TRNG I HC TN C THNGKHOA IN - IN TB MN K THUT IN

TI LIU HNG DN TH NGHIM

MCH IN

Tp.H Ch Minh, thng 8 - 2014

TRNG H TN C THNGKHOA IN IN T---------------------

CNG HA X HI CH NGHA VIT NAMc lp - T do - Hnh phc**************NI QUY

PHNG TH NGHIM IN-IN T

IU I. TRC KHI N PHNG TH NGHIM SINH VIN PHI:1. Nm vng quy nh an ton ca phng th nghim.2. Nm vng l thuyt v c k ti liu hng dn bi thc nghim.3. Lm bi chun b trc mi bui th nghim. Sinh vin khng lm bi chun b theo ngyu cu s khng c vo lm th nghim v xem nh vng bui th nghim .4. n phng th nghim ng gi quy nh v gi trt t chung. Tr 15 pht khng c vo th nghim v xem nh vng bui th nghim .5. Mang theo th sinh vin v gn bng tn trn o.6. Tt in thoi di dng trc khi vo phng th nghim. IU II. VO PHNG TH NGHIM SINH VIN PHI:1. Ct cp, ti xch vo ni quy nh, khng mang dng c nhn vo phng th nghim.2. Khng mang thc n, ung vo phng th nghim.3. Ngi ng ch quy nh ca nhm mnh, khng i li ln xn.4. Khng ht thuc l, khng khc nh v vt rc ba bi.5. Khng tho lun ln ting trong nhm.6. Khng t di chuyn cc thit b th nghimIU III. KHI TIN HNH TH NGHIM SINH VIN PHI:1. Nghim tc tun theo s hng dn ca cn b ph trch.2. K nhn thit b, dng c v ti liu km theo lm bi th nghim.3. c k ni dung, yu cu ca th nghim trc khi thao tc.4. Khi my c s c phi bo ngay cho cn b ph trch, khng t tin sa cha.5. Thn trng, chu o trong mi thao tc, c thc trch nhim gi gn tt thit b.6. Sinh vin lm h hng my mc, dng c th nghim th phi bi thng cho Nh trng vs b tr im th nghim.7. Sau khi hon thnh bi th nghim phi tt my, ct in v lau sch bn my, sp xp thit b tr v v tr ban u v bn giao cho cn b ph trch.IU IV.1. Mi sinh vin phi lm bo co th nghim bng chnh s liu ca mnh thu thp c v np cho cn b hng dn ng hn nh, cha np bo co bi trc th khng c lm bi k tip.2. Sinh vin vng qu 01 bui th nghim hoc vng khng xin php s b cm thi.3. Sinh vin cha hon thnh mn th nghim th phi hc li theo quy nh ca phng o to.4. Sinh vin hon thnh ton b cc bi th nghim theo quy nh s c thi nhn im ktthc mn hc.IU V.1. Cc sinh vin c trch nhim nghim chnh chp hnh bn ni quy ny.2. Sinh vin no vi phm, cn b ph trch th nghim c quyn cnh bo, tr im thi. Trng hp vi phm lp li hoc phm li nghim trng, sinh vin s b nh ch lm th nghim v s b a ra hi ng k lut nh trng.

Tp.HCM, Ngy 21 thng 08 nm 2014KHOA IN-IN T( k)

TS. V HONG DUY

MC LCBi 1: Mch in DC, Mch Thevenin-Norton v mng mt ca2Bi 1.1: Mch in DC2Bi 1.2: Mch Thevenin-Norton v mng mt ca10Bi 2: Th nghim mch in AC 1 pha v 3 pha15Bi 2.1: Mch in xoay chiu 1 pha15Bi 2.2: Mch in xoay chiu 3 pha22Bi 3: Mng 2 ca tuyn tnh khng ngun v mch cng hng RLC31Bi 3.1: Mng 2 ca tuyn tnh khng ngun31Bi 3.2: Mch cng hng RLC42Bi 4: Qu trnh qu mch tuyn tnh-mch phi tuyn49Bi 4.1: Qu trnh qu mch tuyn tnh49Bi 4.2: Mch phi tuyn57Ti liu hng dn th nghim mch in

1B mn K thut in, Khoa in in t

H V TN: ................................LP:...................................MSSV:...................................

Nhn xt ca gio vin hng dn..............................................................................................................

Bi 1MCH IN DC, MCH THEVENIN-NORTON V MNG MT CABi 1.1MCH IN DCI/. MC CH o dng in, o in p, o cng sut. Kho st mch in ni tip, mch in song song. Kho st nh lut K1, nh lut K2 v nguyn l xp chng. Kho st hin tng ngn mch, h mch.II/. PHN L THUYTnh lut OhmQuan h gia dng v p trn mt phn t in tr R:U = R.ICng sut trn in tr R:P = U.I

nh Lut K1Tng i s cc dng in bt k ti mt nt bng khng.

I k 0nt

Qui c: Cc dng in c chiu dng i vo nt th ly du (+), cn i ra khi nt th ly du (-), hoc ngc li.nh Lut K2Tng i s cc in p trn cc phn t dc theo cc nhnh trn mtvng bng khng.

Uk 0vng

Nguyn l xp chngp ng to bi nhiu ngun kch thch tc ng ng thi th bng tng ccp ng to bi mi ngun kch thch tc ng ring r.III/. DNG C TH NGHIM Bng mch th nghim: bn s 1 MODULE 1A v MODULE 1B. Ngun p v ngun dng DC iu chnh c. VOM. Dy ni.IV/. PHN TH NGHIM1/. Kho st nh lut K1 v K2 cho mch in ni tip v song song: a/. Mch ni tip Mch in th nghim nh hnh 1: R1=2.8 , R2=2.2 , R3=2.2 , R4=2.4 .(1)R2 A JV1R3R1R4A

Hnh 1J

S dng mch (a),MODULE 1A Cc bc thc hin:1. t ngun dng v tr OFF.2. Kt ni ngun dng vo s mch.3. Mc Ampe k o dng in chy qua in tr R1.4. Bt ngun dng ON, chnh ngun dng gi tr 2A.5. c ch s dng in qua R1.6. Dng mt Vn K ln lt o cc hiu in th gia hai u ngun dng, R1, R2, R3, R4. Ghi cc gi tr o c vo bng s liu.7. Bt ngun dng v tr OFF.8. Qu trnh lm tng t nh cc bc trn, o dng in qua cc in tr R2.9. c v ghi cc ch s in p, dng in o c vo bng.

3B mn K thut in, Khoa in in t

Phn tNgun dngR1R2R3R4

Dng21.4380.5220.5520.552

p3.9653.671.2421.6431.17

Cng sut7.935.2770.6480.9060.645

Cu hi:1. T bng s liu, hy tnh cng sut ca mi phn t (in vo bng).2. Hy kim chng nh lut K1 v K2. nh lut K1 ti nt 1:Ti nt 1 :Tng dng in vo bng tng dng in raIR1=IR2+IR3+IR4 nh lut K2 cho vng V1: Tng i s cc in p ca cc phn t trn vng mt bng khng.

Uk 0

3. Nhn xt v mi quan h gia cng sut ca ngun dng vi tng cngsut tiu tn trn cc in tr.`Tng cng sut ca ngun bng tng cng sut ca ti trn ton mch Pngun Pti

b/. Mch song song Mch in th nghim nh hnh 2: R1=10 , R2=22 , R3=50 , R4=50

S+ dng mch (b), MODULE 1A. Cc bc thc hin:1. t ngun p v tr OFF.2. Kt ni ngun p vo s mch.3. Mc Ampe k o dng in chy qua in tr R1.

5B mn K thut in, Khoa in in t

4. Bt ngun p ON, chnh n gi tr 20V.5. c ch s dng in qua R1.6. Dng mt Vn K ln lt o cc hiu in th gia hai u R1, R2, R3, R4. Ghi cc gi tr o c vo bng s liu.7. Bt ngun p v tr OFF.8. Qu trnh lm tng t, ln lt o cc gi tr dng in chy qua cc in tr R2, R3 v R4.9. c v ghi cc ch s in p, dng in o c vo bng sau

Phn tNgun pR1R2R3R4

Dng0.930.9180.480.2160.222

p209.1210.6910.710.7

Cng sut18.68.3725.1312.3112.375

Cu hi:1. T bng s liu, hy tnh cng sut ca mi phn t (in vo bng trn).2. Kim chng nh lut K1 v K2 nh lut K1 ti ntTng i s cc dng in ti nt bng khng. Tng dng in vo bng tng dng in ra

I k 0nt nh lut K2 cho vng V1Tng i s cc in p trn cc phn t dc theo vng mt bng khng.

Uk 0vng

3. Nhn xt v mi quan h gia cng sut ca ngun p vi tng cng suttiu tn trn cc in trGia cng sut ca ngun p vi tng cng suttiu tn trn cc in tr gn nh nhau.

2/. Nguyn l xp chng Mch in th nghim nh hnh 3: R1=33 , R2=5 , R3=10 , R4= 3.2 .

6B mn K thut in, Khoa in in t

R1R3

VIR4R- 2A

Hnh 3

S dng s trong MODULE 1B. Cc bc thc hin:1. t ngun p v ngun dng v tr OFF.2. Kt ni ngun p v ngun dng vo s mch.3. Mc Ampe k o dng in chy qua in tr R2.4. ng kho K1, h kho K2. iu ny cho php kch thch mch bng ngun p v trit tiu ngun dng (ngun p ON, ngun dng OFF),5. Bt ngun p ON v ngun dng OFF, chnh ngun p gi tr 20V.6. c v ghi vo bn s liu ch s dng in qua R2, gi l dng I21.7. Dng Vn K o hiu in th gia hai u R2, gi l U21. Ghi cc gi tr o c vo bng s liu.8. M kho K1, ng kho K2. iu ny cho php kch thch mch bng ngun dng v trit tiu ngun p (ngun p OFF, ngun dng ON), chnh ngun dng gi tr 2A.9. c v ghi vo bn s liu ch s dng in qua R2, gi l dng I22.10. Dng Vn K o hiu in th gia hai u R2, gi l U22. Ghi cc gitr o c vo bng s liu.11. ng c 2 kho K1 v kho K2. iu ny cho php kch thch mch bng c hai ngun dng v ngun p (c hai ngun ON).12. c v ghi vo bn s liu ch s ca Ampe k, ch dng in qua R2,gi l dng I2.13. Dng Vn K o hiu in th gia hai u R2, gi l U2. Ghi cc gi tr o c vo bng s liu.

Ngun p ONNgun dng OFFNgun p OFFNgun dng ONNgun p ONNgun dng ON

I21U21I22U22I2U2

0.3731.90.251.2740.6273.186

Cu hi:1. T bng s liu trn, so snh I2 vi I21 v I22, U2 vi U21 v U22.

Tng: I21+I22=I2Tng: U21+U22=U2

2. Nguyn l xp chng:

p ng to bi nhiu ngun kch thch tc ng ng thi th bng tng ccp ng to bi mi ngun kch thch tc ng ring r.3. Nhn xt v nguyn l xp chng cho mch in trnTng dng ngun p v tng dng ngun dng trong mch bng tng dng trong mch.4. Sinh vin hy t mc mch kho st nguyn l xp chng trn cc in tr R1, R3, R4 theo trnh t cng vic ging nh trn.

in tr R1

Ngun p ONNgun dng OFFNgun p OFFNgun dng ONNgun p ONNgun dng ON

I11U11I12U12I1U1

0.52117.91-0.036-1.2840.48316.63

Nhn xt:Tng: I11+I12=I1Tng: U11+U12=U1

in tr R3

Ngun p ONNgun dng OFFNgun p OFFNgun dng ONNgun p ONNgun dng ON

I31U31I32U32I3U3

0.141.544-0.285-2.996-0.142-1.478

Nhn xt:............. Tng : I31+I32=I3.......... Tng : U31+U32=U3

in tr R4

Ngun p ONNgun dng OFFNgun p OFFNgun dng ONNgun p ONNgun dng ON

I41U41I42U42I4U4

0.1380.3561.7054.351.8494.72

Nhn xt:...............Tng : I41+I42=I4.......... Tng : U41+U42=U4........Nguyn l xp chng dng in3/. Hin tng ngn mch, h mch Mch in th nghim nh hnh 4.

S dng s trong MODULE 1BCc bc thc hin: Ngun p cung cp cho mch o l 20V, ngun dng l 2A1. t ngun p v ngun dng v tr OFF.2. Kt ni ngun p v ngun dng vo s mch. Kho K1 v kho K2ng.3. H mch gia hai im E-M bng cch h mch hai im E-F, dng dydn ni tt cc cp im sau li vi nhau: A-B, C-D, G-H.

8B mn K thut in, Khoa in in t

4. Bt ngun p v ngun dng ON.5. o hiu in th h mch gia hai im E-M, gi l Uhm .6. o dng in qua cc in tr R1 v R3. Gi l IR1, IR3.7. Ghi cc gi tr o c vo bng s liu.8. Bt ngun p v ngun dng v tr OFF.9. Ngn mch gia hai im E-M bng cch mc gia hai im E M mtAmpe k .10. Bt ngun p v ngun dng ON. c ch s ca Ampe k, gi l Inm.11. Bt ngun p v ngun dng v tr OFF.12. Ni tt gia E-M bng mt si dy dn, mc Ampe k gia hai im A-B o dng IR1.Tng t, o dng IR3.13. Lu , mi ln thay i v tr mc Ampe k nn bt ngun p v ngundng v tr OFF.

H mch hai im E-MNgn mch hai im E-M

UhmIR1IR3InmIR1IR3

9.30.310.3070.9510.577-0.377

T bng s liu, hy nhn xt cc kt qu o c.................................................................................................................................................

9B mn K thut in, Khoa in in t

10B mn K thut in, Khoa in in t

H V TN: ................................LP:...................................MSSV:...................................

Nhn xt ca gio vin hng dn..............................................................................................................

Bi 1.2MCH THEVENIN V MNG MT CA

I/. MC CH Minh ha cc nh l Thevenin v Norton. S dng s tng ng Thevenin v Norton kho st mng mt ca. Kho st s phi hp tr khng gia ngun v ti truyn cng sut cc i.II/. PHN TM TT L THUYTnh L TheveninC th thay th tng ng mt mng mt ca tuyn tnh bi mt ngun in p bng in p trn ca khi h mch mc ni tip vi tr khng Thevenin ca mng mt ca.nh L NortonC th thay th tng ng mt mng mt ca tuyn tnh bi mt ngun dng in bng dng in trn ca khi ngn mch mc song song vi tr khng Thevenin ca mng mt ca.Zo+Zo

UhmInm

a/S Theveninb/S Norton

Trong :Uhm: in p h chng mch trn phn mng mt ca.

Inm: dng in ngn mch qua ca phn mng mt ca.

Zo: tr khng vo mng mt ca khi cho n tr nn th ng.

Mt khc, cc gi tr trong cc s tng ng trn cn c lin quan vi nhau theo cng thc sau:Cng thc ny gip ta xc nh c Zo thng qua vic xc nh Uhm v Inm .Nguyn l truyn cng sut cc i:Theo nh l Thevenin, bt k mt mng mt ca no ni vi ti cng c th a v s tng ng nh hnh sau:ZnE+ZtMch tng ngmng mt ca

c/ Mng mt ca

Trong E v Zn c xc nh theo nh l tng ng Thevenin.Ti Zt s nhn c cng sut cc i nu:

y l iu kin phi hp tr khng gia ngun v tiTrong bi th nghim ny, Zn = Rn v Zt = Rt u l thun tr nn iu kin phi hp tr khng s l:Rt = RnIII/. DNG C TH NGHIM Bng mch th nghim: bn s 1 MODULE 1C. Ngun p v ngun dng DC iu chnh c. VOM. Dy ni.IV/. PHN TH NGHIM Mch th nghim nh hnh 5:

11B mn K thut in, Khoa in in t

R5R1R3AR2RLR4B

VI

Hnh 5Mc ch: Xy dng s tng ng Thevenin v Norton ca mch nhnt 2 cc A v B. S dng s trong MODULE 1C.1/. Xy dng m hnh mch tng Thevenin - Norton Cc bc thc hin:1. t ngun p v ngun dng v tr OFF.2. Kt ni ngun p v ngun dng vo s mch.3. ng 2 kha K1 v kha K2.4. H mch hai im A-B (hai u in tr R2).5. Bt ngun p v ngun dng ON, chnh ngun p 20V v ngun dng 2A.6. Dng Vn k o hiu in th h mch gia hai im A-B, gi l Uhm .7. Tt ngun p v ngun dng.8. Ngn mch A-B bng cch gn Ampe k gia A v B.9. Bt ngun p v ngun dng ON.10. c gi tr trn Ampe k, gi l dng ngn mch Inm.

UhmInmRo

3.4381.013.40396

Cu hi:1. T bng s liu hy tnh in tr tng ng Thevenin/Norton Ro R 0 =2. V mch tng ng Thevenin v mch tng ng Norton. Ghi y cc gi tr ca cc phn t trong mch.

2/. Phi hp tr khng gia ngun v ti (truyn cng sut cc i)

Cc bc thc hin1. Tt ngun p v ngun dng. Bt kho K1 v tr ON, kho K2 ON.2. Ni vo hai im A-B mt bin tr Rt bng cch ni tt hai im B-D vmc Ampe k gia hai im A-C.3. Bt ngun p v ngun dng ON.4. iu chnh bin tr Rt khong 10 gi tr t khng n gi tr cc i.5. ng vi mi gi tr ca Rt hy ghi li gi tr dng in v in p trn Rtri in vo bng sau:

STT12345678910

Rt102030405060708090100

U2.562.933.093.173.223.263.273.293.3013.313

I0.250.140.0960.0720.0570.0470.0420.0370.0340.03

P0.6400.4100.2970.2280.1840.1530.1370.1220.1120.099

Cu hi:1. Tnh ton cng sut tiu tn P trn Rt ng vi tng gi tr ca Rt (in gi tr tnh ton vo bng trn).2. T bng gi tr o c, v th cng sut tiu tn P trn Rt theo gi tr ca Rt, P = f(Rt).

P12345678910Rt

3. T th hy xc nh gi tr ca Rt sao cho cng sut tiu tn trn Rt lln nht. Rtmax =Theo l thuyt gi tr ca Rt l bao nhiu. Rtmax (l H V TN: ................................LP:...................................MSSV:...................................

Nhn xt ca gio vin hng dn..............................................................................................................

15B mn K thut in, Khoa in in t

Bi 2TH NGHIM MCH IN AC 1 PHA V 3 PHABi 2.1MCH XOAY CHIU 1 PHA

I/. MC CHKho st cc thng s c trng ca mt on mch c kch thch bng ngun xoay chiu hnh sin nh: p, dng, tr khng, cng sut, lch pha.V gin vect dng, p ca mt on mch.II/. TM TT L THUYTTrong ch xc lp iu ho (ngun kch thch hnh sin), cc thng s trongmi nhnh c mi quan h vi nhau nh sau:U = Z*IhayI = Y*U = u - iP = U x I x Cos() = Re[Z] x I2 III/. DNG C TH NGHIM Bng mch th nghim: bn s 2 MODULE 2B. Ngun p xoay chiu 3 pha Y/ - 41/24 VAC. VOM s. Watt k. Dy ni.IV/. PHN TH NGHIM1/. Mch thun tr, thun dung, thun cm Mch th nghim nh hnh 7:

VacZ**WAAVB

Hnh 7 Da vo hnh 7, sinh vin ln lt mc mch vi cc ti Z l: R, L, C. Vi trng hp ti R, sinh vin t v mch th nghim vi y dng c o, th nghim c tin hnh theo cc bc sau:1. Bt cng tc ngun v tr OFF.2. Mc mch nh hnh v, lu cc dng dng v p ca Watt k.3. Bt cng tc ngun v tr ON.4. c cc tr s o cc gi tr in p, dng in, cng sut v ghi vobng sau tng ng vi tng trng hp. Qu trnh lm tng t cho cc ph ti L v C. Cc gi tr o c ghi vo bng sau

U (V)I (A)P(kW)

R24.640.53813

L24.460.9758

C25.30.5470

T bng s liu o c, trong mi trng hp hy:1. Xc nh gc lch pha gia hiu in th v dng in gia hai imA-B (in gi tr tnh ton c vo bng trn).2. V gin vect dng v p ca on mch.

3. Tnh cng sut tc dng trn mi phn t mch theo l thuyt. So snhvi cng sut o c..............................................................................................................................................................................................................................................................................................................................................................................................4. Nhn xt...............................................................................................................................................................................................................................................................2/. Mch R, L, C ni tip Mch th nghim nh hnh 8:

VacL * ARC*WAVC

BDHnh 8 Th nghim c tin hnh theo cc bc sau:1. Bt cng tc ngun v tr OFF.2. Mc mch nh hnh v 8, lu cc dng dng v p ca Watt k.3. Bt cng tc ngun v tr ON.4. c cc tr s o tr in p, dng in, cng sut v ghi vo bng sau:

U(V)I(A)P(kW)

on mch A-B25.720.3458

R15.550.348

L20.340.348

C18.160.348

Lu : o cng sut trn cc phn t R, L, C ta ch vic mc cun p caWatt k mc ti: A-C, C-D, D-B. T bng s liu o c hy:1. Xc nh gc lch pha gia hiu in th v dng in gia hai imA-B v tng phn t mch (gi tr tnh ton in vo bng trn).2. V gin vect dng v p ca on mch.

3. Tnh biu thc UAB.I.Cos(). So snh vi cng sut o c...........................................................................................................................................................................................................................................................Xc nh tr khng phc ZAB t gi tr U, I va o c trong bng s liu..............................................................................................................................................................................................................................................................4. Nhn xt..............................................................................................................................................................................................................................................................................................................................................................................................

3/. Mch R, L, C song song S mch th nghim nh hnh 9: * A*WAVRL

VacC

BHnh 9 Th nghim c tin hnh theo cc bc sau1. Bt cng tc ngun v tr OFF.

2. Mc mch nh hnh v 9, lu cc dng dng v p ca Watt k.3. Bt cng tc ngun v tr ON.4. c cc tr s o tr in p, dng in, cng sut v ghi vo bng sau:Lu : trc khi chuyn v tr cc Ampe k ti cc v tr cc nhnh o dngin nhnh th phi bt cng tc ngun v tr OFF.

U(V)I(A)P(kW)

on mch A-B251.25421

R250.5513

L25118

C250.550

T bng s liu o c hy:1. Xc nh gc lch pha gia hiu in th v dng in gia hai im A-B v tng phn t mch bng gin vect (gi tr tnh ton in vo bng trn).2. V gin vect dng v p ca on mch

3. Tnh biu thc UAB.I.Cos(). So snh vi cng sut o c...............................................................................................................................4. Xc nh tr khng phc ZAB t gi tr U, I va o c trong bng s liu..............................................................................................................................5. Nhn xt................................................................................................................................

4/. Mch R, L, C hn hp S mch th nghim nh hnh 10:

*A * WAR

C VacVC

L

BHnh 10

Th nghim c tin hnh theo cc bc sau:1. Bt cng tc ngun v tr OFF.2. Mc mch nh hnh v 10, lu cc dng dng v p ca Watt k.3. Bt cng tc ngun v tr ON.

20B mn K thut in, Khoa in in t

4. c cc tr s o tr in p, dng in, cng sut v ghi vo bngsau:Lu : trc khi chuyn v tr cc Ampe k, Volt k v Watt k vi cc v tr cc nhnh o dng in nhnh th phi bt cng tc ngun v tr OFF.

U(V)I(A)P(kW)

on mch A-B25.770.4062

R10.330.2292

L18.710.2280

C25.760.5640

T bng s liu o c hy:1. Xc nh gc lch pha gia hiu in th v dng in gia hai im A-B v tng phn t mch bng gin vect (gi tr tnh ton in vo bng trn).2. V gin vect dng v p ca on mch.

3. Tnh biu thc UAB.I.Cos(). So snh vi cng sut o c.............................................................................................................................................................................................................................................................4. Xc nh tr khng phc ZAB t gi tr U, I va o c trong bng s liu..............................................................................................................................5. Nhn xt..........................................................................................................................................................................................................................................................................................................................................................................................

21B mn K thut in, Khoa in in t

H V TN: ................................LP: ................................................MSSV: .............................................

Nhn xt ca gio vin hng dn..............................................................................................................

Bi 2.2MCH XOAY CHIU 3 PHA

I/. MC CHKho st vic u dy cho ngun ti v cch mc cc dng c o trong mt mch in ba pha.Kho st vic o cng sut trn mt mch ba pha.II/. TM TT L THUYTMach ba pha l h thng gm ba sc in ng, ba ti v cc dy ni chng. Mt h thng ba pha thng dng nht l h thng ba pha c ngun ba pha i xng. C hai cch mc ti l mc hnh sao v mc tam gic. Trong tng trng hp, cn lu n cc cng thc tnh ton dng, p pha v dy. o cng sut ti ba pha, ngi ta ln lt o cng sut tng pha (phng php dng ba Watt k) hay dng phng php hai Watt k. Mi phng php o c tng c im ring v phm vi kho st ring m sinh vin cn lu .Sinh vin cn n li kin thc mch ba pha trong gio trnh mch in 1 trckhi th nghim bi ny.III/. DNG C TH NGHIM Bng mch th nghim : bn s 2 MODULE 2B, MODULE 2C Ngun p xoay chiu 3 pha, tn s 50 Hz. VOM s. Watt k. Dy niIV/. PHN TH NGHIM1/. Mch 3 pha ti hnh sao - c dy trung tnh S mch th nghim nh hnh 11. Ti Z1, Z2, Z3 l thun tr.

V1Z1IpV2Z2IpV3Z3Ip It

0Hnh 11

t cc cng tt ngun V1, V2, V3 v tr OFF trc khi tin hnh mc mch. Sinh vin t mc mch nh hnh 11 v cc ng h o lng thch hp thc hin o cc gi tr dng in pha, p pha, p dy v dng dy trung tnh ri ghi kt qu vo bng sau:

Up1Up2Up3Ud1Ud2Ud3Ip1Ip2Ip3Itt

25.9424.925.643.243.844.40.550.550.5830.017

o cng sut trn cc ti t suy ra cng sut ca mch ba pha.

P1P2P3Pt

131314

Cu hi:1. V th vect dng p pha v dy. Nhn xt mi quan h gia cc i lng.

.............................................................................................................................................................................................................................................................................................................................................................................................2. Lm li th nghim trn vi mi ti Z1, Z2, Z3 l R-C mc ni tip.Up1Up2Up3Ud1Ud2Ud3Ip1Ip2Ip3Itt

25.3525.5125.474343.243.80.3850.3850.390.4

o cng sut trn cc ti t suy ra cng sut ca mch ba pha.

P1P2P3Pt

55515

V th vect dng p pha v dy. Nhn xt mi quan h gia cc i lng

................................................................................................................................................................................................................................................................................................................................................................................................3. Lm li th nghim trn vi ti Z1, Z2 l thun tr R, Z3 l R, L, C mc ni tip (ti bt i xng).

Up1Up2Up3Ud1Ud2Ud3Ip1Ip2Ip3Itt

2524.825.342.843.543.80.3240.330.3340.42

o cng sut trn cc ti t suy ra cng sut ca mch ba pha.

P1P2P3Pt

99927

V th vect dng p pha v dy. Nhn xt mi quan h gia cc i lng

.................................................................................................................................................................................................................................................................................................................................................................................................................................2/. Mch 3 pha ti hnh sao khng dy trung tnh (i xng v bt i xng) S mch th nghim nh hnh 12. Ti Z1, Z2, Z3 l thun tr.

V1Z1Z3Ip1V2Z2Ip2V3 Ip3

0

Hnh 12 Bt cc cng tt ngun V1, V2, V3 v tr OFF trc khi tin hnh mc mch. Sinh vin mc mch nh hnh 12 v thc hin o dng pha, p pha, p dy ri ghikt qu vo bng sau:Up1Up2Up3Ud1Ud2Ud3Ip1Ip2Ip3

24.7624.1224.74242.643.10.5380.5470.568

o cng sut trn cc ti t suy ra cng sut ca mch ba pha:

P1P2P3Pt

12121236

Cu hi:1. V th vect dng p pha v dy. Nhn xt mi quan h gia cc i lng.

..................................................................................................................................................................................................................................................................................................................................................................................................................2. o in p im trung tnh. Nhn xt. Utt =3. Lm li th nghim trn vi mi ti Z1, Z2, Z3 l R, L mc ni tip.Up1Up2Up3Ud1Ud2Ud3Ip1Ip2Ip3

25.224.7254243.543.40.210.210.217

o cng sut trn cc ti t suy ra cng sut ca mch ba pha:

P1P2P3Pt

2226

V th vect dng p pha v dy. Nhn xt mi quan h gia cc i lng

......................................................................................................................................................................................................................................................................................................................................................................................................4. Lm li th nghim trn vi ti Z1, Z2 l thun dung C, Z3 l R, C mc ni tip (ti bt i xng).

Up1Up2Up3Ud1Ud2Ud3Ip1Ip2Ip3

25.0624.32542.443.2430.3850.3840.39

o cng sut trn cc ti t suy ra cng sut ca mch ba pha:

P1P2P3Pt

55515

V th vect dng p pha v dy. Nhn xt mi quan h gia cc i lng

....................................................................................................................................................................................................................................................................3/. Mch 3 pha ti hnh tam gic (i xng v bt i xng) S mch th nghim nh hnh13. Ti Z1, Z2, Z3 l thun tr.

V1Id1Ip1Ip3V2Z1Id2Ip2V3Z2 Id3

Z3

0Hnh 13

Bt cc cng tt ngun V1, V2, V3 v tr OFF trc khi tin hnh mc mch. Sinh vin mc mch nh hnh v 13 v thc hin o dng pha, dng dy, p pha ( p dy) ri ghi kt qu vo bng sau:

Ud1Ud2Ud3Ip1Ip2Ip3Id1Id2Id3

40.740.8420.8930.9310.961.61.581.62

o cng sut trn cc ti t suy ra cng sut ca mch ba pha:

P1P2P3Pt

373940116

Cu hi:1. V th vect dng p pha v dy. Nhn xt mi quan h gia cc i lng.

......................................................................................................................................................................................................................................................................................................................................................................................................2. Lm li th nghim trn vi mi ti Z1, Z2, Z3 l R, L mc ni tip.

Ud1Ud2Ud3Ip1Ip2Ip3Id1Id2Id3

41.642.8430.5290.5540.56711.061.0441.077

o cng sut trn cc ti t suy ra cng sut ca mch ba pha:

P1P2P3Pt

20222365

V th vect dng p pha v dy. Nhn xt mi quan h gia cci lng

......................................................................................................................................................................................................................................................................................................................................................................................................3. Lm li th nghim trn vi ti Z1, Z2 l thun dung C, Z3 l thun tr R(ti bt i xng).

Ud1Ud2Ud3Ip1Ip2Ip3Id1Id2Id3

43.243.242.810.9740.960.5831.651.847

o cng sut trn cc ti t suy ra cng sut ca mch ba pha:

P1P2P3Pt

224246

V th vect dng p pha v dy. Nhn xt mi quan h gia cci lng

...........................................................................................................................................................................................................................................................................................................................................................................

4/. Kho st hin tng mt pha a/. Ti mc hnh sao Bt cc cng tt ngun V1, V2, V3 v tr OFF trc khi tin hnh mc mch. Sinh vin mc mch nh hnh v 11 vi Z1, Z2, Z3 l thun tr R Bt cc cng tt ngun V1, V2, v tr ON, V3 v tr OFF. Thc hin o dng pha (dng dy), p pha, p dy, dng dy trung tnh ri ghikt qu vo bng sau:

Up1Up2Up3Ud1Ud2Ud3Ip1Ip2Ip3Itt

25.324.6043.22525.40.550.55

So snh kt qu vi kt qu o phn 1. Nhn xt khi xy ra hin tng mt pha.....................................................................................................................................................................................................................................................................b/. Ti mc tam gic Bt cc cng tt ngun V1, V2, V3 v tr OFF trc khi tin hnh mc mch. Sinh vin mc mch nh hnh v 13 vi Z1, Z2, Z3 l thun tr R Bt cc cng tt ngun V1, V2, v tr ON, V3 v tr OFF. Thc hin o dng pha, dng dy, p pha, p dy ri ghi kt qu vo bng sau:Up1Up2Up3Ip1Ip2Ip3Id1Id2Id3

41.820.920.890.9130.4750.4711.4051.4150

Ud1Ud2Ud3

41.420.920.86

So snh kt qu vi kt qu o phn 3. Nhn xt khi xy ra hin tng mt pha...................................................................................................................................................................................................................................................4. thuyt) =5. Nhn xt v cc gi tr ny....................................................................................................................................................................................................................................................................................................................................................................................................

31B mn K thut in, Khoa in in t