CHNG I: O IN P V DNG IN

O LNG IN

BI TP

CHNG I: O IN P V DNG IN

1.1 Mt ampe-k dng c cu o t in c in tr c cu o R(m) =99 v dng lm lch ti a Imax = 0,1mA. in tr shunt Rs = 1. Tnh dng in tng cng i qua ampe-k trong cc trng hp:a) kim lch ti ab) 0,5Dm; (FSD = Imax, full scale deviation)

c) 0,25Dm

Hnh B.1.1Gii:a) kim lch ti a Dm:in p hai u c cu o:

Vm=Im.Rm=0,1mA.99=99mV

IsRs = Vm => Is = == 9,9mADng tng cng:

I = Is + I = 9,9 + 0,1 = 10mA

b) 0,5Dm:Im = 0,5 . 1mA = 0,05mA

Vm = Im.Rm = 0,05mA.99 = 4,95mV

Is =

I = Is + Im = 4.95mA + 0,05mA=5mA

c)0,25mA:Im = 0,25.0,1mA = 0,025mA

Vm = ImRm = 0,025mA.99 = 2,475mVIo=

EMBED Equation.3 1.2 Mt c cu o t in c I= 100A, in tr ni khung quay R= 1K. Tnh in tr shunt mc vo c cu o tr thnh mt ampe-k tng ng vi hnh 1.1.

a) Dm = 100mA = tm o 1

b) Dm = 1A = tm o 2

Gii:

a) tm o 100mA

Vm= ImRm = 100.1 = 100mV

It = Is+ Im => Is = It Im = 100mA 100A = 9,9mA

Rs =

b) tm o 1A: Vm = ImRm = 100mV

Is= It Im = 1A- 100A= 999,9mA

Rs=

1.3 Mt c cu o t in c ba in tr shunt c mc theo kiu shunt ayrton s dng lm ampe-k. Ba in tr c tr s R1=0,05, R2=0,45, R3=4,5, Rm= 1k, Imax= 50A, c mch o nh hnh sau, tnh cc tr s tm o ca ampe-k

Hnh B.1.3Gii:

Ti lch 0,5 Dm

Vs= Imax.Rm= 50A.1k = 50mVIs=

It=Is+Im=50A+10mA = 10,05mA; I=10mA.Kha in C:

Vs= Im(Rm+R3) = 50A.(1k+4,5) = 50mV

Is=

Kha in D:

Vs= Im(Rm +R2 +R3) = 50A(1k + 4,5 +0,45) =50mV

Is = 1A.I = 50A+1A=1,00005A = 1A1.4 Mt c cu o t in Imax =100A,in tr dy ni (dy qun) Rm = 1K c s dng lm vn k DC. Tnh in tr tm o vnk c Vtd= 100V. Tnh in p V hai u vn k khi kim c lch 0,75Dm; 0,75Dm v 0,25Dm ( lch ti a Dm)

Hnh B.1.4Gii:V = IM (Rs + Rm) => Rs =

Khi V= Vtd=100V => IM = Imax =100A

Rs = -1K =999KTi lch 0,75 DmIm = 0,75.100A = 75A

V= Im(Rs+ Rm) 75A(999k +1k)=75V

Ti lch 0,5 DmIm = 50 A

V= 50 A(999 k+1k)=50V

Ti lch 0,25 Dm

V= 25A(999 k+1k)=25V

1.5 Mt c cu o t in c Imax=50 A; Rm =1700 c s dng lm vn k DC c tm o 10V, 50V, 100V. tnh cc in tr tm o theo hnh sau:

Hnh B.1.5Gii

Theo hnh a:

Theo hnh b:

1.6 Mt vnk c tm o 5V, c mc vo mch, o in p hai u in tr R2 nh hnh sau: a) Tnh in p VR2 khi cha mc Vnk.

b) Tnh VR2 khi mc vn k, c nhy 20k/V. c) Tnh VR2 khi mc vn k, c nhy 200k/V

Hnh B.1.6Gii:a) VR2 khi cha mc Vnk.

b)Vi vn k c nhy 20k/V.Rv=5V.20k/V=100k

Rv//R2=100k//50k=33,3k

VR2=

c)Vi vn k c nhy 200k/V

Rv=5V.200k/V=1k

Rv//R2=1M//50k= 47,62k

=4,86V

1.7 Mt c cu o t in c Ifs= 100A v in tr73 c cu o Rm =1k c s dng lm vnk AC c V tm o = 100V. Mch chnh lu c dng cu s dng diode silicon nh hnh v, diode c VF(nh) =0,7Va) tnh in tr ni tip Rsb) Tnh lch ca vnk khi in p a vo vnk l 75V v 50V (tr hiu dng-RMS).

c) Tnh nhy ca vn k. Tn hiu o l tn hiu xoay chiu dng sin.

Hnh B.1.7Gii:

a) Tnh Rs:

y l mch chnh lu ton k nn ta c quan h:

IP(tr nh)= Itb/0,637

Vm (tr nh)=

C cu o c:

nhy=

1.8 Mt c cu o t in c Ifs = 50A; Rm = 1700 kt hp vi mch chnh lu bn k nh hnh sau. Diod silicon D1 c gi tr dng in thun If (nh) ti thiu l 100 A. Khi in p o bng 20% Vtm o , diode c VF = 0,7V, vn k c Vtm o = 50V. a) Tnh Rs v RSH b) Tnh nhy ca Vnk trong hai trng hp: c D2 v khng c D2

Hnh B.1.8Gii:

a)Tnh Rs v RSH

y s dng chnh lu bn k nn ta c:

Ip=Itb/(0,5.0,673): tr nh trong trng hp chnh lu bn k

C cu o c Ifs = Itb = 50 A=> Im= 50 A/(0,5.0,673) = 157 A(tr nh)

Khi V= 20% Vtd, IF(nh) c gi tr 100 A. Vy khi V= Vtd, IF(nh) c gi tr:

b)Tnh nhy:

C D2 trong bn k dng, dng qua D1 c gi tr nh: IF=500 A

Trong bn k m, dng qua vnk c gi tr nh:

Khng c D2:

Trong bn k dng:IF(nh) = 500 A. Trong bn k m: I = 0

Trong chu k ca tn hiu:

Ihiu dng =0,5I F(nh) Vi I l dng in mch chnh chy qua Rs trong bn k dng.

1.9 Mt ampe k s dng c cu o t in c cu chnh lu v bin dng nh hnh v. Bit rng c cu o c Ifs = 1mA v Rm = 1700. Bit dng c Nth = 500; Ns = 4. Diode cVF(nh) = 0,7V; Rs=20k. Ampe k lch ti a khi dng s cp Ip = 250mA. Tnh gi tr RL.

Hnh B.1.9

Gii:

Chnh lu ton k nn ta c:

Im(tr nh)

in p Em hai u cun th bin dng(tr nh):

Em = (Rm+Rs) + 2VF = 1,57mA(20k + 1700) + 1,4V= 35,5V

Es(tr hiu dng) = (0,707.35,5V) = 25,1V

Dng lm lch ti a c cu o c tr hiu dng I:

I = 11,1Itb = 11,1.1mA=11,1mA

Ta c:

(vi Iq=Iqua c cu o)CHNG II: O IN TR

2.1 Cho Eb = 1,5; R1= 15k; Rm =1k; R2 = 1k; Imax = 50A. Xc nh tr s c ca Rx khi Ib = Imax; Im = Imax; Im =3/4 Imax .

Gii:Ti Im =Imax = 50A; Vm = Imax Rm = 50A 1k = 50mA.

Do : . Nh vy dng in: Ib = 100A.Vy Nu .

# Rx +15k = 15k; Rx = 0.

Khi Im =1/2 Imax = 25A; Vm = 25mV I2 = 25A.Suy ra Ib = 50A. Vy Rx + R1 # ; Rx # 15k.

Tng t nh cch tnh trn. Im = 3/4 Imax = 37,5A.

Ib = Im + I2 = 37,5A + 37,5A = 75A.

Rx + R1 = = 20k, Rx = 5k.

2.2 Mt ohm-k loi ni tip c mch o (Hinh di y). Ngun Eb = 1,5V, c cu o c Ifs = 100A. in tr R1 + Rm = 15k.

a)Tnh dng in chy qua c cu o khi Rx = 0.

b)Tnh tr gi Rx cho kim ch thi c lch bng 1/2 FSD, 1/4 FSD, 3/4 FSD (FSD: lch ti a thang o.)

Hnh B.2.2Gii:a. (FSD).

b. lch bng 1/2 FSD:

Im = (v c cu o tuyn tnh.)

lch bng 1/4 FSD:

;

lch bng 3/4 FSD:

Im = 0,75 100A = 75A;

2.3 Mt ohm-k c mch o nh hnh sau. Bit Eb =1,5V, R1 = 15k; Rm = 50; R2 = 50; c cu o c Ifs = 50.

Tnh tr gi Rx khi kim ch th c lch ti a: (FSD); 1/2 FSD v 3/4 FSD.

Hnh B.2.3

Gii:

Khi kim lch ti a (FSD):

Im = 50A; Vm = Im.Rm = 50A50 = 2,5mV.

Dng in mch chnh: Ib = I2 + Im = 50A + 50A = 100A.

Rx = ( Rx + R1) R1 = 15k - 15k = 0

Kim lch 1/2 FSD:

Im = 25A; Vm = 25A 50 = 1,25mV;

Ib = 25A + 25A = 50A.

; Rx = 30k - 15k = 15k.

Kim lch 3/4 FSD:

Im = 0,75 50A = 37,5A; Vm = 37,5A50 = 1,875mV.

; Ib = 37,5A + 37,5A = 75A.

.

2.4 Mt ohm-k co mch o nhiu hnh bai 3. c ngun Eb giam xung ch cn 1,3V. Tnh tr gi mi ca R2 ?.?li cc gi tr Rx tng ng vi lch ca kim: 1/2 FSD, 3/4 FSD.

Gii:

Rx = 0;

Im = 50A (FSD); I2 = Ib Im = 86,67A 50A = 36,67A.

Vm = ImRm = 50A 50 = 2,5mV;

Khi kim lch 1/2 FSD:

Im = 25A; Vm = 25A 50 = 12,5mV

Ib=Im + I2 = 25A + 18,3A =43,33A

Khi kim lch 3/4 FSD:

Im = 0,75 50A = 37,5A; Vm = 37,5A 50 = 1,875mV.

Ib =37,5A + 27,5A = 65A.

2.5 Tnh dng in chy qua c cu o v lch ca kim ch th ca ohm-k c mch d nh hnh v khi ta s dung tm o R1 trong hai trng hp:

a)Rx = 0

b) Rx = 24

Hnh B.2.5Gii:

Mch tng ng ca ohm- k khi ta s dng tm o R1 trong hai trng hp Rx = 0 v Rx = 24 nh sau:

Rx = 0;

Dng Im chy qua c cu o:

Im = 37,5A = Ifs: Khi kim lch ti a.

Rx = 24:

kim lch 1/2 FSD.

2.6 Tnh dng in chy qua c cu o v lch ca kim ch th ca ohm-k c mch nh bi 5, khi s dng tm o R100 va R10k trong trng hp Rx = 0.

Hnh B.2.6Gii: Mch tng ng ca Ohm-k khi ta s dng tm o R100 v R = 0.

: kim ch th lch ti a.

Mch tng ng ca ohm-k khi ta s dungj tm o R10k v Rx = 0.

: Kim ch th lch ti a.

2.7 Ta o in tr bng cch dng phng php V v A c mc r di. Ampe-k ch 0,5A,vn k ch 500V.Ampe k c Ra = 10,10k/V. Tnh gi tr R.

Hnh B.2.7Gii:E + EA = 500V; I = 0,5A

R = 1000 - Ra = 1000 - 10 =990.

2.8 Cc ampe-k, vn k v in tr R bi 2.7 c mc r ngn. Hy tnh ch ca vn k v ampe-k (ngun cung cp vn l 500V).

Hnh B.2.8Gii:

Ni tr ca vn k :

Rv = 1000V 10k/V =10M

Rv // R = 10M // 990 = 989,9

ch ca vn k :

ch ca ampe-k: .

CHNG III:

O IN DUNG, IN CM V H CM

3.1.Cho cu o nh hnh v , bit C1 =0.1F v t s R3/R4 c th chnh c thay i trong khong : 100/1 v 1/100 . Hy tnh CX m cu c th o c.

Hnh B.3.1Gii:Ta c: Cx = C1R3/R4 . Vi : R3/R4 =100/1=>CX = 0,1F(100/1) =10FVi : R3/R4 =1/100 => 0,1F(1/100) =0,001FVy cu c tm o : t 0,001F10F3.2. Cho cu in dung nh hnh sau, thnh phn mu C1 =0,1F ;R3 =10k. Bit rng cu cn bng khi ngun cung cp co f = 100Hz; R1 =125 v R4 = 14,7 . Hy tnh gi tr Rs , CS v h s tn hao D ca t?

Hnh B.3.2

Gii:Ta c : Cs =C1R3/R4;CS == 0.068F ;RS == =183.3D =CSRS = 2 . 100Hz 0,068F183,8 = 0,0083.3. Cho cu in dung nh hnh sau, thnh phn mu C1 =0,1F ;R3 =10k. Bit rng cu cn bng khi ngun cung cp co f = 100Hz; R1 =125 v R4 = 14,7 . Hy tnh gi tr Rs , CS v h s tn hao D ca t?

Hnh B.3.3Gii:Ta c : Cs =C1R3/R4;CS == 0.068F ;RS == =183.3D =CSRS = 2 . 100Hz 0,068F183,8 = 0,008

3.4.Cu Maxwell o in cm dng thnh phn mu C3 = 0,1F, ngun cung cp c tn s f=100Hz. Cu cn bng khi R1 =1,26k; R3= 470 v R4 =500 .Tnh tr gi in cm LS, in tr RS v h s phm cht Q ca cun dy.

Hnh B.3.4Gii:Ta c :LS =C3R1R4 =0,1F

RS =

Q=

3.5. Cu c ngun cung cp f= 100Hz cn bng khi C3 =0,1F, R1 =1,26k , R3 =75 v R4 =500. Tnh in cm LP ,in tr RP v h s phm cht Q ca cun dy?

Hnh B.3.5

Gii:L0,1F

R

Q =

3.6. Hy tnh thnh phn tng ng LS,RS ca cun dy c :LP =63Mh ; RP = 8,4k ( f =100Hz).Gii:

RS =;th: RP = 8,4k ; R; X

=>XP =2

EMBED Equation.3 100Hz63mH =39,6

X=1,57=7,056 107

RS = ; XS = 39,6

LS === 63mH

3.7. Hay tinh thanh phn tng ng CP ,RP cua tu in co RS =183,8 va CS =0,068f (f=100Hz).

Gii:

Ta co: RP =( RS2 +XS2 )/RS ; RS2 = (183,8)2 =33,782

XS =1/2X=5,478.108RP =( 33,78.103 +5,4782,99M

XP ==23,41.103

CP = 1/(2.100Hz.23,41k)= 0,068F

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