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Kontur toklar usulidan foydalanib tarmoqdagi barcha toklarni aniqlaymiz
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| bet | 8/10 | | Sana | 14.05.2024 | | Hajmi | 0,69 Mb. | | #232732 |
Bog'liq KURS ISHI (4)2. Kontur toklar usulidan foydalanib tarmoqdagi barcha toklarni aniqlaymiz:
Bunda tarmoqdagi barcha toklar yuq, konturdan faqat bita kontur toki o`tadi deb faraz qilinadi.
Kirxgofning II – qonuni asosida har bir kontur uchun tenglamalar tuzamiz:
I – kontur uchun: E1 – E2 = II · (R1 + R4 + R2) – III · R2 – IIII · R4 (4)
II – kontur uchun: E2 + E3 = III · (R2 + R5 + R3) – II · R2 – IIII · R5 (5)
III – kontur uchun: 0 = IIII · (R4 + R5 + R6) – II · R4 – III · R5 (6)
Endi, masalaning shartida berilganlarni (4), (5) va (6) tenglamalarga qo`yamiz:
72 – 12 = II · (6 + 4 + 1) – III · 1 – IIII · 4 (4I)
12 + 4 = III · (1+10+12) – II · 1 – IIII · 12 (5I)
0 = IIII · (4 + 12 + 4) – II · 4 – III · 12 (6I)
(4I), (5I), (6I) larni ixchamlashtirib, quyidagi tenglamalar sistemasini hosil qilamiz:
60 = 11 · II – III – 4 · IIII (4 II)
16 = – II + 23 · III – 12 · IIII (5 II)
0 = – 4 · II – 12 · III + 20 · IIII (6 II)
(4 II), (5 II) va (6 II) lar uch noma`lumli uchta tenglamalar sistemasidir. Bu tenglamalar sistemasini determinant usulidan foydalanib yechamiz:
Unda
= 60 · 23 · 20 + (–1) · (–12) · 0 + (–4) · 16 · (–12) –
– (–4) · 23 · 0 – 60 · (–12) · (–12) - (–1) · 16 · 20 = 27600+0+768+0 –
– 8640 + 320 = 20048.
= 11 · 16 · 20 + 60 · (–12) · (–4) + (–4) · (–1) · 0 – – (–4) · 16 · (–4) – (11) · (–12) · 0 – 60 · (–1) · 20 = 3520 + 2880 + 0 – – 256 – 0 + 1200 = 7344.
= 11 · 23 · 0 + (–1) · 16 · (–4) + 60 · (–1) · (–12) – –60 · 23 · (–4) – 11 · 16 · (–12) – (–1) ∙ (–1) · 0 = 0 + 64 + 720 + 5520 + 2112 – 0 = 8416
Kontur toklarining qiymatlari:
bo`ladi.
Endi haqiqiy I1, I2, I3, I4, I5 va I6 toklarini aniqlaymiz:
konturning tashqarisidagi qarshiliklar (R1, R6, R3) dan o`tayotgan tok kuchlari (I1, I6, I3) o`sha kontur toklariga teng bo`ladi;
konturning abv qismidan faqat II kontur toki o`tadi va bu tok zanjirning shu qismidan o`tayotgan haqiqiy I1 tokiga teng bo`ladi, ya`ni II = I1= 6,7 A. Xuddi shunday II va III - kontur toklari III = I3 = 2,45 A; IIII = I6 = 2,8 A bo`ladi. Qolgan I2, I4, I5 toklari Kirxgofning I – qonuni asosida aniqlanadi, ya`ni:
I2 = II – III = 6,7 – 2,45 = 4,25 A.
I4 = II – IIII = 6,7 – 2,8 = 3,9 A.
I5 = IIII – III = 2,8 – 2,45 = 0,35 A.
Izlanayotgan barcha toklar aniqlandi.
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