Skkn Hoi Giang Vat Ly 9

Nhng kin thc c bn

I. T VN

Do thc t trnh nhn thc ca hc sinh THCS cha cao, c bit l i vi vng nng thn, thi gian tip thu trn lp cn t so vi lng kin thc v kh nng t duy, nhn dng, phn loi bi ton xc nh c yu cu ca bi ton l ht sc kh khn i vi phn ln hc sinh.

Bn cnh , do nhu cu ham hc, ham hiu bit ca s hc sinh c trin vng, do mc quan trng ca vt l 9 i vi vic thi hoc sinh gioi cac cp v tip tc hc ban khoa hoc t nhin cc lp trn nn yu cu t ra l phi chn la, sng lc v phn loi bi tp hng dn cho hc sinh l cng vic v cng quan trng i vi mi gio vin dy bi dng. Thc t cho thy: kin thc l v hn, cc loi, cc dng bi tp ni chung, bi tp v mch in ni ring l rt phong ph v a dng:

- Mch in mc ni tip, mc song song.

- Mch in hn hp tng minh.

- Mch in hn hp khng tng minh.

- Mch cu, mch i xng, mch tun hon, mch bc thang...Trong qu trnh bi dng vt l THCS cho hc sinh, nu ta ch phn ra cc phn c, nhit, in, quang; mi phn lm mt vi bi hc sinh quan st, ghi chp v ghi nh my mc theo kiu ti hin th rt kh c th ghi nh bn vng v p dng khi cn thit.

Vic bi dng hc sinh c trin vng i hi gio vin phi nh hng c v phn loi tng dng bi tp cho hc sinh, vi mi dng trc ht cung cp cho hc sinh h thng kin thc c bn, nhng im cn lu , cung cp cch gii c th, chn la bi tp cho hc sinh luyn gii nm vng phng php vi mc t n gin n phc tp. Trong cc dng bi tp th vic hc sinh bit phn tch cch mc cc b phn trong mch in phc tp th mi c th bt tay vo vic gii cc bi tp khc.Trong qu trnh bi dng cho hc sinh mi nhn, hc sinh thuc i tuyn d thi hc sinh gii, iu m ti nhn thy hu ht hc sinh l i vi nhng mch in phc tp, cc em u b lng tng, b tc khng tm ra hng phn tch mch in.Song do iu kin c hn v thi gian, iu kin v phng tin, dng, vt cht.. nn khng th nghin cu k trnh by cc cho cc dng bi tp v cc loi mch in m y ti ch a ra mt vi kinh nghim nh gip hc sinh bin i t mch in hn hp khng tng minh tr v mch in hn hp tng minh c th thc hin gii mt cch n gin v nh vy, khi hc sinh bit cch v li mch in th khi hc sinh s c s hng th bt tay vo vic khai thc nhiu dng ton, bi ton v mch in.

Vy gip hc sinh c kh nng gii ton vt l phn nh lut m, bi dng hc sinh c trin vng chn i tuyn hc sinh gii... t kt qu cao, ti la chn chuyn NHNG KINH NGHIM GIP HOC SINH G MACH IN V TNH GI TRI IN TR THOA MN YU CU BI TON VT L LP 9 trong chuyn nay ti chi nu ra cc bi ton v mch in hn hp khng tng minh cung cp cho hc sinh c thm gii php gii bi ton loi mch in ny.

II. GIAI QUYT VN 1. C S LI LUN

Chng I . Nhc li mt s kin thc c bn

Mt mch in c th gm nhiu on mch in. Mi on mch in gia hai im ca on mch in c th gm mt hay nhiu b phn, cc b phn c th mc ni tip hoc mc song song vi nhau.

1. nh lut m:

U = I.R v

2. nh lut m i vi cc loi on mch

a/ on mch ni tip:

* Tnh cht:

Hai in tr R1 v R2 c mt im chung l C.

I = I1 = I2.

(1a)

U = U1 + U2.

(2a)

R = R1 + R2.

(3a)

.

(4a)

*Ch :

U1 = I1.R1 = I.R1 = .R1 = U..

(5a)

U2 = I2.R2 = I.R2 = .R2 = U. .

Chia U thnh U1 v U2 t l thun vi R1 v R2.

.

- Nu R2 = 0 th theo (5a) ta thy :

U2 = 0 v U1 = U.

Do trn s (H.1). Hai im C v B: UCB = I.R2 = 0. Khi im C coi nh trng vi im B (hay im C v B c cng in th).- Nu R2 = (rt ln)

U1 = 0 v U2 = U.

b/ on mch mc song song: * Tnh cht:

Hai in tr R1 v R2 c hai im chung l A v B.

U = U1 = U2 .

(1b)

I = I1 + I2.

(2b)

.

(3b)

.

(4b)

*Ch :

(5b)

Chia I thnh I1 v I2 t l nghch vi R1 v R2 :

- Nu R2 = 0 th theo (5b) ta c:

I1 = 0 v I2 = I.

Do trn s (H.2). Hai im A v B c :UAB = 0. Khi hai im A v B c th coi l trng nhau (hay hai im A v B c cng in th).

- Nu R2 = (rt ln) th ta c : I2 = 0 v I1 = I.

(Khi R2 c in tr rt ln so vi R1 th kh nng cn tr dng in ca vt dn l rt ln. Do ta c th coi dng in khng qua R2.)

3. Mt s im lu :

- Trong mt mch in, cc im ni vi nhau bng dy ni (hoc ampe k) c in tr khng ng k c coi l trng nhau. Khi ta chp cc im li v v li mch tnh ton.

- Trong cc bi ton, nu khng c ghi ch g c bit th ta c th coi:

RA 0 v RV .- Khi gii bi ton vi nhng s mch in mc hn hp tng i phc tp, nn tm cch a v mt s tng ng n gin hn. Trn s tng ng, nhng im c in th nh nhau (bng nhau) c gp li (chp li) lm r nhng b phn phc tp ca on mch c ghp li to thnh on mch n gin hn.

Chng II. Mch in hn hp khng tng minh.1/ Nhn xt chung:

- Mch in hn hp khng tng minh cng l mt loi mch in mc hn hp, song cch mc kh phc tp, khng n gin m phn tch cch mc cc b phn trong mch in c ngay. V vy, thc hin c k hoch gii, bt buc phi tm cch mc li a v mch in tng ng n gin hn.

Nh rng, gia cc im ni vi nhau bng dy dn, ampe k... c in tr khng ng k l nhng im c cng in th, ta gp li (chp li). Khi v li mch in, ta s c mch in tng ng dng n gin hn.

- Phn tch cch mc cc b phn trong mch in l bc kh quan trng, n gip ta thc hin yu cu ca bi ton trnh c nhng sai st.

Cui cng, ta p dng cc tnh cht v h qu ca nh lut m i vi tng loi on mch ni tip v song song.

2/ Cc bi tp th d c th

2.1 Bi tp th d 1:

Cho s mch in c mc nh s hnh v 3.

Bit R1 = 6; R2 = 3; R3 = 8; R4 = 4.

Khi on mch c mc vo mt ngun in,

ampe k ch 3A.

a/ Tnh hiu in th ca ngun in.

b/ Tnh dng in i qua R1 v R2.

Hng dn hc sinh thc hin gii

Vi vic ln u tin gii bi ton mch in hn hp nh th ny, hc sinh lng tng trong vic phn tch mch in. V vy, sau khi c gio vin cung cp vic chp cc im ni vi nhau bng dy dn, ta yu cu hc sinh quan st k s v nhn xt cch mc.

Bc 1: Nhn xt:

Ta thy cc im A v D c ni vi nhau bng dy dn c din tr khng ng k, nn chng c cng in th v ta chp l thnh mt im. Nh vy th gia hai im A v B c mt on mch mc song song gm 3 mch r. Mch r th nht cha R1, mch r th hai cha R2, mch r th ba cha R3 v R4. Bc 2: Thc hin bi gii:

- Mch in c v li tng ng nh sau:

- Mch in c mc:

R1 // R2 // (R3 nt R4 )

Gi I1, I2, I3,4 l cc dng in i qua cc in tr R1, R2, R3 v R4.a/ Hiu in th gia hai c ca ngun in cng chnh l hiu in th gia hai mch r cha R3 v R4.

Ta c:

UAB = I34.R34 = I34(R3 + R4)

= 3(8 + 4) = 36(V)

b/ Cng dng in qua R1 v R2 ln lt l :

I 1 =

I2 =

S: U = 36V; I1 = 6A; I2 = 12A.

2.2 Bi tp th d 2:Cho mch in c s cch mc nh hnh v 4.Bit: R1 = 6,5; R2 = 6; R3 = 12; R4 = 10;

R5 = 30. Ampe k ch 2A. Tnh:

a/ Hiu in th 2 cc ca ngun in.

b/ Cng dng in qua mi in tr.


Khi hc sinh quan st s mch in, rt kh c th phn tch c cch mc cc b phn trong mch in, ta yu cu hc sinh quan st v nhn xt s cch mc.

Bc 1; Nhn xt

Ta thy hai im B v C c ni vi nhau bng dy dn c in tr khng ng k. Do , ta chp hai im ny li vi nhau. Khi on mch AC v on mch CD l hai on mch mc ni tip, mi on mch li c 2 in tr c mc song song. Nh vy, mch in gm: Hai on mch mc song song AC v CD mvs ni tip vi nhau v ni tip vi in tr R1 mc vo ngun in. Bc 2: Thc hin bi gii:


- Mch in c mc nh sau:

R1 nt {(R2 // R3) nt (R4 // R5)}

a/ in tr tng ng ca mch AC l :

in tr tng ng ca on mch CD l:

in tr ton mch l: R = R1 + RAC + RCD = 6,5 + 4 + 7,5 = 18()

Vy hiu in th hai cc ca ngun in l:

U = I.R = 2.18 = 36(V)

b/ Cng dng in qua R1 l I1:

I1 = I = 2(A)

Cng dng in qua R2 v R3 l I2 v I3 :

Ta c :

(1)

M :I2 + I3 = I = 2A

(2)

Kt hp (1) v (2), ta c :I2 = (A) v I3 = (A)Cng dng in qua R4 v R5 l I4 v I5:

Ta c :

(3)

M:I4 + I5 = I = 2A(4)Kt hp (3) v (4), ta c :

I4 = (A) v I5 = (A).

S:U = 36V; I1 = 2A; I2 = A; I3 = A; I4 = A; I5 = A.

2.3 Bi tp th d 3:

Cho s mch in nh hnh v 5. Cc in tr u bng nhau v c gi tr l r = 15. Dy ni v ampe k c in tr khng ng k. Khi mc mch in vo ngun in th ampe k ch 2A. Tnh:

a/ in tr tng ng ca ton mch AB.

b/ Hiu in th gia hai cc ca ngun in.


Vi mach in nh th ny, nu hc sinh cha tip cn ln no th d gy cho hc sinh s chn nn v b cuc. Song vi vic chp cc im c cng in th m cc em c tip cn th li gy cho cc em s t m mun c th sc.

Bc 1: Nhn xt:

Ta thy gia cc im A, C, D, F, I c ni vi nhau bng dy dn v ampe k c in tr khng ng k nn chng c cng in th. Do , ta chp cc im ny li lm mt v ni vi dng ngun. Tng t nh vy, gia cc im E, G, H, K, B ta chp li lm mt v ni vi m ngun. Nh vy hai u mi in tr ny, mt u ni vi cc dng, mt u ni vi cc m ca ngun in, ngha l mch in AB gm 5 in tr c mc song song vi nhau. Bc 2: Thc hin k hoch gii:


- Mch in c mc:

R1 // R2 // R3 // R4 // R5.

a/ in tr tng ng ca ton mch AB l:

b/ Hiu in th hai cc ca ngun in l:

UAB = I.RAB = 2.3 = 6(V)

S: RAB = ;UAB = 6(V)

2.4 Bi tp th d 4:

Cho s mch in nh hnh v 6. Cc in tr u bng nhau v c gi tr l r = 49. Dy ni c in tr khng ng k. Tnh in tr tng ng ca ton mch.


Vi mch in phc tp ny, hc sinh sau khi lm quen vi phng php quan st nhn ra c gia cc im c ni vi nhau bng dy dn s c chp li lm r cch mc cc b phn trong mch in.

Bc 1: Nhn xt:

Quan st s mch in, ta thy gia cc im A, C, I, E, G. c ni vi nhau bng dy dn c in tr khng ng k. V vy, cc im ny c cng in th, ta chp li lm mt v mc v pha cc dng ca ngun in, tng t nh vy ta cng c th chp cc im B, K, D, H, F li lm mt v mc v pha cc m ca ngun. Bc 2: Thc hin k hoch gii- Mch in c v li tng ng nh sau : - Mch in c mc:R1 // R2 // R3 // R4 // R5 // R6 // R7.

in tr tng ng ca ton mch l:

EMBED Equation.3

2.5 Bi tp th d 5:

Cho s mch in nh hnh v 7. Cc in tr u bng nhau v c gi tr l r = 12. in tr dy ni khng ng k. Ampe k ch 2,4A.a/ Tnh in tr tng ng ca ton mch.

b/ Tnh hiu in th gia hai u mch in.Hng dn hc sinh thc hin gii

n y, hc sinh gp phi mt s mch in phc tp hn, khng ch n gin l chp cc im c ni bng dy dn m hc sinh cn phi xc nh cc yu t ca nh lut m (I. U, R) v dng in a vo mch nh th no. T mi nh gi c in th ti cc im, khi nhng im no c cng in th ta chp li lm mt.Bc 1: Nhn xt:

Ta nhn thy:

- Cc in tr c mc vo cc cnh ca hnh lp phng.

- Theo bi cc in tr ny c cng gi tr.

- Dng in c a vo nt A, i ra nt C(hai u ng cho ca hnh lp phng).

Nh vy, cc im B, D. A c cng in th ta chp li lm mt. Tng t nh vy, cc im C, B, D cng c cng in th ta chp li lm mt. Do mch in thc cht gm 3 on mch mc ni tip nhau. Trong on mch AB c 3 in tr R1, R2, R3 mc song song, on mch BC c 6 in tr mc song song, on mch CC c 3 in tr mc song song.Bc 2: Thc hin k hoch gii


- Mch in c mc:(R1//R2//R3) nt (R4//R5//R6//R7//R8//R9) nt (R10//R11//R12)a/ in tr tng ng ca on mach AB l :

in tr tng ng ca on mch BC l :

in tr tng ng ca on mch CC l :

Vy in tr tng ng ca ton mch AC l :

RAC = RAB + RBC + RCC = .

b/ Hiu in th gia hai u mch in l :

UAC = I.RAC = 2,4.10 = 24 (V)

S:

RAC = 10;

UAC = 24V.2.6 Bi tp th d 6:

Cho mch in c s nh hnh v 8:

Bit R1 = 600; R2 = 500; R3 = 700; U = 100V. Dy ni v kho K c in tr khng ng k.

a/ Gi s vn k c in tr RV = 2000. Tm s ch ca vn k khi kho K ng, kha K m.

b/ Gi s vn k c in tr rt ln RV = . Tnh cng dng in chy trong mch khi kho K ng.c/ Nu tho b in tr R3 v thay vn k bng mt ampe k c in tr khng ng k th ampe k ch bao nhiu ?

Hng dn hc sinh thc hin k hoch gii

Sau khi hc sinh thc hin tt vic xt in th cc im chp li v v li mch th gio vin tip tc cho hc sinh lm quen vi dng mch in c xt thm vai tr, chc nng ca vn k trong mch khi vn k c in tr gii hn xc nh v khi c in tr v cng ln.Bc 1: Nhn xt:

Vi mch in ny, gio vin s nhc li cho hc sinh chc nng ca vn k v ampe k:

- Nu vn k c in tr l mt gi tr gii hn no khng i th vn k lc trong mch cho dng in chy qua v xem n nh mt in tr khi tnh in tr tng ng trong mch.

- Nu vn k c in tr v cng ln (tnh cn tr dng in ca vt dn ln) th dng in qua n coi nh khng ng k (c th tho ra khi tnh in tr tng ng).

- Ampe k c in tr khng ng k, c th chp li nhng im c cng in th lm r cch mc cc b phn trong mch in.Bc 2: Thc hin k hoch gii:

a/ Nu vn k c i tr xc nh l RV = 2000

* Khi kho K ng, mch in c mc:

R1 nt {(RV nt R3) // R2)

Ta c :

=

in tr tng ng ca mch l: Rt = R1 + R2V = 600 + 421,87 = 1021,87 ()Cng dng in chy trong mch l:

Vy s ch ca vn k l:

UV = U2V = I.R2V = 0,097.400 = 38,8 (V).

* Khi kho K m, mch in c mc:

R1 nt RV nt R3.

in tr tng ng ca mch l :

Rt = R1 + R3 + RV = 600 + 700 +2000 = 3300 () Cng dng in chy trong mch l:

Vy s ch ca vn k trong trng hp ny l:UV = I.RV = 0,03.2000 = 60 (V)

b/ Nu vn k c in tr rt ln (RV = ), coi nh khng c dng in chy qua vn k v R3 (c th tho ra). Khi kho K ng, mch in lc ny ch gm c 2 in tr R1 v R2 mc ni tip:R1 nt R2.Cng dng in chy mch l:

c/ Khi b i tr R3 v thay vn k bng ampe k (do ampe k c in tr khng ng k nn mch in c mc: Khi s ch ca ampe k l :

S:a/ K ng: UV = 38,8V;

K m:UV = 60V

b/ I = 0,09A

c/ IA = 0,166A

3/ Mt s bi tp p dngBi 1: Tnh in tr tng ng ca on mch AB nh hnh v 9, nu:a/ K1, K2 m.

b/ K1 m, K2 ng.

c/ K1 ng, K2 m.

d/ K1, K2 u ng.

Cho R1 = 2; R2 = 4; R3 = 6; R4 =12; in tr cc dy ni l khng ng k.

S: a/ K1, K2 m: RAB = 12;

b/ K1 m, K2 ng: RAB = 4. c/ K1 ng, K2 m: RAB = 1,2;d/ K1, K2 u ng: RAB = 1.

Bi 2: Tnh in tr RAB, v RAG theo mch in c v H. 10a v H. 10b. Bit mi on u c in tr l R.S:RAB = ;

RAG =

Bi 3: C mch in nh hnh v 11:Bit R1 = R3 = R4 = 4; R2 = 2; U = 6V.

a/ Khi ni gia A v D mt vn k th vn k ch bao nhiu ? Bit vn k c in tr rt ln.

b/ Khi ni gia A v D mt ampe k th ampe k ch bao nhiu ? Bit in tr ca ampe k rt nh. Tnh in tr tng ng ca mch trong trng hp ny.

S: UV = UAD = 5,14V;IA = 2,25APhn III - KT QU T C V BI HC KINH NGHIM

1/ Kt qu t c* Trong qu trnh dy hc sinh cc lp bi dng, trc khi hng dn cho hc sinh kinh nghim ny, khi gp bi tp v mch in khng tng minh th hc sinh thng lng tng, ch c s t l thc hin c, cn li l thc hin c nhng cha t yu cu, thm ch l c hc sinh khng c nh hng gii. iu lm cho hc sinh c tm l chn nn, ngi hc vt l. Kt qu c th: Kt qu

S HS kho stHS khng thc hin cHS thc hin cha t yu cuHS thc hin t yu cu

SL%SL%SL%

4018451742,5512,5

* Vi kt qu nh vy, ti nhn thy khi hng dn hc sinh lm bi tp v mch in phc tp th trc ht ti phi dy kinh nghim gii ton v mch in khng tng minh, c nh vy th hc sinh mi c c s khai thc tip cc dng bi tp khc v mch in. Sau khi hng dn cho hc sinh kinh nghim ny, phn ln hc sinh thc hin bi ton l t yu cu, s t l thc hin cha t yu cu, ch cn li mt vi hc sinh khng thc hin c. T gy cho hc sinh nim am m, yu thch b mn vt l hn. Kt qu c th:Kt qu

S HS kho stHS khng thc hin cHS thc hin cha t yu cuHS thc hin t yu cu

SL%SL%SL%

4037,5922,52870

- Kt qu trong nhng nm bi dng hc sinh gii gn y, ti lun c hc sinh t gii hc sinh gii cp huyn v hc sinh c chn vo i tuyn d thi hc sinh gii cp tnh vi kt qu kh quan.2/ Bi hc kinh nghim

Nh ti trnh by phn u, ti ch nghin cu trong phm vi mch hn hp khng tng minh nhm cung cp cho hc sinh phng php gii i vi mch in loi ny.

gip hc sinh c hng th v ny sinh tnh hung c vn khi hc tp th gio vin c cho hc sinh gii cc bi ton v mch in hn hp tng minh vi 2, 3 ri 4 in tr. Sau , gio vin mi a ra loi mch hn hp khng tng minh dng n gin, khi hc sinh s bt u gp kh khn khi thc hin gii, lc ny gio vin hng dn cho hc sinh phn kin thc mc 3 ...mt s im lu v cng hc hc sinh tin hnh gii ri mi nng dn ln mch hn hp khng tng minh mc t n gin n phc tp nh ti trnh by trong ti.

Vic ti a vo th d 6 trong phn cc gii php ci tin l mun sau khi hc sinh c lm quen vi vic chp cc im c cng in th (2 u dy dn, kho K, ampe k... c in tr khng ng k), gio vin tip tc gii thiu cho hc sinh dng ton v mch in c xt n vai tr, chc nng ca vn k trong mch khi m vn k c nhng gi tr v in tr khc nhau, hoc l trn nhnh cha vn k c mc thm cc b phn tiu th in khc vic tip thu ca hc sinh c lin mch v c s lgc khi chuyn t dng ny sang dng khc, sau khi hng dn cho hc sinh mi dng bi tp trn lp, cn giao thm cc bi tp thuc dng hc sinh p dng lm nh.

y ti thy: Vi cch xt in th ti cc im tm ra nhng im c cng in th v li mch in tng ng n gin hn, khi m hc sinh nm vng kin thc c bn. Vi phng php ny s gip hc sinh trnh c tm l lo s khi gp mch in loi ny, ng thi hc sinh s gii chnh xc v n gin hn nhiu nu nguyn mch in ban u (thm ch c nhiu mch in nu nguyn mch ban u s khng th gii c).

thc hin tt vic xt in th v li mch v phn tch mch in th nht thit hc sinh phi c gio vin cung cp v t nm vng c mc 3 - Mt s im lu . ( chng I Nhc li mt s kin thc c bn).

Vic phn tch mch in l rt cn thit thc hin k hoch gii ton. Song, trong qu trnh lm bi, hc sinh ch nhp m khng cn trnh by phn ny. Trong qu trnh hng dn hc sinh thc hin gii, ti trnh by thnh 2 bc:

Bc 1: Nhn xt

Bc 2: Thc hin k hoch gii.

Trong bi lm, hc sinh ch cn trnh by t bc 2, phn thc hin k hoch gii.

kt qu c chnh xc v sai s l thp nht th cc php tnh nn bin i biu thc ch, ch thay gi tr bng s vo cc i lng biu thc cui cng, sau kim tra li kt qu xem c ph hp vi iu kin bi ton v thc t khng.Mong rng, ti ny s gip hc sinh gii ton vt l phn mch in hn hp khng tng minh c tt hn nhm gp phn nng cao hiu qu, cht lng dy v hc vt l cp THCS. Trong qu trnh bin son ti, chc chn l khng th trnh khi thiu st m c th bn thn ti cha pht hin ra. ni dung v hnh thc ti thm phong ph, ti rt mong c s ng gp kin ca cc HKH, ca bn c v ng nghip. Ti xin chn thnh cm n !MC LC

Ni dungTrang

Phn I - t vn

Phn II Cc gii php ci tin

Chng I - Nhc li mt s kin thc c bnChng II - Mch in hn hp khng tng minh

1 - Nhn xt chung v mch hn hp khng tng minh

2 - Cc th d c th+ Bi tp th d 1

+ Bi tp th d 2

+ Bi tp th d 3

+ Bi tp th d 4

+ Bi tp th d 5

+ Bi tp th d 6

3 Cc bi tp p dng

Phn III - Kt qu t c v bi hc kinh nghim

1 - Kt qu t c2 Bi hc kinh nghim1 - 233 45555 66 88 99 1010 1212 - 141516

1616 - 18

EMBED Equation.3

R1

R2

A

C

B

H.1

R1

R2

I2

I1

I

A

B

H.2

A

R2

R3

R4

R5

R1

A

B

C

D

+

-

-

+

D

C

R1

A

R1

A

R5

R3

R4

R2

B

A

R4

R3

A

R1

D

C

A

A

B

A

R4

R3

A

R2

R1

B

C

D

E

F

G

H

I

K

r

r

r

r

r

+

-

A

+ A

- B

r

r

r

r

r

R1

R2

R3

R4

R5

R6

R7

+ A

C

D

E

FF

G

H

I

K

- B

- B

+ A

R5

R6

R2

R3

R4

R1

R7

A

A

B

B

C

D

D

A

C

H.7

C

R4

- C

+ A

R8

R9

R5

R6

R7

R6

R10

R11

R3

R1

R2

A

B

H.3...

H. 4

H. 5

H. 6

V

+

-

U

R1

R2

R3

K

H. 8

K

K

R3

R2

R1

U

-

+

V

R1

R2

R3

A

R1

U

-

+

V

U

-

+

A

R1

U

-

+

M

N

K2

K1

B

R4

R3

R2

R1

H. 9

A

0

B

C

D

H. 10a

A

B

C

D

E

F

G

H

H. 10b

+

-

U

. A

D

B

C

R1

R2

R3

R4

H. 11

PAGE 19

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