Thng s tc ng c trng cho phn t c kh nng t n (hoc khi c kch thch) c th to ra v cung cp nng lng in ti cc phn t khc ca mch

Bin p l tng 1:1 tng ng vi m hnh mng 4 cc no di y? (h5.44b)

Bin p l tng 1:-1 tng ng vi m hnh mng 4 cc no di y : (h5.45a)Bin p l tng 1:n (vi n l t s vng dy gia cun th cp v cun s cp) l mt mng 4 cc c h phng trnh c trng:

Bit 1octave bng 0,3 decade. Vy tc suy gim 6 dB/octave cng l

20 dB/decade

Biu thc c tuyn bin tng ng hm truyn t:

Biu thc c tuyn bin ng vi h s K ca hm mch l:

Biu thc c tuyn bin ng vi:

Biu thc c tuyn pha tng ng vi:

Biu thc c tuyn pha ng vi h s K ca hm mch l:

Biu thc c tuyn pha ng vi:

Biu thc no c trng cho cch ghp dy chuyn cc M4C?

Biu thc no c trng cho cch ghp ni tip - ni tip cc M4C?

Biu thc no c trng cho cch ghp ni tip - song song cc M4C?

Biu thc no c trng cho cch ghp song song-ni tip cc M4C?

Biu thc no c trng cho cch ghp song song-song song cc M4C?

Biu thc no di y m t ng mi quan h gia cc thng s ca M4C?

Biu thc no di y m t ng tr khng sng ca M4C i xng theo s cu?

Biu thc no m t mch in di y ? (h1.50)

Biu thc no sau y dng cho cc dn np mc ni tip?

Biu thc no sau y dng cho cc dn np mc song song?

Biu thc no sau y dng cho cc tr khng mc ni tip?

Biu thc no sau y dng cho cc tr khng mc song song?

B bin i tr khng m (NIC) thuc mng 4 cc tng h, th ng. Pht biu ny ng hay sai ?

SaiBc phn tch no sau y khng c trong cc bc c bn gii bi ton qu ?

Xc nh ma trn tr khng c trng ZijCc biu thc no sau y biu din h phng trnh dng in vng ca mch nh hnh v ? (h2.51)

Cc im khng ca hm truyn t H(p) ca mch in c th nm trn ton b mt phng phc. Pht biu ny ng hay sai ?

ngCc im khng v cc im cc ca hm mch:

C th l nghim thc hoc nghim phc, nghim n hoc nghim bi.Cc iu kin u ca bi ton qu tun theo lut ng ngt ca cc phn t qun tnh. Pht biu ny c ng khng ?

ngCc phn t th ng dn in hai chiu R, L, C u c tnh cht tng h. Pht biu ny ng hay sai ?

ng

Cc tn s ct trn v ct di ca mt mch RLC ni tip tng ng l 20 kHz v 5kH, th bng thng BW s l:

15 kHz

Cho mch in nh hnh v vi: Eng1=16V; Ing=5mA; Eng2=32V; R1=2,4K; R2=R3=1,6K. in p trn R2 c xc nh bng phng php xp chng l: (H2.53)

Cho mch in nh hnh v, biu thc no sau y l ng: (h2.32) VR1+VR2=5V

Cho mch in nh hnh v, nu gi tr in tr R2 tng, th dng in qua ngun: (h2.33)

GimCho mch in nh hnh v, vi cc s liu: Z1=Z2=10 ; Z3=Z4=20 ;Ing1=3 A; Eng4=30 V. Dng in qua Z3 c xc nh bng phng php xp chng l: (H2.57)

(chiu dng ra khi nt B)

Cho mch in nh hnh v, vi cc s liu: Z1=Z2=20 ; Z3=Z4=10 ;Eng1=60V; Ing4=6A. Dng in trn Z3 c xc nh bng phng php xp chng l: (H2.54)

(chiu t B sang A)

Cho mch in nh hnh v, vi cc s liu: Z1=Z2=20 ; Z3=Z4=10 ;Eng1=60V; Ing4=1,5A. Dng in trn Z2 c xc nh bng phng php xp chng l: (H2.55)

(chiu dng ra khi nt A)

Cho mch in nh hnh v, vi cc s liu: Z1=Z3=10 ; Z2=Z4=20 ;Ing1=3 A; Eng4=30 V. Dng in qua Z3 c xc nh bng phng php xp chng l: (H2.56)

(chiu t A sang B)

Cho mch in nh hnh v. Bit Eng=20V; Ing=2A; R1=24. in p v tr khng trong ca ngun tng ng trong mch Thevenine l: (H2.66)

Cho mch in nh hnh v. Dng in ngun ca mch Norton tng ng c xc nh theo biu thc: (H2.67)

Cho mch in nh hnh v. Dng in qua ngun v cc dng in trn R1, R2, R3 ln lt l: (h2.44)

20 mA; 10 mA, 5 mA v 5 Ma

Cho mch in nh hnh v. Gi s rng kha K v tr 1 rt lu. Ti thi im t=0, kha K chuyn sang v tr 2. Vi e(t)=100 V, R1=R2=50 . Cc iu kin u IL(0) v UC(0) c xc nh l : (h3.40)

Cho mch in nh hnh v. Hm truyn t in p K(p)=Vout(p)/Vin(p) l: (h4.40)

Cho mch in nh hnh v. Hm truyn t in p K(p)=Vout(p)/Vin(p) l: (h4.41)

Cho mch in nh hnh v. Hm truyn t in p K(p)=Vout(p)/Vin(p) l: (h4.42)

Cho mch in nh hnh v. Hm truyn t in p K(p)=Vout(p)/Vin(p) l: (h4.43)

Cho mch in nh hnh v. H phng trnh in p nt ca mch l: (h2.50)

Cho mch in nh hnh v. Ing=5A; Eng2=8V; Eng3=6V; R1=2; R2=3; R3=1. Biu thc no sau y biu din phng trnh in p nt ca mch. (h2.47)

Cho mch in nh hnh v. Ing1=2A; Ing4=3A; R1=5; R2=3; R3=4; R4=6. Biu thc no sau y biu din phng trnh in p nt ca mch. (h2.48)

;

Cho mch in nh hnh v. Ni tr ngun (Zt) ca mch Thevenine tng ng c xc nh theo biu thc: (H2.64)

Cho mch in nh hnh v. Ni tr ngun ca mch Norton tng ng c xc nh : (H2.68)

Cho mch in nh hnh v. Sc in ng (Et) ca mch Thevenine tng ng c tnh theo biu thc: (H2.63)

Cho mch in nh hnh v. Sc in ng (Et) v ni tr ngun (Rt) ca mch Thevenine tng ng c xc nh theo biu thc: (H2.65)

Cho mch in nh hnh v. Ti thi im t=0, kha K ng. Bit e(t)=50 V, R=100. Cc iu kin u IL(0) v UC(0) c xc nh l (h3.39)

Cho mch in nh hnh v. Vi in p ngun bng E=10V, R1=10; R2=R3=20, th tr khng tng ng Rt khi chuyn sang mch Thevenine l: (h2.46) Bng 5Cho mch in nh hnh v. Vi in p ngun E=10V, R1=10; R2=R3=20, th in p ngun tng ng khi chuyn sang mch Thevenine l: (h2.45) Bng 5VCho mch in nh hnh v. Vi E1=10V; R1=10. Dng in ngun ca mch Norton tng ng l: (H2.69)

Cho mch in nh hnh v. Vi R1=10; R2=40. Ni tr ngun ca mch Norton tng ng l: (H2.70)

Cho mch in nh hnh v. Xc nh cng sut tiu tn trn ti? H1.63

Cho mch in nh hnh v: Eng=20V; Ing=2A; R1=24; Rt=16. Dng in trn ti Rt c xc nh bng phng php xp chng l: (H2.52)

(chiu t B n A)

Cho mch in nh hnh v: Ing1=2A; Ing2=0.5A; Ing3=2A; R1=20; R2=40; R3=30. Cc biu thc no sau y biu din phng trnh in p nt A, B ca mch? (h2.49)

;

Cho mch in nh hnh v: Z1=10 ; Z2=Z3=20 ; Ing1=3A; Eng3=30 V. Sc in ng Et) ca mch Thevenine tng ng l: (H2.62)

Cho mch in nh hnh v: Z1=20 ; Z2=Z3=10 . Ni tr ngun (Zng) ca mch Norton tng ng l: (H2.61)

Cho mch in nh hnh v: Z1=20 ; Z2=Z3=10 ; Eng1=20 V; Ing3=1A. Dng in Ngun (Ing) ca mch Norton tng ng l: (H2.60)

Cho mch in nh hnh v: Z1=Z2=10 ; Z3=20 . Ni tr ngun (Zt) ca mch Thevenine tng ng l: (H2.59)

Cho mch in nh hnh v: Z1=Z2=10 ; Z4=20 ; Eng1=60 V; Ing4=3 A. Sc in ng (Et) ca mch Thevenine tng ng l: (H2.58)

Cho mch nh hnh v. in p trn Rt c tnh theo biu thc:(h1.59)

Cho mch nh hnh v. Xc nh trng hp no di y khng phi l dng c tuyn bin hm truyn t in p ca mch? (h4.48)(h4.48b)

Cho mng 4 cc i xng nh hnh v. Hy xc nh cp tr khng cu ZI, ZII ca mng 4 cc i xng cu tng ng (h5.78)

Cho mng 4 cc hnh T nh hnh v vi cc s liu Z1=1; Z2=2; Z3=3. Xc nh cc thng s tr khng h mch Zij (h5.74)

Cho mng 4 cc hnh nh hnh v vi cc s liu Z1=2; Z2=4; Z3=1. Xc nh cc thng s dn np ngn mch Yij (h5.75)

Cho mng 4 cc nh hnh v. Hy xc nh cc thng s tr khng h mch Zij ca mng 4 cc (h5.81)

Cho mng 4 cc nh hnh v. Xc nh iu kin ca Zng v Zt c s phi hp tr khng trn c 2 ca ca mng 4 cc? (h5.60)

C bao nhiu h phng trnh c tnh c trng cho bn cc tuyn tnh, bt bin, tng h? 6 h phng trnh c tnhC bao nhiu loi ngun ph thuc ?4

C s chnh ca phng php phn tch mch bng in p nt da vo nh lut Kirchhoff v dng inC s chnh ca phng php tch mch bng dng in vng da vo nh lut Kirchhoff v in pC s ca phng php Heaviside l: Tnh tuyn tnh ca bin i LaplaceC s phn tch mch tuyn tnh bng phng php xp chng l

Tnh tuyn tnh ca mchCng dng ca th Bode l: Nghin cu c tnh tn s ca hm mch.Cng hng trong mch dao ng n ni tip cn c gi l Cng hng in p

Cng hng trong mch dao ng n song song cn c gi l:

Cng hng dng in

Cng hng trong mch RLC song song xy ra khi

Cng sut phn khng ca mch th ng c trng cho s tiu tn nng lng di dng nhit. Nhn xt ny ng hay sai?Sai

Cng sut tc dng P chnh l cng sut ta nhit trn cc thnh phn in tr ca mch. Pht biu trn ng hay sai ?ng

c trng ca phn t thun cm l:in p trn n t l vi tc bin thin ca dng in

c trng ca phn t thun dung l:Dng in trong n t l vi tc bin thin ca in p

c tuyn bin thnh phn ng vi im khng gc ta l: Mt ng thng c dc 20dB/Dc tuyn bin ng vi h s K ca hm mch l: Mt ng song song vi trc honh (trc decade)

c tuyn pha ca thnh phn ng vi: l: Mt ng song song vi trc decade, ct trc tung ti gi tr / 2 radDn np ca phn t thun cm l

Dn np ca phn t thun dung l

Dn np tng ng ca mch in:(H1.54) Y = 2 - j3 (S)

Dng tng qut th bin ng vi h s K ca hm mch l: (h4.25c) chn di tn s t 1 kHz n 30 kHz v loi b cc tn s khc, phi s dng loi mch lc no ?

Thng di loi b cc thnh phn tn s nh hn 30 kHz, phi s dng: Mch lc thng cao lc ly di tn Audio (t 0 kHz n 20 kHz) v loi b cc tn s khc, phi s dng loi mch lc no ?

Thng thp tm hm gc f(t) t nh F(p), theo Heaviside, cn phi xt: Cc im cc ca F(p) xc nh in p ngun tng ng Eng trong mch Thevenine, th cn: H mch ti xc nh tr khng trong ca ngun Thevenine tng ng, th cn: Loi b ti v ngunim cc ca hm mch l cc im pj tha mn : H2(pj)=0im khng ca hm mch l cc im pi tha mn : H1(pi)=0in p m ngun p l tng cung cp cho mch ngoi s khng ph thuc vo ti. Pht biu ny ng hay sai ?ng

in dung (C), in cm (L), h cm (M) thuc loi: Thng s qun tnh

in tr thuc loi:

Thng s khng qun tnh

iu kin u ca bi ton qu ni ln c tn ti nng lng ban u hay khng. Nhn xt ny ng hay sai ? ngiu kin c s phi hp tr khng c hai ca ca M4C l :

iu kin mng 4 cc sau l i xng? (h5.21) Za = Zb

iu kin mng 4 cc sau l i xng? (h5.22) Ya = Yciu kin n nh ca cc mch in tuyn tnh, bt bin, c thng s tp trung l mi im cc ca hm truyn t H(p): Nm bn na tri ca mt phng phc (khng bao gm trc o)nh lut Kirchhoff 1 cp i vi Dng in ti cc nt mchnh lut Kirchhoff 2 cp i vi in p trong cc nhnh mch th bin ca hm truyn t: v: L ging nhau th Bode c v vi trc Decade c nh ngha (vi 0l tn s chun dng chun ha)

th Bode tng hp ca hm truyn t: c tng hp t cc c tuyn thnh phn theo nguyn tc: Cng th th pha ca hm truyn t: v L i xng nhau qua trc honh (trc decade)i vi cc mch in nhn qu v n nh, ta lun c th tnh ton trc tip p ng tn s H(j) t hm truyn t H(p) bng cch: Thay th p=jDng in ca ngun c gi tr bng dng h mch ca ngun. Pht biu ny ng hay sai ?Sai

Dng in m ngun dng l tng cung cp cho mch ngoi s khng ph thuc vo ti. Pht biu ny ng hay sai?ng

Hm truyn t in p ca M4C theo cc thng s aij?

Hm truyn t in p ca M4C theo cc thng s yij?

Hm truyn t in p ca M4C theo cc thng s zij?

Hm truyn t H(p) ca mch tng t-tuyn tnh-bt bin v nhn qu c nh ngha trc tip t t s gia p ng v tc ng trong min p vi iu kin u ca mch bng khng. Pht biu ny ng hay sai ? ngHy xc nh hm truyn t ca mch in sau. Gi thit KTT l l tng, lm vic ch tuyn tnh. (h5.79)

Hy xc nh hm truyn t ca mch in sau. Gi thit KTT l l tng, lm vic ch tuyn tnh. (h5.80)

H s phm cht ca mch RLC ni tip (ti tn s cng hng) c xc nh:

H s phm cht ca mch RLC song song (ti tn s cng hng) c xc nh:

H s phm cht Q ca mch RLC ni tip c th tng bng cch: Gim R.

H s phm cht Q ca mch RLC song song c th gim bng cch: Gim R.

H s truyn t ca ca mng 4 cc th ng lun tha mn:

H s truyn t G(p) v hm truyn t in p K(p) t l thun vi nhau. Pht biu ny ng hay sai ? SaiH s truyn t phc ca mng 4 cc th ng c tnh theo biu thc

Hnh v th bin tng ng vi:

(h4.30) Hnh v th pha thnh phn tng ng vi im khng gc ta : (h4.32d)

Hnh v th pha ng vi h s K ca hm mch l (h4.27d)

Hnh v di y m t s tng ng ca tranzito lng cc theo cc tham s vt l trong cch mc no? (h5.57)

Emitter chung Hnh v di y m t s tng ng ca tranzito lng cc theo cc tham s vt l trong cch mc mc no? (h5.58)

Base chung

Hnh v di y m t s tng ng ca tranzito lng cc theo cc tham s vt l trong cch mc mc no? (h5.59)

Collector chung H cm c cng bn cht vt l vi:in cm

Khi p dng cc nh lut Kirchhoff, cc du i s l: Cn thitKhi bin i thnh mch Thevenin, Et v Zt khng ph thuc gi tr ti v cc thng s ny: c thc hin khi ti h mch.Khi cc im cc ca H(p) nm bn na tri mt phng phc, ngoi tr tn ti mt vi im cc khng lp nm trn trc o, mch s: bin gii n nhKhi chuyn sang mch Thevenine, nu thay i gi tr ti Rti th: C Et v Rt u khng thay iKhi chuyn sang mch Thevenine, th chiu ca ngun tng ng: c xc nh theo chiu in p h mch tiKhi gii cc bi ton qu , khng cn quan tm n cc iu kin u ca mch. Pht biu ny ng hay sai ? SaiKhi mi im cc ca hm mch F(p) nm bn na tri mt phng phc (khng bao gm trc o), th p ng f(t) s: Hi t v 0 khi

Khi phn tch mch bng phng php in p nt, th s phng trnh c lp to ra l: Nn-1Khi phn tch mch bng phng php dng in nhnh cn s dng: C hai nh lut KirchhoffKhi phn tch mch bng phng php dng in nhnh, chiu ca dng in trn cc nhnh c th chn ty . Pht biu ny ng hay sai ? ngKhi phn tch mch bng phng php dng in nhnh, s phng trnh to ra l: NnhKhi phn tch mch bng phng php dng in vng hoc phng php in p nt, khng nht thit phi quan tm n chiu ca dng in trn cc nhnh. Pht biu ny ng hay sai ? ngKhi phn tch mch bng phng php dng in vng, th s phng trnh c lp to ra l: Nnh- Nn+1Khi phn tch mch in p dng phng php xp chng, th: Ln lt ch gi li mt ngun, cc ngun khc cn c loi bKhi tn s tn hiu vo mch lc thng thp tng vt ra ngoi di thng, in p li ra s:

GimKhi thay i mc in p ngun ca mch ban u th thng s no ca mch tng ng Thevenine s b thay i theo? Sc in ng ca ngunKhi tn ti im cc ca hm mch F(p) nm bn na phi mt phng phc, p ng f(t) s: Tin n v hn khi

Khng p dng c nh l Thevenine-Norton cho mt phn mch khi: N c ghp h cm vi phn mch ti

K hiu no sau y l ca ngun p c lp? (h1.16)

K hiu no sau y l ngun p ph thuc? (h1.18)

K hiu no sau y l ngun dng c lp? (h1.17) K hiu no sau y l ngun dng ph thuc? (h1.19)

K thut chun ha qua cc gi tr tng i da vo nguyn tc chn cc gi tr chun thch hp, nhm tng hiu qu tnh ton. Pht biu ny ng hay sai ?ng

Loi mch lc no s loi b di tn s t 3 kHz n 30 kHz ?

Chn diLa chn dng c tuyn hm truyn t in p ca mch in nh hnh v: (h4.47)(h4.47d) La chn Hm truyn t ca h thng nu th Bode ca n c dng nh hnh v: (h4.44) La chn Hm truyn t ca h thng nu th Bode ca n c dng nh hnh v: (h4.45)

La chn Hm truyn t ca h thng nu th Bode ca n c dng nh hnh v: (h4.46)

Lng truyn t c vit di dng logarit t nhin ca h s truyn t g()=lnG. Pht biu ny ng hay sai? ngMa trn tr khng h mch v ma trn dn np ngn mch l nghch o ca nhau. Pht biu ny ng hay sai ? ng

Mch in di y c bao nhiu nt, nhnh?(h1.40)

4 nt, 7 nhnh

Mch in nhn qu v n nh lun tn ti p ng tn s H(j). Pht biu ny ng hay sai ? ngMch in s lm vic ch tuyn tnh nu: Tt c cc phn t ca mch u lm vic ch tuyn tnh

Mch in tuyn tnh, bt bin truyn thng trong min thi gian c c trng bi: Mt h phng trnh vi phn tuyn tnh h s hng

Mch khuch i thut ton l mng 4 cc khng tng h, tch cc. Pht biu ny ng hay sai ? ngMch l mt h n nh khi mi im cc ca hm truyn t H(p) nm bn na tri ca mt phng phc,. Pht biu ny ng hay sai ? ng

Mch lc loi k (nh hnh v) tha mn: (h5.62) C ( trong k l mt hng s thc)Mch nh hnh v. Dng in trn Rt c xc nh:(h1.60)

EMBED Equation.DSMT4 Mch RLC ni tip, L = 1 mH v C = 10 mF. Tn s cng hng fch l xp x : 1,59 kHz

Mch RLC song song mang tnh cm khng khi: BL ln hn BC

Mch RLC song song mang tnh dung khng khi: BL nh hn BC

Mng 4 cc c cha ngun iu khin l mng 4 cc khng tng h. Pht biu ny ng hay sai ? ng

Mng 4 cc i xng hnh hc tng ng vi mt mng 4 cc i xng cu (nh hnh v) theo mi quan h no di y? (h5.46) ZI= z11- z12; ZII= z11+ z12

Mng 4 cc i xng hnh hc tng ng vi mt mng 4 cc i xng cu (nh hnh v), vi mi quan h: (h5.47) z11 =(Z1 + ZII)z12 =(ZII - ZI)Mng 4 cc i xng v mt in nu: Z11=Z22 v Z12=Z21Mng 4 cc khng tng h, tch cc c th biu din thnh s tng ng c cha ngun iu khin. Pht biu ny ng hay sai ? ng

Mng 4 cc tuyn tnh, tng h c th tng ng vi mt mng 4 cc n gin hnh T nh hnh v, vi cc biu thc quan h: (h5.39)

Z1=Z11-Z12; Z2=Z22-Z12; Z3=Z12=Z21;

Mng 4 cc tuyn tnh, tng h c th tng ng vi mt mng 4 cc n gin hnh T nh hnh v, vi cc biu thc quan h: (h5.40)

Z11=Z1+Z3; Z22= Z2+Z3; Z12=Z21=Z3Mng 4 cc tuyn tnh, tng h c th tng ng vi mt mng 4 cc n gin hnh , vi cc biu thc quan h: (h5.41) Y1=Y11+Y12; Y2=Y22+Y12; Y3= -Y12= -Y21Mng 4 cc tuyn tnh, tng h c th tng ng vi mt mng 4 cc n gin hnh , vi cc biu thc quan h: (h5.42) Y11=Y1+Y3; Y22=Y2+Y3; Y12=Y21= -Y3Mng 4 cc tuyn tnh, tng h c th tng ng vi mt mng 4 cc n gin hnh . Biu thc no sau y khng phi dng m t mi quan h tng ng trn? Y1=Y11-Y12Mng 4 cc tuyn tnh, tng h c th tng ng vi mt mng 4 cc n gin hnh T. Biu thc no sau y khng phi dng m t mi quan h tng ng trn?

Z2=Z22+Z12

Mng 4 cc tuyn tnh, tng h, th ng c th khai trin thnh: 2 s tng ngMng 4 cc tuyn tnh, tng h, th ng c th khai trin thnh s tng ng: Hnh T hoc hnh Mng 4 cc tuyn tnh, tng h, th ng, i xng c th khai trin thnh s tng ng no? Hnh T, hnh cu, hoc hnh PMng bn cc c cha Diode l loi M4C: Khng tng hMng bn cc c cha Transistor l loi M4C:

Khng tng h, tch cc.

M hnh ca in cm trong min tn s phc p l : (H3.26D)

M hnh ca in dung trong min tn s phc l: (H3.27B) Mi ngun in bt k u c th khai trin tng ng thnh mt ngun p hoc mt ngun dng. Pht biu ny ng hay sai ?ng

Mt 4 cc i xng v mt in th ng nhin i xng v mt hnh hc. Pht biu ny ng hay sai ? SaiMt 4 cc i xng v mt hnh hc th ng nhin i xng v mt in. Pht biu ny ng hay sai ? ng

Mt mch lc thng cao c tn ct fC=10 kHz. Khi gim tn s, bt u t fC, in p li ra s:

GimMt mch vng c th c nh ngha L mt ng mch in khp kn

Mt mng 4 cc i xng hnh hc th c th thay th bng s mng 4cc i xng cu tng ng. Pht biu ny ng hay sai ? ngMt mng 4 cc i xng hnh hc th c th thay th bng s mng 4 cc i xng cu tng ng, mi quan h gia chng tun theo: nh l Bartlett- Brune

Mt mng 4 cc thun khng c ni vi ngun v ti thun tr nh hnh v. Khi khng c s phn x tn hiu trn cc ca ca M4C, th Cng sut tc dng trn ti: (h5.61)

t cc iMt mng 4 cc tuyn tnh, bt bin, tng h th tha mn: y12= y21

Nu cng tc S1 trong hnh v chuyn t v tr A sang B th: (h2.43) Dng trn R1 s gimNu dng phng php dng in vng phn tch cc mch c cha ngun p th cn chuyn sang ngun dng. Pht biu ny ng hay sai ? SaiNu khi gii mch in mt chiu thu c dng trong mt nhnh mch c gi tr m th: Chiu thc t ca n l ngc li chiu bn quy c.

Nu mi im cc ca hm nh F(p) l cc nghim n pk, th hm gc f(t) c dng:

Nu ni tr ca ngun in l Rng , cng sut trn ti ln nht ng vi trng hp: Rng = Rt

Nu ni tr ca ngun in l Rng=100, cng sut trn ti ln nht ng vi trng hp:

Rt=100Nu tn s ca in p ngun t vo mch RLC song song thay i, i lng no sau y s khng thay i ?

IRNu thay i cc gi tr tr khng bn phn mch c cha ngun ban u th thng s no ca mch tng ng Thevenine s b thay i theo? Sc in ng v ni tr ngunNgun in l tng l khng c tn hao nng lng. Pht biu ny ng hay sai ?ng

Ngun ph thuc cn gi l ngun c iu khin. Pht biu ny ng hay sai ?ng

Nhc im chnh ca mch lc loi K ? Kh phi hp tr khng vi ngun v ti.Ni tr trong ca ngun tng ng Thevenine v Norton: L bng nhauPhn tch mch bng phng php in p nt dng n s trung gian l in p ti cc nt. Pht biu ny ng hay sai ? ngPhn tch mch bng phng php dng in vng dng n s trung gian l dng in gi nh trong cc vng kn. Pht biu ny ng hay sai ? ngPhn tch mch bng phng php ngun tng ng thng p dng trong cc trng hp khng i hi phi xc nh tt c cc dng v p ca tt c cc nhnh. Pht biu ny ng hay sai ? ngPhng php in p nt khng thun li cho nhng mch c ghp h cm. Pht biu ny ng hay sai ? ngPhng php Heaviside thc cht l: Phn tch F(p) thnh tng cc nh c bnPhng php ngun tng ng da vo: nh l Thevenine- NortonPhng php xp chng ch dng phn tch cc mch c cha cc ngun tc ng c cng tn s. Nhn xt ny ng hay sai ? SaiPhng php xp chng ch dng phn tch cc mch khng c ghp h cm. Nhn xt ny ng hay sai ? SaiPhng php xp chng c th dng phn tch cc mch phi tuyn. Nhn xt ny ng hay sai ? SaiQu trnh qu ca mch xy ra khi trng thi cn bng nng lng trong mch b ph v. Pht biu ny ng hay sai ? ngQu trnh qu trong mch in l qu trnh mch chuyn t trng thi ban u ti mt trng thi xc lp mi di mt tc ng kch thch no . Pht biu ny ng hay sai ? ngS tng ng ca on mch c dn np Y=2-j9 (S):(H1.58) S tng ng ca on mch c dn np Y=5+j7(S) l(H1.57)

S tng ng ca on mch c tr khng Z= 4-j3 (H1.56) S tng ng ca on mch c tr khngZ=5+j6 l (H1.55c)

S tng ng hnh T ca M4C tng h thng c xc nh theo cc thng s: ZijS tng ng hnh ca M4C tng h thng c xc nh theo cc thng s:

YijS bin i mt mch in thnh mch Norton l to ra: Mt mch song song n gin tng ng vi mch in ban uS dng phng php ngun tng ng bin i thnh mch Thevenine tng ng l to ra mt mch ni tip n gin tng ng vi mch in ban u. Pht biu ny ng hay sai ? ngSut in ng ca ngun c gi tr bng in p ngn mch ca ngun. Pht biu ny ng hay sai ?Sai

Sc in ng ca ngun in thuc loi:Thng s tc ng

Ti im cng hng ca mt mch RLC ni tip: Mch c tnh thun tr, dng vi p cng pha

Ti tn s cng hng ca mch RLC ni tip, dng in cng hng Ich l: Cc i

Ti tn s cng hng ca mch RLC song song, dng in tng qua ngun l: Cc tiu

Tn s ct ca mch lc thng cao loi K ?

Tn s ct ca mch lc thng thp loi K?

Tn s cng hng ca mch RLC ni tip hoc song song l:

Tn s cng hng trong mt mch RLC ni tip c th gim bng cch: Tng C

Tn s no sau y l b hn 2 decade so vi 1 kHz?

10 Hz.

Theo phng php th Bode, c tuyn bin ca hm mch F(p) c v da trn nh ngha:

Theo phng php th Bode, c tuyn pha ca hm mch F(p) c v da trn nh ngha:

Thng s ca ngun p tng ng vi ngun dng c Ing=2A; Rng=15 l:E=30V; R=15

Thng s ca ngun dng tng ng vi ngun p c Eng=50V; Rng=100V Ing= 5A; R = 10

Thng s in cm

Thuc loi thng s qun tnh

Thng s in dung:Thuc loi thng s qun tnh

Thng s no ca ngun tng ng Thevenine khng b ph thuc vo mc in p ngun ca mch ban u? Ni tr ngunThng s tc ng c trng cho phn t c kh nng t n (hoc khi c kch thch) c th to ra v cung cp nng lng in ti cc phn t khc ca mch. Pht biu ny ng hay sai ?

ng

Tm ra mt mnh sai trong s cc mnh sau: M4C phi hp tr khng c kt cu i xngTm ra mt mnh sai trong s cc mnh sau: M4C suy gim l tng c kt cu thun khngTnh n nh ca mch lin quan ti v tr cc im khng ca hm truyn t H(p) ca mch trn mt phng phc. Pht biu ny ng hay sai ? SaiTng i s cc dng in trn cc nhnh ti mt nt th: Bng khng

Transistor l loi mng 4 cc

Khng tng h, tch ccTr khng ca cc phn t qun tnh trong min tn s phc p ch c tnh bng biu thc Z=U(p)/I(p) khi nng lng ban u trong phn t bng khng. Pht biu ny ng hay sai ? ngTr khng ca mch RLC ni tip ti tn s cng hng l Cc tiu

Tr khng ca mch RLC song song ti tn s cng hng l Cc i.

Tr khng ca phn t thun cm l:

Tr khng ca phn t thun dung l:

Tr khng ca vo b bin i tr khng m (NIC) theo Zt ca ra? (h5.56)

Tr khng tng ng ca on mch nh hnh v(H1.80)

Tr khng tng ng ca on mch nh hnh v: (H1.79)

Tr khng tng ng ca mch in:(H1.53)

Z = 3 - j4

Tr khng v dn np ca cc phn t th ng trong min tn s thng hon ton c th suy ra t cch biu din trong min tn s phc p bng cch thay th p=j. Pht biu ny c ng khng ? ngTr khng v dn np ca in cm trong min p c dng

Tr khng v dn np ca in dung trong min p c dng:

Tr khng v dn np ca in tr trong min p c dng:

Trong cch biu din dn np di dng Y=G+jB, th: G l in dn, B l in np

Trong cch biu din tr khng di dng Z=R+jX, th: R l in tr, X l in khng

Trong cun dy c t bin dng in. Pht biu trn ng hay sai ? SaiTrong mch in, nu Vs1= Vs2 v cc gi tr R1, R2v R3 l bng nhau, th: (h2.31) C VR3 v IR3 u bng 0Trong mch RLC ni tip, nu dng in nhanh pha hn so vi in p t vo, tng ng vi trng hp no sau y l ng ? XC ln hn XL

Trong mch RLC ni tip, nu dng in tr pha hn so vi in p t vo, tng ng vi trng hp no sau y l ng ? XL ln hn XC

Trong mch th ng, cng sut phn khng c th c gi tr dng hoc m. Nhn xt ny ng hay sai?ng

Trong mch th ng, cng sut tc dng P c th c gi tr m. Nhn xt ny ng hay sai?Sai

Trong min tn s phc, quan h gia in p v dng in trn in tr l:

Trong min tn s phc, quan h gia in p v dng in trn in cm l:

Trong min tn s phc, quan h gia in p v dng in trn in dung l:

Trong m hnh phn t in cm min p, thnh phn L.iL(0) ng vai tr: Mt ngun sut in ng v ngc chiu vi U(p)Trong m hnh phn t in dung min p, thnh phn: ng vai tr: Mt ngun sut in ng v cng chiu vi U(p)Trong mt mch RLC ni tip, nu VC ln hn VL , th dng in Nhanh pha so vi in p t vo mch

Trong mt mch RLC ni tip, nu VC ln hn VL . Nhn xt no sau y ng ? Mch mang tnh dung khng

Trong mt mch RLC ni tip, nu VL ln hn VC . Nhn xt no sau y ng ? Mch mang tnh cm khng

Trong mt mch vng khp kn, tng i s cc st p :Bng khng

Trong mt vng khp kn, tng i s cc st p trn cc nhnh th ng Phi bng vi in p ngun cung cpTrong t in c t bin in p. Pht biu trn ng hay sai ? SaiT biu thc hm mch: c tuyn bin s l:

T biu thc hm mch: c tuyn bin s l:

T nh lut Kirchhoff 1 ch c th vit c Nn-1 phng trnh c lp, vi Nn l s nt trong mch. Pht biu ny ng hay sai ? ngT nh lut Kirchhoff 2 c th vit c Nnh phng trnh c lp, vi Nnh l s nhnh ca mch. Pht biu ny ng hay sai ? SaiV mt kt cu, mch in c hi tip ni tip in p (tn hiu hi tip ni tip vi tn hiu vo v t l vi in p u ra) ph hp vi kiu ghp no ?

Ghp ni tip- song songV mt kt cu, mch in c hi tip song song in p (tn hiu hi tip song song vi tn hiu vo v t l vi in p u ra) ph hp vi kiu ghp no ? Ghp song song- song songV mt kt cu, mch in c hi tip song song dng in (tn hiu hi tip song song vi tn hiu vo v t l vi dng in u ra) ph hp vi kiu ghp no ? Ghp song song- ni tipV mt kt cu, mch lc tn s l tng l mt mng 4 cc c suy gim c tnh tha mn: a()=0 trong di thng; trong di chnVic thay th dng trong cc nhnh bng cc n s trung gian trong phng php in p nt v dng in vng nhm mc ch: Lm gim s phng trnh cn thit phi thnh lp.Vi bn cc tuyn tnh, bt bin, tng h, ta lun c: Z12=Z21Vi mt s bi ton khng chnh, trong cun dy c th c t bin dng in, trong t in c th c t bin in p. Nhn xt ny ng hay sai ? ngVi ngun in p Eng=10V, in tr ngun Rng=100. Xc nh cng sut cc i m n c th cp cho ti?

Pmax=0.25 Xc nh cc thng s dn np ngn mch Yij ca mng 4 cc (h5.83)

Xc nh cc thng s tr khng h mch Zij ca mng 4 cc (h5.82)

Xc nh cc thng s tr khng h mch Zij ca mng 4 cc sau (h5.76)

Xc nh cc thng s truyn t Aij ca mng 4 cc (h5.84)

Xc nh hm gc f(t) nu bit nh ca n l:

Xc nh hm gc f(t) nu bit nh ca n l:

Xc nh hm gc IL(t) nu bit nh ca n l

Xc nh hm gc IL(t) nu bit nh ca n l:

Xc nh hm gc UC(t) nu bit nh ca n l

Xc nh hm gc UC(t) nu bit nh ca n l:

Xc nh hm gc UC(t) nu bit nh ca n l:

Xc nh hm gc UC(t) nu bit nh ca n l:

Xc nh hm gc UC(t) nu bit nh ca n l:

Xc nh s mng 4 cc tng ng hnh T ca mng 4 cc v di y: (h5.77)

Xc nh tnh cht ca mch nh hnh v? (h5.63) Mch lc thng thp loi K

Xc nh tnh cht ca mch nh hnh v? (h5.65) Mch lc thng cao loi K

Xc nh tnh cht ca mch nh hnh v? (h5.66) Mch lc chn di loi K

Xc nh tnh cht ca mch nh hnh v? (h5.67) Mch lc thng di loi K Xc nh tnh n nh ca khu c hm truyn t sau y:

Khng n nhXc nh tnh n nh ca khu c hm truyn t sau y: bin gii n nhXc nh tnh n nh ca khu c hm truyn t sau y: n nhXc nh tnh n nh ca khu c hm truyn t sau y:

Khng n nh

Xt hai thnh phn: Hj(p) v , th Bode (bin v pha) ca hai thnh phn ny: L i xng nhau qua trc DecadeXt mch RLC ni tip tn s cng hng, pht biu no sau y l sai ? Mch hot ng nh mt mch cm khng.

Xt mt ngun c tr khng . iu kin phi hp cng sut tc dng trn ti t cc i l:Tr khng ti bng lin hp ca tr khng ngun ()

Xt mt ngun pht c ni tr thun Zng=R0 v mt ti thun tr Zt= R0 , khi nhn xt no sau y l sai? Cn thm mt khu phi hp tr khng gia ngun v ti.Xt tn s ca mch ti f0=1 kHz. Tn s ln hn 1decade so vi f0 s l ti: 10 kHzXt tn s ca mch ti f0=10 kHz. Tn s thp hn 1decade so vi f0 s l ti: 1 kHz ngha ca vic phc ha cc thng s mch in truyn thng l: Chuyn cc h phng trnh vi tch phn trong min thi gian thnh h phng trnh i s trong min tn s.

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