1-misol: iii nimaga teng?

3-misol: ushbu taqsimlangan tugundagi i_3i3​i, start subscript, 3, end subscript ning qiymati nimaga teng?

Bitta javobni tanlang:

Bitta javobni tanlang:

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-6\,\text{mA}−6mAminus, 6, start text, m, A, end text

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\phantom{+}0\,\text{mA}+0mAempty space, 0, start text, m, A, end text

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+6\,\text{mA}+6mAplus, 6, start text, m, A, end text

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+12\,\text{mA}+12mAplus, 12, start text, m, A, end text

Tekshirish

[Javobni koʻrsatish]

Kontur kuchlanishi

Quyida toʻrtta rezistor va kuchlanish manbaidan tashkil topgan zanjir berilgan. Biz uni Om qonunidan foydalanib yechamiz. Keyin natijaga qarab mulohazalar qilamiz. Masalani yechishdagi birinchi qadam – bu tok kuchini topish. Keyin esa alohida rezistorlardagi kuchlanishlarni hisoblaymiz.

Bu zanjirning elementlari ketma-ket ulanganligi bois beshta elementning barchasida faqat bitta \blueD iistart color #11accd, i, end color #11accd toki oqadi. \blueD iistart color #11accd, i, end color #11accd ni topish uchun toʻrtta ketma-ket ulangan rezistorlarni bitta ekvivalent rezistor holatiga keltirish mumkin.

R_{ketma-ket} = 100 + 200 + 300 + 400 = 1000\,\OmegaRketma−ket​=100+200+300+400=1000ΩR, start subscript, k, e, t, m, a, minus, k, e, t, end subscript, equals, 100, plus, 200, plus, 300, plus, 400, equals, 1000, \Omega

Om qonuni orqali tokni topamiz:

\blueD i = \dfrac{V}{R_{ketma-ket}} = \dfrac{20\,\text V}{1000\,\Omega} = 0{,}020\,\text A = 20 \,\text{mA}i=Rketma−ket​V​=1000Ω20V​=0,020A=20mAstart color #11accd, i, end color #11accd, equals, start fraction, V, divided by, R, start subscript, k, e, t, m, a, minus, k, e, t, end subscript, end fraction, equals, start fraction, 20, start text, V, end text, divided by, 1000, \Omega, end fraction, equals, 0, comma, 020, start text, A, end text, equals, 20, start text, m, A, end text

Endi bizga tok kuchi ma’lum. Keyingi qadamda toʻrtta rezistordagi kuchlanishni topamiz. Dastlabki sxemaga qaytib, 555 ta elementning barchasidagi kuchlanishlarni yozib chiqamiz.

Har bir rezistordagi kuchlanishni topish uchun Om qonunini toʻrt marta qoʻllaymiz.

v\phantom{_{\text{R1}}} = \blueD i\,\text RvR1​=iRv, empty space, equals, start color #11accd, i, end color #11accd, start text, R, end text
v_{\text{R1}} = 20\,\text{mA} \cdot 100\,\Omega = +2\,\text{V}vR1​=20mA⋅100Ω=+2Vv, start subscript, start text, R, 1, end text, end subscript, equals, 20, start text, m, A, end text, dot, 100, \Omega, equals, plus, 2, start text, V, end text
v_{\text{R2}} = 20\,\text{mA} \cdot 200\,\Omega = +4\,\text{V}vR2​=20mA⋅200Ω=+4Vv, start subscript, start text, R, 2, end text, end subscript, equals, 20, start text, m, A, end text, dot, 200, \Omega, equals, plus, 4, start text, V, end text
v_{\text{R3}} = 20\,\text{mA} \cdot 300\,\Omega = +6\,\text{V}vR3​=20mA⋅300Ω=+6Vv, start subscript, start text, R, 3, end text, end subscript, equals, 20, start text, m, A, end text, dot, 300, \Omega, equals, plus, 6, start text, V, end text
v_{\text{R4}} = 20\,\text{mA} \cdot 400\,\Omega = +8\,\text{V}vR4​=20mA⋅400Ω=+8Vv, start subscript, start text, R, 4, end text, end subscript, equals, 20, start text, m, A, end text, dot, 400, \Omega, equals, plus, 8, start text, V, end text

Tok kuchi va kuchlanishlarni aniqladik. Elektr zanjir yechildi.

Endi sxemaga rezistorlardagi kuchlanishlarni va manbani kiritib chiqsak boʻladi. Ushbu beshta kuchlanish element kuchlanishi deyiladi. (Zanjirdagi tugunlarni \greenE{\text a}astart color #0d923f, start text, a, end text, end color #0d923f dan \greenE{\text e}estart color #0d923f, start text, e, end text, end color #0d923f gacha harflar bilan belgilaymiz, bu esa keyinchalik ushbu tugunlar haqida mulohaza qilganimizda asqatadi.)

Keling, endi tezda tekshirib olamiz. Rezistorlar kuchlanishlarini qoʻshib chiqamiz.

2\,\text{V} + 4\,\text{V} + 6\,\text{V} + 8\,\text{V} = 20 \,\text V2V+4V+6V+8V=20V2, start text, V, end text, plus, 4, start text, V, end text, plus, 6, start text, V, end text, plus, 8, start text, V, end text, equals, 20, start text, V, end text

Rezistorlardagi kuchlanishlar yig‘indisi manba kuchlanishiga teng boʻladi. Bu bizning hisob-kitoblarimiz toʻgʻri ekanini tasdiqlaydi.

Endi kuchlanishlarni boshqacharoq tartibda, yaʼni kontur boʻylab qoʻshib chiqamiz. Bu yerda hech qanday nazariya yoʻq, biz faqat bir xil hisoblashni boshqacha tartibda bajaryapmiz, xolos.