530 076 K300T PGS.TS N G U Y E N Q U A N G L A C S N G U Y E N T H I N H I - T S C H U V A N L A N H KI THUAT GIAI NHANH BAI TAP TRAC NGHIEM irl*if TAP! • D O N G LTJC H O C V A T R A N D A OD O N G C O • SONG C O • D A O D O N G V A S O N G D I E N TU" D O N G DIEN XOAY CHIEU • Danh cho hoc sinh Idp 12 chiidng trinh chuan va nang cao • On tap va nang cao ki nang lam bai Bien scan theo npi dung va cau true de thi cua bp GDSfDT BAN D A IH O C QUOC GIA H AN O I fi — ¥ & r N G U Y E N T H I N H l - Tfc> (Jhtu VAJN £ A m r Kl THUAT GIAI N A H HN • BAI TAP TRAC NGHIEM • D O N G L U C H O C VAT RAN • DAO DONG CO • S6NG CO • DAO D O N G VA SONG D I £ N T f D N G DI$N XOAY C H I E U • Danh cho hpc sinh Idp 12 chUcfng trinh chuan va nang cao • On tap va nang cao k i nang lam b^i • Bien soan theo noi dung va cau triic de thi cua bp GD&DT THIJ ViFNTiNHBINMTHUAN DCK] NHA XUAT BAN DAI H O C Q U O C GIAH A NOI CHCJ LOfI N O I D A U DONG L U C HOC VAT RAN Quyen sach giiip cac em hoc sinh lop 12 va cac em dang on luyen de thi CD - D H h^ thong hoa kien thuc ly thuyet va luy^n tap each tra loi cau hoi trac nghiem Kien thuc ly thuyet duoc he thong hoa theo chuan chuang trinh Bai tap trac nghiem dugc phan cac dang phu h^p vai npi dung ly thuyet va mot so'each giai dac bi?t dugc cap nhat mot each kip thai t u cac de thi thu dai hoc ciia cac truong nhu THPT chuyen Phan Boi Chau, THPT chuyen Dai hpc Vinh, THPT chuyen DHSP Ha Noi, va de thi CD - D H cac nam gan day, ke ca de thi CD- D H nam 2013 vua qua Mot so'bai tap kho dugc danh dau * de cac em luu y doc Quyen sach dugc tach hai tap de c6 d i i dung lugng cho vi^c chuyen tai ngi dung ly thuyet va khoi lugng bai tap day dii, phong phii * Chuyen dpng quay ciia vat ran quanh mpt true co dinh Su tuong t u gii>a cac dai lugng goc dac trung cho chuyen dgng quay ciia vat r ^ quanh mgt true co djnh va dai lugng dai dac trung cho chuyen dgng thang: Tap bao gom: - I H E T H O N G H O A K I E N THLfC Mot vat ran quay quanh mot true co'djnh thi mgi diem tren vat ran vach nen nhOng vong tron cac mat phang vuong goc vai true quay, c6 tam nam tren true quay va ban kinh bang khoang each tir diem den true quay Moi diem ciia vat quay cung mot goc cung mgt khoang thai gian N h u v^y, chuyen dgng quay quanh mgt true ciia vat ran la tong hgp nhiing chuyen dgng tron tren vat ran Chu de 1: Dong luc hgc vat ran; - Chu de 3: Song co; - Chu de 4: Dao dgng va song dien tu - Chu de 5: Dong di^n xoay chieu; * Tap bao gom: - Chu de 6: Tinh chat song anh sang; Chuyen dgng quay (chieu khong doi) Chu de 2: Dao dgng ca - Chuyen dgng thang (true quay co djnh, chieu quay khong doi) Tga do: x Tga goc: (p Toe do: v = x' Toe goc : ( = cp' Gia toe: a = v' = x" Gia toe goc: y = co' = cp" Chuyen dgng thSng deu: Chuyen dgng quay deu: V = hang so; a = 0; x = xo + vt 0) = hang so; y = 0; (p = cp,, + cot Chu de 7: Lugng t u anh sang Chuyen dgng thSng bien doi Chuyen dgng quay bien doi deu: - Chu de 8: Thuyet tuong doi hep; deu: - Chu de 9: Vat ly hat nhan nguyen t u a = hang so' - Chu de 10: T u vi mo den vT mo V = vo + at X nen chi dgc quyen sach bang mat nhu doc tieu thuyet ma nen c6 gia'y but = xo + vot + -at^ de ghi lai nhOng dieu tam dac va tinh toan cac dai lugng vat ly ma bai tap yeu v ^ - V o =2a(x-Xo) cau theo ggi y huong dan giai da cho Lam dugc nhu vay, cac em cam chac kha Cong thuc lien he giOa cac dai lugng goc va dai lugng dai: \ • • De C O kien thuc virng chac va luyen dugc each giai bai tap, cac em khong nang lam trgn ven de thi CD - D H sap toi Dau da C O be day kinh nghiem lau nam va da co'gang bien soan quyen sach tot de phuc vu dgc gia nhung chung toi luon vui long don nhan nhirng y kien quy bau t u cac thay c6 giao va cac em hgc sinh Nhom tac gia y = hang so' CO = COQ + yt (p = (P(,+C0(,t + ^yt^ s = r(p, co^-cog =2y((p-(Po) V = rco; at = ry; a„ = rco^ Chii y: Dan vj ciia cac dai lugng hf don vj SL x(m); v(m/s); a(m/s^); (p (rad); CO (rad/s); y (rad/s2) Phuang trinh dgng luc hgc ciia vat tin quay quanh mpt tr\ic co dinh a Moi lien h^ giira gia toe goc va momen lyc > Momen life dot vai mot true quay Momen M ciia luc F doi vai true quay A co Ian bang: M = Fd (1) Cty TNHH T r o n g d o d la canh tay d o n cua luc F (khoang each tie t r y c A den gia cua Ivc F ) • T h a n h d o n g chat, c6 k h o i l u g n g m va c6 tie't d i ^ n n h o so v o i chieu dai 1 = -lml2 C h p n ehieu quay cua vat l a m chieu d u o n g , ta c6 q u y uoe: - M > k h i F eo tac d u n g l a m vat quay theo ehieu d u o n g - M < k h i F eo tac d u n g l a m vat quay theo ehieu ngupe ehieu d u o n g > Mot lien he giita gia toe goc va momen luc T r u o n g h p p v^t ran la m p t qua cau n h o eo kho'i l u g n g m gSn vao m p t dau t h a n h rat nh? va dai r Vat quay tren mat phMng n h a n n a m ngang x u n g q u a n h m g t true A th^ng d u n g d i qua m p t dau eiia t h a n h d u o i tae d y n g a i a lye F , t h i m o m e n lye tac d u n g len vat ran la: M = (mr2)y 12 • V a n h tron d o n g chat eo k h o i l u g n g m , ban k i n h R, t r y c quay A di qua tarn v a n h va v u o n g goc v o i mat p h a n g v a n h tron: I = m R • D i a t r o n m o n g d o n g chat c6 k h o i l u g n g m , eo ban k i n h R, t r y e quay A di qua t a m dia t r o n va v u o n g goc v o i mat dia: I = - m R • Qua cau dac d o n g chat eo k h o i l u g n g m , eo ban k i n h R, t r y c quay A di (2) t r o n g d o y la gia toe goc ciia vat ran m qua tam qua cau: I = - mR o c Phuong trinh dpng l y c hpc a i a vat tin quay quanh mpt tryc co d i n h la: T r u o n g h g p vat ran g o m nhieu chat d i e m c6 k h o i l u o n g Ian l u g t la mi, mj, each true quay A n h u n g khoang n, r\, khac n h a u t h i m o m e n lye tae d y n g len vat ran la: M = ly (5) T r o n g do: I la m o m e n quan t i n h ciia vat ran d o i v o i t r y c quay A; M la m o m e n lye tac d y n g vao vat ran d o i v o i trye quay A; y la gia toe goc ciia vat (3) ran t r o n g ehuyen d o n g quay q u a n h true A Momen dpng lugmg d i n h luat bao toan momen dpng l u g n g a Momen dpng l u g n g L eiia vat rin t r o n g ehuyen d p n g quay q u a n h t r y e la: b M o m e n q u a n t i n h T r o n g p h u o n g t r i n h (3), dai l u g n g Y^mj^ dac t n m g cho m i i e q u a n t i n h i cua vat quay va d u g e g o i la m o m e n quan rinh, k y hi$u la I M o m e n q u a n t i n h I d o i v o i m o t true la dai l u g n g dac tnmg cho m u c quan t i n h ciia vat ran t r o n g ehuyen d g n g quay quanh true ay (4) M o m e n q u a n t i n h c6 d o n vj la kg.m^ > M o m e n quan tir\ ciia m g t vat ran khong chi phy thuoe vao khoi l u g n g a i a vat ran ma eon p h u thuge vao ea sy phan bo khoi l u g n g xa hay gan true quay > D i n h l y ve trye song song: I Q = Vift ciia no, t r y c quay A d i qua t r u n g d i e m ciia va v u o n g goc v o i t h a n h : D o n v j cua m o m e n l y e la N m * MTVDWirKK^g + md^ T r o n g do: d la khoang each tir true quay O den true song song qua t r o n g tam G ciia vat; lo; IG t u o n g l i n g la m o m e n quan h'nh ciia vat d o i v o i t r y c O va trye G > M o m e n q u a n t i n h ciia m g t so vat ran eo trye quay do'i x i i n g : L = lo) (6) T r o n g do: I la m o m e n quan t i n h ciia vat ran d o i v o i t r y e quay; co la toe dp goc eiia vat ran t r o n g ehuyen d p n g quay q u a n h tryc D o n vj eiia m o m e n d p n g l u g n g la kgmVs b Phuong trinh dpng l y c hpc ciia vat tin quay quanh mpt tryc dupe viet duoi dang khac: AL M =At (7) T r o n g do: M la m o m e n lye tae d y n g vao vat ran; L = Ico la m o m e n d p n g •trgng ciia vat ran d o i v o i trye quay; AL la dp bien thien m o m e n d p n g l u g n g ciia vat ran t r o n g t h o i gian At c D i n h luat bao toan momen dpng l u g n g K h i M = t h i L = L(o = hang so N e u t o n g eac m o m e n lye tae d u n g len m g t vat ran (hay he vat) d o i v o i m p t true hang t h i tong m o m e n d p n g l u g n g ciia vat ran (hay h ^ vat) d o i v o i t r y c dupe bao toan • T r u o n g h g p I k h o n g d o i t h i w k h o n g d o i : vgt ran (hay h | v^t) d u n g yen ho§e quay deu Cty TNHH MTV DWH Ki thuatgiai nhanh BTTN Vat Li 12, tap - Nguyen Quang Lac • T r u o n g h g p I thay d o i t h i co thay d o i : vat ran (hay he vat) c6 I g i a m t h i co tang, C O I tang t h i co g i a m (Ico = hang so' hay I^co^ = I2CO2) a D p n g n a n g W d cua v a t r a n quay q u a n h m p t true eo d i n h l a : (8) Vi : M p t b a n h xe quay deu q u a n h true co d j n h v o i tan so goc k h o n g d o i 6000 v o n g / p h i i t T r o n g 2,5s banh xe quay d u p e m p t goc la ^ 1257T rad B 250n rad T r o n g d o : I la m o m e n quan t i n h cua vat ran d o i v o i true quay, co la toe d o goc ciia vat ran t r o n g chuyen d o n g quay q u a n h true C 500 n rad D 750TI r a d Huong dan giai: Ta CO f = 6000 v o n g / p h i i t = 100 v6ng/s T u (6) va (8), d p n g nang Wd ciia vat ran quay q u a n h m o t t r y c co' d i n h co the vie't d u o i d a n g : Suy ra: CO = 27tf = 2007t (rad/s) D o do: cp = cot = 20071.2,5 = VVd=^ (9) T r o n g do: L , I Ian l u g t la m o m e n d o n g l u g n g va m o m e n q u a n t i n h cua vat ran d o i v o i true quay D o n g nang cua vat ran co d o n v i la j u n , k i h i e u J quay q u a n h m p t true co d i n h bang t o n g cong thuc hien b o i m o m e n ngoai luc tac d u n g vao vat Ta co: • AW,=llco2-llco?=i(co^-co2) = A (10) T r o n g do: I la m o m e n quan t i n h ciia vat ran d o i v o i true quay, coi la.to'c dp goc liic d a u ciia vat ran, C02 la to'c dp goc liic sau ciia vat ran, A la t o n g cong ciia cac m o m e n ngoai luc tac d u n g vao vat ran, AWd la d p b i e n t h i e n d p n g nang ciia vat ran Dang B A I T A P L I E N Q U A N D E N C H U Y E N D O N G Q U A Y D E U (rad) => C h p n C 50m, quay deu v o i toe dp 120 v o n g / p h i i t Toe d p d a i tai m p t d i e m n a m a v a n h canh quat bang: A 112,6m/s B 314 m/s C 412 m/s D 538 m/s Huang dan giai: in r — Ta co: co = 120 v o n g / p h i i t = in rad/s; ^ = ^ = 25m D o do: v = co.r = 4.3,14.25 = 314(m / s) Chpn B V i d u 3: M p t vat ran quay q u a n h m p t true co d m h x u y e n qua vat v o i p h u a n g t r i n h tpa d p goc cp = 5,6 + 4,5t, t r o n g cj) t i n h bang r a d i a n (rad) va t t i n h bang giay (s) M p t d i e m tren vat each true quay k h o a n g r = 6em t h i co toe d p dai bang: A 12cm/s VI D V MINH HQA SOOTI V i d u 2: M p t canh quat ciia m a y phat dien chay bang sue gio c6 d u o n g k i n h b D i n h l y b i e n t h i e n d p n g n a n g : D o bien thien d o n g n a n g cua m o t vat ran I I C A C D A N G B A I T A P vA = (vi m i v o n g vat ran quay d u p e m p t goc la 2rc rad) D p n g n a n g ciia v a t r a n q u a y q u a n h m p t true W,=llco2 So' v o n g vat ran quay dupe: " = ^ Khattg Viet B 18 cm/s C 27 cm/s D 35 cm/s Huang dan giai: T u phucmg t r i n h c = 5,6 + 4,5t ta suy ta: co = cp' = 4,5rad / s p Vay V = co.r = 4,5.6 = ( c m / s) => C h p n C CUA VATRAN P h u o n g p h a p g i a i : V a n d u n g cac kien t h i i c ve chuyen d p n g quay deu ciia vat ran: • Dang B A I T A P L I E N Q U A N D E N C H U Y E N D Q N G Q U A Y B I E N To'c d p goc: CO bang hang so' DOI DEUCUAVAT RAN • Gia toe goc: Y = • Tpa d p goc ( p h u o n g t r i n h chuyen d p n g ciia vat ran quay deu q u a n h m p t t r v c c o ' d j n h ) : (P = (PQ + cot ' ' ' L i e n he giCra to'c d p goc va to'c dp d a i : v = cor Gia toe h u o n g tarn: a,., = — = a)^r P h u o n g p h a p g i a i : Sii d u n g cac p h u o n g t r i n h chuyen d p n g quay bien d o i deu ciia vat ran q u a n h m p t true eo d j n h n h u sau: co = coo + y t ; ; g'v; i , ; cp = (p(,+coot + - Y t ; co^-co^ =2y((p-cpy) ^ ^ „"J— ^ ^ Trong c> la toa goc tai thai diem ban dau t = 0; coo la toe dp goc tai }o thai diem ban dau t = 0; cf) la toa dp goc tai thai diem t; co la toe dp goc tai thai diem t; y la gia toe goc (y la hang so) > Khi giai toan, ta neu xet vgt quay theo mpt chieu nhat dinh va chpn chieu quay ciia vat la chieu duong Khi do: • Neu vat ran chi quay theo mpt chieu nhat dinh va toe dp goc tang dan theo thai gian thi chuyen dpng quay la nhanh dan (y > 0) • Neu vat ran chi quay theo mpt chieu nhat dinh va toe dp goc giam dan theo thai gian thi chuyen dpng quay la cham dan (y < 0) > Khi khong de y den chieu quay ciia vat ma vat chuyen dpng quay nhanh dan deu thi co va y cung dau (coy > 0), chuyen dpng quay cham dan deu thi co va y khac dau (coy < 0) v^ , > Thanh phan gia toe phap tuyen a„ co dp Ian: a„ = — = c r o > Thanh phan gia toe tiep tuyen co dp Ian: a, = — = ry > Vec to gia toe a dupe xae dinh: a = a„ + a(; ve dp ion: a = -Ja^+a^ V i dy 1: Mpt banh xe quay nhanh dan deu tir trang thai dung yen, sau 4s banh xe dat van toe goc 24 rad/s Goc ma banh xe quay dupe thai gian la: A.23rad B.37rad C 48 rad D 55 rad Huong dan giai: _ , co-cog - ^ , Ta co: y = = = 6rad / s t-to Tu co^ - co^ = 2yAcp =>Acp = ^ 2y =^ i ^ 2.6 Huong dan giai: j w ,^ A p dung cong thuc: co = cO(, + y t , coy = r a d / s , vi banh xe quay ch|m dan deu nen y = -6rad / s^ Do phuang trinh van toe goc la: CO = 12 - 6t (rad/s) 12 Khi banh xe dung lai thi co = => = 12 - 6t t = — = 2s Vay sau 2s thi banh xe dung lai Chpn moc thoi gian t = tai thai diem banh xe bat dau chju lye ham tac dyng, tpa dp goc ban dau cpg = Chpn chieu duong la chieu quay cua vat ran, do, ap dung cong thuc: cp = (P(, +cOot + i Y t ^ =0 + 12.2-i.6.2^ =12rad ChpnD V i d u 3: Mpt banh xe co duong kinh 54 cm quay nhanh dan deu, giay van toe goc tang t u 180 vong/phiit len 480 vong/ phiit Gia toe huong tam ciia diem M a vanh banh xe sau tang toe dupe 3,5s la A 330,4 m / s l B 411,2 mls\ 456,3 m/s^ D 512,7 m / s l Huang dan giai: Ta co: COQ - 67irad / s va co = 16rcrad/s.^ y= CO-co,) = 167t-6K ^ , , = 27i:rad / s Sau 2s, van toe goc bang: co = COQ + yt = 671 +27i,5 = 137trad / s Gia toe huang tam: a^, = co^R = co^ ^ = (l37r)^ :% = 456,3m / Chon C = 48rad^ChonC V i d\ 2: Mpt vat ran dang quay deu quanh mpt true co dinh vai toe dp goc 12 rad/s thi chju mpt luc ham tac dung va chuyen dong quay cham dan deu voi gia toe goc rad/s^ Chpn moc thoi gian t = tai thai diem banh xe bat dau chju luc ham tac dung Chpn chieu duong la chieu quay eiia vat ran Thoi gian t u banh xe chju lire ham tac dung den liic dirng lai va goc quay khoang thai gian la A t = Is; c = 6,7 rad p B t = 1,5 s; c = 7,5 rad p C t = s; (p = 11 rad D t = s; c = 12 rad p / k Dang B A I TAP LIEN Q U A N DEN M O M E N Q U A N T I N H CUA V A T R A N , HE V A T R A N Phuong phap giai: Su dyng cac kien thuc sau: I = " ^ i ^ ^ > Momen quan tinh ciia mpt so vat ran co true quay doi xung: ^ • Thanh dong chat co khol lupng m va co tie't di?n nho so voi chieu dai ciia no, true quay A d i qua trung diem ciia va vuong goc vai thanh: U-lml^ ^2 ;,, K i thuatgidi nhatih BTTN Vat Li 12, tap - Nguyen Quanx Lac Cty TNHH • Vanh tron dong chat c6 khoi lu^ng m, ban kinh R, tryc quay A di qua tarn vanh va vuong goc voi mat phang vanh tron: I = mR^ • DTa tron mong dong chat c6 khoi lugng m, c6 ban kinh R, tryc quay A di • Qua cau dac dong chat c6 khoi luong m, c6 ban kinh R, true quay A di qua tam qua cau: I =—mR^ ; > Momen quan tinh cua he vat ran: I = Ii +12 + i do: Ii, I2, Ian l u ^ t la momen quan tinh cua eae vat ran m j , m^ V i du 1: Hai qua cau lie dac lam eung b3ng mot loai thep, dong chat eo ban kinh gap Ian R, = 4R2 He thiie lien he giua momen quan tinh cua hai qua cau doi voi true quay di qua tam ciia moi qua cau la =51211 C.I, =512 12 D.l2 = 314I, Huong dan gidi: Goi D la khoi lu-gng rieng cua; V i va V2 la the tich ciia cac qua cau Ta c6: m^ = D.V^ = B 0,27 kg.m^ C 0,54 kg.m^ Viet D 1,2 kg.m^ Huang dan gidi: Theo djnh ly true song song, ta c6: {'^ ) ^ ,: => Chgn C Trong do: d la khoang each vuong goc t u true quay O den true song song qua tam G ciia vat; lo; Ic; tuong ung la momen quan tinh eiia vat doi voi true O va true G B.I2 Khattg I = I c + md^ = | m R + mR^ = | m R = |.4.0,3^ = 0,54(kgm2) > Dinh ly ve true song song: IQ = I,- + md^ A I i = 314l2 DWH V i dy 3: Mot dia tron dong chat c6 ban kinh 30 cm, kho'i lugng m = 4kg Momen quan tinh ciia dla doi voi mot true quay vuong goc voi mat dla tai mot diem tren vanh c6 gia tri nao sau day? A 0,12 kg.m^ qua tam dia tron va vuong goc voi mat dla: I = — mR MTV D.y R^; = D.V2 = ^-^^l Dang 4: ,^ v BAI TAP LIEN Q U A N D E N PHl/ONG TRINH D Q N G H O C C U A V A T R A N Q U A Y Q U A N H M O T TRVC C O Ll/C D I N H Phuong phap giai: • Van dung phuang trinh dong lye hoc ciia vat ran quay quanh mot true CO djnh M = l y va phuong phap dong lue hoe de giai bai toan chuyen dong quay ciia vat rSn • Khi tinh momen lue, ta ap dung eong thuc M = Fd va luu y rang M c6 the nhan gia tri duong hoac am, thuoc vao tac dung lam quay vat theo chieu duong hay am ciia tung lue, sau tinh tong momen M • Neu he gom hai vat chuyen dong quay thi each khao sat moi vat thuc hien nhu tren Neu he vat gom mot vat chuyen dong quay thi ap dung djnh luat Newton va eae eong thue nhu da biet lop 10 cho vat chuyen dong tinh tien (xem vat nhu chat diem) V i du 1: Mot vanh tron dong chat, khoi lugng m = 2kg, ban kinh R = 0,5m, true quay qua tam va vuong goc voi mat phang vanh Ban dau vanh dung U2J = 45=512=>Ii=512l2 ^ C h p n C yen thi chiu tac dung boi mot lue F tiep xiic voi mep ngoai vanh Bo qua moi V i du 2: Mot manh eo khoi luong m = kg, chieu dai 1,2 m Doi voi true quay di qua khoi tam va vuong goc voi thanh, momen quan tinh ciia bang: A 0,3 kgm^ B 0,6 kgm^ C 0,9 kgm^ Huang dan gidi: Ta c6: I = ^ m l ^ = ^ , ^ = 0,6kgm2 =^ Chon B D 0,12 k g m l ma sat Sau 3s vanh tron quay dugc mot goe 36 rad Do Ion eiia lue F la A, I N B N C 4N, D 8N Huang dan gidi: Taco: {p = | t ^ =36(rad)=>y = r a d / s ^ I = mR^ - , ( k g m ^ j Momen lue: M = F.R = I7 = 8.0,5 = ( N m ) Do Ion ciia lye F la: F = ^ = ( N ) => Chon D V i dy 2: Mpt dia dac dong chat, khol lupng 0,4kg, ban kinh 20cm, c6 tryc quay A di qua tam dia va vuong goc voi dla, dang dung yen Tac dung vao dTa mpt momen luc khong doi 0,16 N.m Tinh quang duong ma mpt diem tren vanh dia di dupe sau 5s ke tu liic tac dung momen luc A 50m B 32m C 64m D 128m Huang dan gid i: Momen quan tinh cua dia: I = ^mR^ = ^.0,4.0,2^ = 8.10"^ (kg.m^ Gia toe goc cvia dia: y = y = 20rad/s^ Dang 5: BAI TAP LifiN Q U A N D E N D I N H L U A T BAO T O A N MOMEN D O N G LU'ONG Phuang phap giai: Su dung cac kien thuc sau: ô*ã ã Momen dpng lupng L cua vat ran chuyen dpng quay quanh true la: L = I{o Trong do: I la momen quan tinh cua vat ran doi vai true quay; c la toe dp o goc ciia vat ran chuyen dpng quay quanh true • Tinh Momen dpng lupng cua h^ vat ran doi vai tryc quay theo cong thuc: n # L = Li + L + +Ln= Goc ma dia quay dupe: c - -^yt^ = -^.20.5^ = 250(rad) p Quang duong ma mpt diem tren vanh dia di dupe sau 4s la: s = R(p = 250.0,2 = 50 (m) => Chpn A Vi du 3: Mpt banh xe chju tac dung cua mpt momen luc M i khong doi Tong cua momen M i va momen luc ma sat c6 gia tri bang 24 N.m Trong 5s dau, toe dp goc ciia banh xe bien doi tu rad/s den 10 rad/s Sau momen M i ngung tac dyng, banh xe quay cham dan va dung hSn l^i sau 50s Gia su momen luc ma sat la khong doi suo't thai gian banh xe quay Momen luc M i CO gia trj A M l = 11,3 N.m C M l = 33,4 N.m B M l =26,4 N.m D M i = 48,6 N.m i=i Trong do: Lj=IjCOj , • Neu momen luc bang khong thi ap dyng dinh luat bao toan momen dpng lupng: Ltnrac = Ls.iu • Neu momen luc khac khong thi ap dung djnh ly ve dp bien thien mo men dpng lupng: ^ -M dt Vi dy 1: Mpt cung, nh^, manh dai Im, quay xung quanh mpt true vuong goc vai va di qua tam GSn vao hai dau hai chat diem c6 kho'i lupng la m, = 2kg va = 1,5kg Thanh quay deu quanh tryc, moi qua cau CO toe dp la m/s Tinh momen dpng lupng cua he A 3,55 kgm^s B 4, 65 kgm^s Huang dan gidi: C 6,75kgm^s D 8,75 kgm^s Huang dan gidi: Ban dau: M = M j + M^^ = 24(N.m) Toe dp goc cua moi qua cau: Toe dp goc ciia banh xe liic CO M i : =^ = - lO(rad) Momen dpng lupng ciia h^: CO = COQ + yt -> = 2rad / s^ L = Lj + L = IiCOj + I2CO2 = (ij +12 )to Momen quan tinh ciia banh xe: , M 24 2\ I = - = - = 12(kg.m2) = ^miR^+m2R^)(o = (mi+m2)R^.a) = 3,5.0,5h^l0 = 8,75(kg.m2/sj Khi M l thoi tac dung: co' = to + y't = ' f t' y' = -0,2rad / s^ Suy ra: M^^ = y'l =-0,2.12 =-2,4(Nm) Vay M l = M - M ^ , =26,4(N.m).=>ChpnB => Chpn D Vi dy 2: Mpt nguoi c6 khoi lupng 50kg dung mep mpt san quay hlnh tron CO khoi lupng 200 kg, duong kinh 4m Bo qua ma sat true quay va sue can khong Ban dau san va nguoi deu dung yen Nguoi bSt dau chay voi toe dp khong doi 4,2 m/s (doi voi dat) quanh mep san a) San se quay n h u the'nao? Toe d p goc cua san la bao nhieu? A c u n g chieu v a i chieu chay ciia n g u o i v a i toe dg l , ( r a d / s) B nguoc eiiieu v o i chieu chay eiia n g u a i v o i toe d o - l , ( r a d / s) C eiing chieu v o i chieu chay eiia n g u o i v o i toe d o 3,05(rad / s) V i d u 3: M o t san quay h i n h t r u c6 k h o i l u g n g M = 180kg va ban k i n h R = 1,2m dang d u n g yen M o t n g u o i k h o i l u g n g m = 40kg chay tren m a t dat v a i toe d o 3m/s theo p h u a n g tiep tuyen v d i mep san va nhay len san Bo qua ma sat true quay T i n h toe d o ciia n g u o i k h i o tren san A 0,52 m/s B 0,72m/s lai chay theo p h u a n g tiep tuye'n eiia san tai M v a i toe d o n h u tren H o i toe ' '» ' ' C - , 5 ( r a d / s) B 1,425 (rad/s) L , = 1,(0, = m R ^ ^ = m R v j M o m e n d g n g l u g n g eiia he k h i n g u o i nhay len san: D -1,425 (rad/s) Huang dan gidi: L2=(li+l2)«2 K h i n g u o i chay q u a n h mep san t h : m o m e n d o n g l u g n g ciia he k h i d o la: Hay L | = IjCOj +I2CO2 Do m o m e n d o n g l u g n g duge bao toan, nen: L i = Lo Tir d o : V2 = - ^ ^ ^ i ^ _2m(o^ _ CO, D o 0)1, C02 khae dau nen san se quay theo chieu nguge v o i chieu chay ciia n g u a i v a i toe d o goc: ©2 = = = = - l , ( r a d / s) => C h g n B b) K h i n g u a i chay den d i e m M t h i : L2 = I,o), +12(02 D o m o m e n d g n g l u g n g d u g e bao toan nen: L2 = Lo Suy ra: ^=-M._ ^2 " ' 50 , Thay so: CO2 = - - — , = 2.200 ' Chgn C 14 mv ' 2 i M 40+ ^ Dang B A I T A P y2y T R O N G (0, = ^"^2 4,2 4.R 4.2 m + — M R.v ^ ' D E N D O N G N A N G C U A VAT D O N G RAN QUAY P h u o n g p h a p g i a i : Su d u n g eac kien thuc sau: • D g n g nang eiia vat ran quay q u a n h m o t true co d j n h : Wd = ^Ico^ • D g n g nang ciia vat ran vua quay vira chuyen d g n g tjnh tien: ^ + — mvr- ^ v o i V(^ la van toe k h o i tam mjCOj v 13 CHUYEN m2R^ v2 R = 2^.0,92(m/s) 180 LIEN QUAN W J == -ICO (0, ^ Chgn D r,coj +I2CO2 = Tudo: = m+ = mR2+-MR2 m R , = m + — M R.v Suy ra: + l2«;= ^ = Theo d j n h luat bao toan m o m e n d g n g l u g n g , ta eo: L2 = L i a) Ban dau he d u n g yen nen m o m e n d o n g l u g n g : Lo = 1,0, , M o m e n d g n g l u g n g ciia n g u o i truoc k h i nhay len san: goc cua san k h i d o la bao nhieu? A 0,525(rad / s) D 0,92m/s Hu&ng dan gidi: D nguge chieu v a i chieu chay eiia n g u o i v o i toe d o - , ( r a d / s ) b) N g u o i chay dan vao giua san k h i den d i e m M each true quay I m t h i C 0,82m/s V i d u : M o t k h o i cau dac d o n g chat eo khoi l u g n g m = 80 kg, ban k i n h 0,6m / \ = - , 5 ( r a d / s) K h o i cau quay q u a n h true d i qua tam ciia no T i m d g n g nang ciia k h o i cau k h i toe d o goc ciia khoi cau la 20 rad/s A 1234J B.2304J C 3156J D 4758J Huang dan gidi: Huong dan gidi: Do banh xe quay deu, nen tong momen lyc: Momen quan tinh ciia khoi cau la: I = - m R ^ = -.80.0,6^ = l l , ( k g m ^ ) Dgng nang cua khoi cau la: W j = ijco^ = 11,52.20^ = 2304(j) => Chon B M =M +M ^ =0 M „ , = - M = -60(N.m) fO-CO(, 0-50 r / ' j / \ Gia toe goc ciia banh xe: = — ^ — = = -5^rad / s j V i dy 2: Mot banh da hinh try khoi lugng m = 100kg, ban kinh R = 30cm chju tac dung ciia mot momen lyc khong doi M = 36 N.m , , , , M -60 Momen quan tmh: I = — = —r- = 12kg.m y b a) Can bao nhieu thoi gian de tang toe banh da t u trang thai nghi den toe dp goc bang 12 rad/s Cong cua lyc ma sat: A ^ , = AW, - W , o = - i l o ) ^ - =-15000(j) = -15kJ A 0,5s B 1,5s C.2,5s D 3,5s b) Dpng nang ciia banh da a toe dp goc bang bao nhieu? A- 123J B 231J C 324J D 543J Huang dan gidi: a) Momen quan tinh cua banh da: I = ^mR^ i.100.0,3^ = 4,5|kg.m^j Gia toe goc cua banh da: y = — = — = sf rad / s^) I 4,5 V / Thoi gian can thiet: t - ^^Z^ = => Chpn C V i du Mot banh da dupe lam quay quanh mot tryc thang dung voi toe dp goc khong doi 200 rad/s Mpt dTa kim loai dung yen eo 16 thiing tarn rod nh? xuong banh da dang quay eho tryc xuyen qua 16 thung Ma sat giiia dla va banh da lam cho dia quay nhanh dan deu den toe dp goc 200 rad/s thoi gian 0,5s Momen quan tinh ciia dia doi voi true quay la 6.10"^ kg.m^ Xae djnh: a) Momen lye ma sat tac dung len dia = ] ( ) ^ Chpn B A 0,6Nm b) Dpng nang ciia banh da toe dp goc la: A.40J D 3,6Nm C.80J B.60J D 120J c) Nang lupng toan phan da truyen cho dia A 80J B A I T A P L I E N Q U A N D E N D I N H L Y B I E N T H I E N C 2,4Nm b) Dpng nang dia nhan dupe = i l t o ^ = 1.4,5.122 = 324(J) =^ Chpn C ^ang B l,2Nm B 120J Phuong phap giai: Djnh ly ve dpng nang:: AWd = Wd2 - Wdi = ll(co2 - 0)2) = A = MA(p Trong do: As = rAcp Chu y: Doi voi cong phat dpng c6 gia trj duong (A > 0), doi voi cong can c6 gia tri am (A < ) V i dy 1: Tac dyng vao banh xe mot momen lyc khong doi M = 60N.m thi banh xe quay deu voi toe dp goc co = 50rad/s Thoi tac dung momen lyc M thi sau 10s banh xe se dung lai Tim cong ciia lyc ma sat tac dung vao banh xe ke t u thoi tac dung momen lyc M den banh xe dung lai A -3kJ B.-9kJ C.-15kJ D -23kJ D 240J Huang dan gidi: D Q N G N A N G C U AV A T R A N Q U A Y C 200J a) Gia toe goc cua dia: y = =^ = 400(rad / s^) Momen lyc ma sat tac dyng len dia: M = ly = 6.10"^.400 = 2,4(N.m) =>ChpnC b) Dpng nang dia nhan dupe: W j = ^Ico^ = i.6.10"^2002 = 120(j) • => Chpn D c) Nang lupng de thang c6ng cua lye ma sat can tro ehuyen dpng eua banh da, giir cho banh da quay voi t6c dp goc khong doi (thanh phan chuyen nhi?t): AW = A = AWd - Wd - Wdo = 120 G) Nang lupng toan phan truyen cho dia la: W = W , + A W ^ + _ ^ ( j ) => Chpn D THUV!ENTiNHBINH25^U^ 17 Huong dan D UMN = V ; L = , H C UMN = V ; L = , H M Jh ^^yTNHH B UMN = V ; L = , H A UMN = V ; L = , H gidi: H a i gia t r j cua Z L cho c u n g m o t c u o n g d o hieu d u n g t h i : R L , Ro MTV DWH Khattfj Vi^t N Zc=|(zLi+ZL2)-200i^ Huang dan gidi: D u a vao d p I f c h pha gifi-a cac N D o Z L J < Z L nen Z L J < Z c ^ cpi - (P2 =-| M a t khac, (pi = - 92 Giai he ta co: cpj = - - ; (p^ = ^ 6 dien ap v o i n h a u va v o i d o n g dien ta CO gian d o vecto n h u h i n h ve Theo de U^jvj = Uf^^p nen tarn giac M N P la tarn giac can tai M , Ui M : H ^ - ^ M N I Voi sincpj = — • 6' tan 100-200 R = 10073Q =:> Chon B M H la d u o n g t r u n g tuyen Cau 83 M o t d o a n m a c h d i e n xoay chieu n h u h i n h ben H o p d e n X g o m m o t L P =ULtg30''=15V3V;U MN - u i - - = 30N/3V hoac mQt so l i n h k i ^ n R , L , C mac nol tiep D i e n ap giOa hai d a u h o p X tre pha han c u o n g d o d o n g d i f n t r o n g mach ^ H o p X chua cac p h a n tir „_,o cos 30 MtL A R no'i tiep v o i C UR=UPQ=UMN=30V3V ^BB ^ H C C h i chua p h a n tit L 90V U ^Hp R Huong dan L = 0,0827H => C h o n D d o hi?u d u n g , n h u n g pha ciia hai d o n g d i e n d o khac n h a u m p t goc — R va Zc V3 D Ca A , B, C c u n g sai 364 y, ^ L , r = Oj R B ^ nen d o a n mach t r o n g h o p X p h a i chua phan ti'r C hoac p h a n t u L va C n h u n g (loan mach t r o n g hgp den p h a i co t i n h d u n g khang Suy dap an D la d i i n g Chgn D Biet: UAB = 120 ( V ) ; 200 R - ^ Q gidi: £3u_84 C h o m a c h d i f n n h u h i n h ve: CO gia t r j Ian l u g t bang: B Zc = 200Q; R = 0 V Q ^ H p ) i e n ap hai d a u h o p den tre pha h o n c u a n g d o d o n g d i e n t r o n g mach t r i ciia Z L la Z^^ = lOOQ, Z ^ ^ = 300Q cho ta hai d o n g d i ^ n co c i i n g m p t cuang A Z c = 200Q; R = ~ Q V3 X Z^ Zc C Mach CO R, ZL va Zc voi va ZL < Zc ' B R = 25>/3fi va L = 318mH , Huong dan gidi: vay la tam giac vuong tai N NB 60 tga = AN A R = 25N/3Q vaC = 0,127mF A A RvaR Xi J X? B R va C C L va C r* D L va R Hwang dan gidi: Vi dp l^ch pha ciia di^n ap tren hai dau hai hpp den la = ^ nen chi c6 hai truong hpp sau: Hpp chua R, hpp chua C hoac hop chua R, hpp chua L Ki thuatgiai nhanh BTTN Vat Li 12, t&p - Nguyen Quang Lac cfinh cac p h a n t u mac t r o n g X, Y va gia t r j ciia chiing M a t khac, theo dau bai t h i doan mach k h o n g cho d o n g d i e n mQt chieu chay A X chiia Rx = 30Q, L x = H - H ; Y chiia Rv = C O , Cy = qua nen t r o n g mach phai chua p h a n tit C Suy p h u o n g an B la d i i n g V37t => C h o n B B X chua Rx = 30Q, Cau 88 Cho doan mach A B g o m bien t r o n o i tiep v o i h o p k i n X H o p X chi chua cupn thuan cam L hoac t u C D i e n ap hai dau mach c6 gia t r i hieu durig I U = 200(v), tan so' f = 50Hz K h i bien trcr c6 gia t r i cho cong sua't mach cue C = - ^ F B.XchuaL = - H 2n n 10''* C.Xchua C = - — F , chiia t u C = \/2 D i e n d u n g cua t u : C = = W2 10"^ = 27tfZ^ ; Y chiia RY = 400, Cy = ^ F i /; ' ' , = — H ; Y chua RY = 40Q, Cy = ^ F IOTI 471 gidi: day t h u a n cam Lx V i cupn day thuan cam k h o n g c6 tac d u n g can t r o d o n g D o n g d i ^ n tuc t h a i sam pha h a n dien ap hai dau mach d o d o t r o n g hop D u n g khang ciia mach: ZQ = H bai t h i X chua t r o n g ba p h a n t u nen X phai chua d i ? n t r o t h u a n Rx va cupn d i f n m o t chieu nen: ZQ = = ^ H o p X cho d o n g dien m o t chieu d i qua nen X k h o n g chua t u d i ^ n Theo de gidi: C o n g sua't mach cue dai k h i Zc = R => Z = ^ + C X chua Rx = 30Q, 47371 ,-3 Huong dan n Huong dan 3V3 ^ D.XchuaL = - H 7t 7371 D X chua Rx = 30Q, H o p X chua p h a n t u nao va gia t r j ciia no la bao nhieu? A.Xchiia 30 n "3 = - T ^ H ; Y chua RY = 40Q, Cy = - ^ F IOTI dai t h i I = \/2 ( A ) va d o n g dien tiic t h a i som pha h a n dien ap h a i dau m a c h ~ F 471 42.ZQ j R^ = Hi = ^ = 30Q ^ I, K h i mac A , B vao n g u o n dien xoay chieu: VaihppXtaco: Z A M = ^R^ + ^ =Hi = ^ IT = 100(Q) = Jz^ -'AM v2v2 (F) ^ Chon C ^ T C Cau 89 Cho d o n g dien xoay chieu d i vao m o t hop k i n t h i tha'y mach k h o n g ticu R> thu dien nang va c u a n g d o n g d i ^ n sam pha han d i ^ n ap D i e u khang d i n h ^AM = A H o p k i n chua d i e n t r o - R = ^ - ^ = 30V3D 1071 tg(pAM = nao sau day d i i n g ? ^ = 73 rad Ta ve d u p e gian vec to B H o p k i n chua tu d i ^ n = 60Q A >Au h i n h ben: C H o p k i n chua cuon cam D H o p CO the chua t u dien hoac cuon cam Huong dan URV =UMB-sin^ = 40V ^ Ucy gidi: =UMB-COS^ = 40V3V ^ R Y = - ^ = 40Q UAB B M a c h k h o n g tieu t h u dien nang => mach chua L hoac C M a c u o n g d o n j ' d i e n sam pha h a n d i f n ap => mach chua C => Chon B Cau 90 Cho hai hop k i n X, Y chi chua ba phan t u : R, L (thuan), C m^*^^ noi tiep K h i mac hai d i e m A , M vao hai cue ciia m o t n g u o n dien m o t chieu tin A m p e k e c h i 2(A), V o n ke'Vi chi 60(V) K h i mac hai d i e m A , B vao hai c\fc c u ' CY = 10 ,-3 47371 Zc^ = ^ = 40V3Q F => C h p n C mpt n g u o n d i e n xoay chieu tan so'SOHz t h i A m p e k e c h i 1(A), V o n ke Vi ch' 60(V), V o n ke V2 chi 80(V) va u ^ j ^ lech pha so v o i u ^ g m o t goc lnl3 Xa"^ 369 Cau 91 C h o m^ch di^n xoay chieu n h u hinh ve X la mpt hpp den chua phan A Ux =25%/T4.cos lOOTtt- 41n (V) 180^ L tu R h o § c L holic C biet U ^ B = IO0V2cosl007ct(V); ^ = V ( A ) , P = 0 W , 4171"! B Uy = 25728.cos lOOTlt10-^ - ( F ) , i tre pha hon u^g T i m cau tgo X va gia trj cua phan tu? 37t C Uy=257l4.cos A Hpp den chua phan tU R = 40i7 lOOTtt- B Hpp den chiia phan tu C = 3nF D Uy = 25728 cos lOOTlt- C Hpp den chua cupn day khong thuan cam L = — H ; r = 50Q D Hpp den chua cupn day thuan cam L = — H 1 X A H K = ^(5072)^ +(10072)^ rB B 4071 150, (V) 4071 150 (V) -2.50.100.cos| =50V6 = > U L = UC = 2576 ( V ) Y\g dp dong di|n tre pha hon dif n ap nen hpp X phai chua phan tu L den phai chua cupn day khong thuan cam Vay hpp den la mpt cupn day co r (pxw =50(n) HE 2576 73 OH => tg(px = va R Mat khac theo dau bai hpp den chua phan tu R hoac L ho^c C nen hpp Taco: P = i^T^t = — = N Ta lai c6: H K = U L + Uc = 2UL = 2Uc Huang dan giai: 100 M J , Huotig dan giai: Theo bai ta ve dupe gian nhu hinh ben: Xet tam giac O H K ta c6: c -i X 180, 5o7^ 4l7[, 180 (rad) I Ux = N / O H ^ T H E ^ - ^ + 50^2 = (v) Vgy phuang trinh: ,Ux = U x 72coS(ioo;, _ (px) = 25 728 cos (IOOTI - Mgitkhac: r ^ + ( Z L - Z c ) = — ) (V) 180 => Chpn D ZL-ZC Giai ra: Z, = 80(w) =^ L = ^ = — ^ ^ ' (0 lOOTt = — ( H ) => Chpn C 57t^ ' £3iL?2.Cho mach di?n nhu hinh ve Biet X la hpp chua phan tix L'' Ri,Ci mac noi tiep Dat vao hai dau mach di?n ap xoay chieu c6 tan so C = lOOn (rad) thi di^n ap mgch c6 u^^j = 100.cos(l007tt) (V)' O Cau 93 Khi mac Ian lupt mpt difn trd thuan, mpt ty difn va mpt cupn day thuan cam vao mpt di^n ap xoay chieu (c6 di^n ap hifu dung U va tan so'f) thi cuong dp dong di^n hi^u dung chay cac phan tu c6 gia tri Ian lupt la 5A; 1,25A; 2,5A Neu dat vao hai dau doan mach gom difn tro thuan, ty di?n cupn day mac noi tiep vai di^n ap xoay chieu ke tren thi dong di^n chay 3ng mach c6 gia trj hi^u dyng la A 75A B.275A C.375A Huang dan giai: D.47lOA ' "MB =200 cos l O O T t t - - (V) Phuang trinh di^n ap hai dau hpp X la: k 371 P o i v o l mach RLC: U = ^\^m+{^iv => C h o n A Cau 94 Cho mach di#n xoay chieu n h u h i n h ve D i ^ n ap hai d a u doan mach la • U h a i d a u doan m^ch s6m pha ^ so v o i i V o n ke Vj va v o n ke V2 Ian l u g t chi R A IO0V2V va IO0V2V Va R = Z L = Zc (do h i ^ u dien the tren cac phan t u bang nhau) IL P o i v o i mach RL: P U = ^/u^,, + B = =.U2K=-^ = ^ ^ I ^ M = V^^OO^ + (lOO - lOO)^ = lOOV ' u ^ g = 200sin(l007it)(v), d i ^ n t r o cua cac v o n ke R y * +Uicf iP = U2R ^/2 (do Z j = R ) = 50V^V =>ChonB B 150V va 50V Cau 96 P a t d i ^ n ap xoay chieu c6 gia t r j h i ^ u d u n g 200V va tan so k h o n g d o i C lOOV va lOOV vao h a i d a u A va B cua doan mach mac n o i tiep theo t h i i t u g o m bien t r o R, cuon cam t h u a n c6 d o t u cam L va t u dien c6 d i ^ n d u n g C thay d o i G p i N la D 200V va 200V diem n o i g i i i a cupn cam thuan va t u d i f n Cac gia t r i R, L, C h i i u han va khac khong V o i gia t r j C = C i t h i dien ap h i ^ u d u n g giOa h a i d a u bien t r o R c6 gia Huang dan gidi: tri k h o n g d o i va khac k h o n g k h i thay d o i gia t r i R ciia bien t r o V o i gia t r i Q Ta c6: u AB C = =>UAB Voi t h i d i e n ap hieu d u n g g i i i a A va N bang = " A M+ " M B = U A M A 200 V + U M B U A B = U L + U C ; B IO0V2V K h i C = C j t h i d i e n ap hieu d u n g g i i i a hai dau R la: tarn giac v u o n g can nen: U R = I R = 200 A M - '-'MB - 4i ' 1^ , ^ I >^.(z,-Zcf = 100V Pe U p k h o n g p h y thupc vao R t h i Chpn C Cau 95 P a t vao mach d i f n R L C h i ^ u dien the xoay chieu c6 hieu d i ^ n the hieu t C 60V2 V Huang dan gidi: I P 9o72 V t h i Zc^ =2Zc^; \5\^-l'Z\^ = ^/R27Z[ = ^ / R ^ ^ (1) UAB=I'ZkB v o i bang va bang TOO V Ne'u lam ngan mach t u d i ^ n (noi tat hai b a n cue cua no) t h i h i ^ u d i e n the hieu d u n g tren d i ^ n t r a thuan R la K h i C^^C^ = ^ v o i z;,^ d u n g k h o n g d o i Biet h i ^ u d i ^ n the h i ^ u d u n g tren cac p h a n t u R, L, C deu B ^ V = ZQ_^ UAM S o c h i ( V i ) = So chi (V2) =100V A 50V D.200V2V Huang dan gidi: U A M - U R T u gian d o ta c6, tarn giac O A M la U C lOOV ZAB = +(ZL - )' = +(ZL - 2Zcj)' = ^ Z2 ^ (2) \r (1) va (2) ta thay Z;^^ = Z^^B nen U;^^ = ^ A B = U = 200V => C h p n A T a c o : U = ^t i ( U , - U c f ^ 97 M o t m a y bien ap c6 so v o n g cupn so cap la 3000 vong, cupn t h u cap ^OO vong, d u p e mSc vao m a n g d i ^ n xoay chieu tan so 50Hz, k h i d o c u o n g d p '^ong d i ^ n qua c u p n t h i i cap la 12A C u o n g d p d o n g d i ^ n qua cupn so cap la : V l A B.2A C.3A P.4A Chia ve vol ve'ciia (2) va (3) ta dup'c: Huang dan gidi: ' i , , Ap dyng cong thirc may bien ap: ^ DO UT I, N, II N -y^ = -rr- 500 ~ 12 = 2A => Chpn B Ui" ^ N , ^ 3000 Cau 98 Mot may bien the c6 ti so so' vong day ciia cupn so cap va thu cap la B Giam hi$u di^n the va dong di^n 10 Ian , C Tang hi^u di^n the 10 Ian va giam cuang dong dien 10 !an L • D Giam hi?u difn the 10 Ian va tang cuong d dong di^n 10 Ian O l2_U, I1 u N Tac6-A dyng giiia hai dau cugn thii cap de ho la 100 V cuon thu cap, neu giam hot n vong day thi di§n ap hifu dung giua hai dau de cua no la U, neu tang them n vong day thi di?n ap la 2U Neu tang them 3n vong day cuQn thu cap thi di?n ap hi^u dyng giua hai dau de ho ciia cupn bang C 220V D 120V -i'i Ul_N2 Ui N =>•< N, vai U2=U: _u_ N2 Cong suat m^ch t h i i cap: P2 = U2I2 = 96(W) => Chpn C Cau 101 Di^n nang mpt tram phat di?n dupe truyen d i voi di^n ap 2kV, hi?u suat qua trinh truyen tai la H = 80% Bie't cong suat truyen tai khong doi Muo'n hifu suat truyen tai dat 95% thi ta phai lam gi? A Tang dien ap len k V U2=2U: C Tang di?n ap len den k V N2 +n " N, B Giam di?n ap xuo'ng I k V D Tang di?n ap len k V =-=^U2=2U,=4kV =^ Chpn C duong day tai) la 20 kV, hi^u suat ciia qua trinh truyen tai di^n la H = 82% Khi N cong suat di$n truyen di khong doi, neu tang di^n ap (a dau duong day tai) len den 60 kV thi hi?u suat ciia qua trinh truyen tai d i f n se dat gia trj (2) A H = 92% 2U • Cau 102 Diqn nang o mpt tram di?n dupe truyen di duoi di?n ap (6 dau N j - n U2 Neu tang n vong day cugn thii cap thi: voi Hi n V I , = 16(A) N Tac6:4 = ^ = ^0,2.p U2 Ap2 (1) u, D 240V - 96W :6(V) Ni I, C V - W Huang dan gidi: voi U = 0 V Neu giam n vong day a cupn thii cap thi: Chpn A Huang dan gidi: yinN, Huang dan gidi: Ta c6: = U = 200V B 120V-4,8W I1 phi) mOt dien ap xoay chieu c6 gia tri h i | u dyng khong doi thi di?n ap hieu (4) " N., ^ 0,8A thi hi^u di^n the hieu dyng va cong suat mach t h u cap la: '1 U, l ^ = i L ^ChpnC U, 10 B.300V Ni rong va 100 vong Hi?u di$n theva cuang dp hi?u dyng a mach sa cap la 120V ^ = 10Ui Cau 99 Pat vao hai dau cupn so cap ciia mpt may bien ap l i tuong (bo qua hao A 200V " :au 100 Mpt may bien the c6 so vong day ciia cupn so cap va t h u cap la 2000 A 6V-4,8W Huong dan giai: N N, _ N2 + N _ 2N2 H i _ -,^2 ^ U'1 ^'1 A Tang hi?u dien theva dong di?n len 10 Ian .U2=U, = =i> N , = 3n T u (4) va (1) ta thay: 1:10 Bo qua hao phi may bien the Dung may bien the c6 the: Ta co: ^ - — i N, U., N j - n Neu tang 3n vong day a cugn thu cap thi: _ N2 +3n N2- do: I , = —-ly U, N2 f n _N2+n U, N (3) A my' B H = 94% •'• C H = 98% D H = 96 % Huong dan gidi: Gpi P la cong sua't truyen di Cong suat hao phi tren duong truyen la: AP = R cos^ cp H i e u suat cua qua t r i n h t r u y e n tai d i # n nang la: C a u 105 M o t m a y bien ap c6 so v o n g cupn so cap va t h i i cap Ian l u p t la 2200 v o n g va 120 v o n g Mac cuon so cap v o i m a n g dien xoay chieu 220V - 50Hz, k h i P • P Khi U = U i - ( k V ) , d i ^ n ap hieu d u n g g i i i a hai dau cuon t h u cap de h o la: :.,;V C h p n C C a " ^06 M p t t r a m phat dien t r u y e n d i m p t cong suat lOOkW tren day d a n c6 dien t r o R = 8Q D i e n ap t u t r a m phat dien la U = lOOOV C o n g suat hao p h i =>ChonC C a u 103 M o t m a y bien ap c6 hi?u suat bang 1, so v o n g day cua cuon so cap va t h u cap Ian l u g t la 1000 v o n g va 50 v o n g D i e n ap hieu d u n g cupn so cap 220V, c u o n g d p d o n g d i e n la 0,18A B.4A C.6A Huong A.80kW dan Huang dan D A A.^y = R r A ^ = gidi: ,1:, gidi: = 80000 = k W £a» 107 D u o n g day tai dien c6 dien t r o f i d a n dien t u A den B D i e n ap hieu d y n g A la 5000V, cong suat la 500kW H e so cong suat tren d u o n g day tai la U J i coscpi , ^ R = P, => U , ! , coscp, = U J , coscpi => I2 = -rH = ' 0,8 H i e u suat tai d i ^ n la: # A 87,5% =>ChpnB C a u 104 D i ^ n n a n g cua m p t t r a m phat d i ^ n dupe t r u y e n d i d u o i d i ? n ap 2kV va cong suat 200kW H i e u so chi cua cong to t r a m phat va cong to tong noi AP _ P D 40kW;95% gidi: C o n g suat hao p h i tren d u o n g day bang hieu so c h i cac cong t o t r o n g \h, Vixc C 79,5% Huang dan d u o n g day va h i ^ u suat cua qua t r i n h t r u y e n tai di?n Ian l u p t la C 20kW; 90% B.88% A T a c : I = — A P = H Ucosip tieu t h u sau m o i d e m chenh l^ch 480 k W h C o n g suat hao p h i tren Huang dan D.70W > Chpn A H i ? u suat cua m a y bien ap bang nen: B.83kW;85% C.70KW I C o n g suat hao p h i tren day d a n la cong suat toa nhiet tren day dan: Tac6::^ = ^=>U,=U,-^-^ = llV A 0 k W ; % B.80W so cong suat ciia mach so cap va mach t h u cap Ian l u p t la va 0,9 C u o n g dp h i ^ u d y n g mgch t h u cap la: A A ^en day d a n la gidi: Rl2=_5^ RP D 77,5% ^ ^ U^cos^cp "u^cos^cp'S P-AP T u do: H = = 0,875 = 87,5% => C h p n A ^ u J C o n g suat t r u y e n d i cua m p t t r a m phat dien la 200kW H i ^ u so chi cac cong to d i e n a t r a m phat va n o i t h u sau m p t d e m I f c h n h a u la bang: , ^ = 20(kW) C o n g suat den n o i tieu t h u la: 200 - 20 = 180 (kW) H i ^ u suat qua t r i n h t r u y e n tai: H ^ 1^.100 = % => C h p n C 200 ^hem k W h H i ^ u suat tai dien la ,A.70% B.80% C.90% Huang dan D 95% gidi: l?ch cua so chi ciia cong to d i ^ n giCra n o i t h u va t r a m p h a t la cong hao J k h i t r u y e n tai t r o n g m p t d e m la: jCtyTNHH m A = 480kWh ^ A:-y^ = A t = 20000W P — AP AP Hi?u sua't truyen tai di?n nang: H = — - — = - -— 24.3600 V|y hi?u suat truyen tai di?n la: H = ^ — ^ 0 % MTV DWH = 90% w Tu(l)va(2)=:> - H = — = P-!^ P ^ Chpn C Cau 109 Dung rngt hi^u di$n the 6kV de tai di mot cong sua't lOOkW den mot Suy ra: - = noi each xa 7,5km Cong suat ton hao tren duong day bang 3% cong suat car, P-i-; - = Khang Vm (2) ' ) P-V > truyen di Tim kho'i lugng dong can thiet de lam day dan biet di?n tro suat cua dong bang 1,7.10 Tim, khoi lugng rieng cua dong D = 8,8 tan//w ? A m = 6,117 tan C m = 4,117 tan D m = 3,117 tan : CMML Mot may phat di^n xoay chieu mpt pha c6 roto quay voi toe dp 300 B m = 5,117 tan ^ * M Huang dan gidi: Cuong dp dong d i f n tren day tai bang: I - I - I - V> -50 ^phat A ^ ••>••'•• Cong sua't hao ton tren duong day la: Pj = 3?/oPphat = 0,03.100 = 3kW p Di?n tra cua day tai: R = -|- = 10,8Q ' l2 The tich cvia day: V = SI = p — ^ R 1^ , Kho'i lugng ciia dong can diing: m = VD - p—D = 3,117 ta'n R => Chpn D Cau 110 Di^n nang a mpt tram phat di^n dupe truyen d i v6i di?n ap dau duang day tai la 5kV, hi|u sua't qua trinh truyen tai la 80% Muo'n hi^u sua't qua trinh truyen tai tang den 95% thi phai A tang di$n - ap len den lOkV j' B tang di^n ap len den 15kV Huang dan gidi: Cong sua't hao phi tren day dan toa nhi?t: ^ " ' l ? y vong/phiit, may c6 10 c^p eye t u thi dong di^n phat c6 tan so: B 10s-i A 50S-' C 20s' D 100s' k Huang dan gidi: P Ta c6: f = np = — 10 = 50(Hz) = 5o(s"^) =i> Chpn A 60 ' ' *' ^ Cau 112 Noi hai cufc cua mpt may phat di?n xoay chieu mpt pha vao hai dau doan mach AB gom di^n tro thuan R mac noi tiep voi cupn cam thuan Bo qua dien tro cac cupn day cua may phat Khi ro to eiia may quay deu n vong/phiit thi cuong dp dong di^n hi^u dyng doan mach la l A Khi roto cua may quay deu vai to'c dp 3n vong/phut thi cuong dp dong di?n hi^u dyng doan m^ch la 3A Ne'u roto cua may quay deu voi to'c dp 2n vong/phiit thi cam khang ciia doan m^ch AB la A 2RV3 B ^ C RN/3 D V3 V3 Huang dan gidi: Tan so' bie'n thien va sua't d i f n dpng eye dai may phat di^n xoay chieu ipt pha tao la: f = —-; EQ = (oNBS = 27tfNBS => Di^n ap hi?u dyng hai dau 60 B>9n mach la: U = E = ^ = V27ifNBS P* Khi roto ciia may quay deu vai toe dp nj = n va n j = 3n thi f2=3fi=>ZLi=3Z^;U2=3Ui C giam di^n ap xuong 2,5kV D giam di$n ap xuong 0,85kV Theo bai ra: Ij = A; I j = N/3A r=> I2 = -Tsii = ^ = N/3 ^-^ •=> Z2 > Z2 3> R2 + ( Z , J ' = 3(R2 + ZL J'=:> Z^ = Z, if Cau 116 Cho mach d i # n R, L, C mac no'i tie'p D o n g d i # n qua m^ch d u p e c u n g cap b o i may phat d i e n xoay chieu m p t pha K h i t r o n g mach xay h i ^ n t u p n g cpng h u o n g dien t h i tan so'ciia d i f n ap dat vao hai dau doan mach la 50 H z T i n h so v o n g quay d u p e ciia roto ciia may phat dien t r o n g phiit? Biet rang phan cam ciia may phat d i ^ n xoay chieu m p t pha c6 cap cue? Kt thudt xiai nhanh BTTN Vcit Li 12, tap - Nguyen Quang Lac K h i roto cua m a y quay deu v o i toe d p nj = 2n t h i f3=2fi=>ZL3=2Zn=^^ChonB A 1200 v o n g Cau 113 M o t m a y phat d i ^ n xoay chieu m p t pha c6 cap eye, roto quay v o j B 1500 v o n g Huong dan toe d p 900 vong/phiit M a y phat dien t h u hai c6 cap cue De hai d o n g d i ^ n d o cac may phat hoa d u p e vao eiing m p t mang d i ^ n t h i m a y phat d i ^ n t h u hai B 1100 vong/phiit C 600 vong/phiit D 800 vong/phiit '" Huang dan gidi: ^au 117 Ilia m o i cupn day ciia p h a n l i n g xa'p xi bang 60 A 50 v o n g B 100 v o n g i n, = P2 = 600 vong/phiit ^ C h p n C Cau 114 Phan u n g ciia m p t may phat d i ^ n xoay chieu c6 100 v o n g day giong n h a u T u t h o n g qua m p t v o n g day c6 gia t r j eye dai la m W b va bien thien dieu hoa v o i tan so 50Hz Suat di?n d p n g ciia may c6 gia t r i h i ^ u d y n g la : A 111,25V B 133,28V Huang dan C 145,11V D 157,21V T u tinh duoc: E = V2 D 140 v o n g gidi: T u t h o n g qua k h u n g c6 N v o n g day: ^ft ())^ N.B.S.cos((ot + (p) = (pQ,.cos(o)t + (p) V o i (J)Q^ la t u thong cue dai qua mpt v o n g day Suat dien d p n g cam l i n g tiic thai bang dao ham bac nha't ciia t u thong theo thoi gian n h u n g trai dau: e = -(()' = NBSa)sin(cot + cp) = EQ sin(a)t + (p) Eo = NBSco = NcDo27tf Oo = m W b = ->'•- C 120 v o n g Huong dan gidi: Suat d i ^ n d p n g cue d a i giCra hai d a u cupn day p h a n l i n g la: voi M p t m a y phat dien ma phan cam g o m hai cap cue t u quay v o i toe dp p n g hi^u d u n g 220V, t u t h o n g cue d a i qua m o i v o n g day la m W B So' v o n g fi = b ^^^^ 60 :r 500 vong/phiit va phan l i n g g o m hai cap cupn day mac n o i tie'p, c6 suat d i ^ n De hai d o n g d i ? n cac m a y phat hoa dupe vao eiing m p t m ^ n g di^ n thi Do d o : = 60 gidi: Vay so v o n g quay dupe t r o n g m p t p h i i t la: 25.60 = 1500 v o n g =>ChpnB Tan so' f ciia d o n g d i ^ n may phat d i ^ n xoay chieu m p t pha c6 p cap cue va roto quay v o i toe d p n vong/phiit: f = D 2100 v o n g Ta CO f = np, suy n = — = 25 vong/giay p phai CO toe d p quay ciia roto la A 550 vong/phiit C 1800 v o n g 6.10-m ' Voi E , , - E V = NBSco = " NBS^^ 60 So v o n g day ciia m o i cupn day ciia phan u n g (c6 cupn day) la: = 133,28V Ev'^ = NBSco = N B S ^ 60 =:> C h p n A => C h p n B Cau 115 Phan l i n g ciia m p t may phat d i | n xoay chieu c6 200 v o n g day giong = 4.(t)oj.27inp = 49,5 50 v o n g * Cau 118 Phan c i i n g ciia m p t may phat d i ^ n xoay chieu c6 200 v o n g day g i o n g n h a u T u t h o n g qua m p t v o n g day c6 gia t r j cue d a i la , m W b va b ie n thien T u t h o n g qua m p t v o n g day c6 gia t r i eye dai la m W b va bien thien d i e u hoa voi tan so 50 H z Suat d i ? n d p n g m a y p h a t c6 gia t r j hi?^ dieu hoa v o i tan s6'50Hz Suat dien d p n g ciia may c6 gia t r j cue dai la: d u n g la A 8,88 V B 11,54 V Huang dan Taco: ^ = ^ ^ C 14,36 V D 18,66 V gidi: A 120V B 125,7V C 120N/2V D 220V Huong dan gidi: Tu" thong cue dai qua vong day ciia cupn day phan l i n g : ^ ^ ^ ^ ^ - ^ " - ^ =8,88(v) 381 cpo = m W b = 2.10-3 wb C&u 121 M p t m a y p h a t d i ? n ma phan cam g o m hai cSp eye t u quay v o i toe d p 1500 v o n g / p h i i t va p h a n l i n g g o m hai cupn day mac no'i tiep, eo suat d i ^ n d p n g Suat d i ? n d p n g c\tc d g i giua hai dau cupn day p h a n u n g : hi$u d y n g 220V, t u t h o n g cue dai qua m o i v o n g day la m W b M o i cupn day Eo = N < o CO =125,7V => Chpn B P Cau 119 M p t m a y phat di?n xoay chieu c6 p h a n cam g o m hai cap eye va quay v a i v a n toe 1500 v o n g / p h i i t Phan u n g g o m cupn day g i o n g n h a u mac noi g o m CO bao n h i e u vong? A 74 v o n g B 97 v o n g A 21 v o n g B 27 v o n g Ta c6: f = C 44 v o n g D 78 v o n g Huong dan gi&i: " " i ; v.-?.- : : ' - 60 = 50Hz 220V2 CO 9oô A ã N =j^^^^^ = 60 , g EQ = (oN(po = 27tfN(Po = 27tfpnN(po n p _ 1500.2 Eo = N (PQ Taco: vong D 198 v o n g Huang dan gidi: tiep T i r t h o n g eye dai qua m o i v o n g day la m W b Suat d i ? n d p n g h i § u d y n g may tao la 120 V So v o n g day cua m o i cupn day la C 156 v o n g = 198 v o n g 5.10-^271.50 •ChpnD Cau 122 M p t m a y phat d i f n ba pha mac h i n h c6 diC^n ap pha 127V N g u o i d u a d o n g ba pha vao ba d i f n t r o giong m5c h i n h t a m giac, m o i tai c6 N :i?n t r o lOOfi C o n g suat tieu t h y ciia cac tai la Suy so v o n g day tren m o i cupn la: N ' = — = 27 v o n g A 1452 W B 1600W C.3210W D.4800W Huang dan giai: => C h p n B Cau 120 M o t m a y phat d i ^ n ba pha mac h m h c6 dien ap h i f u d u n g p h a Cac di?n trcic g i o n g nen c u o n g dp d o n g dien qua c h i i n g la n h u bang 127V, tan so 50Hz N g u o i ta d u a d o n g ba pha vao ba tai n h u mac V i cac d i ^ n t r o d u p e mac h i n h t a m giac nen d i ^ n ap dat vao m o i tai bang dien h i n h tarn giac, m o i tai c6 dien t r o a va cam khang 32Q T o n g cong suat cac ap day C o n g suat tieu thy bang cong suat tieu thy tren tai tai tieu t h y la P = 3UIcoscp = U I = - ^ = 1452W A.3121W B.2178W ^ C 4242W D 1116W =>Chpn A Huang dan giai: Cau 123 M p t m a y p h a t dien xoay chieu tren stato c6 b o n c^p eye quay v a i toe H i ^ u di?n the dat vao m o i tai la: dp 750 v o n g / p h i i t tao d o n g d i ^ n eo tan so f De dat d u p e tan so' tren v a i may Ud=UpN/3= 127V3«220(V) phat d i ^ n c6 sau cap cue phai quay v o i toe dp bao nhieu ? T o n g t r o cua m o i tai la: Z = ^R^+Zl A 200 v o n g / p h u t = ^24^ + 32^ = 40(fi) D o n g dien tren m o i tai la: Z , 40 ^ B 300 v o n g / p h u t C 500 v o n g / p h i i t ' :;; C o n g suat tieu t h y tren m o i tai la: Pi=Rl2=24.(5,5f =726(W) T o n g cong suat cac tai tieu t h y la: t n i ;jr! Huong dan giai: t*^ ' D 600 v o n g / p h i i t v; , j, 750 Ta eo: f = n i p i = = 50Hz • •}• ^ V o i m a y phat d i ? n eo e | p eye, ta eo: f = n2p2 = 6n2 T u do: n2 = f 50 vong/giay P2 P = 3Pi=2178(w) Toe d() quay cua may bang: —.60 = 500 v o n g / p h i i t => C h p n C => C h p n B 383 Can 124 lA)n;; dK.-a d o M u i t n i i } p l M i d u i i xoay chieu phat sau tang ap du(?c truyen di xa bSng mpt duong day c6 di#n tra lOO Coi h$ so cong sua't bang 1, dien ap dua len duong day la 35kV, cong suat cua may phat la HOOkW.Cong sua't hao phi tren duong day la A IIOOOW B 14000W I ' C 15000W D 16000W Huong dan gidi: ' Ta c6: R = lOfi; costp = ', ^ ' Dien ap dua len duong day: U = 35kV = 35.10^ V Cau 127 Mpt dong ca khong dong bp ba pha c6 cac cuon day mac theo kieu hinh vao mang di?n xoay chieu ba pha c6 difn ap day 380V Dpng co c6 cong sua't 5kW va h? so cong sua't 0,8 Cucmg dp dong difn chay qua dpng cola t Cau 125 Dien ap hieu dung giira hai dau mot pha ciia mpt may phat dien xoay chieu ba pha la 220V Trong each mac hinh sao, dien ap hieu dung giua hai day pha la: C.381V D 513V B Ba cuon day ciia may phat m5c theo hinh tam giac, ba cupn day ciia dpng C O theo hinh tam giac C Ba cupn day ciia may phat mac theo hinh sao, ba cupn day ciia dpng co theo hinh llOV Tai ciia cac pha giong nhau, mac hinh tam giac, moi tai c6 di^n tra thuan va dung khang 12Q (mac noi tiep) Cong sua't tieu thu cua D Ba cupn day ciia may phat mac theo hinh sao, ba cupn day ciia dpng co theo hinh tam giac dong ba pha la: C 1831W D 2904W Huang dan gidi: : Tong tro cua m6i tai: Z = + [Z^ - Z^f A Ba cuon day ciia may phat mac theo hinh tam giac, ba cuon day ciia dpng C O theo hinh => Chon C Cau 126 Mot may phat dien xoay chieu ba pha mac hinh c6 dien ap pha in B 1567W = 220V Cau 128 Mpt dpng co khong dong bp ba pha hoat dpng binh thuong dien ap hieu dung gii>a hai dau moi cuon day la 220V Trong chi co mpt mang dien xoay chieu ba pha mpt may phat ba pha tao sua't di?n dpng hieu dung a m6i pha la 127V De dpng co hoat dpng binh thuong ta phai m^c iheo each nao sau day? ' Trong each mic hinh c6: U j = VsUp = 220N/3 = 381V A 1231W Huang dan gidi: Cong suat ciia dpng co bang tong cong sua't tren ba pha ciia dpng co: P = 3UpI.cos(p^ I = « 9,5A => Chpn C > ' '' ' ^ 3UpCos(p Huong dan gidi: 24Q, cam khang D 5,9 A V3 Chon D B 142V C 9,5 A dpng C O bang dien ap pha ciia mang dien ba pha: U = ^ Cong sua't dien hao phi tren duong day: AP = RI^ = 16.10^ W = 16kW A 120V B 6,5A Cac cuon day dupe mac theo kieu hinh nen dien ap dat vao moi pha cua ^ ^ ; " Cong sua't dong dien tren day phat: P = 1400kW = 1,4.10^ W p Cuone done dien tren day: I = = 40A " • Ucoscp A 5,6 A A Huang dan gidi: • B a cupn day cua may phat theo hinh thi dien ap hieu dung giua hai day - ^jlA^+{30 - uf = 30Q Do cac tai mac hinh nen dien ap hieu dung tren moi tai: Wiala: - VsUp = 12/73 = 220V Ba cupn day ciia dpng co theo hinh tam giac thi di?n ap hieu dung dat vao i6i cupn day ciia dpng co la 220V, dpng co hoat dpng binh thuong Ud = V U p = i i o V v => Chpn D _Ud_110V3 11 Cuong dp dong dien qua moi tai: I = -— = — = Cong suat tieu thu cac tai: P = 3Pi = RI^ = 3.24 => Chpn D 384 riio^' = 2904VV 129 Mpt dpng C O xoay chieu hoat dpng binh thuong voi di^n ap hieu g 220V thi sinh cong sua't co hpc la 170W Bie't dpng co co he so cong i't 0,85 va cong sua't toa nhiet tren day qua'n dpng co la 17W.B6 qua cac hao i khaccuong dp dong dien cue dai qua dpng co la A 72 A B 2A C 3A D 373 Huang dan gidi: Cong sua't toan phan bSng tong cong suat co ich va cong suat hao phi: va di?n ap hi^u dyng hai dau cupn day la 135V Gia trj cua Uo gan gia trj nao nhat sau day? A 95V Ptp=Pci+Php=187W B.75V C 64V.; D 130V Huang dan gidi: Cong suat tieu thu dien nang ciia dpng co: P(p = UIcoscp Taco => = 187 Ucos(p = lA=>Io = =>l2 = 3Ii =^Z,=3Z2 =45(V); U^^ = ( V ) ^ ^ =IV2-V2(A) 220.0,85 => Zl= 9Zl => Chon A Cau 130 Stato cua mot dong co khong dong bo ba pha gom cuon day, cho Hay dong di^n xoay chieu ba pha tan so 50Hz vao dong co Tit truong tai tam cua R2 + ( z ^ - Z^^ f = 9R2 + z, — ^ stato quay voi van toe bang bao nhieu? A 1000 vong/phiit B 1500 vong/phiit C 2000 vong/phvit D 2500 vong/phiit = > ( R + z ) = ZLZe^ Vi Huang dan gidi: = —- n (Pi nen tan cp, tan ^2 = - (vi (pj < 0) Trong stato co cuon day tuong ung p = cap cue, t u truong t^i tam ciia stato quay voi toe do: n = Zc = 1500 vong/phiit => Chon B Ma tancpj = P R —i-;tan(pj=- R R Cau 131 Mot dong co dien xoay chieu mot pha tao mot cong sua't co hoc Zr 630W va CO hieu suat 90% Dien ap hieu dung hai dau dpng co la U = 200V, Suy R hf so'cong sua't ciia dong co la 0,7 Cuong dong di?n qua dong co la A 2,5 A B.5A C 7,5 A ZL-ZC R ^ ^ J D l O A Hay 3R2 + SZL^ - 4ZLZCI + Z^^ = Huang dan gidi: =^ 3(R2 + ZL2 ) - 8(R2 + Zi,2) + Z^^ = Dua vao dinh nghla hif u sua't ciia dong co va cong thiic tinh cong sua't ta dugc: =^z2^ = 5(R2 + ZL^)r) H =i = 90% P(p = U.I cos (p P H.Ucoscp 630 90%.200.0,7 = 5A T u (*) va ta suy Z^^ = 2,5ZLZCI => Zci = 2,5ZL 2(R2 + Z ^ ) = ZLZCI = 2,5 Z^ : ^ Z L = 2R va Z o = 5R (***) Ch(?n A \ Suy ra:Zi2 = R + ( Z L - Z c J Cau 132*: Dat dif n ap u = UQ COS cot ( V ) (voi UQ va co khong doi) vao hai dau doan mach gom cuon day khong thuan cam mac noi tie'p voi tu dien co dien =>ZI = R V I O =10R2 va Zj^ = ^ R + Z ^ -RVS dung C (thay doi duQ'c) Khi C = CQ thi cuong dp dong di^n mach som pha hon u la (p < (Pi < - '1 _ U va di^n ap hi^u dung hai dau cupn day la 45V ^ J Khi C = 3Co thi cuong dp dong dien mach tre pha han u la 92 ~ ^ ' il Zdj Zjj U =U,,—L '1 z = UH,V2 Do Uo = U N/2 = 2Udi = 90V => Chpn A 3^ Cau 133*: Dat di#n ap u = 120A/2 cosZnft (V) (f thay doi dugc) vao hai dau Cau 135: Dat dien ap u = 220^2 cosl007tt(v) vao hai dau doan mach mac noi doan mach mic noi tie'p gom cuon cam thu an c6 dp t u cam L, dien tro R va tu dien C O di^n dung C, voi CR^ < L Khi f = fi thi dien ap hieu dung giua hai dau tie'p gom di^n tro O , cupn cam thuan c6 dp t u cam tu dien dat cue dai Khi f = f2 = fiV2 thi di^n ap hieu dung giua hai dau dien tro dat cue dai Khi f = b thi dien ap hieu dung gii>a hai dau cupn cam dat cue dai ULmax 10"^ Gia tri cua Ui max gan gia trj nao nhat sau day? ' A 173 V B.57V C 145 V dung - g ^ F - Khi c = p Cuong dp dong di?n cue dai IQ = = TooTT ^'^^ i = 2,2cos 1007tt + -4 j v( A ) r ^ C h p n C • A = 4 V => Chpn B Huang dan gidi: ZL , - (V) => Chpn C Cam khang = 207^Q; l , = ^ = ^ = 11(A) Z 20V2 ^{Z^-Z^f U Q R = I Q R = 11.20 = 220V; U Q L = I Q • Z ^ = n = 8 V = 80N/3V = , V * 145 C i = 2,2cos 1007tt + - ( A ) H va t\ di^n c6 di^n Theo gia thie't U , = 0 V = Hi > =^ (i) Ki hi^u may bie'n ap M c6 so' vong day moi cupn tuong ung la N j , Di?n ap hieu dung giira hai dau so cap va thu cap la U j / U j t Khi noi hai dau so ca'p may => — L = _ J L 12,5 M2 vao hai dau thu cap may M i : (2) ^ ' Khi noi hai dau ciia cugn thii- cap cua M2 voi hai dau cupn thii cap cua M i thl di%n ap hi^u dyng hai dau cupn so cap cua M de ho bang 50V 50 =^ (3) N; MUC L U C Chii de 1: Dpng luc hpc vat ran Tir (2) va (3) CO =^ U i = N / I A = V H? thong hoa kien thuc Thayvao(l)c6Hl = ^ Cae dang bai tap va v i du minh hpa =8 Bai tap t u luy^n c6 huong dan each giai => ChQn C Cau 137: Mpt khung day dan phSng, d?t, hinh chu nhat c6 di?n h'ch cm^, quay deu quanh mpt tryc dol xung (thupc mat ph^ng cua khung) t u truang deu c6 vecto cam ung t u vuong goc vai true quay va c6 dp Ion 0,4 T Tu thong cue dai qua khung day la A 2,4.10-3 Wb B l,2.10-3Wb C 4,8.10-3Wb Huang dan gidi: Ta CO t u thong cue dai OQ = BS = 0,4.60.10"'' = , - ^ W b => Chpn A D 0,6.10'^Wb 19 Chii de 2: Dao dpng ca 68 thong hoa kien thuc 68 Cac dang bai tap va v i dy minh hpa Bai tap t u luy|n eo huong dan each giai Chu de3: Song ca 73 » 99 173 He thong hoa kien thiic 173 Cac dang bai tap va v i dy minh hpa 180 Bai tap t u luyen eo huong dan each giai 194 Chu de 4: Dao dpng va song di$n tir 231 He thong hoa kien thuc 231 Cac dang bai tap va vi du minh hpa 235 Bai tap t u luyen c6 huong dan each giai 247 Chu de 5: Dong dif n xoay chieu H ^ thong hoa kien thuc 291 291 Cac dang bai tap va vi du minh hpa 295 Bai tap t u luy^n c6 huong dan each giai 324 www.nhasachkhangviet.vn L w S H ® •••••• % CONG TY TNHH MOT THANH VIEN DWH KHANG VIET • Nha Sach • KHANG VIET • Oia chi:71 OinhTien Hoang - P.Oakao-Quan -Tp HoChf Minh oien thoai: (08) 39115694 - 39105797 - 39111969 - 39111968 Fax: (08) 39110880 Email: khangvietbookstore@yahoo.com.vn Website: www.nhasachkhangviet.vn 935092 Gia: 89.000 525594 ... i tac d u n g TAT - - m v , Wd2=-m2V +-mv B^B mgt dia tron k h o i l u g n g m2 = 1kg Ian k h o n g t r u g t tren mat p h i n g ngang v o i + W^g 1 Wj3 =-Ico^ = - - m R 2 xr- D.2,3J gidi: W... =1400(4.1 0-2 )'' =32.1 0-2 J fChi d i qua vj t r i X = 2cm , the nang ciia 16 xo: ; '' = i k x =1400(2.1 0-2 )^ =8.1 0-2 J N a n g l u g n g ciia 13c W = ^ k A '' ^ D o n g nang: W j = W - W, = 32.1 0-2 - 8.1 0-2 ... cua vat k h i no each v j t r i can bSng 2cm la A 43.1 0-3 J va 34.1 0-3 J B 32.1 0-^ J va 24.1 0-2 J C 65.1 0-3 J va 13 1 0-3 J D 51.1 0-3 j T2 - + •| T '' =f2.f2 2 0)^ = (Oj + CO2 C h u y : M o t 16