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Phương pháp giải các bài tập thí nghiệm vật lí ở trường trung học cơ sở

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HU''''ONG DAN QUAN LI su''''DUNG PHtfCiNC PHAP CIAI CAC BAI TAP THI NCHICH VAT LI fli TRtfCiNC TRUNC HOC CCf Sd TS Nguyin Dinh Hung Trudng Cao ddng Hdi Duong Vgt li Trung hpe ca sd (THCS) g^m cdc ph4n CO, n[.]

HU'ONG DAN QUAN LI su'DUNG PHtfCiNC PHAP CIAI CAC BAI TAP THI NCHICH VAT LIfliTRtfCiNC TRUNC HOC CCf Sd TS Nguyin Dinh Hung Trudng Cao ddng Hdi Duong V gt li Trung hpe ca sd (THCS) g^m cdc ph4n CO, nhipt, dipn, quang Ddy la nhihig kidn thiic \h nhChig hipn tupng ty nhidn CO bdn, rdt gan gui vd gdn lien vdi ddi song cua moi HS Kien thiic vgt li d moi bdi hpe thudng dupe hinh thdnh thdng qua 2, 3, 4, thi nghipm Do dd, vipc dgy hpe Vat li khdng chi hinh thdnh d HS kidn thiic khoa hpe vd nhimg hipn tuong ty nhien, nhihig irng dung kT thugt ma cdn gdp phdn hinh thdnh vd phdt tridn d HS nang luc tu duy, kl nang phan tich, tdng hpp, k! ndng thyc hdnh vd kT thugt Tuy nhien, sdch gido khoa (SGK) Vgt li, sdch bdi tap vdt li THCS dii CO nhidu bdi tap tu dinh tinh ddn dinh lugng nhung cd rdt it bai tap thi nghipm vgt li (BTTNVL) Qua trinh gidi BTTNVL doi hdi HS thyc hipn mpt chudi thao tdc tu duy, thyc hanh nhu chu trinh ciia nhd nghidn ciiu Vi vdy nd khdng nhiing cd tdc dyng hinh kidn thiic mdi, ndng cao chdt lugng ndm vung kidn thiic ciia HS, hinh thdnh vd phdt trien d HS nang luc tu duy, ndng lye thyc hdnh ma cdn hinh vd phdt trien HS nhixng kT nang ca bdn cua nha nghien ciiu Bdi tdp thi nghifm Vat h' Ngay nhgn bdi 15/12/2012; Ngdy duyel ddng 36 BTTNVL Id nhiing bdi tgp phuang dn thi nghiem kiem vgt li md gidi nd ngudi ta tra); phdi phdn tfch, vdn dyng kiln B3: Tien hdnh thi'nghiem, thiic dl ldp lu^n tir dd de xuat thu thgp vd xii li thong tin thu mpt thi nghiem vd tiln hanh dupe qua thi nghiem; thi nghipm dd nit kdt lugn B4: Kit lugn (dua Idi gidi bdi todn) Vi dy: Cho hidu didn thi U Co thi chia BTTNVL = 12 V, mpt bdng den 6V, mpt bdng den 4,5V - 4,5W, mpt thdnh: Bai tgp djnh tinh: Kdt bien trd, ampe kd, von ke, day qud Idi gidi cua bai tgp thinoi, khda K nghipm djnh tinh Id sii dyng a) Xdc djnh tri so cua cdc thi nghiem tao mpt sy kipn, phan bien trd, dipn trd ddy tdc mdt hipn tugng vgt li cy the tir moi bdng den vd cong suat gidi quyet dupe vdn dd bdi cua den 6V cdc den sdng todn ddt binh thudng - Bdi tgp djnh lugng: Kdt b) Cho them mdt dm ke qud Idi giai la nhihig sd cy Xdc dinh dipn trd day tdc bdng the vd phdp mpt dai lugng den nd khong dugc mac vgt li hay sii dyng ket qud ciia vao ngudn dien So sdnh vdi phep cdc dgi lugng vat li dd kdt qud d cdu a va gidi thich nghipm Igi mpt cong thiic, mgt Birdc I : Tun hidu bdi toan dinh luat (tim dii kipn da cho, phdi lim), - Bdi tgp hpp den cd the coi tdm tdt bai toan Id mpt bai tgp thi nghipm vi Cho:U = 12V;U„=6V;U^ gidi bdi tgp hpp den ngudi = 4,5V, Pjj = 4,5 W; bidn trd, vdn hpe phdi Igp lugn dd dua dy kd, ddy noi, , den sang binh doan ve ddi tupng, hien tugng thudng vd diing thi nghipm dl kidm Tim: tra dy dodn a ) I ^ , ^ M,R.,.,R.«vaP^, -' lit([roagin^hX' d l ' d2 , dl Phuffng phdp gidi b) Rj,, Rj2 bang dm kd den BTTNVL khdng dugc c ^ ngudn dien, giai Phuong phdp chung dl gidi thich ket qua mdt BTTNVL cd thi duoc Birdc 2: Lgp ludn thyc hipn qua bon budc sau: Mudn tim R^^ ,R^ can BI: Tim hilu bdi todn; dugc hipu dien thd ddt vdo hai B2: Lap lu§n (dl xdc dinh ddu bidn trd, hipu didn thd dat mdi lidn he giiia cac yeu td da vao hai ddu ddy toe bdng den cho, phdi tim vd xay dung mdt cho den sdng binh thudng' 25/01/2013 • TAP CHI THIFT 81 GIAO DUC - SO 90 - 02/2013 H U Q N G DAN QUAN LI SI/DUNG va cudng dp ddng didn qua chiing dd Ap dyng cdng thiic: R = U/I ta tinli dugc R Rieng den D^, bilt U^, Pj™ nen nlu dilu chinh dl \J^ = 4,5V thi cd thi xac dinh dugc dien trd ddy tdc den D2 theo cdng thdc R^ = U^dj/Pj; (Chii y: Bdi todn khong yeu cdu tinh theo cdng thiic Do dd viec sir dyng cdng thiic dd tinh chi de kilm tra kdt qud do) Theo bdi ra, dd D,, D^ sdng binh thudng thi U^, = 6V, U^ = 4,5V Cd thd xdy mpt sd trudng hpp sau: Ndu mdc Idn lugt timg den vdo hipu dien the 12V thi cac den phdi dugc mdc ndi tilp vdi bidn trd rdi dieu chinh biln trd cho hipu didn thd giua hai ddu mdi bdng den bang hieu didn the dinh miic cua chiing hinh 3) Hinh So dd mgch dien thi nghiem R^ R Neu hai den mdc ndi tidp vd mdc vdo hieu dipn the 12V thi chung phai dupe mdc ndi tidp vdi qua bidn trd, dieu chinh bidn trd de U^,, = 6v, U^ - 4,5V (hinh 4) Tuy nhien, Idii den D^ sdng binh thudng thi I^ = I A, I, = lA vd P, = 6W NghTa Id ndu den Dj ed cdng sudt djnh muc khdc 6W thi hai den khdng thd sdng binh thudng va khdng the su dung so dd mgch dipn d hinh Hmh Sa mgch dien thi nghiem R^ R^, hai den mdc ndi tiep - Ndu hai den mdc song song xh\ mdc vdo hipu dipn thi 12V thi chiing phdi dugc mac theo so dd hinh Didu chinh cdc bien trd de hipu dipn thd Uj^, = 6V (hay U-, = 6V),U,, = 4,5V Tuy nhien, bai toan chi cho dimg mpt bidn trd ndn khdng the dung so thi nghiem nhu hinh Hinh So mgch dien thi nghiem R^ 7?^^ hai den mdc song song Nhu vdy sau budc lap ludn GV giiip HS xdc dinh dupe sa dd maeh didn thi nghiem can sir dyng Id so cho d hinh (vdi mdi den can tidn hdnh thi nghiem it nhat lan) Budc 3: Tiln hdnh thi nghiem CJ budc nay, tatienhdnh cdc Ihao tac sau: Lya chpn dung cy thich hpp (nhu yeu cdu bdi todn); Lap bang ghi sd lieu U, I, dugc; Ldp rap mgch didn thi nghiem nhu sa dd hinh 3; Kiem tra su an toan; Tidn hdnh thf nghiem: Ddt hirgt cua bien trd tai vi tri cho R^^ cd gid tri eye dai -> ddng khda K -> didu chinh bien Ird (ddng thdi quan sat sd chi cua vdn ke) cho U = U^^ -> dpc sd chi cua ampe kd -> ^ i kit qua U, I dupe vdo bang so lieu (tiln hdnh thi nghipm mdc ddn vdo mgch dipn rdi 1^ Igi thi nghi^ vdi den 2); - Ap ( ^ g cdngfliuctinh U (U = U^ ± AU), tlnh I a = la, ± AI) vd cdng thirc tinh R (R = U/I), tinh P(P=U.I) Birdc 4: Ket lugn Phan tich ket qud thi nghiem va nit ket lugn: Thi nghipm chung td, ddn sdng binh thudng thi a)R„- ,R„= b) Dimg dm ke R^ thdy gia tri dupe nhd hem gid tri xdc dinh dupe phdn (a) Gidi thich: Khi ddt vdo hai ddu day tdc bdng den mpt hidu didn thd, cd mdt dong dien chgy qua ddy tdc bdng den, ddy toe bdng den ndng sdng, nhiet dp ddy tdc bdng den cao nen dien trd ciia nd Idn hon nd cd nhiet dp phdng (R phy thupc nhiet dp) Mgt s6 bdi tgp thi nghiem phan dien hgc - Vgt li9TBCS Chiing tdi xin gidi thieu mdt sd bai tap thi nghipm da suu tdm vd xdy dyng sau: 1) Bdng thi nghiem hay chung td rdng cd sy phy thupc lan giiia hidu didn thd d hai ddu mpt doan maeh dien vd cudng dp ddng dien maeh 2) Bdng thi nghiem hay chiing td rdng didn trd day dan phu thudc bdn chdt, kich thudc ciia day 3) Cho cac dien Ord, bdng den, ngudn didn (mpt chidu), TAP CHI THIET BI GIAO DUC-SO 90-02/2013 • 37 HUaNG DAN QUAN Li SLT DUNG khda K, ddy nli Hdy chiing td rdng: - Trong mgch dipn gdm cac dipn trd mdc ndi tiep: + Dipn trd tuong duang Idn ban cdc di^n trd thdnh philn; + Cudng dp ddng di§n todn mgch bdng cudng dp ddng dipn di qua moi dipn trd; + Hipu dipn the hai dau dogn mgch Idn ban hi^u di^n the giita hai dau moi dipn trd - Trong mgch dif n g6m cdc dipn trd mdc song song: + Didn trd tuong duong nhd hon cac dipn trd thdnh phdn; •I- Hieu didn thd hai dau dogn mgch bang hipu didn the giiia hai ddu mdi didn trd; + Cudng dp ddng dien todn mgch Idn ban cudng dp ddng dien qua mdi didn trd 4) Cho ngudn diea (mdt ehieu), mdt bidn trd, bdng den, khda K, day ndi Hay chiing to rdng bien trd Id dipn trd cd tri sd thay ddi dugc 5) Cho ngudn didn (mdt ehieu) cd hieu di?n thi thay ddi dugc, mpt bien h-d, khda K, didn frd, ddy ndi, cac vdn ke, ampe ke Hay nghipm Igi dinh lugt 6m 6) Cho ngudn dipn (mpt chilu) 4.5V, mpt biln trd, khda K, cdc day dien trd, day ndi, vdn kd, ampe kd Hay nghiem Igi cdng thiic R = p l/S 7) Cho mpt ngudn dien (mpt ehieu), mdt bien trd, mdt dien trd, ampe ke, vdn ke, day ndi, khda K Hay xac dinh tri sd phan biln trd ndm hong dogn mgch 38 dipn 8) Cho biln trd, von kl, ampe kl, cdc dipn trd, khda K, ddy ndi vd nguon dipn (mgt chilu) Chiing td rdng cd thi dimg bilnft-ddk Idm thay dli hipu di^n thi giiia hai dau mpt dogn mgch di$n vd cudng dp ddng dipn chgy qua m^t dogn mgch dd 9) Cho biln trd, vdn kl, ampe kl, dipn trd, khda K vd nguon dipn (mpt ehieu), ddy ndi Chiing td rdng cd the dimg bien trd dd phan dp, phdn ddng cho mdt doan mgch didn 10) Mpt nguon dien (mpt chidu), mpt day dipn trd cd ehieu ddi AB = 0,5 m, didn tro suat 1,2 Qm, dudng kinh 0,2mm, day ndi Hay xac dinh dien trd ciia cdc doan AC = 0,15m, AD = 0,25m, AE - 0,4m vd AB bang ampe ke, von ke va bdng dm kd rdi dimg cdng thiic R = p l/S de nghipm lai kdt qud Kit luan Viec gidi BTTNVL ddi hdi HS phdi van dyng kidn thiic dd hpe, sy hieu biet, sy trdi nghidm cimg nhiing pham chat kT thuat, kT ndng thyc hdnh cua bdn than Viec gidi BTTNVL se giiip cimg co, khdc sdu kiln thiic, hidu kien thirc Idm cho chat lugng ndm viing kiln thiic dugc ndng cao, giiip hinh thdnh vd phdt triln d HS ndng lye tu duy, tinh can than, ti mi; hmh vd phat triln d HS ndng luc cimg kT ndng thyc hdnh, kl thugt tra ddng gid c ^ phai dua vdo nhung BTTNVL mdt cdch thich hgp Tdi llfu tham khao Dodn Duy Hinh (chd bidn) - Nguyen Van Hda Ngd Mai Thanh - Nguyen Due Tham V§t U NXB Gido dye, H 2010 Dodn Duy Hinh (chii bidn) - Nguydn Van Hda Vii Quang - Ngd Mai Thanh - Nguyen Diic Tham Bdi tdp Vat li 9, NXB Gido dye, H 2010 Summary The laboratory exercises in physics requires students to apply knowledge leamed, the knowledge, the experience and the technical qualities and skills of practice itself The laboratory exercises in physics will help strengthen, deepen knowlet^, understanding and knowledge for quality mastering advanced knowledge, help shape and develop in students the capacity to think, the careful meticulous; formation and capacity development in students with practical skills and techniques Therefore, in the teaching process as well as in worthwhile checking Vi vdy, qua trinh should be included in physics dgy hpe cung nhu kilm laboratory exercises properly • TAP CHI THIET BI GIAO DUG-SO 90-02/2013

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