🔙 Quay lại trang tải sách pdf ebook Giáo trình lý thuyết mạch điện Ebooks Nhóm Zalo PGS. TS. l £ VAN b a n g Giao trinh L Y T H U Y E T M A C H O I E N H ■ Sach dung cho cac trudng Cao dSng va he Trung hoc chuyfen nghiep 4 (Tai ban Ian th u ba) (D A IH ftlH /fN G W fN [ T K P N G T A M H O C L l l g NHA XU AT BAN GIAO DUC Cong ty Co phan sach Oai hoc - Day nghe - Nha xuat ban Giao due giff quyen cong bo tac pham. Moi to chirc, ca nhan muon sti dung tac pham dwd moi hinh thirc phai dwoc si/ dong y cua chu so huv quyen tac gia. 04 - 2009/CXB/58 - 2117/GD M a so : 6 H 1 4 5 y 9 - D A I L o i g i o i t h i e u Nam 2002, Vu Giao due Chuyen nghiep - Bo Giao due va Dao tao da phoi hdp vdi Nha xuat ban Giao due xuat ban 21 giao trinh phuc vu cho dao tao he THCN. Cac giao trinh tren da dildc nhieu triidng sit dung va hoan nghenh. De tiep tue bo sung nguon giao trinh dang eon thieu, Vu Giao due Chuyen nghiep phoi hap eung Nha xuat ban Giao due tiep tuc bien soan mot so giao trinh, sach tham khao phuc vu cho dao tao d cac nganh : Bien -B ien tit, Tin hoc, Khai thac cd khi. Nhitng giao trinh nay tritdc khi bien soan, Vu Giao due Chuyen nghiep da giti de citdng ve tren 20 triidng va to chiic hoi thao, lay y kien dong gop ve noi dung de cutting cac giao trinh noi tren. Tren cO sd nghien citu y kien dong gop cua cac triidng, nhom tac gia da diiu chinh noi dung cac giao trinh cho phii help vdi yeu cau thitc tien hefn. Vdi kinh nghiem giang day, kien thitc tich luy qua nhieu nam, cac tac gia da co gang de nhitng noi dung ditdc trinh bay la nhiing kien thitc cd ban nhat nhitng van cap nhat ditdc vdi nhitng tien bo cua khoa hoc ky thuat, vdi thitc te san xuat. Noi dung cua giao trinh con tao sit lien thong tit Day nghe len THCN. Cac giao trinh ditdc bien soan theo hitdng md, kien thitc rong va cd gang chi ra tinh itng dung cua noi dung dittfc trinh bay. Tren cd sd do tao dieu kien de cac triidng sit dung mot cach phii hdp vdi dieu kien cd sd vat chat phuc vu thitc hanh, thitc tap va dac diem cua cac nganh, chuyen nganh dao tao. De viec doi mdi phitdng phap day va hoc theo chi dao cua Bo Giao due va Dao tao nham nang cao chat litdng day va hoc, cac tritdng can trang bi dii sach cho thit vien va tao dieu kien de giao vien va hoc sinh co dii sach theo nganh dao tao. Nhitng giao trinh nay cung la tai lieu tham khao tot cho hoc sinh da tot nghiep can dao tao lai, nhan vien ky thuat dang trite tiep san xuat. Cac giao trinh da xuat ban khong the tranh khoi nhitng sai sot. Rat mong cac thay, co giao, ban doc gop y de Ian xuat ban sau ditdc tot hcfn. Moi gop y xin giti ve : Cong ty CP sach Dai hoc - Day nghe, 25 Han Thuyen, Ha Noi. VU GIAO DUO CHUYEN NGHIEP - NXB GIAO DUC Cong ty Co phan sach Dai hoc - Day nghe - Nha xuat ban Giao due giff quyen cong bo tac pham. Moi to chirc, ca nhan muon sir dung tac pham dwdi moi hinh thde phai dwoc si; dong y cua chu sohiru quyen tac gia. 04 - 2009/CXB/58 - 2 1 17/GD M a so : 6 H 1 4 5 y 9 - D A I C huang 1 CAC KHAI NIEM VA DINH LUAT CO BAN VE MACH DIEN 1.1. MACH DIEN. CAC BIEN CO BAN : DONG DIEN VA DIEN AP De ti'nh toan mot cach g£n dung cac qua trinh dien tu trong cac thiet bi dien ngudi ta dung mot mo hinh toan hoc don gian hoa goi la mach dien. 0 day, nguoi ta chi quan tarn d6'n mot so (him han) cac dai luong dien la cac ddng dien va cac dien dp. Dong dien (hay ci/dng do dong dien) i(t) Trong H. 1.1.1 ve mot doan day dan g AB co tiet dien la S. Goi q(t) la luong : —----------- ^ dien tich duong chuyen dong qua tiet J dien S theo chieu tu A den B (chi boi mui a B ten tren hinh ve) o thoi didm t va Hinh l.l.l Aq = q(t + At) - q(t) la luong dien tich chuyen dong qua S theo chiSu tu A den B trong khoang thdi gian At thi tri so trung binh cua cudng do dong dien theo chi6u AB trong At la : i A B Irung binh Trj so trung binh nay cang chinh xac neu ta lay At cang nho. Cho At —> 0, ta dinh nghla cuang do dong dien theo cliieu AB tai thoi diem t la dao ham cua luong dien tich q(t) theo thoi gian : At—>o At at Theo quy uoc, nguai ta lay chieu cua dong dien la chi^u chuyen dong cua cac dien tich duong. Chieu duong cua dong dien trong m6i phan tu cua mach dien duoc ve bang mot mui ten. Vi du 6 H. 1.1.2a, trong doan mach giua hai diem A va B nguoi ta chon chieu di tir A den B lam chi£u duong cua dong dien. Chieu nay duoc ve bang mot mui ten hudng tu A den B di kem voi ki hieu i(t). Ta cung co the dung chi oA' AR Ad r'Ki nh,s„ H„rtno Ha r hon cho dong dien i(t), tuc la : i(t) = iAB(t). 5 1 Chitang 1 CAC KHAI NIEM VA DINH LUAT CO BAN VE MACH DIEN 1.1. MACH DIEN. CAC BIEN CO BAN : DONG DIEN VA DIEN AP De ti'nh toan mot cach gin dung cac qua trinh dien tu trong cac thiet bj dien nguoi ta dung mot mo hinh toan hoc don gian hoa goi la mach dien. O day, nguoi ta chi quan tarn ddn mot so (huu han) cac dai luong dien la cac ddng dien va cac dien dp. Ddng dien (hay ci/dng do dong dien) i(t) Trong H. 1.1.1 ve mot doan day din 5 AB co tiet dien la S. Goi q(t) la luong —-----------— dien ti'ch duong chuyen dong qua tiet dien S theo chi6u tu A den B (chi boi mui a B ten tren hinh ve) 6 thoi didm t va Hinh l.l.l Aq = q(t + At) - q(t) la luong dien tich chuyen dong qua S theo chi6u tu A den B trong khoang thoi gian At thi tri so trung binh cua cudng do dong dien theo chi6u AB trong At la : i = ^ - AB trung binh a . Trj sd trung binh nay cang chinh xac neu ta lay At cang nho. Cho At —> 0 , ta dinh nghla cuong do dong dien theo chieu AB tai thoi diem t la dao ham cua luong dien ticli q(t) flieo tlioi gian : Aq _ dq(t) >o At dt Theo quy udc, nguoi ta lay chieu cua ddng dien la chi6u chuyen dong cua cac dien tich duong. Cliieu duong cua ddng dien trong mdi phan tu cua mach dien ducrc ve bang mot mui ten. Vi du 6 H. 1.1.2a, trong doan mach giua hai diem A va B ngudi ta chon chieu di tu A den B lam chidu duong cua ddng dien. Chieu nay duoc ve bang mot mui ten hudng tu A den B di kem vdi ki hieu i(t). Ta cung cd the dung chi sd AB d6 chi chieu duong da chon cho ddng dien i(t), tuc la : i(t) = iAB(l). 5 Hinh 1.1.2 Nguoi ta thudng chi chieu duong cua dong dien bang m6t mui ten dat nhu 6 H. 1.1.2b. Don vj cua cudng do dong dien trong he don vj quoc te SI la ampere, viet tat la A. Dong dien noi chung co tri sd thay ddi theo thdi gian. Vi du / n \ IabCO = Kt) = 10>/2sin lOOTCt + — V 6 A la bieu thurc cua mot ddng dien hinh sin (thudng goi la ddng dien xoay chi^u), cd bidn do la 10\/2 A. Dudng bieu di6n cua i(t) duoc ve tren H. 1.1.3. Hinh 1.1.3 Vi i(t,) = 12.25A > 0 nen tai thdi diem t, dong dien i(t) co chieu thuc trung vdi chidu duong cua no, tire la tai thdi diem t| dong dien co trj so la 12,25A va chay theo chieu tu A den B. Tai thdi diem 1 1 S + T ^ r r S = 600 ta co *2 600 cot, = 1007t 100 7 In 600 ~6 / va i(t2) = 1 0 V 2 sin = 10%/2 cot1 + ^r 1= 10\/2 sin ^ - = -12,25A Vi i(t2) = -12,25 A < 0 nen tai thdi diem t2 dong dien i(t) co chieu thuc nguoc vdi chieu duong cua no, tire la tai thdi diem t2, ddng dien i(t) cd tri sd tuyet doi la 12,25A va chay theo chidu tu B den A. Trong trudng hop don gian nhat, ddng dien cd chieu va tri sd khong doi theo thdi gian i(t) = I = hang sd, duoc goi la ddng dien khong doi (thudng goi la ddng dien mot chieu). i(t) = I A>B t i(t) I a) b) Hinh 1.1.4 Dien ap u(t) Ngudi ta goi hieu so dien the tu diem A den diem B cua mot mach dien la dien dp uAB(t). uab(0 = vA(t) - vB(t) (1.1.2) O day, vA (hoac vB) la dien the cua diem A (hoac diem B) duoc ti'nh doi vdi dien the cua mot diem O nao do cd dien the chon bang khong v0(t) = 0. Don vi cua dien the va dien ap trong he don vi quoc te SI la volt, viet tat la V. Chidu duong cua dien ap uAB giCra hai di6m A va B duoc chi bang mot mui ten hudng tu diem A den diem B ve tren doan mach kem theo kf hieu u(t) nhu ve d H.1.1. 5. 7 Khi chieu duong cua dien ap giSa hai diem da duoc chi ro bang mot mui ten tren hinh ve thi ngudi ta co the bo qua chi so AB B U ( t ) ma chi viet u(t) thay cho uAB(t). Hinh 1.1.5 Trong trudng hop chung, dien ap giua hai diem cua mach dien la mot ham so cua thdi gian t. Vi du u(t) = uAB(t) = 220V2sin 1007it + — V la mot dien ap hinh sin co bien do la 220V. C) thdi diem t,= — staco u ( t, ) = 2 20\f2sin— = 269,44V ouu Vi u(t,) = 269,44V > 0 nen tai thdi diem t, dien ap u(t,) co chieu thuc trung vdi chieu duong da chon, tuc la tai thdi diem t| dien the diSm A cao hon dien the diem B la 269,44V. Tai thdi diem t2 = s ta cd V / Vi u(t2) = - 269,44V < 0 nen tai thdi diem t2 dien ap u(t2) co chieu thuc nguoc vdi chieu duong da chon. Turc la, tai thdi diem t2 dien ap u(t2) cd tri sd tuyet doi la 269,44V nhung hudng tu B den A, noi cach khac dien the diem B cao hon dien the diem A la 269,44V. Trong trudng hop don gian nhat, mot dien ap u(t) cd chieu va tri sd khong thay doi theo thdi gian : u(t) = U = hang sd duoc goi la dien dp khong doi. thudng goi la dien dp mot chieu. Ddng dien i(t) va dien ap u(t) la hai bien co ban cua mach dien. Cac dai luong khac deu cd the dinh nghla thong qua hai bien co ban nay. Dudi day ta se xet mot sd phan tu co ban nhat cua mach dien. 1.2. NGUON DIEN AP (HAY NGUON SL/C DIEN DONG) De tao ra dien ap dat vao mach dien, ngudi ta dung cac nguon dien. Vf 0 trong (1.4.1) la thong sd dac trung cua phan tir dien trd duoc goi la dien trd (resistance) cua phan tu dien iro. He thuc (1.4.1) cd ten la dinh luat Ohm. Don vj cua dien trd trong he SI la ohm, viet tat la Q . He thiic (1.4.1) con cd the viet dudi dang sau : Ur U) 1r (t) - = G.uR (t) (1.4.2) Hinh 1.4.1 R Trong do : G = — duoc goi la dien dan. Don vi cua dien dan trong he SI la siemens, viet tat la S. Vdi chieu duong cua uR(t) giong vdi chieu duong cua iR(t) thi * tieu thu bdi dien tro R la : C° nS su3t n.4.3) 10 pK(t) = u R(t).iR(t) = R.iR(t) Don vj cua cong suat trong he SI la watt, viet tat la W. Vi R > O v a iR(t)> O n e n : pR(t)> 0 (1.4.4) Dieu nay co nghla la dien trd R luon tieu thu dien nang. Dien nang nay duoc bien doi thanh nhiet nang, tieu tan vao moi trudng xung quanh. Vi the, ngudi ta noi ring dien trd R dac trung cho hieu ting nhiet ciia ddng dien (hieu ung Joule). Nang luong toa nhiet tren dien trd trong thdi gian tir thdi diem t() = 0 den thdi diem t duoc tinh theo cong thuc : (1.4.5) Don vj cua nang luong trong he SI la joule, viet tat la J. Ta cd : 1 joule = lwatt x lgiay hay 1J = 1 Ws Khi dien ap UR va ddng dien IR trong dien trd la dien ap va ddng dien kh6ng doi theo thdi gian thi djnh luat Ohm duoc viet dudi dang : (1.4.6) hay (1.4.7) Vi du 1.4.1 : R Xet mach dien d H. 1.4.2 gom mot dien trd R = lOQdat vao ngudn sue dien dong E = 12V. Tinh dien ap UR, ddng dien IR tren dien trd. Tinh cong suat toa nhiet tren dien trd va nang luong tieu thu bdi dien trd trong mot gid. Hinh 1.4.2 Ta cd : UR = E = 12V Theo djnh luat Ohm (1.4.7) ta cd : Cong suat toa nhiet tren dien trd : PR = RIr = 10 x (1,2)2 = 14,40W hay pR = U RIR = 12x1,20= 14,40W 1 1 Difin nang tieu thu bdi dien trd na> trong thdi gian T = 1 gid la W = Pr T = 14,40 x 60 x 60 = 51,84 x l 0 3J Vi du 1.4.2: Dien trd R = 10Q duoc noi vao mot ngudn ddng dien cd tri sd Int; = 5A (H.l.4.3). Tinh dien ap UR tren dien trd va cong suit toa nhiet PR tren R dien trd Gicii : Ddng dien chay qua dien trd R chi'nh la ddng dien cua ngudn ddng da cho : I r = I„g = 5 A Diep ap tren dien trd UR bang : UR = R.Ir = 10Q x 5A = 50V Cong suat toa nhiet tren dien trd la : P R = U RIR = 50 x 5 = 250 W Hinh 1.4.3 hay : p = RI2 = 10 x (5) = 250W 1.5. DIEN CAM L. D|NH LUAT LENTZ Phan tu dien cam (inductor) thudng goi la ciion day dien cam la phan tu co ban cua mach dien ma dien ap tren no ti le vdi toe do bien thien theo thdi gian cua ddng dien chay qua no. Neu goi uL la dien ap giua hai cuc cua ph£n tu dien cam va iL la ddng dien chay qua no, vdi quy udc chieu duong cua uL cung chieu vdi chieu duong cua iL nhu d H. 1.5.la ta cd phuong trinh (vi phan tuyen ti'nh cap 1) : di, u, = L dt L -/'VTX. B Ae a ) Hinh 1.5.1 b) (1.5.1) B Trong (1.5.1), he sd ti le L > 0 duoc goi la dien cam (inductance) cua phan tu dien cam. Don vi cua dien cam trong he SI la henry, viet tat la H. 12 He thuc (1.14) la he qua cua djnh luat cam ung Lentz ve sue dien dong tu cam, thudng vie't la : e, =■di d /. . \ ' d r = -dTL la tu thong, eL la sue dien dong tu cam, vdi chieu duong chon giong vdi chieu duong cua iL nhu d H. 1.5.1b. That vay, vi cac chieu duong cua uL va eL chon nhu d H. 1.5.1 b, ta cd : V B = v a + £ l Tu do : eL = vB - vA Trong khi do : u( = uAB =vA - vB Tu (1.5.4) va (1.5.5), ta cd : u l - - e L Tu (1.5.6) va (1.5.2) ta cd u L = - eL= - -Ldidt (1.5.3) (1.5.4) (1.5.5) (1.5.6) = L i rdt(1.5.6) di, ,——la dinh luat Lentz va dien dt cam L la thong sd dac trung cho hieu tuang tu cam. Vdi chieu duong cua uL va iL chon giong nhau nhu d H. 1.5.la thi cong suat dua nang luong vao phan tu dien cam la : dt dt I 2Li =dw M dt 6 day : P l = ‘l ul =» l Ldt = Li. diL _ 1 (1.5.7) 1 T -2 W M = 2 L la nang lucmg tu trudng tich luy trong phan tu dien cam. (1.5.8) Do he thurc (1.5.7) ma dien cam L cung dac trung cho hien tuong tich tru nang luong tu trudng trong mach dien. vr du 1.5.1 : Xet mach dien d H.l.5.2 gom mot cuon day dien cam cd dien cam L = 0,1H noi vdi mot ngudn ddng ing(t)= 1 0 ^ 2 sin(lOOOt) mA. Hay tim dien ap uL tren cuon day dien cam, cong suat pL va nang luong tu trudng wM trong cuon day ? Ve cac dudng cong iL(t), uL(t), pL(t). 13 Gioi : Dong dien iL chay qua phin tu dien cam chfnh la dong dien cua nguon dong : iL(t) = ing(t) = 10V2sin(lOOOt) mA Chon chieu duong cua dien ap uL tren dien cam cung vdi chieu duong cua dong dien iL nhu 0 H.l.5.2. Theo dinh luat Lentz © I (1.5.1) ta co : Hinh 1.5.2 uL = L ^ = 0,1x A ( ioV 2 x 10“3 sin lOOOt) = 0,1 x 10V2 x 10~3 x lOOOcos lOOOt = V2 cos lOOOt V Cong suat dua nang luong vao tir trudng cua phan tu dien cam la : Pl = u l *l = \/2cos(lOOOt) x |l0V 2.10-3 sin lOOOt j = 20x1 O'3 sin lOOOt cos lOOOt W = 20 x sin lOOOt cos lOOOt mW „.sin2000t „ hoac pL = 20 ----- — - = 10sin2000t mW Nang luong tu trudng trong cuon day dien cam : WM = ^ LiL = | x 0Tx(l0>y2sinl000t)2 = i x 0,1 x ( 10V2 )" sin2 lOOOt - cos2000tN = 10sinz lOOOt = 10 = 5(l - cos2000t) J Cac dudng bidu diln cua iL(t). uL(t) va pL(t) duoc ve tren H .l.5.3 vdi Im = 10\/2 mA Um = sl2 V Pm=10 mW va 0) = 1000 rad/s. 14 -p„ Hinh 1.5.3 1.6. DIEN DUNG C. DONG DIEN CHUYEN D|CH Phan tu dien dung (capacitor) thucfng goi la tu dien la phin tu co ban cua mach dien ma dong dien qua no ti le thuan vdi toe do bien thien theo thdi gian cua dien ap tren no. Neu ic(t) la ddng dien chay qua tu dien va uc (t) la dien ap giua hai cuc cua tu dien vdi chieu duong cua ic va chieu duong cua uc chon giong nhau nhu d H. 1.6.1 thi phuong trinh vi phan tuyen tinh cap mot lien he giua ic(t) va uc(t) la : , duc (t) ic (t) - C- dt( 1.6 . 1) Trong do he sd ti le C > 0 duoc goi la dien dung (capacitance) cua tu dien. Don vi cua dien dung trong he SI la farad, viet tat la F. He thuc (1.6.1) la he qua cua gia thiet cua Maxwell ve ddng dien chuyen dich (displacement current) qua moi trudng dien moi giua cac ban cuc cua tu dien. dq d duc ic = - r - ^ “ (Cuc) = c ~ r " dt dt dt A+q Hinh 1.6.1 ( 1.6.2) 6 day ±q la dien tich tren cac ban cuc cua tu dien. Do (1.6.2) ma dien dung C dac trung cho hien tuong ddng dien chuyen dich. Vdi chieu duong cua uc va ic chon giong nhau nhu d H. 1.6.1 thi cong suat dua nang luong vao dien trudng trong khong gian giua hai ban cuc cua tu dien la : 15 duc _ d P c - > c uc - u c C dt - — ± C u J ] = d ^ . W_ldt 6 day w E = la nang luong dien trudng tich lay trong tu dien. (1.6.3) (1.6.4) Do he thiic (1.6.4) ma dien dung C cung dac trung cho hien tuong tich luy nang luong dien trudng trong mach dien. Vi du 1.6.1 : A Mot tu dien co dien dung C = 10|aF duoc noi vao mot ngudn ap hinh sin : u(t) = l(W2sinlOOOt V Hay tim dong dien qua tu ic va cong suat dua nang luong vao dien trudng cua tu dien. Giai : Hinh 1.6.2 Ta cd : uc = u(t) = 100\/2 sin lOOOt V Chon chieu duong cua ic(t) trung vdi chieu duong cua uc(t) nhu d H. 1.6.2 ta co dur ic = = lOx 10 6 x — ^ 100^2 sin lOOOtj = 10 x 10 ~6 x 100 V2 x 1000 cos 1 OOOt = V2 cos 1 OOOt A Cong suat tich luy nang luong vao dien trudng cua tu dien : pc = uc .ic =(l00V 2 sin lOOOt) x(V2 cos lOOOt) = 200( sin2QQOt^ = 100sin2000t W Nang luong dien trudng tich luy trong tu dien : we = ^ Cuc = 10*10"6 x(l00V2 sin 1000t)2 = — x 10 x 10~6 x 2(l00\/2 j2 sin2 lOOOt = 0 ,lsin 2 lOOOt = 0,11 - cos2000t= 0 .0 5 (1 -cos2 0 0 0 t) J 16 1.7. 0|NH LUAT KIRCHHOFF CHO CAC DONG (HAY D|NH LUAT KIRCHHOFF THLT NHAT) H. 1.7.1 ve m6t mach dien phiic tap. Ta goi nhanh cua mach dien la bo phan cua mach dien giua hai diem bat ki trong do cd cung mot ddng dien chay qua. 3 Hinh 1.7.1 Hinh 1.7.2 Mach dien d H. 1.7.1 gdm co 3 nhanh. Nhanh thu nhat (nhanh 1) gdm co ngudn sdd e(t) noi tiep vdi dien trd R,. trong hai phan tu nay cd cung mot ddng dien i] chay qua. Nhanh thu: hai (nhanh 2) gdm cd dien trd R2 noi tiep vdi dien cam L, trong hai phan tu nay cd cung ddng dien i2 chay qua. Nhanh thu ba (nhanh 3) gdm cd dien trd R-* noi tiep vdi dien dung C, trong hai phan tu nay cd cung mot ddng dien i? chay qua. Ta goi diem ma d do cac nhanh noi vdi nhau la nut cua mach dien. Trong mach dien d H. 1.7.1, cac nhanh 1, 2 va 3 duoc noi vao hai nut G) va ® . De thay ro cac nhanh va cac nut cua mot mach dien va su noi cua cac nhanh vdi nhau tai cac nut, ngudi ta cd the dung mot loai so do don gian hoa cua mach dien goi la graph cua mach dien. Vi du graph cua mach dien d H. 1.7.1 duoc ve nhu d H.l.7.2. Mdi nhanh cua graph tuong ung vdi mot nhanh cua mach dien va duoc ve bang mot doan dudng cong hay thang. Chieu cua nhanh tren graph tuong ung vdi chieu duong cua ddng dien cua nhanh. Ojnh luat Kirchhoff cho cac dong (hay djnh luat Kirchhoff thdf nhat) Su rang buoc giua cac ddng dien cua cac nhanh trong mot mach dien duoc thiet lap theo dinh luat Kirchhoff ve cac ddng (hay dinh luat Kirchhoff thu nhat). Djnh luat nay cd the phat bieu nhu sau : O mot thdi diem bat ki va tai mot nut bat ki cua mach dien, tong cac ddng dien co chieu duong di toi nut bang long cac ddng dien cd chieu duong rdi klioi nut. 17 Tai nut 1 tren H. 1.7.3 ta co i|(t) + i2(t) = i3(t) + i4(t) (1.7.1) Phuong trinh (1.7.1) co nghla la : ^ L + ^ T 1 = ^ 1 + ^ l h a y d q i + d q 2 = d 0 thi d thdi diem t xet nhanh k thuc su tieu thu nang luong. Ne'u pk(t) < 0 thi d thdi didm t xet nhanh k lai cung cap nang luong cho phan con lai cua mach. 23 Xet m6t mach di6n co m nhanh va n nui. 1 u cac puuung uum iap dinh luat Kirchhoff cho cac dong cho (n - 1) nut cua mach didn nut va cac phuong trinh lap theo dinh luat Kirchhoff cho cac ap cho (m - n + 1) vong co ban cua mach : X uk(l ) = 0 vong ta co the churng minh djnh ly Tellegen ve sir can bang c6ng suat trong mach dien sau day : "Trong mot mach dien bat ky, d mot thdi diem bat ky, tong cac cong suat tieu thu boi tat cd nhanh cua mach bang khong" m m X P k (t)= X Uk(t)ik(t)= 0 (1'9 1 ) k=l k=l hay noi cach khac, t6ng cong suat phat ra boi cac ngudn trong mach bang tdng c6ng suat tieu thu boi cac phan tu khac trong mach. Ta hay churng minh su can bang cong suat cho mach dien o H .l.9.3. Hinh 1.9.3 Ta viet phuong trinh theo dinh luat Kirchhoff cho cac dong cho nut l : - i , - i 2 + i3 = 0 (1.9.2) va cac phuong trinh theo dinh luat Kirchhoff cho cac ap cho cac vong I va II : U] - u2 = 0 ( 1.9.3) u2 + u3 = 0 (1 9 4 ^ 24 Cac phuong trinh (1.9.3) va (1.9.4) cung tucmg ducfng voi viec bieu dien dien ap cac nhanh qua dien the' cac nut (D va (D sau day: u, = v 2 - v , U 2 = v2 -V , U 3 = V , - v2 Tdng cong suat tieu thu boi ta't ca cac nhanh ciia mach la : £ P k = U 1M 1 ! + U 2 * 2 + U 3*3 k=l = (v2 - Vj)i| + (v2 - v,)i2 + (v, - v2)i3 = v ,(- i, - i2 + i3) + v2(+ ij + i2 - i3) Chu y den (1.9.2) ta co : Z P k =0 k=i Vay dang thuc (1.9.1) ciia dinh ly Tellegen da duoc churng minh. Vdi cJU true ciia cac nhanh 1, 2 va 3 nhu d H .l.9.3 ta cd u, = - e , + R,i| u2 = — e2 + R2i2 U3 R3t3 va J ] p k = u,i, + u 2i2 + u 3i3 = ( - e , + R ,i|) i, + ( - e 2 + R 2i2)i2 + ( R 3i3)i3 = 0 k = l Tirdo : e, ij + e2i2 = Rj if + R2i2 + R^i^ (1.9.5) O ve trai ciia phuong trinh (1.9.5) la tdng cong sua't phat ra ciia cac nguon sire dien dong e, va e2, con d ve phai ciia (1.9.5) la tong cac cong suat tieu thu tren cac dien trd R,, R2 va R3 ciia mach dien. 25 CAU HOI KIEM TRA 1.1. Mach dien la gi ? Cac bien cd ban cua mach dien la gi ? 1.2. Dong dien la gi ? Chieu duong cua dong dien la gi ? 1.3. Dien ap la gi ? Chieu duong cua dien ap la gi ? Cho mot so vi du. 1.4. Nguon dien ap la gi ? Quan he giufa dien ap va silfc dien dong ? Cho mot so vi du ve nguon dien ap. 1.5. Nguon dong dien la gi ? Cho mot so vi du ve nguon dong dien. 1.6. Dien trd R la gi ? Phat bieu djnh luat Ohm .Viet bieu thdc cua cong suat va nang lUdng tieu thu tren dien trd. 1.7. Dien cam L la gi ? Phat bieu djnh luat Lentz. Viet bieu thdc cua dien ap tren dien cam uL. Viet bieu thCrc cua cong suat va nang li/dng tu1 trudng trong dien cam theo L va iL. 1.8. Dien dung la gi ? Viet bieu thifc cua dong dien chuyen djch ic theo C va uc. Viet bieu thifc cua cong suat va nang lUdng dien trudng trong dien dung theo C va uc . 1.9. Phat bieu dinh luat Kirchhoff ve cac ddng hay dinh luat Kirchhoff thif nhat. Neu mach dien co n nut thi so toi da cac phuong trinh doc lap tuyen tinh co the lap dUdc theo dinh luat Kirchhoff thd nhat la bao nhieu ? Cho mot sd vi du ve he phUdng trinh doc lap tuyen tinh viet theo dinh luat Kirchhoff thCr nhat cho mot sd mach dien. 1.10. Phat bieu dinh luat Kirchhoff ve cac ap hay djnh luat Kirchhoff thtf hai. Neu mach dien co n nut va m nhanh thi so toi da cac phuong trinh doc lap tuyen tinh co the lap duoc theo dinh luat Kirchhoff thu1 hai la bao nhieu ? Bang cach nao cd the xac dinh duoc cac vdng doc lap cho mot mach dien? Cho mot sd vi du ve he phuong trinh doc lap tuyen tinh viet theo dinh luat Kirchhoff thCr hai cho mot sd mach dien. 1.11. Phat bieu djnh ly Tellegen 1.12. Chdng minh sir can bang cong suat trong mot sd mach dien don gian. 26 BAI TAP BT.1.1. LUdng dien tich q(t) chuyen dong qua tiet dien S cua doan day d in AB theo chieu tU A den B (H .B T.1.1a) co do thi theo thdi gian cho d H.BT.1.1b. Hay tinh dong dien iAB(t) va ve do thi cua dong dien theo thdi gian. a) b) H. BT.1.1 B T.1.2. Hoi nhu bai tap B T.1.1 vdi q (t) = I0 ”3V2 sin lOOOt C. BT.1.3. Cho biet dien the cua cac diem A, B, C va O nhu sau: V 0 = 0. VA = 100V, V B = 80V, V c = 50V. Hay ti'nh cac dien ap: UAB, UBC, UAC, U BA, UCB va UCA. B T.1.4. Hoi nhu bai tap B T.1.3 vdi: v0 = 0, vA = 100V2 sin lOOOt V v b ( 0 = 8OV2 sin lOOOtV » vc (t ) = 50>/2 sin lOOOtV B T.1.5. Mot dien trd R = 100Q dUdc noi vdi nguon dien ap u(t) = e(t) nhu d H.B T.1.5a. Do thi cua e(t) duoc ve d H.BT.1.5b. Tmh dong dien iAB(t) va iBA(t) trong dion trd. Ve dudng cong iAB(t) va iBA(t). Tinh cong suat va dien nang tieu thu bdi dien trd trong khoang thdi gian 0 < t < 20ms. H.BT.1.5 27 BT. 1.6. Giai bai tap BT. 1.5 trong tri/dng hop e(t) = 100n/2 sin IOOtu V B T.1.7. Mot dien trd R = 10 Q di/dc noi vao mot nguon dong dien j(l) d H.BT.1.7a. Di/dng cong j(t) difdc cho d H .BT.1.7b. Tinh dien ap uAB(t) va uBA(t) tren dien trd. Ve di/dng cong uAB(t) va UbaM - Tinh cong suat va dien nang tieu thu tren dien trd trong khoang thdi gian 0 < t < 20 ms. j(t) a) H.BT.I.7 BT. 1.8. Giai bai tap BT. 1.7 trong tri/dng hdp j(t) = 5\[2 sin lOOrct A. B T .1.9. Mot phan tif dien cam co L = 100 mH di/dc ndi vao mot ngudn dong dien j(t) nhtf d H .B T .1.9. Do thi cua j(t) cho d H .B T.1.7b. Tinh dien ap uAB(t) va uBA(t) tren dien cam. Ve di/dng cong uAB(t) va uBA(t). Tinh cong suat pL(t) va nSng li/dng w M difa vao tu1 trirdng cua dien cam trong khoang thdi gian 0 < t < 20 ms. u L id) H.BT.1.9 BT. 1.10. Giai BT. 1.9 trong trifdng hop j(t) = 5\f2 sin lOOnt A B T .1 .11. Mot tu dien co dien dung C = 10 difoc ndi vdi ngudn dien ap u(t) = e(t) nhir d H .B T .1 .1 1. Do thi cua e(t) di/dc cho d H .B T.1.5b. Tinh dong dien iAB(t) va iBA(t) qua dien dung. Ve e ( t ) ( ^ ) di/dng cong iAB(t) va iBA(t). Tinh cong suat pc(t) va nang li/png wE difa vao dien tru’dng cua tu dien trong khoang thdi gian 0 < t < 20 ms. u(t) B H 1 11 28 B T .1.12. Giai bai tap B T .1.11 trong tri/dng hdp e(t) = 100V2sinl007it V. B T .1.13. Cho mach dien d H .B T.1.13 vdi R = 100 Q, C = 5 ^F va e(t) = 100\/2sin lOOOttt V. Tinh iR(t), ic(t) va i(t). Ve cac di/dng cong cua iR(t), ic(t) va i(t). B T .1.14. Cho mach dien b H .B T.1.14 vdi R = 100 Q, L = 0,2 H va j(t) = V2 sin 10007tt A. Ti'nh uR(t), uL(t) va u(t). Ve cac di/dng cong cua uR(t), uL(t) va u(t). j(t)i -------- p A i H.BT.1.13 1---------- R u(t) r H.BT.1.14 L ; i B T .1.15. Cho mach dien b H .B T .1.15 vdi Ri = R2 = R3 = 1 Q, R4 = R 5 = R6 = 20 Q. Biet l4 = 1A va i5 = 2A, l3 = 3A. Hay tim I,, l2, l6, E 1t E2 va E3. H. BT.1.15 B T .1 .16. Tinh cac cong suat tren cac dien trd va cong suat cua cac nguon sdd trong mach dien cua B T .1 .15. Kiem tra can bang cong suat trong toan mach. 29 Chitting 2 DONG DIEN HiNH SIN TRONG CAC MACH DIEN DON GIAN NHAT 2.1. DONG DIEN HiNH SIN. TRj SO HIEU DUNG Dien nang thudng duoc cung cap cho cac thiet bi ky thuat dudi dang dien ap va dong dien hinh sin, thudng goi la dien ap va dong dien xoay chieu (alternating current, viet tat la a. c hay AC). Dong dien hinh sin co bieu thuc toan hoc la : i(t) = Im sin(cot + y j ) (2 . 1.1) trong do lm la bien do cua dong dien; cot + \\ii la goc pha cua dong dien tai thdi difim t. Tai t = 0 thi goc pha bang ij/, duoc goi la goc pha dau cua ddng dien, co duoc goi la tan so goc cua ddng dien. Ddn vi cua co la rad/s. Dai ludng T = — la chu ky cua ddng dien hinh sin. Ngudi ta con dung dai co luong f = ^ goi la tan i'd'cua ddng dien hinh sin. Ddn vi cua tan so f trong he SI la hertz, viet tat la Hz. TSn so cua ddng dien hinh sin dung trong cong nghiep 6 da so cac nude la 50 Hz, d My la 60 Hz. Dudng bieu dien ddng dien hinh sin dude ve tren H .2.1.1 30 Nguoi ta goi tri so hieu dung I cua dong dien hinh sin la tri so trung binh cua i2(t) trong mdt chu ky I = M J i 2 ( t ) d l = J l J , s i n 2 ( c o t + V , ) d l ( 2 . 1 - 2 ) Ta co T, . Tf|~ l-c o s 2 ,(cot + i|/i )1dt s i r , 2 ( o ) t + v , ) d t * \2W J - -------------------- K- ------------ — m 2Jdt - Jcos2(cot + \j/( )dt o o i vi tich phan thu hai Jcos2(cot + y, )dt = 0 . V 2(2.1.3) v a y I = -® r tuc la tri so hieu dung I cua dong dien hinh sin bang bien do Im cua no chia cho 'Jl . He thuc (2.1.2) duoc rut ra tu su tuong duong ve nang luong toa nhiet cua dong dien hinh sin tren mot dien tro R trong mot chu ky T va nang luong toa nhiet cua mot dong dien co tri so khong ddi 1 tren dien tro R trong thdi gian T: T R Ji2(t)dt = RI2T hay I2 = ^ j*i2 (I)dt o Su dung he thuc (2.1.3) ngudi ta thudng viet bieu thuc cua ddng dien hinh sin dudi dang i(t) = \/2Isin(cot + vy,) (2.1.4) f N Vi du, vdi i(t) = 20^2 sin 1 00 TCt + —3A thi trj sd hieu dung la I = 20 A va bien do la Im = 20^2 A, tan sd goc la to = 100rc — , goc pha dau vy,=— rad, tan sd s 3 f = = ■■^ 7t = 5 0 H z va chu ky T = j = — = 0.02s . 2n 2n f 50 31 dien Tuong tu vdi (2.1.1), ngudi ta cun g lumi'ii aiijgaiiiiiCi ana su uuiism. ap hinh sin la : U [u2(t)dt (2.1-5) j U msin2(cot + H/u)dt va ta cung co U = U , (2.1.6) 2.2. BIEU D llN DONG DIEN HiNH SIN BANG VECTO VA SO PHlfC Da’ thuc hien cac phep tinh (cong, tru, dao ham va tich phan) doi voi cac dong dien hoac dien ap hinh sin duoc de dang, nguoi ta c&n dung cach bieu dien cac ham so' hinh sin bang vecto va so' phuc. 2.2.1. Bieu dien dong dien hinh sin bang vecto Xet m6t dong dien hinh sin co bieu thuc cua tri so' tuc thoi la: i(t) = Im sin(cot + v|/) = V2Isin(cot + v|y) trong do Im la bien do, I la tri so hieu dung, co la tan so goc va i\i la goc pha d5u. Ta hay lay mot he true toa do doc cuc co true Ox dat nam ngang (H 2.2.1). Ta ve mot vecto OM vdi do dai OM=|OM| bieu dien cho tri sd hieu dung I cua dong dien theo mot ty le xfch rnj da chon sao cho do dai OM co tri sd thfch hop. a) Vi du, vdi I = 10A, neu ta chon ty la xich la m, = 2 — thi : cm 7^77 I 10A OM = — = ----------= 5cm m, 2A /cm 32 Ta dat vecto OM d vj tri sao cho goc hop boi OM va true Ox ti'nh theo chi6u tir true Ox de'n OM bang goc pha d£u i\i, tuc la (Ox,OM) = h/ ■ Theo quy uoc, ngudi ta chon chieu ducrng cua goc la chieu nguoc vdi chieu quay cua kim ddng ho. Vecto OM xac dinh nhu tren duoc goi la vecto ddng dien hieu dung I , hoac m6t each van tat hon, la vecto ddng dien I . Vi' du hai dong dien hinh sin co cung tan sd goc co nhung co cac trj sd hieu dung va goc pha dSu khac nhau i, (t) = I| V2 sin(cot + V|/|) = 10^2 sin^cot + 30° j A 12 (t) = I2 \/2 sin (cot + \\i2) = 20^2 sin (cot + 45° J A se tuong ung duoc bieu dien bang hai vecto dong dien hieu dung I, v a l2 nhu o H.2.2.IB. 2.2.2. Tinh tong cua hai dong dien hinh sin co cung tan so bling phuong phap do thj vecto Gia su ta can tim tdng cua hai dong dien hinh sin noi tren : 13 (t) = ij (t) + i2 (t) = I, \f2 sin (cot + ) + I2%/2 sin(cot + v|/2) Ta dat i3(t) = I3V2 sin(cot + \|/3) . De tim trj sd hieu dung I3 va goc pha dau v*/3 ta chi cin thuc hien phep cong vecto doi vdi cac vecto dong dien (hieu dung):M I, I3 = I , + I 2 hay OM, = OM, + OM2 bang phuong phap hinh hoc nhu ve d H.2.2.2. De biet do ldn cua I3 ta la'y do dai OM3 nhan vdi ty le xfch mj: I3 - nri|. OM3 De xac dinh tri sd cua vj/3 ta chi can do goc V);3 tren hinh ve. Hinh 2.2.2 duoc goi la do thi vecto cho cac dong dien. Can chu y rang, viec cong vecto cac dong dien nhu tren chi ap dung cho cac dong dien hinh sin co cung mot t5n sd goc co (hoac cung tan sd f). 33 Tir do thj vecto d H.2.2.2 ta tha'y ro rang noi chung I3 * I, + I2, ' a 1- I2 chi khi vj/, = vj/2. 2.2.3. Bieu dien dong dien hinh sin bSng so phurc = Ii + Phuong phap bieu dien cac dong dien hinh sin bang vecto noi tren co uu diem la no cho ta mot hinh anh ro rang v i do Ion (tuong doi) giua cac tri so hieu dung khac nhau cua cac dong dien va anh hudng quan trong cua sir khac nhau ve pha cua cac dong dien hinh sin co cung tan so. Nhung phuong phap nay co nhuoc diem la viec do va ve thudng kem chinh xac va khong tien dung cho trudng hop c£n tinh vdi nhieu dong dien hinh sin (vi du tinh tdng ciia 10 dong di6n hinh sin co cung t^n sd) nhung co cac trj sd hieu dung va cac goc pha dSu rat khac nhau. De tranh nhuoc diem nay ngudi ta can dung cach bieu dien dong dien hinh sin bang sd phuc. Bay gid ta hay bieu dien vecto I bang sd phufc I = IeJV (2.2.1) la sd phuc co moduyn (module theo tieng Phap, hoac modulus theo tieng Anh) va ac - guy - men (argument theo tieng Phap hoac argument hay angle theo tieng Anh) la \\i. Trong (2.2.1) e la co sd ciia logarit tu nhien (e = 2,71828...). Ngudi ta cung thudng dung cach viet tat: i = IeJM' = IZi|; (2.2.2) (doc la I goc i(/). Chu y rang, theo cong thuc Euler iVeJ = cosij; + jsini|/ (2.2.3) vdi j = la don vj ao nen (2.3.4) I = IeJV = l(cosij/ + jsinvj/) = = I cos V|y + jl sin vj/ = Ix + jly nen sd phuc I con duoc bieu dien bang 2 thanh phan trong he toa do chu nhat Oxy la Ix = Icosvj/ I = I sin nhu d H.2.2.3. Ngudi ta goi true Ox la true cac so thuc hay true thuc 0 +1 va true Oy la true cac so ao hay true ao 0 +j hay Oj.Hinh 2.2.3 34 Ngucri ta goi Ix la phan thuc cua sd phtic I : Ix = Icosvj/ = R e jij (2.2.5) vdi Re la ky hieu vie't tit cua partie reelle hay real part co nghla la phan thuc. Ngudi ta goi Iy la phSn ao cua sd phric i : Iy = Isinvj/ = Im jij (2.2.6) vdi Im la ky hieu viet tat cua partie imaginaire hay imaginary part co nghla la phin ao. Noi chung, mot sd phirc co dang a + jb duoc goi so phuc o dang toa do chu nhat hay dang dai so (algebraic form), mot sd phtic d dang re-*9 = rZGduoc goi la sd phirc d dang toa do doc cuc (polar form), con mot sd phirc d dang rej9 duoc goi la sd phurc d dang ham mu (exponential form). Ta co quan he sau day giua ba dang cua cung mot sd phuc (H.2.2.4) : re-*9 = r(cosG + j sin 0) = rcosG + rjsinG = a + jb (2.2.7) voi va a = rcosG b = rsinG r = Va2 + b 2 tg e= —a (2 .2 .8) (2.2.9) (2.2. 10) (2.2. 11) Hinh 2.2.4 hay G=arctg - 1 — = tan — < a J (2.2.12) Ta co the’ bo qua budc trung gian la vecto I va coi rang dong dien hinh sin co bieu thiic la i(t) = Imsin((ot + v|/) duoc bieu diln bang mot anh (image) cua no la mot sd phuc I , goi la ddng dien hieu dung phuc hay mot cach van tat hon la ddng dien phuc : i = iejM/ =IZV|/ = IX + j l y (2.2.13) Vi du, cac dong dien hinh sin: i, (t) = 10V2 sin^cot + 30° j A va i2 (t) = 20V2sin(cot 4-45° j A 35 duoc bieu diin bang dong dien (hieu dung) phUc i, = I,eJV| = 10ej3°° = 10Z30° A 1!rvs . 0 0 = 10(cos30°+ jsin30°) = 10 = (8,66025 + j5,00000) A va dong dien (hieu dung) phuc T + J 2 z, Z \ / I2 = i 2ejM/2 = 20ej45° = 20Z450 A = 20(cos45° + j sin 45°) = 20 y f l .y f c ' ~ r + j - z - = (14,14214 + j l 4 ,14214) A 2.2.4. Tinh tong cua hai dong dien hinh sin cung tan so b&ng phuong phap so phurc Mudn tim tdng cua hai dong dien hinh sin co cung tan so: i, (t) + i2 (t) = I, y/2 sin (cot + \\il ) + 12V2 sin (cot + vj/-,) ta se tim tdng ciia cac anh phirc ciia chung dudi dang dai so: i, + i 2 = I ,e jV| + I 2eJV2‘ = I, (cosvj/| + jsin vj/,) + 12 (cos+ jsin\j/2 ) = (l, cosy, + I2 c o s y 2) + j ( 11 sinvj/, + I2 sinvj/-,) roi chuy6n ket qua nay sang dang so' mu hoac dang toa do doc cuc de duoc: I , e « = I 3Z v 3 = i 3 Vdi cac may tmh bo tui (scientific calculators) viec ti'nh toan nay co the thuc hien duoc rat nhanh chong va co do chfnh xac cao (xem Phu luc I). Sau cung, vdi I3 va V|/3 da tim duoc ta se viet bieu thirc tri sd tuc thoi: i3(t) = I3\/2sin(cot + v|/3) Vf du 2.2.1 : Tmh tdng cua 2 dong dien hinh sin : i, (t) = 10V2 sin^cot + 30° j A va i2 (t) = 20\/2sin^cot + 45° ) A 36 Giai : Taco i,= (l0 Z 3 0 ° )A i 2 =(20Z 45°)A va I3 = i, + i 2 = 10Z30° + 20Z450 = (8,66025 + j5,00000) + (14,1421 + j 14,1421) = (22,8024 + j l 9 , 1421) A = (29,7720Z40,0128°) A = (29,7720ej40 0128 ) A Vay i3 (t) = 29,7720 V2 sin (cot + 40,0128° j A Tren thuc te, nguoi ta thudng viet ket +j qua vdi bon chu sd co nghla: i 3 =(22,80 + j l 9 , 14) A = (29,77Z40,01°) A va i3 (t) = 29,77 sin (cot + 40,01° j A Do thi vecto cho cac dong dien i , , i2 va i3 duoc ve d H.2.2.5. Hinh 2.2.5 Nhung dieu noi tren doi vdi cac dong dien hinh sin cung tan sd cung ap dung duoc doi vdi cac dien ap hinh sin cung tan sd. Vi du, dien ap hinh sin u(t) = 220^2 sin( cot + 6 0 ° j V se duoc bieu dien -► o bang vecto di£n ap U co mdduyn la 220V va ac - guy - men la 60 , hoac duoc bi6u didn bang dien dp hieu dung phuc: U = 220ej60° = (220Z 60°) V = 220^cos60° + jsin60° j = 220 1 . S ' 2 + JT= (ll0 ,0 0 + jl90,53) V 37 Vf du 2.2.2 : Tinh tdng cua hai dien ap hinh sin U| (t) = 220^2 sin (cot + 60° j V va u2 (t) = 180\/2 sin^cot — 30°^ V G ia i: Ta co cac dien ap (hieu dung) phirc U, = 220ej60° = 220Z600 V va u 2-j30° _ Hinh 2.2.6 = isue J ~ = 1 8 0 Z -3 0 V Ta tmh tdng cua hai di6n ap (hieu dung) phirc U 3 = U, + U 2 = (220Z60°) + ( l 8 0 Z - 3 0 ° ) = = (l 10,00 + jl 90,53) + (155,88 -j90,000) = = (265,88 + j l 00,53) V = = (284,25Z20,71°) V vay u3 (t) = 284,25V2sin(cot + 20,71° j V . Do thi vecto cho cac dien ap U, , U2 va U 3 duoc ve d H.2.2.6. 2.3. MACH DIEN CHI CO DIEN TRO R 2.3.1. Dong dien va dien ap. Djnh luat Ohm dirdi dang phurc ' r(^) Xet mach dien don gian nhat chi co mot phan tu co dien tro R ndi vao mot ngudn dien ap hinh sin (H.2.3.1) Ta cd uR (t ) = u (t ) = U msin(cot + H'u) (2.3.1) Tir biSu thuc cua dinh luat Ohm cho dien trd U R (0 = R i R ) 38 u(t) Hinh 2.3.1 uR(t) (2.3.2) ta rut ra ‘R (t) = = ^ j p - s i n ^ t + m/ Ur ) U , \ = - ^ - s i n ( c o t + y u) = iRm sin(cot + (2. 3. 3) vdi i lta i= i W = ] k (2.3.4) va V i = y UR=Vu (2.3.5) Vay dong dien hinh sin trong dien tro R co cung tan so va cung pha vdi dien ap hinh sin dat vao dien trd. Ta ky hidu IR, UR la cac tri sd hieu dung cua iR (t) va uR ( t ) , tuc IR = % (2.3.6) v2 U R = - ^ - (2.3.7) \l2 He thuc (2.3.4) co the viet la H iR Ta dinh nghla cac dien ap hieu dung phirc U, U Rva dong dien hieu dung phurc iR nhu sau U = UejVu = UZ\|/U U R = U ReJ,," ' = U RZ V „B va i R = IRe IR = l R^V iR He thirc (2.3.8) tuong ling vdi he thuc sau i R = % = -H (2.3.9) R R R He thuc (2.3.9) duoc goi la bieu thurc cua dinh luat Ohm dudi dang phuc cho dien trd R. No tuong duong vdi tap hop 2 he thirc (2.3.4) va (2.3.5). Cac dudng cong ciia iR(t)va uR(t) = u(t) duoc ve tren H. 2.3.2 va do thi vecto cua cua IR va U R = U duoc ve d H. 2.3.3. 39 1 • • I r U r = U Hinh 2.3.2 Hinh 2.3.3 2.3.2. Cong suat tCrc thdi va cong suat tac dung Vdi cac chieu ducrng c u aiR (t)va uR(t) chon giong nhau nhu 6 H. 2.3.1 thi cong suat tuc thoi pR (t) tieu thu boi dien tro R se bang P r ( 0 = u r ( 1) » r ( 0 = = U Rm sin (cot + v|/UR ) IRm sin(cot + V lR ) = = R I Rmsin2 (cot + H/,) (2.3.10) ( d d a y y UR = y iR = m/s) Ta nhan tha'y pR (t)>0 vdi moi t. Duong cong cua pR(t) duoc ve d H.2.3.2. Ngudi ta dinh nghla cong suat tac dung (active pow er) PR la tri sd trung binh cua cong suat tuc thdi pR (t) trong mot chu ky cua dong dien 1 T PR = — JpR (t) dt (2.3.11) Ta de thay rang 1 T PR = - j R I 2Rmsin2(cot + M/,)dt = = R[ ^ J !Rm sin2(wt + v|/i ) dt] tutc la Pr = R I r (2.3.12) 40 Vf dy 2.3.1 : Cho mdt dien trd R = 100 Q ndi vao ngu6n dien ap u(t)=220-\/2sincot V. Hay tim iR (t), I r . I r , pR (t) va PR . Gi6i : u(t) 220\f2 sin cot = Ta colR(l) R 100 =2,2V2sincot A. I ^ = S ? = 2 , 2 A. R 100 va U 220Z0 o Ir R ' 100- = 2,2Z0° A. Cong suat tutc thdi : P r (t) = uR (t)iR (t)= R iR(t) = 100x(2,2V2)2sin2cot= 968sin2 cot W C6ng suat tac dung : PR = R I r = 100x(2,2)2 =484 W Ta cung co the tmh duoc PR tu cong thuc dinh nghla cua no theo (2.3.11) : T T 1 1 PR = — JpR (t) dt = — j968sin2 cot dt = 4968 (1 - c o s 2cot) dt = o T =■— |484 d t—— |484cos2cot dt Chu y rang tich phan thu hai bang khong 484 . , —— (sin47i-sin0) = 0 2(0 nen ta co 484cos2cotdt = 484sin2cot 2co PR = 1 |484dt = 484 W 41 2.4. MACH DIEN CHi CO DIEN CAM L 2.4.1. Ddng dien va dien ap. Djnh luat Ohm di/di dang phurc Ta xet mach dien dan gian chi gom co mot phan tir co dien cam L ndi vao mot nguon 0- dien ap hinh sin (H.2.4.1). iL(t) u(t) voi Ta co uL(t) = u(t) = U m sin (cot + iyu) (2.4.1) di. (t) UL(l) 0- = (2.4.2) Hinh 2.4.1 Vdi nguon dien ap hinh sin thi d che do xac lap dong dien trong mach cung la mot dong dien hinh sin co cung t5n sd. Dat iL ( t ) = ILm sin(cot + V(/iL) uL ( l ) = U Lm sin(cot + M/UL) va thay vao (2.4.2) va (2.4.1) ta duoc (2.4.3) (2.4.4) _ILm sin(wt + ViL )] = U Lm sin (cot + Vul ) = U m sin (cot + V u ) hay 71 1 sin wt + ViL + 2 r ULm + M y ) = U m sin(cot + y u)(2.4.5) Tir (2.4.5) ta rut ra: U m = U Lm=(coL)lLm (2.4.6) va (2.4.7) hay t _ U Lm U m Lm coL coL (2.4.8) va71 71(2.4.9) Vdi cac ky hieu IL, UL la cac tri sd hieu dung cua iL(t) va uL(t): I, t _ Lm L _ S ‘ u , = ULm 42 he thurc (2.4.8) cho ta : I. = H i, (2.4.10) L /v\T wL Ta djnh nghla cac dien ap hieu dung phirc U L = U LejM,uL = U lZ vj/ Ul (2.4.11) U = U e ^ u = UZvj/u (2.4.12) va dong dien hieu dung phuc i L = I LeM i - = I l Z V,l (2.4.13) Tu (2.4.9), (2.4.10) va (2.4.11) ta co U L = U LeJM,uL = (o ,L )lLe V ,L + 2 = (a)L ) ( I LeJM''L )e J hay U L =;jcoL )iL (2.4.14) .71 (C huyrang e 2 = COs - + j s ;n - = 0 + jl = j). He thurc (2.4.14) con co the viet la: IL =-.J- f (2.4.15) j
    dang i U L U L U L l L = - ^ = ^ r = — r (2.4.18) /JL J^L J00^ hoac U L = ZLi L = ( j X L) i L -(jc o L )iL (2.4.19) 43 Cac he thuc (2.4.18) va (2.4.19) duoc goi la cac bieu thuc cua dinli I not Ohm dudi dang phuc cho dien cam. Tir (2.4.9) ta thay rang dong dien hinh sin trong dien cam cham ve pha so vdi dien ap tren no mot goc la ^ , hay noi cach khac, dien ap hinh sin tren dien cam vuot trudc \ i pha so vdi dong dien trong no mot goc la ^ . Dieu nay duoc .71 the hien bdi thira sd j = e 2 trong bieu thurc cua tOng trd phirc cua dien cam: Z L = jcoL = jX L = coLZ^ = X LZ j (2.4.20) D6 thi vecto cho U L va i L duoc ve d H.2.4.2. Bieu thurc ciia tri sd tuc thdi ciia dong dien iL(t) UL = U la : J sin cot + y uL 2 coLsin U r cot + - — Hinh 2.4.2 coLsin « t + H/UL- | (2.4.21) Cac duong cong cua uL(t) va iL(t) cho trudng hop vj/u = \\i = 0 duoc ve d H.2.4.3. 44 2.4.2. Cong suat ttfc thdi pL(t) va cong suat phan khang QL Vdi cac chi6u duong ciia uL(t) va iL(t) chon giong nhau nhu 6 H.2.4.1 thi cdng suat tire thdi pL(t) trong di£n cam bang P l (0 = u l (0 *l (0 = U Lm s i n ( Wt + V„L ) * JLm s in ( fflt + V iL ) = X Ll [ msinj^cot + v(/iL + ^ s in ( c o t + v)/iL) = X LILmc o s ( wt + ViL )sin(cot + v|/iL ) hay P l ( 0 = ( x l j l ) sin2(wt + ViL ) = Q l sin2 (cot + vjy ) (2.4.22) Ta nhan thay rang cong suat tire thdi Pl(0 dao dong hinh sin vdi tan sd goc 2co (ga'p doi tan sd goc cua nguon dien ap), vdi bien do bang : QL = X LI* (2.4.23) Ngudi ta dinh nghla cong suat tac dung PL la tri sd trung binh cua cdng suat turc thdi pL(t) trong mot chu ky cua dong dien: p L 4 . f P L ( t ) d l ( 2 A 2 4 ) Thay (2.4.22) vao (2.4.24) ta d6 thay rang 1 T PL = Y | , X LI2 sin2(cot + ¥ iL )d t = 0 (2.4.25) Ve mat vat ly dieu nay co nghla la d dien cam khong co su tieu hao nang luong ma chi co sir trao doi nang luong giua nguon va tir trudng. Trong m6i chu ky luong nang luong do nguon dua vao tir trudng cua dien cam cung bang luong nang luong ma tir trudng tra lai nguon. Dieu nay th£ hien d su bang nhau cua cac phan dien tich duong (gidi han giua dudng cong pL(t) va true thdi gian) va cac phan dien tich am (gidi han giua dudng cong pL(t) va true thdi gian) trong mot chu ky T d tren H. 2.4.3 45 De dac trirng cho cuong do cua su trao d6 i nang luong giura nguon va tu trudng cua dien cam nguoi ta dung bien do ciia cong suat tuc thoi p ^ t) tuc: Ql = Xl I2 (2.4.26) goi la cong suat phan khang (reactive power) cua dien cam. Cong suat phan khang QL co don vj la volt - ampere phan khang (volt - ampere reactif) viet tat la VAr. Vf du 2.4.1 : Mot dien cam L = 200 mH duoc noi vao ngu6n dien ap hinh sin u(t) = 220^2 sin wt V vdi f = 50 Hz. Hay tim ZL, i L, i^ t), P l(0 va QL. G ia i: Tong trd phuc cua dien cam bang Z L = jcoL = j(2nfL) = j(2 n x 50 x 0,2) = j62.832 Q Dong dien hieu dung phuc trong dien cam bang o • _ U 220Z0 L " Z L " j62,832= — j3,5014 A = 3 .5 0 1 4 Z - — A Dong dien hieu dung IL = 3,5014 A. Bieu thuc cua dong dien tuc thdi trong dien cam la iL (t) = 3,5014^2 sin Cong suat tuc thdi P L ( 0 = ul ( 0 x * l( 0 220V2 sin ( 100m) 3,5014%/2 sin 1 0 0 n - ^ = (770.3l)sin2 100 m - j v Cong suat phan khang cua dien cam q l = X l I l = (62.832)(3.5014)2 =770.31 VAr 46 2.5. MACH DIEN CHI CO DIEN DUNG C 2.5.1. Dong dien va dien ap. Djnh luat Ohm dudi dang phurc Ta xet mach dien don gian chi gom co mot ic (i) ; ' , ' 0-----------^ phan tir co dien dung C noi vao mot nguon dien ap hinh sin (H.2.5.1) Ta co: u(» u c (t) uc ( t ) = u (t) = U msin(wt + Vu) (2.5.1) 0- duc (t) va ic (t) - C dt(2.5.2) Hinh 2.5.1 Vdi nguon dien ap hinh sin thi d che do xac lap dong dien trong mach cung la mot dong dien hinh sin co cung tan so. Dat U c ( t ) = U C m s i n ( « t + V u c ) ic ( t ) = I C m s i n ( ® t + ^ic ) va thay vao (2.5.2) va (2.5.1) ta duoc i c ( t ) = c ^ [ u aV,sin(wt + M v ) = c ^ [ U msin( wt + V u )] hay ICm sin(cot + My ) = (odC)UCmcos(cot + v iy ) = (2.5.3) (2.5.4) = (coC)UCmsin[ o)t + \j/Uc + y ) (2.5.5) Tir (2.5.5) ta rut ra va hay Ic™ ~ coCUCm 7t MV +2 ^Cm ( 1 1 vcoCy 71 2 U, v coCy (2.5.6) (2.5.7) (2.5.8) 47 Vdi cac ky hieu Ic , Uc la cac tri sd hieu dung cua ic(t) va uc (t) I — ^Cm ' " " T T U -H em U c _ J 2 he thurc (2.5.8) co the: viet la I - Uc l c - / , Ta djnh nghla cac dien ap hieu dung phurc Uc = U c eJVuc = U c Zv/Uc U = UeJVu = U Z y u va dong dien hieu dung phurc Ic = Ic eJV'c = Ic Zvj/'c Tir (2.5.9) ta co (2.5.9) (2.5.10) (2.5.11) (2.5.12) i - I eJM/ic - - H e AC - 1C e _ / J coC = (coC)( J| Vuc hay Uc eJH,uc V 2 ic = (jcoC)Uc .71 (Chu y rang e 2 = co s—+ jsin —= 0 + jl = j ) 2 J 2 (2.5.13) He thurc (2.5.13) con co the’ viet la: . Uc _ Uc (2.5.14) Lc ~ r - n “ ( 1 vj®c , . 1 -J coC (2.5.15) 48 hoac Ur = jcoC Ic -_ l j ~J coC Cac hd thurc (2.5.14) va (2.5.15) cho ta quan he giua Uc va ic co dang gidng vdi cac bidu thurc cua djnh luat Ohm dudi dang phiic cho dien trd va cho didn cam. Ngudi ta dua vao ky hieu Xr = — c coC goi la dien khang cua dien dung (capacitive reactance) va ZG - JXC - (2.5.16) (2.5.17) goi la tdng trd phuc (complex impedance) hay tong trd (impedance) cua dien dung C. Ta co the viet (2.5.14) dudi dang i -He ir —-JXC vcoCy hoac _ z cic - ( - jx c)ic - -j coC (2.5.18) (2.5.19) Cac he thurc (2.5.18) va (2.5.19) duoc goi la cac bieu thurc cua dinh luat Ohm dudi dang phuc cho dien dung. Tu (2.5.19) ta thay rang, dong dien hinh sin trong dien dung vuot trudc ve 7t pha so vdi dien ap tren no mot goc la — , hay noi cach khac, dien ap hinh sin tren 7t dien dung cham sau ve pha so vdi dong dien trong no mot goc la —. Dieu nay .n duoc the hien d thira sd - j = e 2 trong bieu thuc cua tdng trd phuc cua dien dung : f 1 ^ 2 c 2(2.5.20) Zc ~ ~j coC ~~ j X c _ v(OCyZ - £ = Xc Z - £ Do thi vecto cho Uc va Ic duoc ve d H.2.5.2. Bidu thurc cua trj sd turc thdi cua dong dien ic(t) la: ic (t) - 1C m sin ®t + Vu + \ UC m vcoC, U r '_ 1 _' vcoC, sin sin « t + Vuc + 2 cot + y u + - (2.5.21) 71 2 uc = u Hinh 2.5.2 49 Cac dudng cong cua uc(t) va ic(t) cho trudng hop v|/u = V U(_ = 0 duoc vc d H.2.5.3. Hinh 2.5.3 2.5.2. Cong suat turc thdi pc(t) va cong suat phan khang Qc Vdi cac chieu duong cua uc(t) va ic(t) chon gidng nhau nhu d H.2.5.1 thi cong suat tire thdi pc(t) trong dien dung bang pc (t) = uc(l) ’c(t) = UCm sin (cot + y uc ) x ICm sin (cot + V|/ic ) = x c Im sin(cot + y Uc)cos(o)t + v|/Uc) hay pc (t) = Xc I2 sin2(cot + \(/Uc) = Q c sin2(cot + vj/Uc) (2.5.22) Ta nhan thay rang cong suat turc thdi pc (t) dao dong hinh sin vdi tan sd goc 2co (gap doi tan sd goc cua nguon dien ap) vdi bien do bang: Q c = x c l2 (2.5.23) Ngudi ta djnh nghla cong suat tac dung Pc la tri sd trung binh cua cong suat tifc thdi pL(t) trong mot chu ky cua dong dien: 1 rT pc = x P c ^ d t (2.5.24) I Jo Thay (2.5.22) vao (2.5.24) ta de thay rang: Pc = Tf l x c Ic s in 2 (cot + V Uc) dt = 0 ' 2.5.25) 50 Ve mat vat ly, diiu nay co nghla la d didn dung khong co su tieu hao nang luong ma chi co su trao d<5i nang luong giua ngu6n va dien trudng. Trong mdi chu ky, luong nang luong do nguon dua vao dien trudng cung bang luong nang luong ma dien trudng tra lai ngudn. Di6u nay the’ hien d su bang nhau cua cac phan dien tich duong gidi han giua dudng cong pc(t) va true thdi gian va cac phan dien tich am gidi han giua dudng cong pc(t) va true thdi gian trong mot chu ky T d tren H. 2.5.3. De dac trung cho cudng do cua su trao doi nang luong giua nguon va dien trudng cua dien dung, ngudi ta dung bien do ciia cong suat tuc thdi p c (t) tuc la: QC = X CI2 (2.5.26) goi la cdng suat phan khang (reactive power) cua dien dung. Cong suat phan khang Qc co don vi la volt - ampere phan khang (volt - ampere reactif), viet tat la VAr. Vi du 2.5.1 : Mot dien dung C = 20 pF duoc ndi vao ngudn dien ap hinh sin u(t) = 220\/2 sincot V, vdi f = 50 Hz. Hay tim Zc , ic , ic (t), pc(t) va Qc . Giai : Tdng trd phirc cua dien dung bang = -J- 7 ■ 1 1 c J coC J 2jifC 1 271 x 50 x 20 x 10= — j 159,15 Q Dong dien hieu dung phuc trong dien dung bang Ic = U 220Z0= jl, 3823 = 1,3823Z— Zc — jl 59,15 Ddng dien hieu dung Ic = 1,3823 A. Bieu thurc cua ddng dien tire thdi trong dien dung la % \ .(t) = (l,3823)V2 sin Cdng suat tire thdi pc(t) = uc(t) ic(t) 1007114,- A = (1,3823)72cos(lOOTTt) A V ■J = 220V2 sin(100m )x 1,3823 72 cos(100nt) = 304,11 sin(2007tf) W 51 i Cong suat phan khang Q c = X c I q = (159,15) (l,3 8 2 3 )2 = 304,11 V A r 2.6. MACH RLC 2.6.1. Djnh luat Ohm dang phurc va tong trd phurc cho mach RLC Bay gid ta xet mot mach dien don gian gom cac phan tu dien trd R, dien cam L va dien dung C ndi tiep, goi tat la mach RLC (H.2.6.1). Khi dien ap dat vao mach u(t) la dien ap hinh sin u(t) = U m sin(cot + y u) thi d u R(t) u L(t) u(t) i(t) u c(t) che do xac lap dong dien trong mach cung la dong dien hinh sin co cung tan sd goc co. Hinh 2.6.1 Neu dat i(t) = Im sin(cot + v|/j) la dong dien trong mach (d che do xac lap) ta can tim Im va \jtheo Um, V)/u, R, L, C va co da cho. Ap dung dinh luat Kirchhoff vd cac ap (hay dinh luat Kirchhoff thu hai) cho vong RLC ta co (2.6 .1) VI cac dien ap uR, uL va uc va u(t) deu la cac ham hinh sin co cung tdn sd goc co nen ta co the’ dung cac anh phurc U R, U L, Uc va U cua chung va viet phuong trinh theo dinh luat Kirchhoff cho cac ap dudi dang phuc: UR+UL+UC=U (2.6.2) Theo dinh luat Ohm dang phurc cho cac phan tu R. L va C ta lan lum co: Ur =RI (2.6.3) (2.6.4) (2.6.5) ( 2 .6 .6 ) (2.6.7) 52 UL = jcoL I= jX Li = Z Li vOi Z L = JX L = J wL Zc = —j x c = _ J1 1 coC jcoC Tl> (2.6.2) va (2.6.3) * (2.6.7) ta co : Ri + j X Li - j X c i = U (2.6.8) hay (R + j X L - j X c ) i = [ R + j ( X L - X c ) ] i = ( R + j X) i = U (2.6.9) hay Zi = U (2.6.10) voi Z = R + jX = R + j( X L - X c ) = R + j coL -1 coC(2.6. 11) He thufc (2.6.2) duoc goi la dinh luat Ohm dang phuc cho mach RLC va Z xac dinh boi (2.6. 11) duoc goi la tong trd phifc (complex impedance) hay tdng trd (impedance) ciia mach RLC. Ta cung co the viet (2.6.10) dudi dang: I U ( 2 .6 . 12) Dat ta co _ , „ j

    Xc thi X = XL - Xc > 0 va cp > 0. Trong trudng hop nay, dong dien cham pha sau dien ap. Neu XL < Xc thi X = XL - Xc < 0 va cp < 0. Trong trudng hop nay, dong dien vuot trudc dien ap. Trong trudng hop dac biet XL = Xc thi X = XL - Xc = 0 va cp = 0 tuc la dong dien cung pha vdi dien ap. 53 Cac do thi vecto ve cho mach RLC cho 3 trudng hop nay nhu 6 H.2.6.2. Trudng hop dac biet (XL = Xc ) se duoc xet ky hon trong muc 2.7. U, UL U, a) b) U, U = U , c) Vf du 2.6.1 Hinh 2.6.2 Cho mach dien RLC (H.2.6.1) vdi R = 100 Q; L = 200 mH; C = 20 ^F. Biet u(t) = 1 2 0 ^ sin co t V va f = 50 Hz. Hay tmh Z, i, U R, U L va U c . Giai : Ta co co = 2 n f = 2rtx50 » 314,16 rad/s. XL = coL = 314,16x0,2 = 62,832 Q. 1 X c =- = 159,15 Q 314,16 x 20 x 10-6 X = XL - Xc = 62,832 - 159,15 = - 96,318 Q. Vay tdng trd phuc ciia mach la : Z = R + jX = 1 0 0 - j96,318 = 138,84Z - 43,93° Q Dong dien hieu dung phuc trong mach bang : 120Z0 o= 0,6225+ j0 .5996 = 0,8643Z43.93° A Z 100 — j96.318 Dien ap hieu dung phuc tren cac phan tu : U R = R i = 100x(0,8643Z43,93°) = 86,43Z43,93° V U L = jcoLI = (j62,832)(0,6225 + j0,5996) = -37.674 + j3 9 ,113 = 54.306Z133,93 V 54 uc = -J1coCI = ( - j l 59,15 )(0 ,6225 + jO, 5996) = 95,426-j99,071 = 137,55Z - 46,07 V Bieu thuc cua dong dien tire thoi: i(t) = 0,864372 sin (cot + 43,93°) A Bieu thurc cua dien ap tutc thdi tren cac ph£n tir: uR = 86,4372 sin (cot + 43,93° ) v uL = 54,30672 sin (cot+ 133,93°) V uc = 137,5572 sin (cot - 46,07°) V Do thi vecto bieu diin cac dong dien va dien ap phirc trong mach duoc vedH.2.6.3. Chu y rang vi XL < Xc hay X = XL - Xc < 0 nen cp < 0, dong dien i(t) vuot pha trudc dien ap u(t). 2.6.2. Cong suat turc thdi p(t). Cong suat tac dung P. Cong suat phan khang Q va cong suat bieu kien S tron g mach dien a) Cong suat ttxc th&i p(t) Cong sutft tire thdi, ky hieu la p(t), cua mot bo phan mach dien giua hai cuc A va B la tich sd ciia dien ap tuc A thdi dat vao mach u(t) = uAB(t) va dong dien i(t) = iAB(t). u(t) Vdi chi6u duong ciia u(t) va i(t) chon giong nhau B (hudng tir A den B) nhu d H.2.6.4 thi p(t) la cong suat tiic thdi tieu thu boi bo phan mach dien giua 2 cuc A va B. Trong trudng hop mach RLC d H.2.6.1 ta co: p(t) = u (t) i(t) = ( u R(t) + uL(t) + uc ( t))i(t) = u R (t).i(t) + uL(t).i(t) + uc (t).i (t) i(t) Hinh 2.6.4 = PR (t ) + PL(t) + P c ( 0 (2.6.18) 55 Gia su i ( t ) = Im sin cot ta co uR = R Imsincot u l = ojLImsin f n N cot + — v 2 J = XLIm coscot uc =\COCy I m s in = - X c Im coscot va Pr = uRi(t) - Rim sin2 cot = RI1 -c o s 2cot = RI2 (l - cos2cot) (2.6.19) / \ \ v .2 • sin2cot Pl = ul (t) i(t ) = X LI msin“ tc oscot = XLi ; - — — = X LI sin2cot (2.6.20) va / \ •/ \ xr t2 t 2 sin2cot ,7 Pc = uc (t) ‘(t) = - x c l m sincot coscot = - X c Im— - v =- X c I sin2cot (2.6.21) ,2 Vay ta co P (t ) = PR(t) + PL(t ) + P c ( t ) = RI2 (l -cos2cot) + (X L -X c )l 2 sin2cot = RI2 (1 - cos2cot) + XI2 sin 2cot (2.6.22) Ta nhan thay cong suat tire thdi p(t) cua mach RLC gom hai thanh phan co tmh chat khac nhau. Thanh phan thu1 nhat pR (t) = RI2 (l-cos2cot) luon khong am pR(t)>0va co tri so trung binh trong mot chu ky bang P = i f pR dt = ^ f R I2(l-c o s 2 c o t)d t = R I2 (2.6.23) 1 •'0 1 JO Thanh phdn thu hai dao dong vdi tan so 2co Pl c = P l + P c = xi sin2wt (2.6.24) co bien do bang Q = XI2 va co tri so trung binh trong mot chu ky bang khong 1 rT 1 fT ? PLCdt = ^ XI sin2cotdt = 0 T Jo T Jo (2.6.25) 56 Thanh phSn thur nha't pR (t) cua cong suat tuc thdi la cong suat tieu thu thuc sir tren dien trd R tuong irng vdi hieu urng Joule (bien ddi dien nang thanh nhiet nang). Thanh phan thu hai pLC (t)ciia cong suat tuc thdi la cong suat "tieu thu" boi dien cam L va dien dung C tuong ung vdi su trao doi nang luong giua tir trudng va dien trudng cua mach dien va nguon dien. b) Cdng suat tac dung P Ngudi ta dinh nghla cdng suat tac dung (active power) P la tri sd trung binh cua cong suat tuc thdi trong mot chu ky T cua dong dien. P = Y j j p ( t ) d t (2.6.26) Tir (2.6.22), (2.6.23) va (2.6.25) ta de thay rang 1 f T j 1 f T T Jo T Jo Don vi cua cong suat tac dung la watt (viet tat la W). De’ dac trung cho cong sua't trao doi nang luong giua dien trudng va tir trudng ciia mach vdi nguon dien, ngudi ta dung bien do Q = XI2 (2.6.28) ciia thanh phan pLC(t)- Dai luong Q nay duoc goi la cdng suat phan khang (reactive power) va co don vi la VAr (volt - ampere reactif). Chu y rang, tir do thi vecto d H.2.6.2 ta co: U R = RI = U coscp .(2.6.29) U LC = XI = Usin(p (2.6.30) nen P = RI2 = Ul coscp (2.6.31) va Q = XI2 = Ulsincp (2.6.32) Ta dinh nghla S = Vp2 + Q 2 = UI (2.6.33) la cdng suat bieu kien (apparent power) vdi don vi la VA (volt - ampere) ta co P = Scoscp (2.6.34) va Q = Ssincp (2.6.35) Thira sd coscp trong (2.6.34) duoc goi la he so cdng suat (power factor). He sd cdng sua't cd y nghla rat quan trong va se duoc xet ky hon d muc 2.10. 57 c) Cdng suat phuc De tinh P, Q va S theo cac dien ap hieu dung phuc U va dong dien hieu dung phirc, nguoi ta dinh nghla cdng suat (bieu kien) phirc (apparent complex power) S = U I (2.6.36) * trong do I la so' phirc lien hop ciia dong dien hieu dung phirc. Giathiet i = I.eJM' 1 va U = U.eJ'|/u ta co S = UI = (U eJ'|,u )x (Ie “JM'1) =UIej(Vu~Vi) = UIeJlp (2.6.37) hay S = Ul (cos (p + j sin cp) = Ul coscp + jUI sin cp = P + jQ (2.6.38) Vay P = Re jsj = Re ju I j (2.6.39) Q = Im jsj = Im ju i j (2.6.40) O day Re { } la phan thuc (real part) va Im { } la phan do (imaginary part) ciia mot sd phirc. Vi du 2.6.2 : Tmh cac cdng sua't tire thdi p(t), cdng suat tac dung P, cdng suat phan khang Q, cdng suat bieu kie'n S va S trong mach RLC d vf du 2.6.1. Gidi : Ta cd u(t) = 120V2 sin cot V i(t) = 0,8643^2sin(cot + 43,93°) A Vay cdng suat turc thdi p(t) = u(t)i(t) = (l20 V2sincot)( 0,8643V2j sin (o)t + 43,93' = 2(103,7)(sincot)sin(cot + 43,93° j = 103,7 co s(-43,93°)-cos(2cot + 43,93°)J = 103,7 cos (-43,93° j - 103,7 cos (2cot + 43,93° j = 74,7 - 103.7 cos(2cot + 43,93° j W 58 Cdng sua't tac dung P = RI2 = 100 x (0,8643)2 = 74,70 W hay P = Ul coscp = 120 x 0,8643 x cos (-43,93° ) = 74,7 W Cong sua't phan khang Q = XI2 = -96,32 x (0,8643)2 = -71,95 VAr hay Q = Ul sin cp = 120 x 0,8643 sin (-43,93° j =-71,95 VAr Cdng sua't .bieu kien S = UI = 120x0,8643 = 103,716 VA hay S = Vp2 + Q 2 = y(74,70)2 + (-7 1 ,9 5 )2 =103,7 VA He sd cdng sua't Cdng sua't (bidu kien) phuc S = U I= (l2 0 Z 0 ° )(0 ,8 6 4 3 Z -4 3 ,9 3 ° ) = 103,7Z-43,93° = ( 7 4 ,7 0 - j71,95) VA. 2.7. CONG HlTdNG DIEN AP TRONG MACH RLC Trong mach dien cd dien trd R, dien cam L va dien dung C noi tiep, goi tat la mach RLC (H.2.7.1) cd mot tinh trang dac biet khi dien khang XL cua dien cam bang dien khang Xc ciia dien dung, tuc la X, = X r hay coL = —— L coC Luc nay, dien khang ciia toan mach triet tieu X = X L - Xc = 0 va tdng trd ciia mach bang dien trd Z = R + jX = R Ddng dien trong mach se ciing pha vdi dien ap ciia ngudn (2.7.1) (2.7.2) (2.7.3) (2.7.4) 59 UL U c U L 71 • U , •1 j U _= RI R 2 Uc Hinh 2.7.1 Dien ap hieu dung phirc tren dien cam U L va dien ap hieu dung phirc tren dien dung Uc co moduyn bang nhau nhung nguoc pha nhau: U L = j X Li (2.7.5) Uc = - j X c i (2.7.6) tuc va do do UL = - U C U L + U C = 0 (2.7.7) (2.7.8) Do (2.7.7) va (2.7.8) ngudi ta goi tinh trang nay la cong huang dien dp. Do thi vecto tuong urng vdi tinh trang cong hudng dien ap duoc ve d H.2.7.1. Gia thiet rang dien trd R. dien cam L va dien dung C la cac hang so, tri sd hieu dung U ciia dien ap nguon khong ddi nhung tan sd goc co ciia ngudn co the bien thien tir 0 den oo. Ta hay khao sat sir bien thien ciia ddng dien hieu dung I theo vdi tan sd goc co. Ta co i - b u R + j coL -coC (2.7.9) va I =U \2 (2.7.10) IR2 + coL- — coC T in sd goc co0 a day xdy ra cong hudng dien ap duoc xac dinh tir he thuc woL = — r (2 .7 . 1)) COnC 60 Tir do va 0)0 = LC co„ =■ " V lc (2.7.12) (2.7.13) Nguoi ta goi co0 la tan so goc cong huong cua mach RLC, con tan so f() duoc xac dinh boi 2tiVlC duoc goi la tan so cdng liudng cua mach RLC. (2.7.14) Tir (2.7.10) va (2.7.11) d i dang tha'y rang, khi co cong huong dien ap thi dong dien hieu dung dat tri so cuc dai va bang U Id = R Tri sd I() duoc goi la ddng dien cong hudng. (2.7.15) Ky hieu UL0 la dien ap hieu dung tren dien cam khi cong hudng, Uco la dien ap hieu dung tren dien dung khi cong hudng, ta cd: (2.7.16) Chu y rang, khi cong hudng U = Rl() Ta dinh nghla he so pham chat Q bang : Q =Ulo _ Uco “ oL _ 1 1 L U U R OfiCR R V C Neu J — > R thi Q > 1 va UL0 = UC() > U Neu J — < R thi Q < 1 va UL0 = Uco < U (2.7.17) (2.7.18) (2.7.19) (2.7.20) D i thay ro y nghia cua he so pham chat Q ta can khao sat dac tinh In f V“ o ) 6 day ~ la ty sd cua ddng dien hieu dung I vdi ddng dien khi cong hudng 1, In CO con — la ty sd cua tan sd goc a) vdi t 2.765X103 __ U U 220 Ta nhan thay rang, vdi he so pham cha't Q - 12,57 > 1 thi cac dien ap tren dien cam va dien dung khi cong hudng ldn ga'p 12,57 lan dien ap cua ngudn. Dieu nay co the gay nguy hidm doi vdi ngudi hoac thiet bi. Vf du 2.7.2 : Trong mach dien d vi du 2.7.1. gia su cac tri so cua cac thong so L, C va U van giu nguyen nhu cu nhung R = 200 Q. Tinh I„, UR0, UL0 va Uol khi cong hudng dien ap? Gicii : Cong hudng dien ap xay ra d tan sd: 1 f<> =2rtVLC 27iv/400xl0"3 x 2 5 ,3 3 x l0 " 6= 50 Hz Ddng dien khi cong hudng bang: U 220 0 R 200 ’ Cac dien ap tren R, L, C khi cong hudng bang: U R = U = 220 V He sd phaim chat cua mach bay gid la: 1 l 400x10 RV C 200^25,33x 10“= 0,6283 <1 va U L() ='Uco = QU = 0,6283x220 = 138.3 V. Ta nhan thay rang vdi Q < 1 thi khi cong hudng, dien ap tren dien cam va dien ap tren dien dung be hon dien ap cua ngudn. Hai vi du 2.7.1 va 2.7.2 cho ta thay rang khi ndi mach RLC ta can tmh trudc xem d tan sd da cho trong mach cd xay ra cong hudng dien ap khong, va neu xay ra cong hudng thi ddng dien I va cac dien ap U L, Uc khi cong hudng la bao nhieu. 2.8. MACH R, L, C SONG SONG VA CONG HlfCJNG DONG DIEN Xet mach dien gdm 3 nhanh R, L, C ndi song song d che do xac lap hinh sin (H.2.8.1). Hinh 2.8.1 64 Goi IR, IL, ic va I la cac dong dien nhanh, theo djnh luat Kirchhoff ve cac dong, ta co i = i R+ i L+ i c (2 .8.1) Theo dinh luat Ohm cho m6i nhanh ta co U I r _ R u u jcoL= U -J coL l<=~ z ~ / cu\ ( 2 .8 .2 ) (2.8.3) U coC = U (jcoC) (2.8.4) Vay voi 1 =R Y = R - j G =R 1 coL coL - coC U = YU -toC = G - j ( B l - B c ) = G - jB (2.8.5) (2.8.6 ) (2.8.7) B, =coL (2.8.8) va Bc = coC (2.8.9) Y duoc goi la tong ddn phuc (complex admittance) cua mach, G la dien dan tac dung va B la dien dan phan khang. Tu cac do thi vecto, ta nhan tha'y rang neu IL > Ic thi I cham sau U (H.2.8.2a). Neu IL < Ic thi I vuot trudc U (H.2.8.2b). - Trong trudng hop dac biet, khi IL = Ic (2.8.10) thi IL + ic = 0 (2 .8. 11) va i = i R + ( i L ^ i c ) = I r (2.8.12) tuc la i cung pha voi U (H.2.8.2c). Tinh trang dac biet nay duoc goi la cong hudng ddng dien Tir (2.8.6) va (2.8.11) ta suy ra rang dieu kien de co cong huong dong dien trong mach R, L, C song song la: B = - V - cdC = 0 coL (2.8.13) 65 hayconLr _(0°C (2.8.14) Ta can chu y rang khi cong huong dong dien thi dong dien I co tri so nho U nhat va bang IR =R I = Ic b) Hinh 2.8.2 c) 2.9. CONG HUONG DONG DIEN TRONG MACH SONG SONG CO CUON DAY VA TU DIEN Xet mach dien g6m 2 nhanh noi song song (H.2.9.1). O day R L la dien tro bi6u dien cho cac ton hao nang luong trong cuon day dien cam va Rc la mot dien tro bi£u di£n cho ton hao nang luong trong tu dien. Ta co: U U u ( R l - J X l ) / _ \ / \ R ; -jU - %X l , (2.9.1) II Z l Rl + jX l r 2 + x 2 VR 1 + x 2l J ^ R 2+Xlj u u U ( R c + j X c ) = 0 VC> l c Z c R c - j X c R c + X 2 Rr v Rc + xc j (2.9.2) R , R r \ f X, Vay i = I, + ir =U -jU R l + X l R6 + x ^ y Tu (2.9.1) ta nhan thay rang, neu cac thong so cua mach thoa man dieu kien: (2.9.3) Xr R“ + X l R c + x c thi I = U R . R r Rl + Xj^ Rc + Xc lire la I cirng pha vdi dien ap U . 66 (2.9.4) (2.9.5) Tinh trang dac biet nay duoc goi la cong huong dong dien trong mach song song cua cudn day dien cam va tu dien. Dd thi vecto cua cac dong dien cho mach nay a tinh trang khong cong hudng ducrc ve d H.2.9.2a va do thj vecto cho tinh trang cong hudng ddng dien duoc ve d H. 2.9.2b. a) b) Hinh 2.9.2 Dieu kien cong hudng ddng dien (2.9.4) cd the viet la: 1 “ L coC R2l + (coL)2Rc + coC haycoL coC(2.9.6) R^+(coL)2 1 + (coCRc )2 Ndi chung dieu kien (2.9.6) dan tdi mot phuong trinh bac hai doi vdi co, L, C, Rl hoac Rc. Vi du 2.9.1: Xet mach dien d H. 2.9.1 vdi RL = 15 Q, Rc = 30 Q, Xc = 30 Q. Hay tim tri sd cua dien cam de cd cong hudng ddng dien d tSn so f = 50 Hz. Ti'nh trj sd cua ddng dien cac nhanh tuong ung vdi tinh trang cong hudng neu U = 100Z0° V . Giai : Trj sd XL de cd cong hudng ddng dien duoc xac dinh tu he thuc (2.9.4): X, X, r 2l + x 2 Rc + Xc 67 , r X L 30 1 tuc la —-— —- = — -------------- = — 15 + 30 + 30 60 Ta duoc mot phuong trinh bac hai doi voi XL: X^ - 60Xl + 225 = 0 Phuong trinh nay co 2 nghiem la 60 + Veio2"- 900 X L, = ------- — z---------- = 55,98 Q 6 0 - V 6 0 2 -9 0 0 va X L2= --------^ ---------- = 4,019 Q ~2 Cac tri so' cua dien cam tuong ung se la X QR L ' = 2^ = 2 ^ 5 0 = ° J7 8 2 H = 178' 2 m H L! s T 7 = r ^ s | 2 ' 7 9 x i r ’ H = 2 2 7if 2kx 50 = 12,79 mH Vdi I2| = 178,2 mH thi dong dien cac nhanh se la i - R ^ = T ^ b 8 = (° '4466‘ J,' 667)A ‘c = R T q x 7 = « n 3 o = < ,-6 6 7 - j |'667) A va i = i L+ ic =0,4466+1,667 = 2,114 A Vdi L2 = 12,79 mH thi ddng dien cac nhanh se la U 100 , Il " Rl + j X L2 ” 15 + j4,019 _ (6,220_ J1’667) A U 100 . Ic = Rc - j X c - 3 0 - j30 " ^1,667 + J1’667) A va j = i. + ir =6,220+1,667 = 7,887 A 2.10. NANG CAO HE SO CONG SUAT Tai cua he thong dien trong cdng nghiep chu yeu la cac dong co dien khong ddng bo. Dd la cac tai cd tinh chat dien cam. cd the bieu dien bang dien trd Rt 68 ndi tiep vdi dien khang XL (H.2.10.la). Ddng dien IL cung ca'p cho dong co 71 cham sau dien ap U mot goc (pL vdi 0 < (pL < — nhu ve d dd thi vecto H.2.10. lb Hinh 2.10.1 Ddng dien chay vao dong co dien la R + j x L ' =u r 2 + x 2l■JUX, i 0 U ( R - j X , ) , J R Ta phan tich IL thanh 2 thanh phan: \ R R 1 + X (2 . 10. 1) L J Thanh phan tac dung ILtd =U cung pha vdi dien ap U R2 + X2 ; /X, va thanh phan phan khang ILpk = -jU |U v —-------— cham sau dien ap U goc — R + x L J ' ' 2 Chinh do thanh phan phan khang ILpk nay ma he sd cdng suat cos(pLcua dong co thudng nho hon l.v i'd u coscpL= 0,8 (hay cpL =36,87°). Neu goi P la cdng suat tac dung cua dong co thi ddng dien IL cd tri sd (moduyn) bang: P I, U coscpL(2.10.2 ) He thuc (2.10.2) chung to rang, vdi P va U da xac dinh thi IL cang ldn khi coscpLcang nho. Ddng dien IL ldn se gay ra tdn hao dien nang tren dudng day dien. Hon nua neu he sd cdng suat cua ca he thong dien cd tri sd thap thi cac nha may phat dien se khong the phat duoc cdng suat tac dung ldn nhu da thiet ke gay ra lang phi rat ldn. 69 Vi the can phai dat va'n de nang cao he sd cong sua't cua cac tai trong he thong dien. De nang cao he so cdng sua't cua dong co dien khong ddng bo, ngudi ta ndi song song vdi no vdi tu dien co dien dung C nhu ve d H.2.10. la. Khi cd tu dien thi ddng dien tren dudng day se la i = IL + i c , ddng dien I se cham sau dien ap mot goc cp < cpL vdi coscp > coscpL . Tir dd thi vecto (H.2.10.1) ta de thay rang: lc = I LPk “ Ipk = I idtg(PL - I |d t89 = I,d(tg9L - 18 2 + ( X l + Xc )2 u R + X R -jXI) cp = arctgXL - X C k) 1 = U R m)

    ). BT.2.12. Ngifdi ta djnh nghTa giai thong cua mach RLC la giai tan d do vdi I 0 = — la dong dien cong hifdng. Hay tim giai thong cua mach RLC d BT.2.11. BT.2.13. Khao sat va ve difdng cong cong hifdng I / 10 = f(a > /w 0 ) cua cac mach RLC vdi cac he sd pham chat Q = 1, Q = 5 va Q = 50. Tinh cac tri sd l/l0 vdi cd/ coq = 0,8 va w/co0 = 1.2 tifdng Cfng vdi cac trj sd cua Q ndi tren. Cho cac nhan xet ve anh hifdng cua Q doi vdi tinh chon loc tan sd cua cac mach RLC. Hifdng dan : Sir dung chifong trinh may tinh viet bang MATLAB d phu luc PL.2.2 BT.2.14. Cho mach dien d H.BT.2.14 vdi R = 1 kQ, L = 20 nH, C = 20 pF va j(t) = 10\/2 sin cot mA. H.BT.2.14 75 Tim tan so co de co cong hirong dong dien. Tinh dien ap u(t), iR(t). iL(t) va ic (t) khi cong hirdng. BT.2.15. Cho mach dien 0 H.BT.2.15 voi R = 100 n , L = 100 mH va u(t) = 100a/ 2 sin lOOOt V. Tim dien dung C de co cong hifdng dong dien. Tinh iR(t), iL(t), ic (t) va i(t) khi cong huong. BT.2.16. Cho mach dien 0 H.BT.2.16 vdi R = 100 Q, L = 0,1 H u(t) = 100\/2 sin lOOOt V. Tim trj so cua dien dung C de co cong huong dong dien. Tmh iL(t), ic (t) va i(t) khi cong hudng. u(l) u(t) H.BT.2.15 H.BT.2.16 cho BT.2.17. Trong mach dien a H .BT.2.17 ^ i R 1 = 5 Q, R2 = 10 Q, C = 50 F va U = 100Z0° V . Tim tri so cua dien cam L de co cong hudng dong dien d tan so goc co = 1000 rad/s. ^ Tinh i ,, i 2, I khi cong hirdng. Ve do thj vecto. BT.2.18. Cho mach dien d H .BT.2.17 vdi R 1 = 5 Q, R2 = 10 Q, L = 100 mH, C = 50 jiF. H.BT.2.17 Tim tan so goc co0 de co cong hudng ddng dien. Tinh i , , i 2 va I khi cong hirong neu U = 100Z 0° V . Ve do thi vecto cho tri/dng hop cong hi/dng 76 C huang 3 CAC PHUONG PHAP GIAI MACH DIEN TUYEN TINH PHUC TAP Giai mot mach dien la tim cac dong dien hoac cac dien ap cua tat ca cac nhanh hay cua mot so nhanh nao do cua mach dien khi da biet so dd cua mach va cac thong so cua cac phan tir cua mach. Trong chuong nay ta se xet cac phuong phap de giai cac mach dien tuyen tmh phuc tap d che do xac lap hinh sin trong trudng hop mach dien khong co h6 cam. Trudng hop mach dien co h6 cam se duoc xet d chuong tiep theo. Didu nay se giup cho viec tiep thu noi dung cua hai chuong nay trd nen d-*— © 6 © © © b) a) Hinh 3.1.1 Graph ciia mach dien nay duoc ve d hinh 3 .1.lb. Mach dien gom cd m = 6 nhanh va n = 4 nut. Gia thiet da cho tri sd cua sue dien dong hieu dung phuc E,va E2 cua cac ngudn vdi cac chi£u duong da chon, cac tri sd ciia cac tdng trd phirc Z ,, 'Z2, Z,, Z4, Z, va Z6 ciia cac nhanh va can tim tri sd ciia cac dong dien hieu dung phuc cua cac nhanh. 77 Theo phuong phap cac dong dien nhanh ta tien hanh cac budc sau day: - B udc 1: D at ky hieu va chon chieu duong cho ddng dien cac nhanh. Ta ky hieu I, , \ 2, I3, i4 , I5 va I6 la cac dong dien hieu dung phuc trong cac nhanh co tong tro tuong ting la Z,, Z2, Z3, Z4, Z, va Z6. Doi vdi dong dien trong m6i nhanh, ta co the chon mot trong hai chieu co the lam chieu duong. Doi vdi cac nhanh cd nguon sdd, ngudi ta thudng chon chidu duong cua ddng dien nhanh gidng vdi chieu duong da chon cho cac sdd. VI vay. ta chon chieu duong cua i, giong vdi chieu duong cua E[ va chon chieu duong cua I2 gidng vdi chidu duong cua E2 . Doi vdi cac nhanh khong cd ngudn sdd, thi ta cd the chon mot chieu tuy y lam chieu duong cho ddng dien nhanh. Gia sir ta chon cac chiSu duong cho i3, i4 , i5 va i6 nhu cac mui ten ve tren hinh 3.1.1. - B udc 2 : Viet cac phuong trinh doc lap tuyen tinh theo dinh luat K irchhoff thif nhat (cung goi la dinh luat Kirchhoff cho cac ddng). Cac phuong trinh nay cd dang: nut Ta quy udc rang cac ddng dien cd chieu duong di ra khoi nut duoc lay vdi dau duong (+), va cac ddng dien cd chieu duong di vao nut duoc lay vdi da'u am (-). Cac phuong trinh theo dinh luat Kirchhoff thti nhat viet cho cac nut © , @ va © lan luot la: —i, + i3 + i4 = o ( 3 .i .i ) - i 2 - i 3 + i 5 = o (3. 1.2 ) i 2 - i 5 + i6 = o (3.1.3) Chu y rang phuong trinh theo dinh luat Kirchhoff thti nhat viet cho nut @ i, - i4 - i 6 = o (3 . 1.4 ) khong doc lap tuyen tinh vdi he cac phuong trinh (3.1.1), (3.1.2) va (3.1.3). That vay. neu cong cac phuong trinh (3.1.1), (3.1.2) va (3.1.3) lai ta se duoc: —i i + i 4 + i6 =o ( 3. 1.5 ) Dem nhan phuong trinh (3.1.5) vdi (-1) ta se duoc phuong trinh (3.1.4). Dieu nay chting to rang phuong trinh (3.1.5) la mot td hop tuyen tinh cua cac phuong trinh (3.1.1), (3.1.2) va (3.1.3). 78