TINACloud-ni ishga tushirish va ularni Internetda tahlil qilish uchun Interaktiv DC rejimini tanlash uchun Quyidagi misollarni bosing yoki bosing. TINACloud-ga arzon narxlardagi ma'lumotni oling va misollarni tahrirlang yoki o'zingizning davrlarini yarating
Biz allaqachon aylanib o'tgan elektr zanjirini (bir chastotada) Thvenin yoki Norton ekvivalenti bilan almashtirish mumkinligini ko'rdik. Ushbu texnikaga asoslangan va Maksimal quvvat uzatish teoremasi DC kontaktlarning zanglashiga olib kelganda, biz AC kuchlanishining aylanish davridagi maksimal quvvatni olish shartlarini aniqlay olamiz. O'tkazgichning aylanishi uchun Thvenin empedansi ham, yuk ham reaktiv tarkibiy qismga ega bo'lishi mumkin. Ushbu reaktsiyalar hech qanday o'rtacha quvvatni o'zlashtirmasa ham, ular yuk revantsiyasi Thvenin empedansining reaktsiyasini bekor qilmasa, ular kontaktlarning zanglashiga olib keladi. Shunday qilib, maksimal quvvatni uzatish uchun Tefenin va yuk reaktsiyalari kattaliklarda teng bo'lishi kerak, ammo teskari tomonda; Bundan tashqari, shaharning maksimal quvvat teoremasiga muvofiq, qarshilik ko'rsatadigan qismlar teng bo'lishi kerak. Boshqacha qilib aytganda, yuk empedansi teveninning empedansi ekvivalentiga teng bo'lishi kerak. Xuddi shu qoida yuk va Nortonni qabul qilish uchun qo'llaniladi.
RL= Qayta ZTh} va XL = - Im {ZTh}
Bu holatda maksimal quvvat:
Pmax = 
V2Th va men2N sinusoidal tepalik qiymatlarining kvadratini ifodalaydi.
Keyinchalik, ba'zi misollar bilan teoremni keltiramiz.
misol 1
R1 = 5 kohm, L = 2 H, vS(T) = 100V cos wt, w = 1 krad / s.
a) C va R ni toping2 Shunday qilib, R ning o'rtacha kuchi2Ikkita kutupli maksimal bo'ladi
b) Bu holatda maksimal o'rtacha kuch va reaktiv kuchni toping.
v) bu holatda v (t) ni toping.
V, mA, mW, kohm, mS, krad / s, milodiy, H, m F birliklari: v
a) Tarmoq allaqachon Thévenin formasida, shuning uchun biz birlashtirilgan formadan foydalanishimiz va Z ning haqiqiy va xayoliy qismlarini aniqlashimiz mumkin.Th:
R2 = R1 = 5 kohm; wL = 1 /w C = 2 ® C = 1 /w2L = 0.5 mF = 500 nF.
b.) O'rtacha quvvat:
Pmax = V2/ (4 * R.1) = 1002/ (2 * 4 * 5) = 250 mVt
Reaktiv quvvat: avval tok:
I = V / (R.1 + R2 + J (wL - 1 /wC)) = 100 / 10 = 10 mA
Q = - I2/ 2 * XC = - 50 * 2 = - 100 mvarv) Maksimal quvvat uzatishda yuk kuchlanishi:
VL = I * (R.2 + 1 / (j w C) = 10 * (5-j / (1 * 0.5)) =50 - j 20 = 53.852 ga -J 21.8° V
va vaqt funktsiyasi: v (t) = 53.853 cos (wT - 21.8°) V
V: = 100;
om: = 1000;
{a.} R2b: = R1;
C2: = 1 / sqr (om) / L;
C2 = [500n]
{b. /} I2: = V / (R1 + R2b);
P2m: = sqr (abs (I2)) * R2b / 2;
Q2m: = - sqr (abs (I2)) / om / C2 / 2;
P2m = [250m]
Q2m = [- 100m]
{c./} V2:=V*(R2b+1/j/om/C2)/(R1+R2b);
abs (V2) = [53.8516]
c sifatida import cmath
#Kompleksni chop etishni soddalashtiramiz
Shaffoflik uchun #raqamlar:
cp= lambda Z : “{:.8f}”.format(Z)
V = 100
om=1000
#a./
R2b=R1
C2=1/om**2/L
chop etish ("C2 =", cp (C2))
#b./
I2=V/(R1+R2b)
P2m=abs(I2)**2*R2b/2
Q2m=-abs(I2)**2/om/C2/2
chop etish ("P2m =", cp (P2m))
chop etish(“Q2m=”,cp(Q2m))
#c./
V2=V*(R2b+1/1j/om/C2)/(R1+R2b)
print(“abs(V2)=”,cp(abs(V2)))
misol 2
vS(T) = 1V cos w t, f = 50 Hz,
R1 = 100 ohm, R2 = 200 ohm, R = 250 ohm, C = 40 uF, L = 0.5 H.
a.) Yuk ko'taruvchisidagi quvvatni toping
b.) R va L ni toping, shunda RL ikki qutbining o'rtacha kuchi maksimal bo'ladi.
Avval biz Tvenin generatorini topishimiz kerak, uni biz RL yukining tugunlarining chap tomonidagi kontaktlarning zanglashiga almashtiramiz.
Bosqichlar:
1. RL yukini olib tashlash va uning uchun ochiq kontaktlarning zahirasini o'rnating
2. Ochiq elektron kuchlanishini o'lchash (yoki hisoblash)
3. Kuchlanish manbasini qisqa tutashuv bilan almashtiring (yoki oqim manbaini ochiq kontaktlarning zanglashiga olib tashlang)
4. Ekvivalent empedansni toping
V, MA, kohm, krad / s, mF, H, milodiy birliklar!
Va nihoyat soddalashtirilgan sxema:
Quvvati uchun echim: I = VTh /(ZTh + R + j w L) = 0.511 / (39.17 + 250 - j 32.82 + j 314 * 0.5)
½I½= 1.62 ta va R = ½I½2 * R / 2 = 0.329 mWAgar maksimal quvvatni topsak
shuning uchun R '= 39.17 ohm va L' = 104.4 mH.
Maksimal quvvat:
Imax = 0.511 / (2 * 39.17) = 6.52 mA va 
Va boshqalar: = 1;
om: = 100 * pi;
va:=Vs*replus(replus(R2,(1/j/om/C)),(R+j*om*L))/(R1+replus(replus(R2,(1/j/om/C)),(R+j*om*L)));
abs (va) = [479.3901m]
PR: = sqr (abs (va / (R + j * om * L))) * R / 2;
QL: = sqr (abs (va / (R + j * om * L))) * om * L / 2;
PR = [329.5346u]
QL = [207.0527u]
{b. /} Zb: = (replus (replus (R1, R2), 1 / j / om / C)));
abs (Zb) = [51.1034]
VT: = Vs * replus (R2,1 / j / om / C) / (R1 + replus (R2,1 / j / om / C));
VT = [391.7332m-328.1776m * j]
abs (VT) = [511.0337m]
R2b: = Re (Zb);
Lb: = - Im (Zb) / om;
Lb = [104.4622m]
R2b = [39.1733]
c sifatida import cmath
#Kompleksni chop etishni soddalashtiramiz
Shaffoflik uchun #raqamlar:
cp= lambda Z : “{:.8f}”.format(Z)
#Lambda yordamida replusni aniqlang:
Replus= lambda R1, R2 : R1*R2/(R1+R2)
Vs=1
om=100*c.pi
va=Vs*Replus(Replus(R2,1/1j/om/C),R+1j*om*L)/(R1+Replus(Replus(R2,1/1j/om/C),R+1j*om*L))
print(“abs(va)=”,cp(abs(va)))
PR=abs(va/(R+1j*om*L))**2*R/2
QL=abs(va/(R+1j*om*L))**2*om*L/2
chop etish ("PR =", cp (PR))
chop etish ("QL =", cp (QL))
#b./
Zb=Replus(Replus(R1,R2),1/1j/om/C)
print(“abs(Zb)=”,abs(Zb))
VT=Vs*Replus(R2,1/1j/om/C)/(R1+Replus(R2,1/1j/om/C))
chop etish ("VT =", cp (VT))
print(“abs(VT)=”,cp(abs(VT)))
R2b=Zb.real
Lb=-Zb.imag/om
chop etish ("Lb =", cp (Lb))
chop etish ("R2b =", cp (R2b))
Bu erda biz TINA ning maxsus funktsiyasidan foydalandik replus ikki impedansning parallel tengligini topish.
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