 570ES PlusGII NHANH mt s bi tp TRC NGHIM VT L 12!

I.GII TM NHANH MT I LNG CHA BIT TRONG BIU THC VT L:1.Sdng SOLVE ( Ch dng trong COMP: MODE 1 )a)V d 1: Tnh khi lng m ca con lc l xo dao ng, khi bit chu k T =0,1(s) v cng

k=100N/m. Ta dng biu thc 2=m

Tk

Ch : phm gn bin X: ALPHA ) ; SOLVE: SHIFT CALC ; Du = trong biu thc: ALPHA CALC

Phng php truyn thng Phng php dng SOLVE

Ta c : 2=m

Tk

>2 24=

mT

k

Suy ra:

2

2

.

4=

k Tm

Th s:

2

2

100.(0,1 )

4=m

=0,25kg

Vy :khi lng m ca con lc 0,25kg

-Vi my FX570ES:Bm: MODE 1

-Bm: 0.1 SHIFT X10X ALPHA CALC = 2

SHIFT X10X ALPHA ) X 100

Mn hnh xut hin: 0.1 2 100=

X

-Tip tc bm:

SHIFT CALC SOLVE = ( chkhong 6s )

Mn hnh hin th:

Vy : m= 0,25 kg

Bn thdng ng h ln lt o thi gian khi gii c 2 phng php ri rt ra kt lun !

Tv d ny chng ta c th suy lun cch dng cc cng thc khc!!!

b)V d 2:Tnh cng ca con lc l xo dao ng, khi bit chu k T =0,1(s) v khi lng =0,25kg.

.-Ta dng biu thc 2=m

Tk

lm Tng t nh trn, cui cng mn hnh xut hin:0.250.1 2=X

-Tip tc bm:SHIFT CALC SOLVE =

( chkhong 6s ),Mn hnh hin th nh hnh bn :

Vy : k =100N/m

0 .1 21 0 0

=X

X= 0.25

L--R = 0

0 . 2 50 .1 2=

X

X= 100

L--R = 0


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II.GII NHANH TNG HP DAO NG NHMY TNH CASIO fx570ES, 570ES Plus.A.KIN THC:1. Tng hp hai dao ng iu ho cng phng cng tn s nhsau:x1 = A1cos (t + 1) v x2 = A2cos (t + 2) th: x = x1 + x2ta c x = Acos (t + ) . Vi:

Bin : A2=A12+ A2

2+2A1A2cos (2 - 1); Pha ban u :tan =2211

2211

coscos

sinsin

AA

AA

+

+

vi 1 2 (nu 12 )

2.Nu mt vt tham gia ng thi nhiu dao ng iu ho cng phng cng tn s:x1 = A1cos (t + 1),x2 = A2cos (t + 2) v x3 = A3cos (t + 3) ... th dao ng tng hpcng l dao ng iu ho cng phng cng tn s: x = Acos (t + ) .Chiu ln trc Ox v trc Oy trong h xOy. Ta c:

Ax = Acos = A1cos 1+ A2cos 2+ A3cos 3 + ..v Ay = A sin = A1sin 1+ A2sin 2+ A3sin 3 + ..

Bin : : A =

2 2x yA A+ v Pha ban u : tan =

y

x

A

A vi [Min, Max]

3.Khi bit dao ng thnh phnx1=A1cos (t + 1) v dao ng tng hp x = Acos(t +) thdao ng thnh phn cn li l x2 =x - x1 . vi x2 = A2cos (t + 2).

Bin : A22=A2+ A1

2-2A1Acos( -1); Pha tan 2=1 1

1 1

sin sin

cos cos

A A

A A

vi 12(nu 12)

4.Nhc im ca phng php trn khi lm trc nghim:Mt nhiu thi gian biu din gin vct, i khi kh biu din c vi nhng bi ton tng hp t3 dao ng trln, hay i tm dao ng thnh phn. Nn vic xc nh A v ca dao ng tng hp theophng php trn mt nhiu thi gian v d nhm ln cho hc sinh, thm ch ngay c vi gio vin.-Vic xc nh gc hay 2 tht s kh khn i vi hc sinh bi v cng mt gi trtan lun tn tihai gi tr ca (v d: tan=1 th = /4hoc -3/4),vy chn gi tr no cho ph hp vi bi ton!.

B. GII PHP :Dng my tnh CASIO fx 570ES, 570ES Plus hoc CASIO fx 570MS.(Gip cc em hc sinh v h trgio vin kim tra nhanh c kt qu bi ton tng hp dao ng trn).1. Csl thuyt:+Dao ng iu hox = Acos(t + ) c thc biu din bng vectquay

ur

A c di t l vi bin A v to vi trc honh mt gc bng gc pha ban u .+Mt khc cng c thc biu din bng s phc di dng: z = a + bi

+Trong ta cc: z =A(sin +i cos) (vi mun: A= 2 2a b+ ) hayZ = Aej(t + ).+V cc dao ng c cng tn s gc nn ngi ta thng vit vi quy c z = AeJ, trong my tnhCASIO fx- 570ES k hiu di dng l: r (ta hiu l: A ).+c bit gic sc hin th trong phm vi : -1800< < 1800 hay -


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Bm:MODE xut hi n chCMPLX

2.Chn ch thc hin php tnh vsphc ca my tnh: CASIO fx 570ES, 570ES PlusCc bc Chn ch Nt lnh ngha- Kt quCi t ban u (Reset all): Bm SHIFT 9 3 == Reset allHin th1 dng (MthIO) Bm SHIFT MODE 1 Mn hnh xut hin Math.Thc hin php tnh v s phc Bm MODE 2 Mn hnh xut hin ch

CMPLX

Tnh dng to cc: r

(ta hiu l:A )

BmSHIFT MODE 3 2 Hin th s phc kiu r

Tnh dng to cc: a + ib. Bm SHIFT MODE 3 1 Hin th s phc kiu a+biChn n vo gc l (D) Bm: SHIFT MODE3 Mn hnh hin th chDChn n vo gc l Rad (R) Bm: SHIFT MODE4 Mn hnh hin th chR nhp k hiu gc Bm SHIFT (-). Mn hnh hin th k hiu

V d: Cch nhp: My tnh CASIO fx 570ESCho: x= 8cos(t+ /3)sc biu din vi s phc 8 600 hay 8/3ta lm nh sau:

-Chn mode: Bm my: MODE 2 mn hnh xut hin chCMPLX-Chn n vo gc l (D) ta bm: SHIFT MODE3 trn mn hnh hin th chD-Nhp my: 8 SHIFT (-) 60 s hin th l: 8 60-Chn n vo gc l Rad (R) ta bm: SHIFT MODE4trn mn hnh hin th chR

-Nhp my: 8 SHIFT (-) (:3s hin th l: 81

3

Kinh nghim: Nhp vi n v nhanh hn n vradnhng kt qu sau cng cn phi chuyn sang n vradcho nhng bi ton theo n vrad. (V nhp theo n vrad phi c du ngoc n ()nn thao tc nhp lu hn,v d: Nhp 90 th nhanh hn nhp (/2)

Bng chuyn i n vgc: (Rad)=

(D).

180

n v gc () 15 30 45 60 75 90 105 120 135 150 165 180 360n v gc (Rad) 1

12

1

6

1

4

1

3

5

12

1

2

7

12

2

3

9

12

5

6

11

12

2

3.Lu :Khi thc hin php tnhkt quc hin thdng i s: a +bi (hoc dng cc:A).-Chuyn t dng : a + bi sang dng: A , bm SHIFT 2 3 =

V d: Nhp: 8 SHIFT (-) (:3 ->Nu hin th: 4+ 4 3 i , munchuyn sang dng cc A :

- Bm phm SHIFT 2 3 = kt qu:81

3

-Chuyn t dng Asang dng : a + bi : bm SHIFT 2 4 =V d: Nhp: 8 SHIFT (-) (:3 -> Nu hin th: 8

1

3, mun chuyn sang dng phc a+bi :

- Bm phm SHIFT 2 4 =kt qu :4+4 3 i

Bm SHIFT 2 mn hnh xut hin nh hnh bn

Nu bm tip phm 3 = kt qudng cc (r )Nu bm tip phm 4 = kt qudng phc (a+bi )( ang thc hin php tnh )


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4.Tm dao ng tng hp xc nh A v bng cch dng my tnh thc hin php cng:a.Vi my FX570ES:Bm chnMODE 2 trn mn hnh xut hin ch: CMPLX.

-Chn n vo gc l (D) ta bm: SHIFT MODE3 trn mn hnh hin th chD(hoc Chn n vo gc l Rad (R) ta bm: SHIFT MODE4trn mn hnh hin th chR )

-Nhp A1 ,bm SHIFT (-) , nhp 1, bm +, Nhp A2, bm SHIFT (-) ,nhp 2 nhn = hin thkt qu.

(Nu hin th s phc dng: a+bi th bm SHIFT 2 3 = hin thkt qu: A)

b.Vi my FX570MS : Bm chnMODE 2trn mn hnh xut hin ch: CMPLX.Nhp A1, bm SHIFT (-) nhp 1 , bm + , Nhp A2 , bm SHIFT (-) nhp 2 nhn =Sau bm SHIFT + = hin th kt qu l: A. SHIFT = hin th kt qu l:

c.Lu Ch hin th mn hnh kt qu:Sau khi nhp ta n du = c th hin th kt qu di dng sv t, mun kt qu di dng thp phn tan SHIFT = (hoc dng phm SD ) chuyn i kt quHin th.

d.Cc v d:V d 1:Mt vt thc hin ng thi hai dao ng iu ha cng phng, cng tn s c phng trnh:x1 = 5cos( t +/3) (cm); x2 = 5cos t (cm). Dao ng tng hp ca vt c phng trnh

A. x = 5 3 cos( t -/4 ) (cm) B.x = 5 3 cos( t + /6) (cm)C. x = 5cos( t + /4) (cm) D.x = 5cos( t - /3) (cm) p n B

Phng php truyn thng Phng php dng s phc

Bin : 2 21 2 1 2 2 12. .cos( )= + + A A A A A

Pha ban u : tan =2211

2211

coscos

sinsin

AA

AA

+

+

Th s:

A=2 2

5 5 2.5.5.cos( / 3) 5 3+ + = (cm)

tan =5.sin( / 3) 5.sin 0 5. 3 / 2 3

15cos( / 3) 5.cos0 35. 12

+= =

++

=>

= /6.

Vy :x = 5 3 cos( t + /6) (cm)

-Vi my FX570ES:Bm: MODE 2

-n vo gc l (D)bm: SHIFT MODE3

Nhp:

5 SHIFT (-) (60) + 5 SHIFT (-) 0 =

Hin th kt qu: 5 3 30Vy :x = 5 3 cos( t + /6) (cm)

(Nu Hin th dng cc:15 5 3

2 2+ i th

Bm SHIFT 2 3 = Hin th: 5 3 30 )

Bn thdng ng h ln lt o thi gian khi gii c 2 phng php ri rt ra kt lun !

Gii khi dngn vo gc l Rad(R):SHIFT MODE 4Bm chn MODE 2 mn hnh xut hin: CMPLX. Tm dao ng tng hp:

Nhp :5 SHIFT (-). (/3) + 5 SHIFT (-) 0 = Hin th: 5 3 1

6

V d 2:Mt vt thc hin ng thi hai dao ng iu ho cng phng, cng tn s

x1=cos(2t + )(cm), x2 = 3 .cos(2t - /2)(cm). Phng trnh ca dao ng tng hpA. x = 2.cos(2t - 2/3) (cm) B. x = 4.cos(2t + /3) (cm)C. x = 2.cos(2t + /3) (cm) D. x = 4.cos(2t + 4/3) (cm)


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Gii: Vi my FX570ES :Bm chn MODE 2 trn mn hnh xut hin ch: CMPLXChn n vo gc l rad (R): SHIFT MODE 4

-Nhp my: 1 SHIFT(-) + 3 SHIFT(-) (-/2 = Hin th 2-2

3

. p n A

V d 3:Mt vt dao ng iu ha xung quanh v tr cn bng dc theo trc xOx c li

)()2

2cos(3

4))(

62cos(

3

4cmtcmtx

+++= . Bin v pha ban u ca dao ng l:

A. .3

;4 radcm

B. .6

;2 radcm

C. .6

;34 radcm

D. .3

;3

8radcm

p n A

Gii 1: Vi my FX570ES : Bm chn MODE 2 trn mn hnh xut hin ch: CMPLXChn n vo gc l radian(R):SHIFT MODE 4

Nhp my:4

3

SHIFT (-). (/6) +4

3

SHIFT (-). ( /2 = Hin th: 4 1

3

Gii 2: Vi my FX570ES : Chn n vo gc l Degre(D):SHIFT MODE 3

Nhp my:4

3

SHIFT (-). 30 +4

3

SHIFT (-). 90 = Hin th: 4 60

V d 4:Ba dao ng iu ha cng phng, cng tn s c phng trnh ln lt l x1= 4 cos(t - /2)(cm) , x2= 6cos(t +/2) (cm) v x3=2cos(t) (cm). Dao ng tng hp ca 3 dao ng ny c bin vpha ban u l

A. 2 2 cm; /4 rad B. 2 3 cm; - /4 rad C.12cm; + /2 rad D.8cm; - /2 radGii: Vi my FX570ES :Bm chn MODE 2 trn mn hnh xut hin ch: CMPLXChn n v gc tnh rad (R). SHIFT MODE 4Tm dao ng tng hp, nhp my:

4 SHIFT(-) (- /2) + 6 SHIFT(-) (/2) + 2 SHIFT(-) 0 = Hin th: 2 2 /4. Chn AV d 5: Dao ng tng hp ca hai dao ng iu ha cng phng, cng tn s

x1= a 2 cos(t+/4)(cm) v x2 = a.cos(t + ) (cm) c phng trnh dao ng tng hp l

A. x = a 2 cos(t +2/3)(cm) B. x = a.cos(t +/2)(cm)

C. x = 3a/2.cos(t + /4)(cm) D. x = 2a/3.cos(t +/6)(cm) Chn BGii: Vi my FX570ES :Bm chn MODE 2 trn mn hnh xut hin ch: CMPLX

chn n v gc tnh theo (D) Bm : SHIFT MODE3 ( Lu : Khng nhp a)

Tm dao ng tng hp: Nhp my : 2 SHIFT(-)45 + 1 SHIFT(-)180 = Hin th: 1 90,

e. Trc nghim vn dng :Cu 1: Cho hai dao ng iu ha cng phng, cng tn s x1= 3 cos(5t +/2) (cm) v

x2 = 3 cos( 5t + 5/6)(cm). Phng trnh dao ng tng hp lA. x = 3 cos ( 5t + /3) (cm). B. x = 3 cos ( 5t + 2/3) (cm).C. x= 3 cos ( 5t + 2 /3) (cm). D. x = 4 cos ( 5t +/3) (cm) p n B

Cu 2: Mt vt thc hin ng thi hai dao ng iu ha cng phng, cng tn s theo cc phngtrnh: x1 = 4cos(t )(cm) v x2 = 4 3 cos(t + /2) (cm). Phng trnh ca dao ng tng hp

A. x1 = 8cos(t + /3) (cm) B. x1 = 8cos(t -/6) (cm)C. x1 = 8cos(t - /3) (cm) D. x1 = 8cos(t + /6) (cm) p n A

Cu 3: Mt vt thc hin ng thi hai dao ng iu ha cng phng, cng tn s theo cc phng

trnh: x1 = acos(t + /2)(cm) v x2 = a 3 cos(t) (cm). Phng trnh ca dao ng tng hpA. x = 2acos(t + /6) (cm) B. x = 2acos(t -/6) (cm)C. x = 2acos(t - /3) (cm) D. x = 2acos(t + /3) (cm) p n A


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5.Tm dao ng thnh phn ( xc nhA2v 2 ) bng cch dng my tnh thc hin phptr:V d tm dao ng thnh phn x2: x2 =x - x1 vi: x2 = A2cos(t + 2)Xc nh A2 v 2?

a.Vi my FX570ES : Bm chn MODE 2 mn hnh xut hin ch: CMPLX-Chn n vo gc l ta bm: SHIFT MODE3 trn mn hnh hin th chD(hoc Chn n vo gc l Radian ta bm: SHIFT MODE4trn mn hnh hin th chR )

Nhp A , bm SHIFT (-) nhp ; bm - (tr), Nhp A1 , bm SHIFT (-) nhp 1 , nhn =kt qu.(Nu hin th s phc th bm SHIFT 2 3 = hin th kt qu trn mn hnh l: A22b.Vi my FX570MS : Bm chn MODE 2 mn hnh xut hin ch: CMPLXNhp A , bm SHIFT (-) nhp ;bm - (tr), Nhp A1 , bm SHIFT (-) nhp 1 nhn =Sau bm SHIFT + = hin th kt qu l:A2. bm SHIFT = hin th kt qu l: 2

c.Cc v d :V d 6: Mt cht im dao ng iu ho c phng trnh dao ng tng hp x=5 2 cos(t+5/12)(cm)vi cc dao ng thnh phn cng phng, cng tn s l x1=A1cos(t +1) v x2=5cos(t+/6)(cm),Bin v pha ban u ca dao ng 1 l:

A. 5cm; 1 = 2/3 B.10cm; 1= /2 C.52 (cm) 1 = /4 D. 5cm;1= /3Gii: Vi my FX570ES :Bm chn MODE 2 trn mn hnh xut hin ch: CMPLX

-Chn n vo gc l rad (R):SHIFT MODE 4. Tm dao ng thnh phn:

Nhp my : 5 2 SHIFT(-) (5 /12) 5 SHIFT(-) ( /6 = Hin th: 52

3

, chn A

V d 7: Mt vt ng thi tham gia 3 dao ng cng phng, cng tn s c phng trnh dao ng: x1

= 2 3 cos(2t + /3) (cm), x2 = 4cos(2t +/6) (cm) v x2 = A3cos(t +3) (cm). Phng trnh dao ngtng hp c dng x = 6cos(2t - /6) (cm). Tnh bin dao ng v pha ban u ca dao ng thnh phnth 3:

A. 8cm v - /2. B. 6cm v /3. C. 8cm v /6 . D. 8cm v /2. Chn A

Gii: Vi my FX570ES :Bm chn MODE 2 trn mn hnh xut hin ch: CMPLXChn n vo gc l rad(R) SHIFT MODE 4. Tm dao ng thnh phn th 3: x3 = x - x1 x2

Nhp my: 6 SHIFT(-) (-/6) - 2 3 SHIFT(-) (/3) - 4 SHIFT(-) (/6 = Hin th: 8- 1 2

.

d. Trc nghim vn dng :Cu 4: Mt vt ng thi tham gia 2 dao ng cng phng, cng tn s c phng trnh dao ng: x1 =8cos(2t + /2) (cm) v x2 = A2cos(t +2) (cm). Phng trnh dao ng tng hp c dng

x=8 2 cos(2t + /4) (cm). Tnh bin dao ng v pha ban u ca dao ng thnh phn th 2:A. 8cm v 0. B. 6cm v /3. C. 8cm v /6 . D. 8cm v /2.

Cu 5: Mt vt ng thi tham gia 3 dao ng cng phng, cng tn s c phng trnh dao ng: x1 =

8cos(2t + /2) (cm), x2 = 2cos(2t -/2) (cm) v x3 = A3cos(t +3) (cm). Phng trnh dao ng tnghp c dng x = 6 2 cos(2t + /4) (cm). Tnh bin dao ng v pha ban u ca dao ng thnh phnth 3:

A. 6cm v 0. B. 6cm v /3. C. 8cm v /6 . D. 8cm v /2.Cu 6: Mt vt ng thi tham gia 3 dao ng cng phng, cng tn s c phng trnh dao ng: x1 =a.cos(2t + /2) , x2 = 2a.cos(2t - /2) v x3 = A3cos(t +3). Phng trnh dao ng tng hp c dng x

= a 2 cos(2t - /4) (cm). Tnh bin dao ng v pha ban u ca dao ng thnh phn th 3:

A. a v 0. B. 2a v /3. C. a 2 v /6 . D. 2a 2 v /2.


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II. BI TON CNG IN P XOAY CHIU DNG MY TNH FX-570ES1.Cch 1:Phng php gin vc t: Dng phng php tng hp dao ng iu ho.-Ta c: u1 = U01 1os( )c t + v u2 = U01 2os( )c t +

-Th in p tng trong on mch ni tip: u = u1 +u2 = 01 021 2os( ) os( )+ + +U c t U c t

-in p tng c dng: u = U0 sin( )t +

Vi: U02 = U201+ U02

2 + 2.U02.U01. Cos( 1 2) ;01 1 02 2

01 1 02 2

sin .sin

cos cos

U Utg

U U

+=

+

V D 1: Cho mch gm: on AM cha: R, C mc ni tip vi on MB cha cun cm L,r. Tm uAB = ?Bit:

uAM = 100 2 s os(100 )3

c t

(V) 1100( ),3

AMU V

= =

uMB = 100 2 os(100 )6

c t

+ (V) ->UMB = 100(V) v 26

=

Bi gii: Dng cng thc tng hp dao ng: uAB =uAM +uMB

+ UAB =2 2100 100 2.100.100.cos( ) 100 2( )

3 6V

+ + = => U0AB = 200(V)

+

100sin( ) 100sin( )

3 6tan100cos( ) 100cos( 1

3 62)

=

+

= +

+ Vy uAB = 100 2 2 os(100 )12

c t

(V) hay uAB = 200 os(1002

)1

c t

(V)

2.Cch 2: Dng my tnh FX-570ES: uAB =uAM +uMB xc nh U0AB v . ( RT NHANH!)a.Chn ch mc nh ca my tnh: CASIO fx 570ES+ ci t ban u (Reset all), Bm SHIFT 9 3 ==+ My CASIO fx570ESbm SHIFT MODE 1 hin th 1 dng (MthIO) Mn hnh xut hin Math.+ thc hin php tnh v s phc th bm my : MODE 2 mn hnh xut hin CMPLX

+ tnh dng to cc : r (ta hiu l A) , Bm my: SHIFT MODE 3 2-Chn n vo gc l (D) ta bm my : SHIFT MODE3 mn hnh hin th chD-Chn n vo gc l Rad (R) ta bm my: SHIFT MODE4 mn hnh hin th chR

+ nhp k hiu gc ta bm: SHIFT (-).

b.V d: Cho: uAM = 100 2 s os(100 )3

c t

(V)s biu din 100 2 -600hoc 100 2 -1

3

My tnh CASIO fx 570ES : Chn MODE: Bm my: MODE 2 mn hnh xut hin chCMPLX-Chn n vo gc l (D) ta bm: SHIFT MODE3 trn mn hnh hin th chD

Nhp my: 100 2 SHIFT (-) -60 hin th: 100 2 -60-Chn n vo gc l Rad (R) ta bm: SHIFT MODE4trn mn hnh hin th chR

Nhp my: 100 2 SHIFT (-) (-:3hin th : 100 2 -1

3

-Cn chn ch mc nh theo dng to cc r (ta hiu l A )- Chuyn t dng : a + bi sang dng A , ta bm SHIFT 2 3 =(- Chuyn t dng Asang dng : a + bi , ta bm SHIFT 2 4 = )

c.Xc nh U0 v bng cch bm my tnh:+Vi my FX570ES : Bm chn MODE 2 trn mn hnh xut hin ch: CMPLX.

-Nhp U01 bm SHIFT (-) nhp 1; bm +, Nhp U02 , bm SHIFT (-) nhp 2 nhn = kt qu.(Nu hin th s phc dng: a+bi th bm SHIFT 2 3 = hin thkt qu: A

+Vi my FX570MS : Bm MODE 2 trn mn hnh xut hin ch: CMPLX.

HnhuAM

BA R L,r

uMB

MC


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Nhp U01, bm SHIFT (-) nhp 1 , bm + , Nhp U02 , bm SHIFT (-) nhp 2 nhn =Sau bm SHIFT + = hin th kt qu l: A SHIFT = hin th kt qu l:

+Lu Ch hin th kt qu trn mn hnh:Sau khi nhp, n du = hin th kt qu di dng sv t, mun kt qu di dng thp phn ta n SHIFT =( hoc dng phm SD ) chuyn i kt quHin th.

V d 1 trn :Tm uAB = ? vi: uAM = 100 2 os(100 )3

c t

(V) 0 1100 2( ), 3AMU V

= =

uMB = 100 2 os(100 )6c t + (V) -> U0MB = 100 2 (V) , 2 6

=

Gii 1:Vi my FX570ES : Bm chn MODE 2 trn mn hnh xut hin ch: CMPLXChn ch my tnh theo D():SHIFT MODE 3

Tm uAB?Nhp my:100 2 SHIFT (-) (-60) + 100 2 SHIFT (-) 30 = Hin th kt qu :

200-15 . Vy uAB = 2000os( 15 )c t (V) Hay: uAB = 200 os(100 )

12c t

(V)

Gii 2:Chn ch my tnh theo R (Radian):SHIFT MODE 4

Tm uAB? Nhp my:100 2 SHIFT (-). (- /3) + 100 2 SHIFT (-) ( /6 =Hin th kt qu:

200-/12 . Vy uAB = 200 os(100 )

12

c t

(V)

d. Nu cho u1 = U01cos(t + 1) v u = u1 + u2 = U0cos(t + ) .Tm dao ng thnh phn u2: (V d hnh minh ha bn)u2 = u - u1 .vi: u2 = U02cos(t + 2). Xc nh U02 v 2

*Vi my FX570ES : Bm chn MODE 2Nhp U0, bm SHIFT (-) nhp ; bm - (tr) , Nhp U01 , bm SHIFT (-) nhp 1 nhn = kt qu.(Nu hin th s phc th bm SHIFT 2 3 = kt qu trn mn hnh l: U022

*Vi my FX570MS : Bm chn MODE 2Nhp U0 , bm SHIFT (-) nhp bm - (tr), Nhp U01 , bm SHIFT (-) nhp 1 nhn =bm SHIFT (+) = , ta c U02 ; bm SHIFT (=) ; ta c 2

V d 2: Nu t vo hai u mt mch in cha mt in trthun v mt cun cm thun mc ni tip mt in

p xoay chiu c biu thc u = 100 2 cos(t +4 ) (V), th khi in p hai u in tr thun c biu thc

uR=100cos(t) (V). Biu thc in p gia hai u cun cm thun s l

A. uL= 100 cos(t +2

)(V). B. uL = 100 2 cos(t +

4

)(V).

C. uL = 100 cos(t +4

)(V). D. uL = 100 2 cos(t +

2

)(V).

Gii 1: Vi my FX570ES : Bm chn MODE 2 trn mn hnh xut hin ch: CMPLXChn ch my tnh theo D ():SHIFT MODE 3

Tm uL? Nhp my:100 2 SHIFT (-). (45) - 100 SHIFT (-). 0 =

Hin th kt qu :10090 . Vy uL= 100 os( )2

c t + (V) Chn A

Gii 2:Chn ch my tnh theo R (Radian):SHIFT MODE 4Tm uL? Nhp my:100 2 SHIFT (-). (/4) - 100 SHIFT (-). 0 =

Hin th kt qu: 100/2 . Vy uL= 100 os( )2

c t

+ (V) Chn A

Hnhu1

BA X Y

u2

M


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V d 3: Nu t vo hai u mt mch in cha mt in trthun v mt tin mc ni tip mt in p xoay

chiu c biu thc u = 100 2 cos(t -4

)(V), khi in p hai u in trthun c biu thc uR=100cos(t)

(V). Biu thc in p gia hai u tin s l

A. uC = 100 cos(t -2

)(V). B. uC = 100 2 cos(t +

4

)(V).

C. uC = 100 cos(t + 4

)(V). D. uC = 100 2 cos(t + 2

)(V).Gii 1: Vi my FX570ES : Bm chn MODE 2 trn mn hnh xut hin ch: CMPLX

Chn ch my tnh theo (D):SHIFT MODE 3Tm uc? Nhp my:100 2 SHIFT (-). (-45) - 100 SHIFT (-). 0 =

Hin th kt qu :100-90 . Vy uC = 100 os( )2

c t

(V) Chn A

Gii 2:Chn ch my tnh theo Radian( R):SHIFT MODE 4Tm uC ? Nhp my:100 2 SHIFT (-). (-/4) - 100 SHIFT (-). 0 =

Hin th kt qu: 100-/2 . Vy uC = 100 os( )2

c t

(V Chn A

V d 4: on mch AB c in trthun, cun dy thun cm v t in mc ni tip. M l mt im trn trndon AB vi in p uAM = 10cos100t (V) v uMB = 10 3 cos (100t -

2) (V). Tm biu thc in p uAB.?

A. u 20 2cos(100 t) (V)AB = B. ABu 10 2cos 100 t (V)3

= +

C. u 20.cos 100 t V)AB 3(

= +

D. ABu 20.cos 100 t V)3

(

=

Chn D

Gii 1: Vi my FX570ES: Bm chn MODE 2 trn mn hnh xut hin ch: CMPLXChn ch my tnh theo (D):SHIFT MODE 3

Tm uAB? Nhp my:10 SHIFT (-). 0 + 10 3 SHIFT (-). -90 =

Hin th kt qu :20-60 . Vy uAB = 20 os(100 )3

c t (V) Chn D

Gii 2:Chn ch my tnh theo Radian (R):SHIFT MODE 4

Tm uAB ? Nhp my:10 SHIFT (-). 0 + 10 3 SHIFT (-). (- /2 =

Hin th kt qu: 20-/3 . Vy uC = 20 os(100 )3

c t

(V) Chn D

e. Trc nghim vn dng :

Cu 1:t in p xoay chiu vo hai u on mch R, L thun cm , C mc ni tip th in p on mch cha

LC l 1 60cos 100 . ( )

2

u t V

= +

(A) v in p hai u R on mch l ( )2 60cos 100 . ( )u t V= . in p hai u

on mch l:

A. ( )3/.100cos260 = tu (V). B. ( )6/.100cos260 = tu (V)

C. ( )60 2 cos 100 . / 4u t = + (V). D. ( )6/.100cos260 += tu (V). Chn C

Cu 2: Cho mch in xoay chiu nh hnh v . t vo hai u A, B mt in p xoay chiu , in p tc thi gia

cc im A v M , M v B c dng : ( )AMu 15 2 cos 200 t / 3 (V)=

V ( )MBu 15 2 cos 200 t (V)= . Biu thc in p gia A v B c dng :

B

A


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C LA BR M

A. ABu 15 6 cos(200 t / 6)(V)= B. ( )ABu 15 6 cos 200 t / 6 (V)= +

C. ( )ABu 15 2 cos 200 t / 6 (V)= D. ( )ABu 15 6 cos 200 t (V)=

Cu 3: Mt on mch gm tin C c dung khng ZC = 100 v mt cun dy c cm khng ZL = 200 mcni tip nhau. in p ti hai u cun cm c biu thc uL = 100cos(100 t + /6)(V). Biu thc in p hai uon mch c dng nh th no?

A. u = 50cos(100 t - /3)(V). B. u = 50cos(100 t - 5 /6)(V).C. u = 100cos(100 t - /2)(V). D. u = 50cos(100 t + /6)(V). Chn D

Cu 4(H2009): t in p xoay chiu vo hai u on mch c R, L, C mc ni tip. Bit R = 10 , cun cm

thun c L=1/(10) (H), tin c C = (F) v in p gia hai u cun cm thun l uL= 20 cos(100t +

/2) (V). Biu thc in p gia hai u on mch lA. u = 40cos(100t + /4) (V). B. u = 40 cos(100t /4) (V).

C. u = 40 cos(100t + /4) (V). D. u = 40cos(100t /4) (V). Chn D

Cu 5: Hai u on mch CRL ni tip c mt in p xoay chiu: uAB =100 2 cos(100t)(V), in p gia hai

u MB l: uMB = 100cos(100t +4

)(V).

Biu thc ca in p gia hai u on AM l:

A. uAM = 100cos(100t +2 )V. B. uAM = 100 2 cos(100t -

2 )V.

C. uAM = 100cos(100t -4

)V D. uAM = 100 2 cos(100t -

4

)V. Chn C

Cu 6: t vo hai u vo hai u on mch R, L, C mc ni tip . Bit R = 10, cun cm thun c

HL10

1= , tin c FC

2

10 3= v in p t vo hai u cun cm thun c dng

VtuL )2100cos(220

+= . Biu thc in p hai u on mch l:

A. Vtu )4100cos(240

+= B. Vtu )4100cos(40

=

C. Vtu )4

100cos(40

+= D. Vtu )4

100cos(240

=

Chn B

Cu 7: Mt mch in xoay chiu RLC ( hnh v) c R = 100 ;

L=3

(H). in p hai u on mch AM cha R c dng:

u1 = 100 cos100 t(V). Vit biu thc tc thi in p hai u AB ca mch in.

A. 200 2 cos(100 )3

u t

= + (V) B. 200 2 cos(100 )4

u t

= (V)

C. 200cos(100 )3u t

= + (V) D. 200 2 cos(100 )4u t

= (V). Chn CCu 8 : mch in hnh v bn , khi t mt in p xoay chiu vo AB th 120 2 os(100 )AMu c t V = v

120 2 os(100 )3MB

u c t V

= + . Biu thc in p hai u AB l :

A. 120 2 os(100 )4AB

u c t V

= + . B. 240 os(100 )6AB

u c t V

= + .

C. 120 6 os(100 )6AB

u c t V

= + . D. 240 os(100 )4AB

u c t V

= + .

Hnhu1

BA R L

u2

M

M

CA B

R L,r


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III. TM BIU THC i HOC u TRONG MCH IN XOAY CHIU DNG MY FX-570ES1.Phng php gii truyn thng:Cho R , L, C ni tip. Nu cho u=U0cos(t+ u),vit i? Hoc nu cho i=I0cos(t+ i),vit u?

Bc 1: Tnh tng trZ: Tnh LZ L= .;1 1

2CZ

C fC = = v 2 2( )L CZ R Z Z = +

Bc 2: nh lut m : U v I lin h vi nhau biU

IZ

= ; Io =Z

Uo ;

Bc 3: Tnh lch pha gia u hai u mch v i: tan L CZ Z

R

= ; Suy ra

Bc 4: Vit biu thc i hoc u:a) Nu cho trc u=U0cos(t+ u) th i c dng: i =I0cos(t + u -).b) Nu cho trc i=I0cos(t + i) th u c dng: u =U0cos(t+ i +).

2.Phng php dng my tnh FX-570ES: (NHANH V HIU QU CHO TRC NGHIM)a.Tm hiu cc i lng xoay chiu dng phc: Xem bng lin h :I LNG IN CNG THC DNG S PHC TRONG MY TNH FX-570ES

Cm khng ZL ZL ZL i (Ch trc i c du cng l ZL )Dung khng ZC ZC - ZC i (Ch trc i c du tr l Zc )Tng tr:

=LZ L.;1

=CZ.C

;

( )22

L CZ R Z Z = +

( )= + L CZ R Z Z i = a + bi ( vi a=R; b = (ZL -ZC ) )

-Nu ZL >ZC : onmch c tinh cm khng-Nu ZL =u

i u i Z

Z

=> =u

Z

i

Ch : ( )= + L CZ R Z Z i ( tng trphcZ c gch trn u: R l phn thc, (ZL -ZC ) l phn o)b.Chn ci dt my tnh Fx-570ES:-Bm SHIFT 9 3 == : ci t ban u (Reset all)-Bm SHIFT MODE 1: hin th 1 dng (MthIO) Mn hnh xut hin Math.-Bm MODE 2 : Tnh ton s phc, trn mn hnh xut hin: CMPLX.

-Bm SHIFTMODE 3 2 : Ci t dng to cc:( r)-Chn n vo gc l (D) , bm : SHIFT MODE3 trn mn hnh hin th chD(-Chn n vo gc l Rad (R), bm: SHIFT MODE4trn mn hnh hin th chR )

-Bm SHIFT (-) : nhp k hiu gc ca s phc-Chuyn ta + bi sang A , bm SHIFT 2 3 =(-Chuyn t Asang a + bi , bm SHIFT 2 4 = )-Dng phm ENG nhp phn o ib.Lu Ch hin th kt qu trn mn hnh:Sau khi nhp, n du = c th hin th kt qu di dng sv t,mun kt qu di dng thp phn ta n SHIFT =( hoc dng phm SD ) chuyn i kt quHin th.

Phm ENG nhp phn o i


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c. Cc V d 1:

V d 1: Mt mch in xoay chiu RLC khng phn nhnh c R = 100 ; C= 41

10. F

; L=2

H. Cng

dng in qua mch c dng: i = 2 2 cos100 t(A). Vit biu thc in p tc thi ca hai u mch?

Gii:2

100 200LZ L.

= = = ;4

1 1

10100

CZ.C

.

= = = 100 . V ZL-ZC =100

-Vi my FX570ES : Bm chn MODE 2 trn mn hnh xut hin ch: CMPLX.-Bm SHIFTMODE 3 2 : Ci t dng to cc:( r)-Chn n vo gc l (D), bm: SHIFT MODE3 trn mn hnh hin th chD

Ta c : iZZRXIZiu CLi )((... 0 +== 2 2 0 100 100= +> X ( i ) ( PhpNHN hai s phc)Nhp my: 2 2 SHIFT (-) 0 X ( 100 + 100 ENG i ) = Hin th: 40045

Vy biu thc tc thi in p ca hai u mch: u = 400cos( 100t +/4) (V).

V d 2: Cho on mch xoay chiu c R=40 , L=

1(H), C=

6.0

10 4(F), mc ni tip in p 2 u mch

u=100 2 cos100 t (V), Cng dng in qua mch l:

A. i=2,5cos(100 t+ )( )4

A B. i=2,5cos(100 t- )( )4

A

C. i=2cos(100 t- )( )4

A

C. i=2cos(100 t+ )( )4

A

Gii:1

100 100= = = LZ L.

; 41 1

10100

0 6

= =CZ

.C.

,

= 60 . V ZL-ZC =40

-Vi my FX570ES : Bm chn MODE 2 trn mn hnh xut hin ch: CMPLX.

-Bm SHIFTMODE 3 2 : Ci t dng to cc:( r)-Chn n vo gc l (D), bm: SHIFT MODE3 trn mn hnh hin th chD

Ta c : i0

( ( )

= =

+

u

L C

Uu

R Z Z iZ

100 2 0

40 40

=

+.

( i ) ( PhpCHIAhai s phc)

Nhp 100 2 SHIFT (-) 0 : ( 40 + 40 ENG i ) = Hin th: 2,5-45Vy : Biu thc tc thi cng dng in qua mch l: i = 2,5cos(100t -/4) (A). Chn BV d 3: Mt on mch in gm in trR = 50 mc ni tip vi cun thun cm L = 0,5/ (H). t

vo hai u on mch mt in p xoay chiu u = 100 2 cos(100t- /4) (V). Biu thc ca cng dng in qua on mch l:

A. i = 2cos(100t - /2) (A). B. i = 2 2 cos(100t - /4) (A).

C. i = 2 2 cos100t (A). D. i = 2cos100t (A).

Gii:0 5100 50= = = L,

Z L.

; . V ZL-ZC =50 - 0 = 50

-Vi my FX570ES : Bm chn MODE 2 trn mn hnh xut hin ch: CMPLX.

-Bm SHIFTMODE 3 2 : Ci t dng to cc:( r)-Chn n vo gc l (D), bm: SHIFT MODE3 trn mn hnh hin th chD

Ta c : i0

( )

= =

+

u

L

Uu

R Z iZ

100 2 45

50 50

=

+.

( i ) ( PhpCHIAhai s phc)


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Nhp 100 2 SHIFT (-) - 45 : ( 50 + 50 ENG i ) = Hin th: 2- 90Vy : Biu thc tc thi cng dng in qua mch l: i = 2cos( 100t - /2) (A). Chn BV d 4 (H 2009): Khi t hiu in th khng i 30V vo hai u on mch gm in trthun mcni tip vi cun cm thun c t cm L = 1/4 (H) th cng dng in 1 chiu trong mch l 1A.

Nu t vo hai u on mch ny in p u =150 2 cos120t(V) th biu thc cng dng introng mch l:

A. 5 2cos(120 )( )4= i t A

B. 5cos(120 )( )4= +i t A

C. 5 2cos(120 )( )4= +i t A

D. 5cos(120 )( )4= i t A

Gii: Khi t hiu in th khng i (hiu in th 1 chiu) th on mch ch cn c R: R = U/I =30

1120 30

4= = =

LZ L.

; i =

u 150 2 0

(30 30i)Z

=

+ ( PhpCHIAhai s phc)

-Vi my FX570ES : -Bm chn MODE 2 trn mn hnh xut hin ch: CMPLX.

-Bm SHIFTMODE 3 2 : Ci t dng to cc:( r)-Chn n vo gc l (D), bm: SHIFT MODE3 trn mn hnh hin thD

Nhp my: 150 2 : ( 30 + 30 ENG i ) = Hin th: 5- 45Vy: Biu thc tc thi cng dng in qua mch l:: i = 5cos( 100t - /4) (A). Chn D

3. Trc nghim vn dng:

Cu 1: Cho mch in xoay chiu c R=30 , L=

1(H), C=

7.0

10 4(F); hiu in th hai u mch l

u=120 2 cos100 t (V), th cng dng in trong mch l

A. 4cos(100 )( )4

i t A

= + B. 4cos(100 )( )4

i t A

=

C. 2cos(100 )( )4

i t A

= D. 2cos(100 )( )4

i t A

= +

Cu 2: Hy xc nh p n ng .Mt on mch in xoay chiu gm: R = 100 ;cun dy thun cm

L =

1

H; t din c in dung 15,9F , mc vo in p xoay chiu u = 200 2 cos(100t)(V) .Biu thccng dng in l:

A. i = 2cos(100t -4

)(A). B. i = 0,5 2 cos(100t +

4

)(A) .

C. i = 2 cos (100t +4

)(A). D. i =

3

2

5

1cos (100t +

4

)(A) .

Cu 3: Cho on mach xoay chiu gm R, L mc ni tip.0.2

20 ,R L H

= = . on mch c mc vo

in p 40 2 cos100 ( )u t V= . Biu thc cng dng in qua mch l:

A. 2cos(100 )( )4

i t A= B. 2cos(100 )( )4

i t A= +

C. 2 cos(100 )( )2

i t A

= D. 2 cos(100 )( )2

i t A

= +

Cu 4: Mt mch gm cun dy thun cm c cm khng bng 10 mc ni tip vi tin c in

dung 42

.10C F

= . Dng in qua mch c biu thc 2 2 cos(100 )3

i t A

= + . Biu thc in p ca hai

u on mch l:


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A. 80 2 cos(100 )6

u t

= (V) B. 80 2 cos(100 )6

u t

= + (V)

C. 120 2 cos(100 )6

u t

= (V) D.2

80 2 cos(100 )3

u t

= + (V)

IV. XC NH HP EN TRONG MCH IN XOAY CHIU DNG MY TNH FX-570ES( RT NHANH V HIU QU CHO TRC NGHIM)

1.Chn ci dt my tnh Fx-570ES:-Bm SHIFT 9 3 == : ci t ban u (Reset all)-Bm SHIFT MODE 1: hin th 1 dng (MthIO) Mn hnh xut hin Math.-Bm MODE 2 : Tnh ton s phc, trn mn hnh xut hin ch: CMPLX.

(-Bm SHIFTMODE 3 2 : Ci t dng to cc: r )-Bm SHIFT MODE 3 1 : Ci t dng to cc: (a + bi).-Chn n vo gc l (D) , bm my : SHIFT MODE3 trn mn hnh hin th chD-Chn n vo gc l Rad (R), bm my: SHIFTMODE4trn mn hnh hin th chR-Bm SHIFT (-) : nhp k hiu gc ca s phc-Chuyn t dng a + bisang dng A , bm SHIFT 2 3 =-Chuyn t dng Asang dng a + bi, bm SHIFT 2 4 =

2.Xc nh cc thng s( Z, R, ZL, ZC) bng my tnh:

-Tnh Z: =u

Zi

0

0( )

=

u

i

U

I

Nhp my: U0 SHIFT (-) u : ( I0 SHIFT (-) i ) = ( PhpCHIAhai s phc)

-Vi tng trphc : ( )= + L CZ R Z Z i , ngha l c dng (a + bi). vi a=R; b = (ZL -ZC )-Chuyn t dng Asang dng : a + bi: bm SHIFT 2 4 =3.Cc V d:V d 1:Mt hp kn (en) ch cha hai trong ba phn t R, L, C mc ni tip. Nu t vo hai u mch

mt in p xoay chiu u= 100 2 cos(100t+4 )(V) th cng dng in qua hp en l

i= 2cos(100t)(A) . on mch cha nhng phn t no? Gii tr ca cc i lng ?Gii: +Vi my FX570ES: Bm chn MODE 2 trn mn hnh xut hin ch: CMPLX.-Chn n vo gc l (D), bm : SHIFT MODE3 trn mn hnh hin th chD

-Bm SHIFT MODE 3 1 : Ci t dng to cc: (a + bi).100 2 45

(2 0)

= =

uZ

iNhp: 100 2 SHIFT (-) 45 : ( 2 SHIFT (-) 0 ) = Hin th: 50+50i

M ( )= + L CZ R Z Z i .Suy ra: R =50; ZL= 50 . Vy hp kn (en) cha hai phn t R, L.

V d 2:Mt hp kn (en) ch cha hai trong ba phn t R, L, C mc ni tip. Nu t vo hai u mchmt in p xoay chiu u= 200 2 cos(100t-

4

)(V) th cng dng in qua hp en l

i= 2cos(100t)(A) . on mch cha nhng phn t no? Gii tr ca cc i lng ?Gii: +Vi my FX570ES : Bm chn MODE 2 trn mn hnh xut hin ch: CMPLX.-Chn n vo gc l (D), bm : SHIFT MODE3 trn mn hnh hin th chD

-Bm SHIFT MODE 3 1 : Ci t dng to cc: (a + bi).


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200 2 45

(2 0)

= =

uZ

i: Nhp 200 2 SHIFT (-) -45 : ( 2 SHIFT (-) 0 ) = Hin th: 100-100i

M ( )= + L CZ R Z Z i . Suy ra: R =100; ZC = 100 . Vy hp kn (en) cha hai phn t R, C.

V d 3:Mt hp kn (en) ch cha hai trong ba phn t R, L, C mc ni tip. Nu t vo hai u mch

mt in p xoay chiu u= 20 6 cos(100t-

3


i= 2 2 cos(100t)(A) . on mch cha nhng phn t no? Gii tr ca cc i lng ?Gii: +Vi my FX570ES: Bm chn MODE 2 trn mn hnh xut hin ch: CMPLX.

-Chn n vo gc l (D), bm : SHIFT MODE3 trn mn hnh hin th chD


(2 2 0)

= =

uZ

i: Nhp 20 6 SHIFT (-) -60 : ( 2 2 SHIFT (-) 0 ) = Hin th: 5 3 -15i

M ( )= + L CZ R Z Z i .Suy ra: R =5 3 ; ZC = 15 . Vy hp kn (en) cha hai phn t R, C.


mt in p xoay chiu u= 200 6 cos(100t+ 6 )(V) th cng dng in qua hp en l

i= 2 2 cos(100t-6

)(A) . on mch cha nhng phn t no? Gii tr ca cc i lng ?

Gii: +Vi my FX570ES : Bm chn MODE 2 trn mn hnh xut hin ch: CMPLX.-Chn n vo gc l (D), bm : SHIFT MODE3 trn mn hnh hin th chD


(2 2 30)

= =

uZ

i: Nhp 200 6 SHIFT (-) 30 : ( 2 2 SHIFT (-) (-30) =

Hin th: 86,6+150i =50 3 +150i .Suy ra: R =50 3 ; ZL= 150. Vy hp kn cha hai phn t R, L.


mt in p xoay chiu u= 200 2 cos(100t+4


i= 2cos(100t)(A) . on mch cha nhng phn t no? Gii tr ca cc i lng ?Gii: +Vi my FX570ES : Bm chn MODE 2 trn mn hnh xut hin ch: CMPLX.

-Chn n vo gc l (D), bm : SHIFT MODE3 trn mn hnh hin th chD


(2 0)

= =

uZ

i: Nhp 200 2 SHIFT (-) 45 : ( 2 SHIFT (-) 0 =

Hin th: 141.42...45 .bm SHIFT 2 4 = Hin th: 100+100i Hay: R =100; ZL= 100.Vy hp kn cha hai phn t R, L.4.Trc nghim:Cu 1: Cho on mch gm hai phn t X, Y mc ni tip. Trong X, Y c th l R, L hoc C. Cho bit hiu in

th gia hai u on mch l u = 200 2 cos100 t(V) v i = 2 2 cos(100 t - /6)(A). Cho bit X, Y l nhngphn t no v tnh gi tr ca cc phn t?

A. R = 50 v L = 1/ H. B. R = 50 v C = 100/ F.

C. R = 50 3 v L = 1/2 H. D. R = 50 3 v L = 1/ H.


16/16

GV:on Vn Lng -T: 0915718188 - 0906848238 16

Cu 2: Cun dy thun cm c h s t cm L = 636mH mc ni tip vi on mch X, on mch X cha 2 trong 3

phn t R0, L0 , C0 mc ni tip. t vo hai u on mch hiu in th u = 120 2 cos100 t(V) th cng

dng in qua cun dyl i = 0,6 2 cos(100 t - /6)(A). Xc nh 2 trong 3 phn t?A. R0 = 173 v L0 = 31,8mH. B. R0 = 173 v C0 = 31,8mF.C. R0 = 17,3 v C0 = 31,8mF. D. R0 = 173 v C0 = 31,8 F.

Cu 3: Cho on mch in xoay chiu gm 2 phn t mc ni tip. Hiu in th gia hai u on mch v cng

dng in trong mch c biu thc: u = 100 2 cos(100 t - /2)(V) v i = 10 2 cos(100 t - /4)(A).

A. Hai phn t l R v L. B. Hai phn t l R v C.C. Hai phn t L v C. D. Tng trca mch l 10 2

Cu 4: Cho mt on mch xoay chiu gm hai phn t mc ni tip. in p gia hai u mch v cng dng in qua mch c biu thc: u = 200cos(100t- /2)(V), i = 5cos(100t - /3)(A). Chn p nng?

A. on mch c 2 phn t RL, tng tr40 . B. on mch c 2 phn t LC, tng tr40 .

C. on mch c 2 phn t RC, tng tr40 . D. on mch c 2 phn t RL, tng tr20 2 .Cu 5: Cho mt hp en X trong c cha 2 trong 3 phn t R, L, hoc C mc ni tp. Mc hp en ni tip vimt cun dy thun cm c L0 = 318mH. t vo hai u on mch in mt hiu in th xoay chiu c biu thc

u = 200 2 .cos(100 t- /3)(V) th dng in chy trong mch c biu thc i = 4 2 .cos(100 t - /3)(A). Xc

nh phn t trong hp X v tnh gi tr ca cc phn t?A. R = 50 ; C= 31,8 F. B. R = 100 ; L= 31,8mH.

C. R = 50 ; L= 3,18 H. D. R = 50 ; C= 318 F.

Cu 6: Mt on mch xoay chiu gm 2 trong 3 phn t R, L hoc C mc ni tip . Biu thc hiu in th 2 u

mch v cng dng in qua mch l 80cos 100 ( )2

u t V

= +

v 8cos(100 )( )4

i t A

= + . Cc phn t

trong mch v tng trca mch lA. R v L , Z = 10 . B. R v L , Z = 15 . C. R v C , Z =10 . D. L v C , Z= 20 .

Cu 7: Mch in ni tip R, L, C trong cun dy thun cm (ZL < ZC). t vo hai u on

mch mt in p xoay chiu 200 2 cos(100t+ /4)(V). Khi R = 50 cng sut mch t gitr cc i. Biu thc dng in qua mch lc :

A. i = 4cos(100t+ /2) (A) B. i = 4cos(100t+/4) (A)

C. i = 4 2 cos(100t + /4)(A) D. i =42 cos(100t) (A)Gi : Khi R = 50 cng sut mch t gi tr cc i. suy ra R=/ZL-ZC/ = 50 .

Mt khc ZC > ZL nn trong s phc ta c: ZL + ZC = -50i. Suy ra:u 200 2 ( : 4)

i 450 50i 2Z

= = =

Chn A

V. KH NNG VN DNG:-Dng my tnh CASIO fx-570ES hoc CASIO fx570ES Plus nhm rn luyn cho HC SINHthao tc nhanh, chnh xc v rt hiu qumt s bi tp TRC NGHIM VT L 12.

Cc em HS thdng CNG CMY TNH CASIO fx570ES & Fx- 570ES Plus !!!GII NHANH mt s bi tp TRC NGHIM VT L 12.Nguyn tc thnh cng: Suy nghtch c c; Cm nhn am m; Hot ng kin tr !Chc cc em HC SINH thnh cng trong hc tp!

Ngi su tm v chnh l: GV: on Vn LngEmail: [email protected] ;[email protected]; [email protected];in Thoi: 0915718188 0906848238

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Giai Nhanh Vat Li 12 Bang FX570ES