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Quvvatlar balansini tuzish
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| bet | 9/10 | | Sana | 14.05.2024 | | Hajmi | 0,69 Mb. | | #232732 |
Bog'liq KURS ISHI (4)3. Quvvatlar balansini tuzish.
Har qanday elektr zanjirida e.yu.k. manbalarining umumiy quvvati iste`molchilarda sarflanadigan quvvatlar yig`’indisiga teng bo`lishi kerak, ya`ni:
1.1 - rasmda keltirilgan elektr zanjiri uchun quvvatlar balansini tuzamiz:
e.yu.k. manbalarining umumiy quvvati:
= 72 · 6,7 – 12 · 4,25 + 4·2,45 = 441,2 Vt.
Iste`molchilarni umumiy quvvati:
= (6,7)2 · 6 + (4,25)2 · 1 +(2,45)2 · 10 + (3,9)2 · 4 + (0,35)2 · 12 + (2,8)2 · 4 = 441,2 Vt. Demak, 441,2 = 441,2 Vt.
4. Potensiallar diagrammasini qurish.
Potensiallar diagrammasi, asosiy murakkab zanjirning tashqi konturi uchun quriladi (2 - rasm); zanjirda biror nuqta (a) ning potensiali nolga teng deb olinib, shu nuqtaga nisbatan boshqa nuqtalarning potensiallari hisoblanadi. Buning uchun, soat mili harakati yo`nalishida, a nuqtasidan boshlab konturni aylanib chiqiladi
2 – rasm
a nuqtaning potensiali nolga teng:
φa = 0.
b nuqtaning potensiali a nuqtaning potensalidan E1 ga katta:
φb = φa + E1 = 0 + 72 = 72 V.
v nuqtasining potensiali b nuqtasining potensialidan I1 · R1 ga kichik bo`ladi, chunki R1 qarshiligida kuchlanish pasayishi φb – φv = I1 · R1 bo`ladi. Bundan,
φv = φb – I1 · R1 = 72 – 6,7 · 6 = 31,8 V.
g nuqtasining potensiali ham v nuqtaning potensialidan I6 · R6 ga kichik bo`ladi:
φg = φb – I6 · R6 = 31,8 –2,8 · 4 = 20,6 V.
d nuqtasining potensiali g nuqtaning potensialidan E3 ga katta bo`ladi:
φd = φg + Ye3 = 20,6 + 4 = 24,6 V.
yana, a nuqtasining potensialini tekshirsak, bunda u d nuqtasining potensialidan I3 · R3 ga kichik bo`ladi:
φa = φd – I3 · R3 = 24,6 – 2,46·10 = 0.
Endi, masshtabda potensiallar diagramma [φ = f(R)] sini quramiz (3 - rasm).
3 - rasm. Potensiallar diagrammasi.
2- masala bo`yicha variantlar.
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