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Yuqorida aytib o'tganimizdek, sinusoidal qo'zg'alish davri yordamida echilishi mumkin murakkab impedanslar elementlar va murakkab tepalik or Murakkab rms qiymatlari oqim va kuchlanish uchun. Kirchhoff qonunlarining murakkab qiymatlari versiyasidan foydalangan holda, o'zgaruvchan tok zanjirlarini doimiy tok zanjirlariga o'xshash tarzda echishda tugunli va to'rli tahlil usullarini qo'llash mumkin. Ushbu bobda biz buni Kirchhoff qonunlari misollari orqali ko'rsatamiz.
misol 1
Oqimning amplituda va fazoviy burchagini topingvs(T) if
vS(T) = VSM cos 2pft; i (t) = ISM cos 2pft; VSM = 10 V; MenSM = 1 A; f = 10 kHz;
Hammasi bo'lib bizda noma'lum 10 kuchlanish va oqim mavjud, ya'ni: i, iC1, iR, iL, iC2, vC1, vR, vL, vC2 va vIS. (Agar kuchlanish va oqim uchun murakkab cho'qqisi yoki rms qiymatlaridan foydalansak, bizda jami 20 ta haqiqiy tenglama bor!)
Tenglama:
Loop yoki mash tenglashishi: for M1 - VSM +VC1M+VRM = 0
M2 - VRM + VLM = 0
M3 - VLM + VC2M = 0
M4 - VC2M + VIsm = 0
Ohm qonunlari VRM = R *IRM
VLM = j*w* L *ILM
IC1M = j*w*C1*VC1M
IC2M = j*w*C2*VC2M
Nodal tenglama uchun N1 - IC1M - ISM + IRM + ILM +IC2M = 0
seriyali elementlar uchun I = IC1MTenglamalar tizimini yechishda noma'lum tokni topishingiz mumkin:
ivs (T) = 1.81 cos (wT + 79.96°) A
Bunday katta miqdordagi murakkab tenglamalarni echish juda murakkab, shuning uchun biz uni batafsil ko'rsatmadik. Har bir murakkab tenglama ikkita haqiqiy tenglamaga olib keladi, shuning uchun biz echimni faqat TINA Interpreter bilan hisoblangan qiymatlar bilan ko'rsatamiz.
TINA ning Tarjimonidan foydalangan holda echim:
om: = 20000 * pi;
Va boshqalar: = 10;
= 1;
Sys Ic1, Ir, IL, Ic2, Vc1, Vr, VL, Vc2, Vis, Ivs
Vs=Vc1+Vr {M1}
Vr=VL {M2}
Vr=Vc2 {M3}
Vc2=Vis {M4}
Ivs=Ir+IL+Ic2-Is {N1}
{Ohm qoidalari}
Ic1 = j * om * C1 * Vc1
Vr = R * y
VL = j * om * L * IL
Ic2 = j * om * C2 * Vc2
Ivs = Ic1
tugatish;
Ivs = [3.1531E-1 + 1.7812E0 * j]
abs (Ivs) = [1.8089]
fiIvs: = 180 * boshq (Ivs) / pi
fiIvs = [79.9613]
import sympy sifatida s
c sifatida import cmath
cp= lambda Z : “{:.4f}”.format(Z)
om=20000*c.pi
Vs=10
=1
Ic1,Ir,IL,Ic2,Vc1,Vr,VL,Vc2,Vis,Ivs=s.symbols('Ic1 Ir IL Ic2 Vc1 Vr VL Vc2 Vis Ivs')
A=[s.Eq(Vc1+Vr,Vs),s.Eq(VL,Vr),s.Eq(Vc2,Vr),s.Eq(Vis,Vc2), #M1, M2, M3, M4
s.Eq(Ir+IL+Ic2-Is,Ivs), #N1
s.Eq(1j*om*C1*Vc1,Ic1),s.Eq(R*Ir,Vr),s.Eq(1j*om*L*IL,VL),s.Eq(1j*om*C2*Vc2,Ic2),s.Eq(Ic1,Ivs)] #Ohm’s rules
Ic1,Ir,IL,Ic2,Vc1,Vr,VL,Vc2,Vis,Ivs=[complex(Z) for Z in tuple(s.linsolve(A,(Ic1,Ir,IL,Ic2,Vc1,Vr,VL,Vc2,Vis,Ivs)))[0]]
chop etish (Ivs)
print(“abs(Ivs)=”,cp(abs(Ivs)))
chop (“180*c.faza(Ivs)/c.pi=”,cp(180*c.faza(Ivs)/c.pi))
TINA yordamida echim:
Ushbu muammoni qo'l bilan hal qilish uchun murakkab impedanslar bilan ishlang. Masalan, R, L va C2 parallel ravishda ulangan, shuning uchun ularning parallel ekvivalentini hisoblash orqali kontaktlarning zanglashini soddalashtirish mumkin. || impedanslarning parallel ekvivalentini bildiradi:
Son jihatdan:
Impedansdan foydalangan holda soddalashtirilgan kontakt:
Tenglama tartiblangan holda: I + IG1 = IZ
VS = VC1 +VZ
VZ = Z · IZ
I = j w C1· VC1
To'rtta noma'lum - I; IZ; VC1; VZ - va bizda to'rtta tenglama bor, shuning uchun echim mumkin.
ifoda I boshqa noma'lumlarni tenglamalardan so'ng:
Raqamli
TINA Interpreter natijasiga ko'ra.
om: = 20000 * pi;
Va boshqalar: = 10;
= 1;
Z: = replus (R, replus (j * om * L, 1 / j / om / C2));
Z = [2.1046E0-2.4685E0 * j]
sys I
I = j * om * C1 * (Vs-Z * (I + Is))
tugatish;
I = [3.1531E-1 + 1.7812E0 * j]
abs (I) = [1.8089]
180 * boshq (I) / pi = [79.9613]
import sympy sifatida s
c sifatida import cmath
Replus= lambda R1, R2 : R1*R2/(R1+R2)
om=20000*c.pi
Vs=10
=1
Z=Replus(R,Replus(1j*om*L,1/1j/om/C2))
chop etish('Z=',cp(Z))
I=s.symbols('I')
A=[s.Eq(1j*om*C1*(Vs-Z*(I+Is)),I),]
I=[kompleks(Z) kortejdagi Z uchun(s.linsolve(A,I))[0]][0]
chop etish ("I =", cp (I))
print(“abs(I)=”,cp(abs(I)))
chop etish (“180*c.faza(I)/c.pi=”,cp(180*c.faza(I)/c.pi))
Oqimning vaqt funktsiyasi quyidagicha:
i (t) = 1.81 cos (wT + 80°) A
Kirchhoffning amaldagi qoidasini fasorli diagrammalar yordamida tekshirishingiz mumkin. Quyidagi rasm i dagi tugun tenglamasini tekshirish orqali ishlab chiqilganZ = i + iG1 shakl. Birinchi diagrammada parallelogram qoidasi bilan qo'shilgan fazalar ko'rsatilgan, ikkinchisida fazor qo'shilishning uchburchak qoidasi ko'rsatilgan.
Keling, TINA ning fazor diagrammasi xususiyati yordamida KVR-ni namoyish etamiz. Tenglamada manba kuchlanishi salbiy bo'lganligi sababli biz voltmetrni "orqaga" uladik. Fasor diagrammasi Kirchhoff kuchlanish qoidasining asl shaklini aks ettiradi.
Birinchi fazor diagrammada parallelogram qoidasi, ikkinchisida esa uchburchak qoidasi qo'llaniladi.
V shaklida KVRni tasvirlashC1 + V.Z - VS = 0, biz yana voltmetrni kuchlanish manbasiga orqaga bog'ladik. Fazor uchburchagi yopiq ekanligini ko'rishingiz mumkin.
misol 2
Barcha komponentlarning kuchlanish va tokini toping, agar:
vS(T) = 10 cos wt V, iS(T) = 5 cos (w t + 30 °) mA;
C1 = 100 nF, C2 = 50 nF, R1 = R2 = 4 k; L = 0.2 H, f = 10 kHz.
Noma'lumlar "passiv" elementlarning kuchlanishlari va oqimlarining murakkab tepalik qiymatlari, shuningdek kuchlanish manbai oqimining (iVS ) va oqim manbaining kuchlanishi (vIS ). Hammasi bo'lib, o'n ikkita murakkab noma'lum. Bizda uchta mustaqil tugun, to'rtta mustaqil ko'chadan bor (M deb belgilangan)I) va beshta "Ohm qonunlari" bilan tavsiflanadigan beshta passiv element - umuman 3 + 4 + 5 = 12 tenglama mavjud:
Nodal tenglamalar N uchun1 IVsM = IR1M + IC2M
N uchun2 IR1M = ILM + IC1M
N uchun3 IC2M + ILM + IC1M +IsM = IR2M
Loop tenglamalari M uchun1 VSM = VC2M + V.R2M
M uchun2 VSM = VC1M + V.R1M+ V.R2M
M uchun3 VLM = VC1M
M uchun4 VR2M = VIsm
Ohm qonunlari VR1M = R1*IR1M
VR2M = R2*IR2M
IC1m = j *w*C1*VC1M
IC2m = j *w*C2*VC2M
VLM = j *w* L * ILM
Shuni unutmangki, har qanday murakkab tenglama ikkita haqiqiy tenglamaga olib kelishi mumkin, shuning uchun Kirxhoff usuli ko'plab hisob-kitoblarni talab qiladi. Differentsial tenglamalar tizimidan foydalangan holda kuchlanish va oqimlarning vaqt funktsiyalarini hal qilish ancha sodda (bu erda muhokama qilinmaydi). Avval TINA tarjimoni tomonidan hisoblangan natijalarni ko'rsatamiz:
f: = 10000;
Va boshqalar: = 10;
s: = 0.005 * exp (j * pi / 6);
om: = 2 * pi * f;
sys ir1, ir2, ic1, ic2, iL, vr1, vr2, vc1, vc2, vL, vis, ivs
ivs=ir1+ic2 {1}
ir1=iL+ic1 {2}
ic2+iL+ic1+Is=ir2 {3}
Vs=vc2+vr2 {4}
Vs=vr1+vr2+vc1 {5}
vc1=vL {6}
vr2=vis {7}
vr1=ir1*R1 {8}
vr2=ir2*R2 {9}
ic1=j*om*C1*vc1 {10}
ic2=j*om*C2*vc2 {11}
vL=j*om*L*iL {12}
tugatish;
abs (vr1) = [970.1563m]
abs (vr2) = [10.8726]
abs (ic1) = [245.6503u]
abs (ic2) = [3.0503m]
abs (vc1) = [39.0965m]
abs (vc2) = [970.9437m]
abs (X) = [3.1112u]
abs (vL) = [39.0965m]
abs (ivs) = [3.0697m]
180 + radtodeg (arc (ivs)) = [58.2734]
abs (vis) = [10.8726]
radtodeg (boshq (vis)) = [- 2.3393]
radtodeg (arc (vr1)) = [155.1092]
radtodeg (arc (vr2)) = [- 2.3393]
radtodeg (arc (ic1)) = [155.1092]
radtodeg (boshq (ic2)) = [- 117.1985]
radtodeg (boshq (vc2)) = [152.8015]
radtodeg (boshq (vc1)) = [65.1092]
radtodeg (boshq (iL)) = [- 24.8908]
radtodeg (arc (vL)) = [65.1092]
import sympy sifatida s
matematikani m sifatida import qiling
c sifatida import cmath
cp= lambda Z : “{:.4f}”.format(Z)
f = 10000
Vs=10
S=0.005*c.exp(1j*c.pi/6)
om=2*c.pi*f
ir1,ir2,ic1,ic2,iL,vr1,vr2,vc1,vc2,vL,vis,ivs=s.symbols('ir1 ir2 ic1 ic2 iL vr1 vr2 vc1 vc2 vL vis ivs')
A=[s.Eq(ir1+ic2,ivs), #1
s.Eq(iL+ic1,ir1), #2
s.Eq(ic2+iL+ic1+Is,ir2), #3
s.Eq(vc2+vr2,Vs), #4
s.Eq(vr1+vr2+vc1,Vs), #5
s.Eq(vL,vc1), #6
s.Eq(vis,vr2), #7
s.Eq(ir1*R1,vr1), #8
s.Eq(ir2*R2,vr2), #9
s.Eq(1j*om*C1*vc1,ic1), #10
s.Eq(1j*om*C2*vc2,ic2), #11
s.Eq(1j*om*L*iL,vL)] #12
ir1,ir2,ic1,ic2,iL,vr1,vr2,vc1,vc2,vL,vis,ivs=[complex(Z) for Z in tuple(s.linsolve(A,(ir1,ir2,ic1,ic2,iL,vr1,vr2,vc1,vc2,vL,vis,ivs)))[0]]
print(“abs(vr1)=”,cp(abs(vr1)))
print(“abs(vr2)=”,cp(abs(vr2)))
print(“abs(ic1)=”,cp(abs(ic1)))
print(“abs(ic2)=”,cp(abs(ic2)))
print(“abs(vc1)=”,cp(abs(vc1)))
print(“abs(vc2)=”,cp(abs(vc2)))
print(“abs(iL)=”,cp(abs(iL)))
print(“abs(vL)=”,cp(abs(vL)))
print(“abs(ivs)=”,cp(abs(ivs)))
chop etish ("180+daraja(faza(ivs))=",cp(180+m.daraja(c.faza(ivs))))
print(“abs(vis)=”,cp(abs(vis))))
chop etish ("darajalar(faza(ko'rinish))=",cp(m.degrees(c.faza(vis))))
chop etish(“daraja(faza(vr1))=”,cp(m.daraja(c.faza(vr1))))
chop etish(“daraja(faza(vr2))=”,cp(m.daraja(c.faza(vr2))))
chop ("daraja(faza(ic1))=",cp(m.degrees(c.faza(ic1))))
chop ("daraja(faza(ic2))=",cp(m.degrees(c.faza(ic2))))
chop etish ("daraja(faza(vc2))=",cp(m.daraja(c.faza(vc2))))
chop etish ("daraja(faza(vc1))=",cp(m.daraja(c.faza(vc1))))
print(“daraja(faza(iL))=”,cp(m.derece(c.faza(iL))))
chop ("daraja(faza(vL))=",cp(m.degrees(c.faza(vL))))
Endi almashtirish yordamida qo'llar yordamida tenglamalarni soddalashtirishga harakat qiling. Birinchi almashtirish eq.9. ekv 5 ga.
VS = VC2 + R2 IR2 a.)
keyin eq.8 va eq.9. EK 5 ga qo'shildi.
VS = VC1 + R2 IR2 + R1 IR1 b.)
keyin eq 12., teng. 10. va menL eq. 2 ta eX.6.
VC1 = VL = jwLIL = jwL (IR1 - MenC1) = jwLIR1 - jwL jwC1 VC1
Express VC1
v)
Express VC2 ekv.4 dan. va ek.5. va ekv.8 o'rnini almashtiring. va VC1:
g.)
Eq.2 ga almashtiring., 10., 11. va d.) Eq.3 ga. va ifoda etamanR2
IR2 = IC2 + IR1 + IS = jwC2 VC2 + IR1 + IS
e.)
Endi d) va e) eq.4 ga almashtiring va I ni ifoda etingR1
Son jihatdan:
I vaqt funktsiyasiR1 quyidagilar:
iR1(T) = 0.242 cos (wT + 155.5°) Ma
O'lchagan kuchlanish: 
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