V (phdn tich va huang dan gidi 2nv 271.50 = 2cm Buac song:'k= — = 507t f O Vi M va O n a m trenCd u o n g trung true ciia hai nguon nen P h u o n g trinh song tai mpt M la: u ^ = 2acos 507it- 2Kd^ P h u o n g trinh song tai mot O la: 27i.OA~ U Q = 2a cos 507tt - = 2acos(507rt-87:) M va O d a o dong nguoc pha khi: = > A ( P M / O = i - ^ = (2k + l ) t ^ d = 3,5>.-k)L = - k > A O = = ^ k < - , Vgy: d„i„ « k ^ , = - =^ d^in = ^ OM^i„ = ^ d ^ i ^ - O A ^ = 7T7cm C h p n d a p an A C a u : Hai nguon ke't h o p Si,S2 each mpt khoang 50mm tren m a t nuoc phat hai song ke't h o p c6 p h u o n g trinhuj = U = 2cos2007it(mm) Van toe truyen song tren mat nuoc la 0,8 m/s Diem gan nhat d a o dpng ciing pha voi nguon tren d u o n g trung true eua S1S2 each nguon Si bao nhieu: D 24mm A 16nim B 32mm C m m ^hdn tich v>d huang dan gidi Giai 1: P h u o n g trinh song tong quat tong h p p tai M la: U M = 2a cos r d2-di cos 2007lt - 7t Voi M each deu Si, S2 nen di = d2 Khi d o d2 - di = => A = 2a cos De M d a o dpng cung pha voi Si, S2 thi: dij^ d i + d ^ ^ k = ^ d i = d = d = kX Goi X la khoang each t u M den Si va S2: d , = d , = d = Jx'2 + rsiS2 ^ = > , k ' ' - , > = > k > 3,125 => = = > d = 4A = 32 mm ChQH d a p an B Giai n h a n h bai n h u sau: Diem M d a o dong ciing pha voi nguon Si nen phai each Si m p t doan d = kX Nhin tu hinh ve ta tha'y: d = k^>S,0^8k>25:^k>3,125:^k„i„ = => d^,„ = 4X = 4.8 = m m Chgn B ;au : Tren mat mot cha't long, c6 hai nguon song ke't h p p Oi, O2 each / = 24em, d a o dong theo ciing mpt p h u o n g voi p h u o n g trinh "oi = " o = Aeoswt (t tinh bang s A tinh bang mm) Khoang each ngSn nhat tu trung diem O eua O1O2 den cac diem n a m tren d u o n g trung true eua O1O2 d a o dong ciing pha voi O b3ng q = 9cm So diem d a o dong voi bien dp bang O tren doan O1O2 la: A 18 B 16 C 20 D 14 ^hdn tich m huang dan gidi P h u o n g trinh d a o dong tai mpt diem c6 giao thoa: u = A cos d, - d , cos (Ot - 71 P h u o n g trinh d a o dong tai O: u = 2Acos cot 27ta (voi / = 2a) 1° J," P h u o n g trinh d a o dong tai M: u = 2Acos cot 27rd^ Dp l^ch pha ciia M so voi O: Acp = — ( d - a ) ^' X 2?^ M d a o dong ciing pha vai O nen: Acp = — ( d - a) = 2k7i => d - a = k> X Diem M gan O nhat thi: k = =>> = d - a = 7a^+q^ - a = 712^+9^ So cue dai tren O1O2: -12 = 3cm - | < k < i = > - < k < : k = ±8;+7;±6;±5;±4;±3;+2;±l;0 Co p17 cuepdai tren OiCh (ke ea O), vay c616 diem dao dpng voi bien d p hang O C h n d a an B C a u 7: H a i n g u o n song ket hop, dat tai A va B each 20cm dao d g n g theo di2 = (8 + x)2 + 82; d22 = (8 - xf- + 8^ p h u o n g t r i n h u = acos(cjt) tren mat nuoc, coi bien k h o n g d o i , buoc song => di2 - d22 = 32x => d i + d2 = 16x X = 3cm Gpi O la t r u n g d i e m ciia A B M o t d i e m n a m tren d u o n g t r u n g true T u (*) va (**) =>di = x + l A B , dao d p n g c i m g pha v o i cac n g u o n A va B, each A hoac B m p t do^n nho di2 = (8 + x)2 + 82 = (8x + 1)2 => 63x2 = 128 : nhat la B.lOem C.13.5cm 10 cm Bieu thiie song tai M : U ; ^ = 2acos c o t - k^in = 27id ket hop v o i hai n g u o n eung pha nen ON = a = - = - = lem 4 c x y ? /' M / / ( X' a Theo bai thi khoang each giua xx' v o i A B la y = 8cm C a u : T r o n g thf n g h i ^ m giao thoa tren mat chat long v o i n g u o n A , B phat song ket h o p nguoe pha Khoang each g i i i a n g u o n la A B = 16cm H a i song t r u y e n d i c6 buoc song la 4em Tren d u o n g thang xx' song song v o i A B , each A B m o t doan 8cm, goi C la giao d i e m cua xx' v o i d u o n g t r u n g true eiia A B K h o a n g each ngan nhat t u C den d i e m dao d p n g v o i bien d p cue dai n a m tren xx' la B l,50em ^hdn C 2,15em tich v>d hucmg dan = o x = l,42cm K h i : d i - d2 = (k + 0,5) X D i e m M gan C nhat k h i k = (*) B 30cm ^han C 40em tich vd hu&ng ddn de doan A M eo gia trj Ion nhat t h i C X M x' M phai nam tren van eye dai bae n h u h i n h ve va thoa m a n : d - d , = k ? t = 1.20 = 20(cm) (1) di D i e m M dao d p n g v o i bien d p eye tieu d p n g eung pha Bie't song m o i nguon phat eo tan so'f = lO(Hz), v a n toe Do M la m o t eye dai giao thoa nen Cach = IH C a u : Tren be mat chat l o n g c6 hai nguon ket hop A B each 40cm dao T a c o ; = ^ = ^ = 20(em) f 10 ^ ' k + - X o j + ( + x)^-^8^+{8-xf 2) M a t khae, tarn giae A M B la tam A I H = l J l + = l,42em 63 C h p n dap an A A 20em tren xx' t h i M thupe cue tieu t h u nhat k = M = d2; X = C M • = l = > x = a, l + ^ a tai A dao d o n g v o i bien dp cue dai Doan A M c6 gia t r i Ion nhat la: gidi K h o a n g each ngan nhat t u C den d i e m dao d p n g v a i bien dp eye tieu nSm Xet d i e m M : Vi the'tu phuong trinh: truyen song 2(m/s) Gpi M la m o t d i e m n a m tren d u o n g v u o n g goc v o i A B D 2,25cm Cach : G o i M la d i e m thoa man yeu cau va dat C M = x, di - d2 = 0,5> = (cm) X T a c o : b^ =e^ - a ^ - ^ - ^ =63 C h p n dap an A = di; B ,'' k = v o i e la tieu diem va e = OB = O A = — = — = 8em 2 k>3,33 d^i„ =4X = 4.3 = 12cm A l,42em voi d u o n g cue tieu gan t r u n g true nha't D i e m M dao d p n g c i i n g pha v o i n g u o n k h i = k T i = > d = kX = k > ^ =1 a Xet d i e m M tren t r u n g true ciia A B : A M X Chpn D A.12cm — (**) B giae v u o n g tai A nen ta c6: A M = d2 = V ( A B ) + ( A M ) =740^+d,2 (2) D.SOem gidi (2) v a o (1) t a d u o c : Thay H a i n g u o n d a o d o n g k h a c b i e n d o v i t h e t a ti'nh b i e n d o t h e o c o n g t h u c : + d ,^ - d ] = : • d i = ( c m ) SJAO^ Cau 10: T r e n b e m a t c h a t l o n g c h a i n g u o n k e t hgrp A B each n h a u 0 c m d a o d o n g c u n g p h a B i e t s o n g d o m o i n g u o n p h a t r a c t a n so f = 10(H2), v a n toe t r u y e n s o n g 2(m/s) G o i M la m o t d i e m B 2 , c m C 12cm D 30cm In (0)n = 5^ + ^ + C O S (d2 - d , ) - ( ( p -cpi) 4; U = 34 + I5V2 A = V34 + I5V2 c m Chpn dap an B ^hdn tich vd hu&ng dan gidi Taco = 5^ + 3^ + C O S nam t r e n d u o n g v u o n g g o c v o i A B t a i A d a o d o n g v o i b i e n d p c u e d a i D o a n A M c g i a t r j n h o n h a t la: A 5,28cm 2n A ^ = A t + A2 + A , A C o s C h p n dap an B Cau 12: T r e n m a t m o t c h a t l o n g c h a i n g u o n s o n g ke't h o p c i i n g p h a c bien d o m m va m m dao d o n g v u o n g goc v o i mat t h o a n g chat long N e u ;i = ^ = ^ = 20(cm) f 10 cho rSng song t r u y e n d i v o i bien d o k h o n g thay d o i t h i tai m o t d i e m M So v a n c u e d a i t r e n d o a n A B t h o a m a n d i e u k i e n : each h a i n g u o n n h u n g k h o a n g d i = 12,75>i v a d = 7,25X se c b i e n d p d a o - A B < d - d , = kA.< A B d o n g la b a o n h i e u ? Hay : -AB X , AB -100 , 100 _ d hucmg ddn gidi Suy : k = , ± , ± , ± , ± Do h a i n g u o n eo b i e n d o k h a c n h a u n e n ta t i n h b i e n d o s o n g t o n g h p p t a i M Vay theo cong t h u e t o n g h o p dao d o n g d i e u hoa: de doan A M c6 gia t r i be n h a t th'i M p h a i n a m t r e n d u o n g A ^ = A f + A ^ +2A,A2Cos eye d a i bac ( k = ) n h u h i n h v e va thoa m a n : = 7^ +8^ d2 - d j = k ? t = - ( c m ) +2.7.8.COS 2n 2n, ^(d2-d,)-({p2-(pi) {U,75X-7,25X)-0 A M a t khac, d o t a m giac A M B la = 7^ + 8^ + 2.7.8.cosn7: = t a r n g i a c v u o n g t a i A n e n t a eo: => A = l c m M B - d -V(AB2) + (AM2)=7lOO^+di2(2) Chpn dap an B T h a y (2) v a o (1) ta d u o c : 7lOO^ + d i ^ - d , = ^ d i =22,5(cm) Chpn dap an B Cau 11: T a i h a i d i e m A , B tren m a t chat l o n g c6 h a i n g u o n phat song: Uys^ = c o s ( ( o t ) c m ; U g = c o s a)t + — c m C o i b i e n d p s o n g k h o n g d o i k h i 4J t r u y e n d i B i e n d p song t o n g h o p tai t r u n g d i e m ciia d o a n A B A A = 8cm B A = N/34 + 15N/2 c m C A = 15cm D 2cm k = 17 -12 +1 = _ V a y CO gia tri cua tan so cho song dung tren day Chpn dap an D Vl du 5: Trong thi nghi^m ve song dimg, tren mpt spi day dan hoi dai 1,5 m voi hai dau co dinh, nguoi ta quan sat thay ngoai hai dau day CO djnh co n a m diem khac tren day khong dao dpng Biet khoang thoi gian giua ba Ian lien tiep voi spi day duoi thMng la 0,1s V?n toe truyen song tren day la _ A 5m/s B 4m/s C lOm/s D m/s Biquyet on luyjn thi dai hoc dat diem td'i da Vat It, tap - L c Van Vitih p B A I TAP VAN DgNG: C a u 1: M o t sgi day A B = 50cm tree l o l u n g dau A co d j n h , dau B dao d p n g 'Phdn tich vd hiiong ddn gidi voi tan so 50Hz thi tren day co 12 bo scSng nguyen K h i d i e m N each A m o t doan 20em la b u n g hay n u t song t h u may ke t u A A b u n g song t h u C b y n g song t h u = k - + - = k + l- U CO tat ca n u t song => c6 bung song => c6 bo song =i> k = = ^ => v = = - ^ ^ ^ = m / s f 3 — 2013) B la d i e m b u n g thi d i e m nut ^kdn tich vd himng dan gidi A (sai) v i d i e m tren day each dau A mot doan bSng nua buoc song k h o n g phai la b u n g song ma la nut song B (sai) v i d i e m tren day each dau A m p t doan bSng m o t phan t u buoc song k h o n g phai la n u t song ma la b u n g song C (sai) V I d i e m tren day each dau B m p t doan b3ng ba phan t u buoc song k h o n g phai la b u n g song ma la n i i t song D ( d u n g ) V ! d i e m tren day each dau B m o t doan bSng mot phan t u buoc song la d i e m nut H i n h ve tren se cho cai n h i n cu the hon Niit ' ,; , 12 + =Y = - => A N = - = k - => k = => tai N la n u t t h u , C a u 2: M o t spi day A B dai 50em, van toe t r u y e n song tren day la 5m/s, dau A dao d o n g v o i tan so 80Hz Tren day co song d u n g hay khong? so'byng song la : B diem tren day each dau A mot doan bSng mot phan t u buoc song la diem nut D d i e m tren day each dau B m p t doan bang m o t phan t u b u o c song la - Chpn dap an B A diem tren day each dau A mot doan b^ng nua buoc song la diem bung C diem tren day each dau B mot doan bang ba phan t u buoc song la diem bung ^," De bie't d u p e N la n i i t song hay b u n g song t h u may ta xet t i so: C h p n dap an A K h i CO song d u n g tren spi day dan hoi AB v o i dau A la d i e m n u t va dau r Chieu dai sgi day mot dau t u do, mot dau co d j n h co song d u n g thoa man: Ngoai hai dau day co d j n h c6 nam diem khac tren day k h o n g dao V l d u 6: ( T r i c h de t h i t h u chuyen Ha TTnh Ian nam tich vd hu&ng ddn gidi f ; Tren day CO 12 bo song nguyen nen k = 12 At = T = , ] s = > f = 10Hz Chieu dai soi day thoa man: = k - = - = ^ 2 D n u t song t h u ^hdn Khoang t h a i gian giua ba Ian lien tiep soi day d u o i thang la m o t chu ky: dong B nut song t h u A Co, c616 b u n g song f B Co, co 21 b u n g song C K h o n g D Co, CO 15 b u n g song ^hdn tich vd huang ddn gidi Bai toan chua cho bie't song d i r n g tao t r u o n g h o p nao nen ta xet ca hai t r u o n g h p p : i » • T r u o n g h p p 1: H a i dau day e o d m h Chieu dai day thoa man: l = k - = k — ^ k = — = ^-^'^-^^ = e Z 2f V Suy co song d u n g xay u n g v o i 16 b u n g song D o t r u o n g h p p thoa m a n nen ta k h o n g xet t r u o n g h p p nira T r u o n g hpp la song d u n g tao tren day m o t dau co d i n h , m p t dau t u d o C h p n dap an A C a u : M o t spi day dan hoi dai 130cm, co dau A co' d j n h , dau B t u dao d o n g v o i tan 50 H z , van toe truyen song tren day la 20 m/s Tren day co bao nhieu n i i t va b u n g song: A CO n u t song va b u n g song B co n i i t song va b u n g song C CO n u t song va b u n g song D co n u t song va b u n g song ^hdn tich vd huang ddn gidi Nhan xet: song d i r n g tao b o i m p t dau t u do, m p t d a u co d i n h 243 => so niit = so bung = so bo + V i the loai dap so B va D (do bung, nut khong bang nhau) Chieu dai soi day thoa man: 1^ 21f 2.1,3.50 k + - -=> k= = k—+ — = k + 2f V 2 20 So' niit = so'bung = so bo + = + = Cau : Soi day AB = 35cm voi dau B tu Tao tai A mot dao dong ngang CO tan so f Van toe truyen song la 3m/s, muon c6 10 bung song thi tan so dao dgng phai la bao nhieu ? B.41,4Hz C 40,7Hz D 74,lHz phai c6 bo song => k = f= k + 21 ^ 1^ 9+ - 2.0,35 = 40,7Hz D^ng 2: Xac d j n h dieu k i f n de c6 song duTng t r e n d a y Vl d u : Soi day dan hoi AB, dau A gan voi can rung c6 tan so f, dau B dugc giu CO dinh f i va ii la hai tan so lien tiep de tao song dung tren soi day Tim tan so nho nhat de tao duoc song dimg tren sgi day B f =^ = i ± i 21 'min ijA 'min 21 D- mi =^ ^hdn tick m hu&ngf danngidi Z , V fi V / = k — = 2f (k + l ) - k V = f2-fl V f m i n — T T — f9 —i^ 21 Vl dM 2: Sgi day dan hoi AB, dau A gan vai can rung c6 tan so f, dau B t u fi va h la hai tan so lien tiep de tao song dirng tren sgi day Tim tan so nho nha't de tao dugc song dung tren sgi day ,j j • , A W = — = f2 - fi B f =^ = i ± i c =^ = i z i D-fmin=^ = f2-fl f 41 Zt Dieu kien de c6 song dung: (1) ^ ^ Hai tan so fi, k la hai tan so lien tiep de tao song dimg ung voi so bo song Ian lugt la k va k + , Ta CO : — 41 = — - — = 2k + l 2(k + l ) + l A p dung tinh chat: — = — = b d b-d 4/ 2k + l ^2 2k + _ f - f i f,-fi _ (2k + ) - k - l Tan so nho nha't de tao song dirng ung vai k = = f2-fl Z/ - I/ 'Phdn tick va hu&ng dan gidi _ (1) K Hai tan so fi, h la hai tan so lien tiep de tao song dirng ung voi so bo ^ k+ f2-fl Tan so'nho nha't de tao song dung ling voi k = Dieu k i f n de c6 song dung: AB = / = k.-^ = k.-;^ ^ ^ = song Ian lugt la k va k + k _ AB = / = (2k + l ) - = (2k + l ) — — = — ^ ^ ^4 ^ ^ 4f 41 2k + l ChQH dap an A C f = X= 21 h Chpn dap an A Chpn dap an C A 78,4Hz V _ f l _ / = ( k + ) — = (2.0 + 1) ^ Uf ^ V f - V - ^ - f l ^ 4f i n m Chpn dap an C Cach tinh sau van cho dap an tucmg tu: Chieu dai sgi day mot dau tu do, mot dau codinh c6 song dimg thoa man: f2 A p dyng tinh chat: b a -c —= a b-d l=k—+—= k + 2, V Dat: fk = k + 2I k+2 V k+ 21 (1) 94=; f fk+l - I f ^ V J 2T" k + + l kH 1^ Dau t u d o (dau h o cua ong) la b u n g song V T u (1) va (2) ta c6: k+]-f\,= — =k—+—= Tan so n h o nhat de c6 song d u n g tren day u n g v o l k = f = k l _ V _ fk+uJi— Suy dap an van la C C f = 2) Bung => Chieu d a i o n g day thoa man: ' 2y 21 "^21 l A 1^ v k + 2y ~ l 2? k + - v 21" Tan so'nho nhat de co song d u n g o n g u n g v o i k = 340 V l d u : M o t soi day d a n h o i d u o c treo thSng d u n g vao m o t d i e m co d i n h , Al dau lai tha t u d o N g u o i ta tao song d u n g tren day v o i tan so nho 4.0,85 = 100Hz Chpn dap an D Nut nhat la ^ Phai tang tan so them m o t l u g n g nho nhat la bao nhieu de lai CO song d u n g tren day? A Af = ^ ong mot dau kin, mot dau giong truang hap song dimg tren sai day B Af = ^ 3^ C Af = , T»}/ nhien, ncu dcy chung ta sc thai/ dugc trifdng hap giao thoa cua song dimg tao j mot dau CO djnh, mot dau tie VI thechi can dp dung cong thitc tinh tan sonho D.Af = 4^ nhat cua trmrng hap Id co dap dn nhanh nhat ^hdn tick pd hiccmg dan gidi v i de cho van toe truyen song va chieu dai day nen ta co Chieu dai soi day m o t dau t u do, mot dau co dinh k h i co song d u n g thoa man: k.11 l = k - + - = V k + - 3; 2i 2) f = 'min 2) 2\ Tan so n h o nhat de co song d u n g u n g v o i k = 0: V 21 41 m (1) 2) 21 3v B A I T A P VAN DUNG: T~ = r^- Vay phai tang t h e m Af = 2^, t h i tren day lai c6 song d i m g ^ V i d u : N g u o i ta tao song d u n g t r o n g m o t cai o n g m o t d a u k i n m o t dau ho d a i 0,85m chua day k h o n g k h i dieu k i ^ n t h u o n g (van toe a m la 340m/s) H o i tan so n h o nhat de co song d u n g t r o n g o n g la bao nhieu? D f = 100Hz _ ^hdn tick m hu&ng d&n gidi T r u o n g h o p giong t r u o n g h o p song d u n g tao b o i sgi day m o t dau co d i n h , m o t d a u t u D a u CO d j n h (dau k i n cua ong) la n u t D 60 m/s 2f f V K h i f, va (2 la hai tan so lien tiep fj < £3 t h i k j va k j la so n g u y e n lien tiep: k = k| + C h p n dap an B Cf-200Hz C 32 m/s D i e u kien de co song d u n g tren day v o i hai dau day la m i t song (hai d a u co k 21 djnh): l = k - = k — — = — = const 41 B f = 75Hz B m / s ^hdn tick m hu&ng dan gidi _ 41 _ = ^ = 3=>f, = ^ = ^ A f = f i - ^ - ^ A f - 50 H z 4.0.H5 A m / s 3v ( l ) v a (2)=> f : 60 H z Xac d j n h toe truyen song tren day biet hai dau day deu la n u t song? (2) "IT 41 = 100Hz C a u 1: M p t soi day dai = 1,2 m co song d u n g v o i tan so lien tiep la 40 H z va Tan so tiep d o de sg'i day lai co song d u n g u n g v o i k = V _ 340 —= N h u vay ta co: — = — ki k2 _ f2 ki+1 kj , , k 21 Van toe t r u y e n song tren day: - = — ^ f V 40 60 k, ki+1 21fi 2.1,2.40 , v= — - = — = 48 m/s kj C h p n dap an A Tuy nhien, ne'u ndm dugc cong thitc a vi du thi ket qua thu diegc se nhanh han Tan so nho nha't de co song d u n g tren day (hai dau co d j n h ) : ^min=Y^=h-h-^^ = 2/(f2 - f i ) = 2.1,2(60 - 40) = m / s 247 Cty TNHH MTV DWH ciia d p n g co (ti so g i u a cong suat h u u i c h va cong suat tieu t h u toan Pco + PR = lUcoscp Khang Vi$t 20P - 200.0,891 + 320 = 20P - 1781 + 320 = => lOP - 891 + 160 = phan) la A 80% B 90% C 92,5% D 87,5 % Giai p h u o n g t r i n h tren ta co hai nghi^m: I i = 6,4 ( A ) va h = 2,5 ( A ) LU'U y : a vi du ta chpn nghiem ma khong can luang lu gi dung khong cdc 'Phan tich v>d huang dan gidi ban! Tat nhien la chiing ta phdi lam nhu the, vi phuang trinh bac hai theo I chi co C o n g suat tieu t h u toan phan P = UIcoscp = 220.0,5.0.8 = 8 W mot nghiem (nghiem kep) Trong vi du ta cUng gidi phuong trinh bac hai theo I => C o n g suat h u u i c h Phi = P - Php = 88-11 = 77W H i e u suat ciia d p n g co:H=^ P nhung Iqi co hai nghiem I phan biet deu dumg Vqy ta chon nghiem ndo? Sau day la cdc gidi quyei van detren: = — = 87,5% 88 N e u I = I i = 6,4 ( A ) t h i cong suat toa n h i f t Pj = R I j = 20.6,4^ = 819,2W cong C h p n dap an D suat qua I o n so v o i cong sua't co N e n loai t r u o n g h p p Vl d y : M p t d p n g co d i ^ n co g h i 220V-176W, h ^ so cong suat bSng 0,8 N e u I = h = 2,5 ( A ) t h i cong suat toa n h i ^ t Pj = R I i =20.2,5^ = W c n g d u p e mac vao mach dien xoay chieu co dien ap hi?u d y n g 380V D e dpng suat n h o h o n cong suat ciia d p n g co Pco do ta chpn nghiem I = 2,5A CO hoat d p n g b i n h t h u o n g , phai mSc d p n g co no'i tiep v o i m p t d i ^ n tro C hpn dap an C thuan CO gia t r i : A 180Q B.300O C 220n V i d g : (Trich de thi t h u S u ph^m H a Npi Ian nam 2013) D 176fl M p t doan m a c h g o m m p t d p n g co d i ^ n mac n o i tiep v o i m p t c u p n day va 'Phdn tich m hu&ng ddn gidi mac vao n g u o n d i ^ n xoay chieu D p n g co d i ^ n tieu t h u m p t cong suat P = UIcoscp^ l = ^ =^ = ^A^ U.coscp 220.0,8 Va coscp = — = 0,8 Z , , = ^ - ^ = '^'^ I 220a P = 9,53kW, d o n g d i ^ n qua d p n g co co c u o n g d p h i ^ u d u n g bang 40y4 va cham pha m p t goc cpj = — so v o i d i f n ap giiJa hai dau d p n g co D i f n ap g i i i a hai d a u cupn day co gia trj h i | u d u n g la 120y va s o m pha m p t goc r = 0,8.2^^ = 0,8.220 = 176Q V o i d p n g CO ( r , L ) Z ^ ^ = yjr^ + ^ 220 = , / l ^ + Z ^ Z L = 132f^ D o R mSc n o i tiep v o i d p n g co (r, L ) , nen de d p n g co hoat d p n g b i n h thuong (Pj = — so v o i d o n g d i | n chay qua no D i ^ n ap h i ^ u d y n g g i i i a h a i dau doan mach la: thi I qua R phai bang A Ta co t o n g t r o ciia ca d o a n mach B 301 V A.190 V C 384 V D 220 V Phdn tich vd htcang ddn gidi )i?n ap giCia hai dau d p n g co som pha m p t goc R mac n o i tiep v o i d p n g ca (r, L ) ta co Z = ^J{R + T)^ Thay so: 380 = J{R + 176f +132^ ^ +zl j = — so v o i c u o n g d p d o n g d i ^ n , nen d p n g ca R = 180Q C h p n dap an A Vl dy 4: M p t d p n g CO d i ^ n xoay chieu hoat d p n g b i n h t h u o n g v o i d i ^ n ap h i ^ u d u n g 200V t h i sinh cong suat co la 320W Diet d i ^ n t r o thuan cua day quan d p n g co la 20 Q va he so cong suat ciia d p n g co la 0,89 Cuong d p d o n g d i ^ n h i ^ u d u n g chay t r o n g d p n g co la A.4,4A B.1,8A C.2,5A Phan tich va hicang ddn gidi Ta co: P = cong suat co + cong suat n h i ^ t = Pco + PR 6m d i ^ n t r o R v a cupn day thuan L 'Jhu vay theo bai ta co gian vecto r u p t n h u sau: p 953 )i?n t r o ciia d p n g co la: R = - y = — 160 Nen U R = 238.25V Suy D.4A = tancpi-Up = 137.6V A ^ E>i^n ap g i u a hai d a u cupn day s o m pha m p t goc cpj = - s o v o i d o n g di?n day E>i?n giCra chay qua no nen ta co: = Ujcoscpj = 60V Cty TNHH MTV DWH Khang Viet = tan ^ i ^ r = 60N/3V Va ^hdn tick vd hitang ddn gidi Cong suat tieu t h u toan phan: P = Va + \\-, D i e n ap hieu d u n g giua hai dau doan mach la: Hay: U = ^ ( U „ + U , ) +(UL,+UL^ P = U.I.coscp = R, + Php 2 - ^ , = 170+17 C h p n dap an C V i d u : M o t d g n g co dien xoay chieu san cong suat co hoc 7,5kW va c6 hieu suat % Mac d o n g co noi tiep v o i m o t cuon cam roi mSc c h i i n g van m a n g dien xoay chieu Gia tri hieu dien t h e h i ^ u d u n g hai dau d o n g co la U M biet rcing d o n g diC^n qua d o n g co co cuong hieu d u n g I = A va tre pha v o i uvi m o t goc n/6 H i e u dien the hai dau cugn cam Ui = 125V va som pha so v o i d o n g dien qua cuon cam la n/3 T i n h hieu dien the C u o n g d g d o n g dien cue dai I„ = -Jl A C h p n dap an C » U : M o t quat dien ma tren ghi 200V-100W de quat boat d g n g b i n h t h u o n g d u o i dien ap 220V n g u o i ta mac noi tiep v o i no dien t r o thuan R Biet rang he so cong suat cua quat la 0,88 Xac djnh c u o n g d o n g d i ^ n t r o n g mach va d i ^ n t r o R B 35n A 39Q D V ; 39" C V ; 39" B V ; " A V ; 40" p = UI cos(p => I = 'Phdn tick vd hitang ddn gidi P U.cos(p 100 25 U 200 200.0,88 44 I 25 44 = 352Q Va cos (p = — - = 0,88 => r = 0,88.Z., = 31OQ P„=I1.Z5H0.9375(W) 0,8 Mi) P M = U M l.Cosc}) • M UM D 53n C 93Q ^hdn tick vd himng ddn gidi hieu d u n g cua m a n g dien va lech pha cua no so v o i d o n g dien H ^• =384V = Vol d g n g CO r, L ) Z^,^ = A(p U = ^ ^ +zf +ZL « Z^ = Q Do R mac noi tiep v o i dgng co (r, L), nen de dgng co boat d g n g b i n h t h u o n g thi I qua R phai b5ng / 4 ( A ) 9373 l.COSip, 40 Ta CO tong t r o cua ca doan mach Z = A p d u n g djnh ly ham cosin ^ ^ Cho tam giac A N B ta co: i • 2.270,6.125.Cos 1' cosa = n 13 44 = 387i^ R = 390 a u : Cho mach dien g o m bong den day toe mSc ncYi tiep v o i d g n g co xoay chieu pha Biet cac gia t r i djnh miic cua den la 120V - 330W, d\qn ap djnh m i i c cua d g n g co la 220V K h i dat vao dau doan mach dien ap xoay 2.U.U M chieu CO gia trj hieu d u n g 332V thi ca den va d g n g co deu boat d g n g d u n g C h p n dap an C cong suat djnh m i i c Cong suat d i n h m i i c ciia d g n g co la: DUNG: A.583W B.605W C a u 1: M o t d g n g co di?n xoay chieu k h i boat d g n g b i n h t h u o n g v o i dien hieu d u n g 220 V t h i sinh cong suat co hgc la 170 W Biet d g n g co co he cong suat 0,85 va cong suat toa nhiet tren day quan d g n g co la 17 W Bo q^" hao p h i khac, c u o n g d o n g dien cue dai la bao nhieu? A.4,4A 25 C h p n dap an A = 384(V) -.0 a = 9"=>(t) = a + " = ^ CO B A I T A P V A N 220 R mic noi tiep v o i d g n g co (r, L) ta co Z = ^^(R + r)^ + Thay so: 387 = 7(^ + 310)^ +167^ U = VU'M+U^^+2UM.UJ.COSA(P 270,6^+125^ • U B.1,8A C N/2 A C 543,4W D 485,8W ^hdn tick m huang ddn gidi Do boat d g n g cac gia trj djnh miic Nen:I = i ^ =^ = 2,75(A) 120 U dm D.4A Do d g n g co la b^ L - R mSc noi tiep nen ta co mach dien g o m : Bt quye't on luyeii llii tint hoc dat diem toi da Viit If, tap - Le Van Vitih^ Cty TNHH MIX nyviiKhang Vift Chnyto Ah (Rpnt R D C ntL) Ta CO gian vecto trup-t BAI TOAN HOP A p dung dinh ly ham cosin cho tarn giac A M B £>EN PHl/QNG PHAP = o + U^jc -2.UD.UDCCOS(180'' -(PDC) 3 ^ = 2 + 2 ^ - 2 c o s ( ° -(p^c) =>COS(PDC De giai mot bai toan ve hop kin ta thuong s u dung hai phuong phap sau: Phuong phap dai so =0-89818 Bi: C a n c u "dau vao" ciia bai toan de dat cac gia thie't co the xay V i vay: Cong suat dpng co la B2: C a n cii "dau ra" ciia bai toan de loai bo cac gia thie't khong phii hop P = UDC-IDC-™SCPDC Br Gia thie't dupe chpn la gia thie't phu hop voi tat ca cac d i i kien dau = 220.2,75.0,89818 = 543,4(W) ChpndapanC Cau vao va dau ciia bai toan P h u o n g phap s i i dung gian vec to trugt : (Trich de thi thtr S u pham H a Npi Ian nam 2013) Bi: Ve gian vec to (trupt) cho phan da bie't eiia doan mach Phat bieu nao sau day ve dong co khong dong bp ba pha la sai? B2: C a n cii vao di> kien bai toan de ve phan lai ciia gian A Vecto cam ung tu cua t u truong quay dong co luon thay doi ca ve Bx D u a vao gian vec to de tinh cac dai lupng ehua bie't, tu lam huong va tri so' sang to hop kin B Roto cua dpng co quay voi toe dp goc nho hon toe dp goc cua tu truong C Nguyen tac hoat dpng cua dpng co d y a tren hi^n tupng cam ung di^n tu v a s u dung tu truong quay D H a i bp phan chinh ciia dpng co la roto va stato ^hdn tick huang dan giai D a p an A sai v i vec to cam ung tu ciia tu truong quay dpng co thay qua moi phan bang cac vec to A M ; M N ; N B noi duoi theo nguyen tac; R - di ngang; L - di len; B /^^-^ A UR * > M C - di xuong Buoc 3: Noi A voi B thi vec to A B chinh la bieu dien U B A Nhan xet: bp ba pha B Nguoi ta CO the tao tu truong quay bang each cho dong di?n mot + chieu chay qua nam cham di^n C Nguoi ta CO the tao tu truong quay bang each cho dong d i f n xoay ba pha D Nguoi ta CO the tao tu truang quay bang each cho dong di?n xoay chieu chay qua nam cham di^n ^hdn tick vd huang ddn gidi Nguoi ta CO the tao tu truong quay bang each cho dong di?n xoay ehieu ba pha chay qua ba cupn day cua stato ciia dpng co khong dong bp ba pha Cac dien ap tren cac phan tu dupe bieu dien boi cac vec to ma dp Ion ty le voi di^n ap hieu dung cua no + chieu ba pha chay qua ba cupn day ciia stato cua dpng co khong dong bg C h p n dap an C ^/ Buoc 2: Bieu dien Ian lupt hieu dien the Phat bieu nao sau day la diing? chieu mot pha chay qua ba cupn day cua stato cua dpng co khong dong /—^ Uc (do la diem A ) : (Trich de thi t h u S u pham H a Ngi Ian nam 2013) A Nguoi ta CO the tao t u truong quay bang each cho dong di?n xoay / Buoc 1: Chpn true nam ngang la true dong dien, diem dau mach lam goc doi ve huong nhung tri so' khong doi la , 5BQ C h p n dap an A Cau • Cdch ve gian vec ta truat Dp lech pha giua cac hieu dien the la goc hop boi giiia cac vec to tuang ling bieu dien chiing + Dp lech pha giira hieu dien the'va cuong dp dong dien la goc hpp boi vec to bieu dien no voi true i + Giai bai toan la xac dinh cac canh, goc ciia tam giae dua vao cac dinh ly ham so'sin, cosin va cac cong thiic khae: a SinA _ b SinB _ C SinC Bi guy el on luy?n thi diii hgc dat diem tot da Vat li, tap 1-he T r o n g toan hpc m p t tarn giac se giai dugc Van L t y TNHH Vinh 'Phan tich vd hixang ddn neu ba canh) t r o n g sau ye'u to'(3 goc va canh) UAB = « « UAB = U A M + U ^ B v i DU MAU: u =100>/2cos(1007tt)V, C -4 10 - F H Q P k i n X chi chiia m g t phan t u (R hoac cupn day t h u a n cam), 71 d o n g d i e n t r o n g mach som pha n/3 so v o i d i ^ n ap g i u a hai d a u doan WO/Sn B Chiia L ; Z L = 0 / V Q C C h u a R ; R = 10oV3Q 220^ AH' /3 f i va L = m H v a C = l,27|iF D R = Vs Phdn tich vd hu&ng ddn T u h i n h ve ta c6: 100 de i n h a n h pha h o n u m p t goc — t h i C h p n dap an A ( h i n h ve) h o p X chua m p t ba p h a n t u : d i ^ n t r o thuan, cuon u p h a i n a m phia d u o i i , ket h p p v o i gia day, t u d i ^ n K h i dat vao hai dau A B m p t d i ^ n ap xoay chieu c6 gia t r i Weu thiet h o p X chi chua hai p h a n t u nen hpp d u n g 2 V , n g u o i ta dupe U A M = V va UMB = V H o p X chua: X p h a i la d i e n t r a t h u a n R va t u d i ^ n C 'r-' ' rr Ug 40 T o n g t r o : Z = — = — = 5Q IQ D tu dien ^ I ^ A C Q va L = 3,18mH gidi T u gia thiet de cho ta c6 gian d o vecto t r u p t n h u Kmh ve: 1' R C d i ^ n t r o t h u a n VI mac n o i tiep C u o n g d p d o n g di?n t r o n g d o ^ n mach n h a n h pha 7i / so ^ i.:pt goc — t h i u p h a i n a m phia d u o i i , ket B c u p n day k h o n g t h u a n cam ' 260^+120^-220^ = VUMB - U L = 7260^-140^ = V l du 3: t r u p t n h u h i n h ve: de i nhanh pha h o n u A c u p n day t h u a n cam ' the dap an B diing Ttr gia thiet de cho ta c6 gian vecto V l d u 2:6 (v6 ly) v i the dap an A sai (v6 ly) v i the dap an C sai = 120^ + 260^ 2U AM 'PhAn tich vd hit&ng d&n gidi UR = 140 N e u h o p X la c u p n day k h p n g thuan cam (L,r) t h i m a c h g o m C, n o i tiep v o i B D C h u a L ; Z L = 10oV3n 220 = 260 - de B d i i n g mach A B H p p X chiia g i ? d i $ n t r o hoac cam k h a n g c6 gia t r i bao nhieu? tan gidi T i day chi d a p an B v i the chac chan la d i i n g va sau d a y la l y l u a n Cho doan mach n h u h i n h ve, biet ChiiaR; R = Vi?t N e u h o p X la t u d i e n t h i mach g o m C, n o i tiep v o i C v i the ta c6: c2 = a2 + b2 - 2abcos C A - ^AU U i e = U A M + ^u^ b2 = a2 + c2 - 2accos B ; 1: KKang N e u h o p X la d i e n t r o t h u a n t h i mach g o m C, n o i tiep v o i R v i t h e t a c6: a2 = b2 + c2 - 2bccos A ; V l dM mVH N e u hop X la cupn day thuan cam thi mach g o m C, noi tiep v o i L v i t h e t a c6: biet t r u o c ba (hai canh goc, hai goc m p t canh, 03 MIV ^ M B Tir h i n h ve ta cr , cs- = R U Z •R Cty TNHll Ml V nvvil Cac gia trj tuc thoi =>C = 1 coZ^ 10071.2,5 A R = 30Q vaC = (mF) C R = 30Q vaL = UAB=100N/2cosl007it (V); IA =N/2 (A), P = 100 (W), C = — - ( F ) , i tre pha 7tV3 ^ B | => ti so u tuc thoi khong doi Nen u^N,u^iA,u^^ cijng pha u^g cham pha hon i -> X la R, C -4 10 R D.R = 30Q vaL = — ( H ) Ta CO gian vecto truot nhu hinh veuj^^j = 3u^^ = l , u ^ ^ hon UAH Tim cau tao X va gia tri ciia phan tu ^ (H) ^hdn tick m hu&ng ddn gidi 3n ^ B.R = f i vaC = — ( m F ) 7t den chiia phan tir: R hoac L hoac (L, r) hoac C, biet C X la tu dien; C = = 15^3^ Doan mach AB = ^ ' ^ U ^ N ' tro cua cuon cam M A la R = 15Q V i d u : Cho mach dien xoay chieu nhu hinh ve A X la cuon day thuan A / _ cam; L = — ( H ) 57t B X la dien tro thuan R; R = lOOQ = 3UMA C O dien ap vuong pha dien ap dau mach X chua gi? Gia tri ciia no? Biet = 1,27.10-^(F) Chpn dap an A X la mpt hpp U ^ N Khang Viet tancpMA=:^ = ^ = > M A = ? -(F) n sm D X la cuon day khong thuan cam; r = 50Q;L = — ( H ) DTI 'Phan tick vd hucmg ddn gidi R = Theo gia thiet i tre pha hon UAD va mach tieu thu dien suy ra: Hop den la mot cuon day CO r tan Taco: P = rl^ = r = = 50(fi) = + ZL = 30Q 7: v6, u, f - C = J^(mF) Chpn dap an A Vi d u 6: Trong hop X chi c6 chiia nhieu nhat la mpt linh kien: di^n tro Mac khac: ZL-ZC = r^ + {Z^ -Z^f 100^ i - r 80 lOOrt 571 [12]' thuan hoac cuon thuan cam hoac tu di^n Dat vao hai dau mach mpt di^n = ^ -50^ ap xoay chieu tan so 50Hz thoi diem t = t j , dong di^n va di^n ap c6 =50: Z L - = 50 ZL=80Q gia tri Ian lupt la l A va -50\/3V thoi diem t = t j , dong d i f n va dien L Z L - = -50' Z L = -20Q(1) ap CO gia trj Ian lupt la - N / S A va -50V Hop X chua phan t u nao, tinh gia tri phan t u do? ^hdn tick vd hu&ng ddn gidi (H) ChQn dap an D Ta biet rang: i cung pha voi u^ nen tai t V I dy : Mach dien xoay chieu M N gom cupn cam c6 tro hgp X, cuon cam thuan mac theo thu t u A la diem giu-a cuon cam c6 tro va hop X B la diem giiia hop X va cuon cam thuan Trong hop X c6 linh kien khac loai (dien tro thuan, tu dien, cuon cam) de bai t, =*• u >0 i>0 hoac t=> f u , =-50>/3V>0 ' nen m^ch chi c6 the la L ho|c C ii = A < u Z L ^ = ( Q ) • tan(pAM= — X V i t h e t a c6 h ^ p h u o n g t r i n h sau: Ml V UVVH Rhang MB ^ -(F) = 50Q^C = = 80.^-40V3(V) V i d y : Cho hai h o p k i n X , Y chi chua t r o n g ba p h a n tir: R, L (thuan), C mac noi tiep K h i mac hai d i e m A , M vao hai cue ciia m p t n g u o n dien ZT mQt chieu t h i L = 2(A), Uvi = 60(V) K h i mac hai d i e m A , B vao hai cue cua m o t n g u o n dien xoay chieu tan so 50Hz t h i L = 1(A), Uvi = 60v; Uv2 = V , U A M lech pha so v o i UMH m o t goc 120^, xac d j n h X , Y va cac gia t r i cua chiing^ => L v = =4073(Q) 40S 0,4^3 lOOn (H) V I d u : M p t doan mach xoay chieu A B g o m hai phan tir X, Y mac n h u tren ^hdn tich v>d huang dan gidi C u o n g d p dao d o n g mach nhanh Day la mot bai todn c6 su dung den tinh chat cua dong dien chieu doi voi cuon cam va tu dien Khi gidi phai X M pha 7r/6 so v o i h i ^ u d i ^ n the giira hai Y B litu y den vai dong dien chieu thi a = =>ZL = Ovd -Y Zr- = — — = 00 coC V i X cho d o n g dien m p t chieu d i qua nen X k h o n g chua t u di?n Theo de bai thi X chua t r o n g ba phan t u nen X phai chua d i ^ n t r o thuan (Rx) va cupn day t h u a n cam (Lx) C u o n day thuan cam k h o n g c6 tac d u n g v o i d o n g dien d a u doan mach Biet cac bien d p cua h i f u d i ^ n the va c u o n g d p d o n g d i ^ n Ian l u p t la Uo = 220 V2 V va lo = 2,2>/2 A , tan so dao d o n g la f = 50Hz H a i phan tir tren la p h a n tir nao so R, L, C? T i n h gia t r j m o i p h a n tir 'Phan tich vd hu&ng dan gidi Gia sir t r o n g doan mach tren c6 k h o n g c6 p h a n tir R N h u v a y t h i X, Y la hai phan tir L, C , m o t chieu nen: Rx = — ^ ( Q ) K h i mac A , B vao n g u o n d i ^ n xoay chieu Z Gpi cp la goc h o p v o i U ; I => tan(p = bai l i tre pha v a i i goc R) - > Y l a C ~ Z " = ±00 =>(p = ± ^ m a theo dau vay mach dien chac chan c6 R (gia sir X la B l ljuyit oil hiyrii TJiT'Jiu hoc dit cliem tot da Vai It, tap - Le Van tancp = — ^ = tan R V t 6, = -4-=*R = V3Zc Vinh Cty TNHH MTV DWH Khang Viet Cho UAB = U\/2cosa)t Khi khoa K dong thi UAM= 200V, UMN = 150V Khi K ngSt thi UAN = 150V, UNH = 200V Cac phan tu hop X c6 the la: (1) Mat khac: A Dien tro thuan B Cupn cam thuan noi tiep voi tu dien IQ Zc = 50Q => C = 2.10 C Dien tro thuan no'i tiep voi cupn cam 2,1\J2 D Dien tro thuan noi tiep voi tu dien -4 -(F) ^hdn tich vd hu&ng dan gidi Khi k dong mach chi R&C mac noi tiep, ta c6 V i d u : (Trich de t h i t h u Su pham Ha Npi Ian nam 2013) Dat dien ap xoay chieu c6 gia tri hi^u dung U vao hai dau mot hop den X thi dong dien mach c6 gia trj hieu dung 0,25A va som pha nil so voi dien ap hai dau hop den X Cung dat dien ap vao hai dau hpp den Y thi dong dien mach van c6 cuong hieu dung la 0,25 A nhung cung pha voi dien ap hai dau doan mach Ne'u dat di^n ap tren vao hai dau doan mach X va Y mac no'i tiep (X, Y chi chua phan tu) thi cuong dp hi^u dung cua dong di§n mach la c:^A D N/2 A ^hdn tich m huang dan gidi U — = 4U Chpn dap an B J-6U^ U +16U ' thuan R, tu dien c6 di?n dung C, mpt hpp den X Diem M giiia A va C, diem N d giira C va X Ar>o / day thuan cam X A C D I N H H O P D E N X T R O N G MACH D I E N X O A Y C H I E U D U N G MAY T I N H F X - E S Nut l^nh M |SHIFT Bam: SHIFT M O D E I Bam: | M O D E Dang toa dp cue: rZB (AZcp) Bam: SHIFT M O D E Bam: SHIFT Ti'nh dang toa dp de cac: a + ib Chpn don vj goc la dp K Y nghia- Ket q u i mm Bam: Thu'c hien phep tinh ve so phij'c J — ^ 4x/2 R Vay X gom dien tro thuan no'i tiep voi cupn Chi djnh dang nhap / xua't toan V I d u : Doan mach AB gom cac phan t u mac theo thii t u : D i f n tro Hai dau NB c6 mpt day noi c6 khoa K (dien tro ciia khoa K va day noi khong dang ke) 250^=150^+2002 Cai dat ban dau (Reset all): 0,25 Khi mac X,Y noi tiep ta c6 I = Nhan xet quan trpng: ChQti che dg Khi hop den la Y thi i ciing pha voi u nen hop Y la di?n tra Ij K mo R va C no'i tiep voi X : Chpn dap an C Khi hpp den la X : theo bai i nhanh pha hon u lupng ^ nen X la tu di^n U codungkhang Z c = - = UAB = M M + U ^ N = V Hoac chpn dan vj goc Ja Rad (R) m Bam: S H I F T Reset all Man hinh xua't hi^n Math Man hinh xua't hifn chii CMFLX • Hien thj so phuc kieu r ZG ^ODE T Hien thi so phuc a+bi ^ODEJ^ Man hinh hien thj chir D Bam: |SHIFT M0DE\ kieu Man hinh hien thj chir R Cty TNHH MTV DWH Khang Vi?t De nhap ky hi#u goc Z Bam: S H I F T J Chuyen t u dang a + bi Bam: Man h i n h hien thi ky hieu Z y Man |SHIFT sang dang A Z (p , AZ Chuyen t u dang A Z < Bam: p gSB M a Z = R + ( Z L - Z ^ ) ] Suy ra: R = Q ; Z L = Q Vay h o p k i n (den) chiia hai phan t u R, L C h o n dap an A hinh hien thj dang V i d u : M o t hop k i n (den) chi chua hai ba phan t u R, L, C m i c noi tiep (p Man h i n h hien thj dang a Sir dung bp n h d dpc Bam: M + lap V oac I S H I F T M+ Gpi bp nha dpc lap Bam: R C L Xoa bp nho dpc lap Bam: S H I F T Iglb = A d Neia dat vao hai dau mach mot dien ap xoay chieu u = 200 N/2 cos(1007tt - —) Man hinh xua't hien M va (V) thi cuong dong dien qua hpp den la i = 2cos(]007it) Xac dinh hpp den A R = lOOQ; Z L = lOOQ B R = lOOQ; Zc= lOOQ hoac M - ©sang dang a + bi + bi M+ |SHIFT C Zc = 90Q; Zi,= OOQ D R = 120Q; Zi,= 120Q I ^hdn tich vd huong ddn gidi Clear Memory? [=] :Yes ^ " i ~ Xdc dinh cdc thong so( Z, R, Zr, Zc) bang may tinh: TinhZ:Z =^ =.^°^^^ tiep Ne'u dat vao hai dau mach mot dien ap xoay chieu u = 100N/2 cos(1007:t + - ) (V) t h i c u o n g d o n g dien qua hop den la I = 2cos(1007tt)(A) Doan mach chiia n h i r n g phan t u nao? Gia trj ciia cac dai l u g n g do? B R = 50Q; Zc= 50Q Zc = Q ; Z L = Q D R = 60Q; ZL= 600^ 'Phan tick v>d hit&ng dan gidi g f SHIFT (-) H i e n t h i : 100-lOOi ( Phep C H I A hai so p h u c ) A R = Q ; Z i = Q 45 ( Z ) N h a p ^ ^ H I F T J V i d u 1 : M o t hpp k i n (den) chi chua hai ba phan h i R, L, C mac noi C _ u _ 200V2Z - (mat chii M ) \a Z = R + ( Z L - Z c ) i S u y r a : R = 100Q;Zc = 100Q : Vay hpp k i n (den) chua hai phan t u R, C C h g n dap an B V i d u : C u o n day thuan cam c6 he so t u cam L = 6 m H mac n o i tiep v o i doan mach X Dat vao h a i d a u d o a n mach h i ^ u d i ^ n the u = 120 \/2 COSIOOTI t ( V ) t h i c u o n g d o n g dien qua cuon day la i = 0,6V^cos(1007i t - n /6){A) Tim hieu dien the hieu d u n g Ux giira hai dau doan mach X? A 120V Voi may F X E S : Barn chon [MODEJ g tren man h i n h xua't hien chir: C M P L X B 240V C.UOsfzV D 60>/2V ^hdn tich v>d hucmg ddn gidi C h o n d o n v j goc la ( D ) , bam: pHIFT] ^0DE\ tren m a n h i n h hien thi Voi dang todn tinh gidi theo mdy tinh Id mot buac dot pha vegidi nhanh Khi chiiD tinh theo mdy tinh khong nhimg ta tim dmc hieu dien the hieu dung (cue dai) ma Bam B H I F l i [ M O D E | @ g j l j : Cai dat dang toa de-cac: (a + bi) ta tim duac cd pha cm hop X tif ta biel dugc hop X cd chua phan tu ne'u a = Othi mach khong c6 R ndo hay la mach c6 tinh cdm khdng hay dung khdng Day Id phuong phdp gidi cue (hoqc cocdLvaC nC''ua>Ovab>0 thi mach coRvdL + (hoac cd cd R, L, C vd mach cd tinh cdm khdng) Truoc tien tinh Zi = 200Q neua R Q Z, - Z ^ - ZC O R, = l=^Ro = Z L - Z C - Z C O = > Z = J R ^ + ( Z L - Z ^ - Z C „ ) ^ = R O = Z C O V = 5OV2 = Zco = n => C(, = 2.10 -4 -(F) in C h p n dap an B N h a n xet: ccicli lam Ihi't hai doi hoi chung ta phai hien luan dime giai han tren cua dai luang vi the cung gdy cho nhieu han hoi roi Tuy nhien neu lam each cong them v&i phimng phdp hai trie thi cdng dat hieu qua tot han Cudi cimg ta nhan thay, gidi hdng mdy tinh ket qud cho nhanh han vd cQng khong can kien tlu'ec suy luqn hay loqi trie gi het vi ket qud cho la ro rdng Cty TNHll Ca BAI TAP V A N DUNG: Cau 1: Trong mach difn xoay chieu gom phan tu X noi tiep phan tu Y Biet rang M ! i i) \ Cau 3: Nhieu hop khoi giong nhau, nguoi ta noi mpt doan mach gom mo cac hpp khoi mac no'i tiep voi di^n tro R = lOOQ doan mach X va Y chua phan tu (dien tro thuan, tu di?n, cupn day thuan cam) dupe dat vao hi^u di#n the xoay chieu tan so 50Hz thi hi^u di^n the som Dat vao hai dau doan mach mot di?n ap: u = U pha 60° so voi dong dien mach Hop kin chua tu di§n hay cuon cam Tinh dien dung ciia tu hoac dp t u cam ciia cuon cam cosl007tt(V) thi di^n ap hieu U •/ s3 dung tren phan t u X, Y duoc Ian luot la Ux= U — va U Y = y A X la cupn day thuan cam; L = — ( H ) n B X la dien tro thuan R; R = lOOQ Biet u nhanh pha hon i Cac phan t u X, Y Ian lug-t la: A Di?n tro thuan R va cupn day thuan cam B Dien tro thuan R va tu di^n ' ' C Cupn day thuan cam va cuon day thuan cam D CuQn day thuan cam va tu di^n 10 - -(F) D X la cupn day khong thuan cam; r = 50Q; L = — ( H ) i*hd,, iich v>d huang dan gidi Nhan xet: u nhanh pha hon i va Ux + U Y C X l a t u d i e n ; C = 571 nen X la d i f n tro thuan 'Phan tick vd huang dan gidi Y cuon thuan cam Doan mach gom X va R mic noi tiep Theo bai ra: Hi§u dien the som pha Chgn dap an A hon cuong dp dong di^n mach nen mach di?n c6 tinh chat cam Cau 2: Dat vao dau mot hop kin X (chi gom cac phan tix mac noi tiep) mot di?n ap xoay chieu u = 50cos(1007it + 7r/6)(V) thi cuong dong dien qua mach i = 2cos(1007it + 27t/3)(A) Neu thay dien ap tren bang dien ap khac c6 bieu thuc u = 50 N/2 cos(2007it + 27i/3)(V) thi cuong A R = 25 (Q), L = 2,5/7i(H), C 10-^/7:(F) =-i = - Chpn dap an B phan t u chua (F) 'Phdn tich vd hu&ng dan gidi "^^2=ZL2-ZC2 =2Z^-—!-(f2=2fi) Theo bai ta c6: ^^0 = 25n ^ = - ^ = 50n L -(H) Cau 4: Cho mach dien xoay chieu nhu hinh ve chua phan t u mac 10" cua chiing, biet Co = In (Zc^ > ZL^ ) L = —(H) 127t^ Z2=2Z, 7t z - ^ a M X -3 Trong hai truong hg-p u va i vuong pha voi nen R = -Z, lOOTt 73, doan mach X va dp Idn Gia su mach gom phan tu thuan R, thuan L va tu C noi tiep Zi = Z p (0 Xac dinh cac ^hdn tick vd huong ddn gidi =T-7"f 100N/3 = N/3R = IOOA/SQ ampe ke chi 0,8A va he so cong suat ciia dong di^n mach la 0,6 D R = 25 (Q), L = 5/127i(H) ^2=%-^,2 Z, V => Z^, s mach mpt hi^u dien the xoay chieu c6 bieu thuc U = 200 \/2 coslOOTtt (V) thi C L = l,5/n(H), C = l,5.10-V7t(F) - ^ii = f - ^ - - f = ^ R noi tiep Bo qua di^n tro ciia mape ke vao dau noi Dat vao dau doan B L = 5/127t(H), C = l,5.1zO-V7t(F) = Z Vi the ta c6: tancp = — t = tan60° = dong dien i = N/2 cos(2007it + ;r/6)(A) NhCrng thong tin tren cho biet X chua ^1 khang Vay hop chua cupn cam ' - coC0 lOOTt 10 - = 20(Q) 271 Tongtro: Z = - = — = 250Q ^ I 0,8 Tong tro doan mach X: Z^ = Z^^ + Z^ Co Cty TNHH MTV DWH Khang Vift Zx = Jz^-Z^ = N/2502 - 20^ = 30769Q R+ Ta lai c6: > J R — £ - R + > 2Z, V R R R f^/„ Dau = xay va chi R = Zc cosq) = I = 0,6 =^ R = 0,6Z = 0,6.250 = 150Q Tong tro cua toan mach: Z = ^ R ^ + ZQ =ZQ\12 Nhu vay, doan mach X gom R va L hoac R va C O (1) Matkhac: Z = — Truong hop 1: X gom R va L (2) I T u ( ) va (2) =^ Z, = - =^(30^) /r U 200 = Z c V = - = ^ •Zf^=100fi ^ c = — -1502=60VriQ 1 01 - lOOn.lOO n -F Cau 6: Cho doan mach AB nhu hinh ve Moi hpp X va Y chi chua hai ba phan tir: dien tro thuan, cuon day thuan cam va tu di?n mic noi tiep Cac von ke'Vi, V2 va ampe kedo dugc ca dong xoay chieu va mgt chieu, di^n tro cac Trimng hop 2: X gom R va C von ke'rat Ion, dien tro ampe kekhong dang ke Khi mac vao hai diem A va M hai cue cua nguon di^n mpt chieu, ampe kechi 2A, Vi chi 60V Khi mac A va B vao nguon dien xoay Z^ = J Z ^ - R = J(30>/69 )^ -150^ = 60VITQ nhung UAM va UMB lech pha ^ -(F) (H) Cau 5: Cho mach dien nhu hinh ve Dien ap giiia hai dau doan mach c6 bieu thuc la u^,^^, =200x/2sinl007it (V) Cuong dong dien i nhanh pha hon dien ap hai dau doan mach dien R la bien tro Dieu chinh R thay cong ^ suat ciia mnch cue dai \ 4lA Xac ^ | \~\ Cuong dong dien i nhanh pha hon dien ap u hai dau doan mach nen X chua tu dien R^ + Z ^ ^ „ V B I Khi mac hai dau hop X voi nguon dien mot chieu, ampe ke chi 2A =i> mach CO dong dien c6 cuong Ii = 2A, chung to hpp kin X khong c6 R+ (vi Z L = 0) Khi mac A va B vao nguon di^n xoay chieu, ta c6: I2 ^2 Zc I ^ R Theo bat d3ng thuc Co-si, ta c6: R + M nao? Tinh gia trj ciia chung R V Hai hop X va Y chua nhiing phan t u Khi ta c6: Z ^ ^ = R = — = — = Q c = o)Z(^ • ' • R^ + Z? = 60^ Z L = 760^ -30^ = 30V3Q R 30V3 2ni 27t.50 -0,165 H o liiquyc't on luyeu thi liai hoc dat ilicm toi da Viit li, tap - Le Van Viith C t y TNim Ta co: tantp^M = — = = N/3 =^ cp^^ = - rad Ta CO hinh ve nhu ben duoi Theo hinh, UMH tre pha so voi dong di?n nen => hop kin Y chua dien tro R' no'i tiep tu dien C U2 _ 60 _ Doi voi doan mach MB: Z^g = -— - oUia I2 Ma Z^,, = ^ / R ^ ^ = 60Q ^ R ^ + Z^ = 60^ Vi UAM vuong pha UMH nen ta c6: tan(p^^^.tan(p^^B =-1 » - z 'C R = -1 o (1) Z, Z, R • R' 1 Ini.Zr 271.50.30 Khang Viet 'Phan tich vd hu&ng dan gidi •C2 Taco: ^ ^ = 0 ; Z^, =100Q X A D B C, E C2 G Khi dong dong thcri ca Ki va K2 thi hi^u dien thehai dau tu dien dat cue dai va hieu dien the hai dau hop den X nhanh pha hon cuong dp dong dien la Ti/4 (Rad) Nen hop den X gom dien tro R va cupn thuan cam co dp tu cam L ( v'l nC'ii CO L vii C tlu ux viidn;^ pha v&i i; ncii la R va C thi ux chqtn va ta CO R = Z[ (vi ux nhanh pha hon i g o c j ) (2) 30 R' ^ G i a i ( l ) v a ( ) ^ R' = 30V3Q ; Z :30Q =>C = =1 MTV D W J f Khi U c = Uenuv thi Z,- = 2!l^ pha han i) „ = 2Z,=R = Z , = ^ Q =:1,06.10-^(F) V a y h o p X c h u a R = 30Q noi tiep L 0,165H hpp Y chua R'= 30^30 noi tiep C = 1,06.10"''F ,-4 1-3 10 F Hop den X Cau 7: Cho mach dien nhu hinh ve Bie't Ci = — — F , C2 =—— 1571 71 chua linh i JR2 + ( Z , - Zc^ - Z,-^ f - Zcj - Zc2 = - ( Z L + ZLJ +1Z^ =Z + ( Z L + ZL^ - Zc,) - Zcj) - Z L =100+ = ^R2 2,5 2.150-2.75 = 250Q=^L, = — ( H ) 71 Cau 8: Cho doan mach AB gom hop kin X chi chua mpt ba phan tu (di^n tro thuan, tu dien hoac cupn day thuan cam) va bien tro R Dat vao hai dau doan AB mpt hi$u di^n thexoay chieu co gia trj hieu dung 200V va tan so 50Hz Thay doi gia tri ciia bien tro R de cong suat tieu thu mach AB la cue dai Khi cuong dp dong dien qua mach co gia trj hi^u dung bSng 1,414 A (coi b^ng 42 A) Bie't cuong dp dong di^n som pha hon hi^u di§n thegiOa hai dau doan mach AB Hoi hop kin chua cai gi? Tinh gia trj ciia no Bo qua di^n tro cac day noi Bi quyei on luyftt thi dai hqc dat diem tot da Vat It, tap 1-Le VSn Vinh Cty TNHH MTV DWH Khang Viet C a u : Mpt hpp kin (den) chi chiia hai ba phan t u R, L, C mac noi tiep