8-sinf fizika fani mashqlari masalalari yechimlari
TRANSCRIPT
ABDUBANANOV AKRAMJON
1
8-SINF FIZIKA FANI
MASHQLARI
MASALALARI
YECHIMLARI
ABDUBANANOV AKRAMJON
2
I BOB. ELEKTR ZARYAD. ELEKTR MAYDON
1-mashq
1-masala. Litiy atomidagi elektronlar va pratonlar zaryadlari miqdorini aniqlang.
Bu masalani ishlash uchun Mendeleyev jadvalidan foydalanamiz
Litiy atomining yadrosida 3 ta praton orbitasida 3 ta elektron bor ekan.
Berilgan Chizmasi Formulasi Hisoblash
π = 3
π = β1,6 β 10β19πΆ
π = +1,6 β 10β19πΆ
ππ = ππ
ππ = ππ
ππ = β1,6 β 10β19 β 3 =
= β4,8 β 10β19πΆ
ππ = +1,6 β 10β19 β 3 =
= +4,8 β 10β19πΆ
Topish kerak:
ππ =? ππ =?
Javob:
ππ = β4,8 β 10β19πΆ
ππ = +4,8 β 10β19πΆ
2-masala. Uglerod atomidagi jami elektronlarning massasi qancha?
Bu masalani ishlash uchun Mendeleyev jadvalidan foydalanamiz
Uglerod atomining yadrosida 6 ta praton orbitasida 6 ta elektron bor.
ABDUBANANOV AKRAMJON
3
Uglerad atomi chizmasi
Berilgan Formulasi Hisoblash
π = 6
ππ = 9,1 β 10β19ππ
π = πππ
π = 9,1 β 10β31 β 6 =
= 54,6 β 10β31ππ
Topish kerak:
π =?
Javob:
π = 54,6 β 10β31ππ
3-masala. Kislorod atomidagi jami elektronlar zaryadi va massasini hisoblang.
Bu masalani ishlash uchun Mendeleyev jadvalidan foydalanamiz
Uglerod atomining yadrosida 8 ta praton orbitasida 8 ta elektron bor.
Uglerad atomi chizmasi
Berilgan Formulasi Hisoblash
ABDUBANANOV AKRAMJON
4
π = 8
π = β1,6 β 10β19πΆ
ππ = 9,1 β 10β19ππ
ππ = ππ
π = πππ
ππ = β1,6 β 10β19 β 8 =
= β12,8 β 10β19πΆ
π = 9,1 β 10β31 β 8 =
= 72,8 β 10β31ππ
Topish kerak:
π =?
Javob:
ππ = β12,8 β 10β19πΆ
π = 72,8 β 10β31ππ
2-mashq
1-masla. Bir-biridan 5 sm masofada joylashgan sharchalarning biriga β8 β 10β8πΆ,
ikkinchisiga esa 4 β 10β8πΆ zaryad berilgan. Zaryadlangan sharchalar qanday kuch
bilan tortishadi?
Berilgan Formulasi Hisoblash
π = 5 π π
π = 5 β 10β2π
π1 = β8 β 10β8πΆ
π2 = 4 β 10β8πΆ
πΉ = π β|π1| β |π2|
π2
Chizmasi
πΉ = 9 β 109|β8 β 10β8| β |4 β 10β8|
(5 β 10β2)2
=9 β 109 β 32 β 10β16
25 β 10β4=
= 11,52 β 10β3π
Topish kerak:
πΉ =?
Javob:
πΉ = 11,52 ππ
2-masala. Biri ikkinchisidan 5 sm uzoqlikda joylashgan bir xil zaryadlangan
sharchalar 3,6 β 10β4π kuch bilan taβsirlashmoqda. Ular qanday miqdorda
zaryadlangan?
Berilgan Formulasi Hisoblash
F F
r
ABDUBANANOV AKRAMJON
5
π = 5 π π
π = 5 β 10β2π
π1 = π2 = π
πΉ = 3,6 β 10β4π
πΉ = π β|π1| β |π2|
π2
πΉ = π βπ2
π2
Bundan:
π = π β βπΉ
π
π = 5 β 10β2 β β3,6 β 10β4
9 β 109=
= 5 β 10β2 β 2 β 10β7 = 10β8πΆ
Topish kerak:
πΉ =?
Javob:
π = 10β8πΆ
3-masala. Zaryadlari 0,36 ππΆ va 10 ππΆ boβlgan sharchalar qanday masofada 9 mN
kuch bilan taβsirlashadi?
Berilgan Formulasi Hisoblash
π1 = 0,36 ππΆ
π1 = 0,36 β 10β6πΆ
π2 = 10 ππΆ
π2 = 10 β 10β9πΆ
πΉ = π β|π1| β |π2|
π2
π2 = π βπ1 β π2
πΉ
Bundan:
π = βπ β π1 β π2
πΉ
π = β9 β 109 β 0,36 β 10β6 β 10 β 10β9
9 β 10β3
= 0,6 β 10β1π = 0,06 π
Topish kerak:
πΉ =?
Javob:
π = 0,06 π π¦πππ π = 6 π π
4-masala. Elektronlar orasidagi elektr itarishish kuchi ularning bir-biriga
gravitatsion tortishish kuchidan necha marta katta?
Berilgan Formulasi Hisoblash
ππ = 1,6 β 10β19πΆ
ππ = 9,1 β 10β31ππ
πΉπ = π β|ππ| β |ππ|
π2
πΉπ‘ = πΊ βππ β ππ
π2
πΉπ
πΉπ‘=
9 β 109 β (1,6 β 10β19)2
6,67 β 10β12 β (9,1 β 10β31)2
F F
r
F F
r
πΉπ‘ πΉπ πΉπ πΉπ‘
r
ABDUBANANOV AKRAMJON
6
πΉπ
πΉπ‘=
πππ2
πΊππ2 =
9 β 109 β 2,56 β 10β38
6,67 β 10β12 β 82,81 β 10β62=
=23,04 β 10β29
552,3427 β 10β74=
= 4,17 β 1045
Topish kerak:
πΉπ
πΉπ‘=?
Javob: β 4,2 β 1045 ππππ‘π πππ‘ππ
3-mashq
1-masala. Massalari 60 g boβlgan ikkita bir xil sharcha vakuumda bir-biridan ancha
uzoqda turibdi. Sharlar oβrtasidagi gravitatsiya tortishish kuchini muvozanatga
keltirish uchun har bir sharchaga bir xil ishorali qanday zaryad berish kerak?
Berilgan Formulasi Hisoblash
π = 60 π
π = 60 β 10β3ππ
π = 1
πΉπ‘ = πΉπ
πΉπ = π β|π1| β |π2|
π2
πΉπ‘ = πΊ βπ β π
π2
πΉπ‘ = πΉπ
π βπ2
π2= πΊ β
π2
π2
π = πβπΊ
π
π = 60 β 10β3β6,67 β 10β12
9 β 109=
= 6 β 10β2β0,7411 β 10β21 =
= 16,3 β 10β11πΆ
Topish kerak:
π1 = π2 = π =?
Javob: π = 0,16 ππΆ
2-masala. Bir xil manfiy zaryadga ega boβlgan ikki metall sharcha 24 sm masofada
2,5 Β΅N kuch bilan oβzaro taβsirlashmoqda. Har bir qancha ortiqcha elektronlar
boβlgan?
πΉπ‘ πΉπ πΉπ πΉπ‘
r
ABDUBANANOV AKRAMJON
7
Berilgan Formulasi Hisoblash
π1 = π2 = βπ
π = 24 π π
π = 24 β 10β2π
πΉ = 2,5 ππ
πΉ = 2,5 β 10β6 π
π = 1,6 β 10β19πΆ
πΉ = π β|π1| β |π2|
π2
πΉ = π βπ2
π2
π = πβπΉ
π
π =π
π
π = 24 β 10β2β2,5 β 10β6
9 β 109=
= 24 β 10β2 β0,5 β 10β7
3=
= 4 β 10β9πΆ
π =4 β 10β9
1,6 β 10β19= 2,5 β 1010
Topish kerak:
π =?
Javob: π = 25 β 1019 π‘π
3-masala. Uchta bir xil oβlchamli metall sharlarning biri +20 ππΆ, ikkinchisi β8 ππΆ,
uchunchisi zaryadsiz. Sharchalar bir-biriga tekkizilib, avvalgi vaziyatiga qaytarildi.
Uchunchi zaryad qanday zaryad oladi?
Berilgan Formulasi va chizmasi
π1 = +20 ππΆ = +20 β 10β6πΆ
π2 = β8 ππΆ = β8 β 10β6πΆ
Toβqnashishdan avval
ππ’π = π1 + π2
ππ’π = (20 β 8) β 10β6 = 12 β 10β6πΆ
Topish kerak:
π3 =?
Toβqnashishdan keyin
Zaryadni saqlanish qonuniga koβra zaryad 3 ta
sharga sharlar bir xil boβlganligi uchun teng
taqsimlanadi va oβzgarmaydi.
π3 =ππ’π
3=
12 β 10β6πΆ
3= 4 β 10β6πΆ π½ππ£ππ: π3 = +4 β 10β6πΆ
π1 π2 π3
ABDUBANANOV AKRAMJON
8
4-masala. Bir xil ishorali 2q va 10q zaryadlar bilan zaryadlangan ikkita bir xil
sharcha bir biridan r masofada oβzaro taβsirlashib turibdi? Sharchalar bir-biriga
tekkizilib, avvalgi vaziyatiga qaytarildi. Bunda ularning oβzaro taβsir kuchi qanday
oβzgaradi?
Berilgan Izoh
π1 = 2π
π2 = 10π
Bu masala potensial mavzusi oβtilgandan keyin
berilishi kerak edi.
Topish kerak:
πΉ2
πΉ1=?
Ishlash tartibi: 1) Zaryadlar bir-biriga tekkazilganda har ikkisidagi potensillar
tenglashgunga qadar potensiali yuqori shardan potensiali kam sharga zaryad oqib
oβtadi. Masalamizda ikkinchi shardan birinchi sharga oqib oβtadi.
π1 = ππ1
π π2 = π
2
π
Toβqnashishdan oldin Toβqnashishdan keyin
2) Sharlar potensiallarini tenglaymiz
π1, = π2
, ππ1
,
π = π
π2,
π π1
, = π2,
Demak, zaryadlar teng taqsimlanar ekan.
3) Zaryadni saqlanish qonunini qoβllaymiz
π1 + π2 = π1, + π2
, = 2π1, = 2π2
,
2π + 10π = 2π1, = 2π2
, π1, = π2
, = 6π
4) Oβzaro taβsir kuchlarini boβlamiz
π1 π2 π2 π1
π π
π1, π2
, π1,
π π
π2,
ABDUBANANOV AKRAMJON
9
πΉ1 = ππ1π2
π2 πΉ2 = π
π1, π2
,
π2
πΉ2
πΉ1=
π1, π2
,
π1π2=
6π β 6π
2π β 10π= 1,8
Javob: 1,8 marta ortar ekan.
5-masala. Ikki nuqtaviy zaryad bir-biridan r masofada turibdi. Zaryadlar orasidagi
masofa 20 sm ga ortirilganda ular orasidagi oβzaro taβsir kuchi 9 marta kamaygan.
Zarayadlar orasidagi dastlabki masofa qanday boβlgan?
Berilgan Formulasi Hisoblash
π1 π£π π2
π2 = π1 + 20
πΉ2
πΉ1=
1
9
πΉ1 = π β|π1| β |π2|
π12
πΉ2 = π β|π1| β |π2|
π22
πΉ1 = 9πΉ2
π β|π1| β |π2|
π12 = 9π β
|π1| β |π2|
(π1 + 20)2
(π1 + 20)2 = 9π12 kvadrat ildiz
chiqaramiz
π1 + 20 = 3π1 β π1 = 10 π π
Topish kerak:
π1 =?
Javob: π1 = 10 π π = 0,1 π
Chizmasi
Dastlabki holat Ortirilgan holat
6-masala. Ikkita bir xil sharcha bir-biridan 10 sm masofada turibdi. Ular bir xil
miqdorda manfiy zaryadga ega boβlib, 0,23 mN kuch bilan oβzaro taβsirlashadi. Har
qaysi sharchadagi ortiqcha elektronlar sonini toping.
Berilgan Formulasi Hisoblash
π1 = βπ
π2 = βπ
πΉ = π β|π1| β |π2|
π2
πΉ = π βπ2
π2
π = 0,1 β β0,23 β 10β3
9 β 109=
π2 π1
π1
πΉ1 πΉ1
π2 π1
π1
πΉ2 πΉ2
20 π π
ABDUBANANOV AKRAMJON
10
π = 10 π π = 0,1 π
πΉ = 0,23ππ
πΉ = 0,23 β 10β3π
π = π β βπΉ
π
π = ππ β π =π
π
= 0,01598 β 10β6πΆ
π β 0,016 β 10β6πΆ
π =0,016 β 10β6
1,6 β 10β19= 1011
Topish kerak:
π =?
Javob: π = 1011 π‘π
7-masala. Bir-biridan 3 sm masofada turgan har biri 1 nC dan boβlgan ikki zaryad
qanday kuch bilan taβsirlashadi?
Berilgan Formulasi Hisoblash
π1 = 1ππΆ = 10β9πΆ
π2 = 1ππΆ = 10β9πΆ
π = 3 π π = 0,03 π
πΉ = π β|π1| β |π2|
π2
πΉ = π βπ2
π2
Chizmasi
πΉ = 9 β 10910β9 β 10β9
0,032=
=9 β 10β9
9 β 10β4= 10β5π
Topish kerak:
πΉ =?
Javob: πΉ = 10β5π = 10ππ
8-masala. Bir-biridan 1 sm uzoqlikda joylashgan ikkala sharchaga bir xil 10-8C dan
zaryad berilgan. Zaryadlar qanday kuch bilan taβsirlashadi?
Berilgan Formulasi Hisoblash
π1 = 10β8πΆ
π2 = 10β8πΆ
π = 1 π π = 0,01 π
πΉ = π β|π1| β |π2|
π2
πΉ = π βπ2
π2
Chizmasi
πΉ = 9 β 10910β8 β 10β8
0,012=
=9 β 10β7
1 β 10β4= 9 β 10β3π
Topish kerak: πΉ =? Javob: πΉ = 9 β 10β3π = 9 ππ
π2 π1 π
πΉ
π2 π1 π
πΉ
ABDUBANANOV AKRAMJON
11
4-mashq.
1-masala. Zaryadi 4 nC boβlgan nuqtaviy zaryadning 6 sm masofada hosil qilgan
maydon kuchlanganligini toping.
Berilgan Formulasi Hisoblash
π = 4ππΆ = 4 β 10β9πΆ
π = 6 π π = 0,06 π
πΈ = π β|π|
π2
Chizmasi
πΈ = 9 β 1094 β 10β9
0,062=
=36
36 β 10β4= 104
π
πΆ
Topish kerak:
πΈ =?
Javob: πΈ = 104 π
πΆ
2-masala. Elektr maydon kuchlanganligi 3000 N/C boβlgan nuqtada turgan zaryadi
20 nC boβlgan sharchaga qanday kuch taβsir qiladi?
Berilgan Formulasi Hisoblash
π = 20ππΆ
π = 20 β 10β9πΆ
πΈ = 3000 π/πΆ
πΈ =πΉ
π
πΉ = πΈπ
Chizmasi
πΉ = 3000 β 20 β 10β9 =
= 6 β 10β5π
Topish kerak:
πΉ =?
Javob: πΉ = 6 β 10β5π = 60ππ
3-masala. Bir jinsli elektrostatik maydonda 5 β 10β8πΆ zaryadga 8Β΅N kuch taβsir
qilmoqda. Zaryad turgan nuqtadagi elektr maydon kuchlanganligini toping.
Berilgan Formulasi Hisoblash
π = 5 β 10β8πΆ
πΉ = 8ππ = 8 β 10β6π
πΈ =πΉ
π
Chizmasi
πΈ =8 β 10β6
5 β 10β8=
= 160 π/πΆ
π
π
πΈ
π
π
πΈ
πΉ
π
πΈ
πΉ
ABDUBANANOV AKRAMJON
12
Topish kerak:
πΈ =?
Javob: πΈ = 160 π/πΆ
4-masala. Zaryadi 3,6 nC boβlgan nuqtaviy zaryaddan qanday masofada maydon
kuchlanganligi 9000 N/C ga teng boβldi?
Berilgan Formulasi Hisoblash
π = 3,6ππΆ
π = 3,6 β 10β9πΆ
πΈ = 9000 π/πΆ
πΈ = π β|π|
π2
π = βππ
πΈ
Chizmasi
π = β9 β 109 β 3,6 β 10β9
9000=
= 6 β 10β2 π
Topish kerak:
π =?
Javob: π = 6 β 10β2π = 6 π π
5-mashq
1-masala. Elektr maydonda turgan 20 nC zaryadga 8 Β΅N kuch taβsir qilmoqda.
Zaryad turgan joyda maydon kuchlanganligi qancha boβlgan?
Berilgan Formulasi Hisoblash
π = 20ππΆ
π = 20 β 10β9πΆ
πΉ = 8ππ = 8 β 10β6π
πΈ =πΉ
π
πΈ =8 β 10β6
20 β 10β9= 4 β 102
π
πΆ
Topish kerak: πΈ =? Javob: πΈ = 4 β 102 π
πΆ
2-masala. Bir xil zaryadlangan ikki nuqtaviy zaryadlar oβzaro 30 Β΅N kuch bilan
taβsirlashmoqda. Birinchi zaryadning ikkinchi zaryad turgan nuqtada hosil qilgan
π
π
πΈ
ABDUBANANOV AKRAMJON
13
elektr maydon kuchlanganligi 5000 N/C ga teng. Nuqtaviy zaryadlarning qiymatini
toping.
Berilgan Formulasi Hisoblash
πΈ = 5000 π/πΆ
πΉ = 30ππ = 30 β 10β6π
πΈ =πΉ
π
π =πΉ
πΈ
Chizmasi
π =30 β 10β6
5000= 6 β 10β9πΆ
Topish kerak:
π =?
Javob: π = 6 β 10β9πΆ = 6ππΆ
3-masala. Maydon kuchlanganligi 1200 N/C boβlgan nuqtada turgan manfiy
zaryadlangan sharchaga 160 Β΅N kuch taβsir qilmoqda. Sharchadagi ortiqcha
elektronlar soni qancha?
Berilgan Formulasi Hisoblash
πΈ = 1200 π/πΆ
πΉ = 160πππΉ = 160 β 10β6π
π = 1,6 β 10β19πΆ
πΈ =πΉ
π
π =πΉ
πΈ
π =π
π
π =160 β 10β6
1200=
16 β 10β7
12πΆ
π =16 β 10β7
12 β 1,6 β 10β19=
= 8,3 β 1011
Topish kerak:
π =?
Javob: π = 8 β 1011 π‘π
4-masala. Zaryadi 7 nC boβlgan nuqtaviy zaryad kerosin ichida turibdi. U oβzidan
10 sm uzoqlikda qanday kuchlanganligini hosil qiladi? Kerosinning dielektrik
singdiruvchanligi 2,1 ga teng deb oling.
Berilgan Formulasi Hisoblash
π = 7ππΆ = 7 β 10β9πΆ πΈ = π β|π|
π β π2 πΈ =
9 β 109 β 7 β 10β9
2,1 β 0,12=
π πΈ
πΉ
ABDUBANANOV AKRAMJON
14
π = 10 π π = 0,1 π
π = 2,1
Chizmasi = 30 β 102π/πΆ
Topish kerak:
πΈ =?
Javob: πΈ = 3000 π/πΆ
5-masala. Muhit ichida turgan 30 nC va -36 nC boβlgan nuqtaviy zaryadlar bir-biri
bilan 18 sm masofada oβzaro taβsirlashmoqda. Ular orasidagi oβzaro taβsir kuchi 150
Β΅N ga teng boβlsa, muhitning dielektrik singdiruvchanligi qancha boβlgan?
Berilgan Formulasi Hisoblash
π1 = 30ππΆ = 30 β 10β9πΆ
π2 = β36ππΆ
π2 = β36 β 10β9πΆ
π = 18 π π = 0,18 π
πΉ = 150ππ = 150 β 10β6π
πΉ =
= π β|π1| β |π2|
π β π2
π
= π β|π1| β |π2|
πΉ β π2
π =
=9 β 109 β 30 β 10β9 β 36 β 10β9
150 β 10β6 β 0,182
= 2
Javob: πΈ = 3000 π/πΆ
Chizmasi
Topish kerak:
π =?
6-masala. Massasi 80 mg boβlgan moy tomchisi manfiy zaryadlangan. U
kuchlanganligi 1000 N/C boβlgan elektr maydonda muallaq turgan boβlsa, undagi
ortiqcha elektronlarning massasini toping.
Berilgan Formulasi Hisoblash
π = 80ππ = 80 β 10β6ππ
πΈ = 1000 π/πΆ
ππ = 9,1 β 10β31ππ
π = 1,6 β 10β19πΆ
πΉ = π
πΈπ = ππ
π =ππ
πΈ
π =π
π
π =80 β 10β6 β 10
1000= 8 β 10β7πΆ
π =8 β 10β7
1,6 β 10β19= 5 β 1012
π = 9,1 β 10β31 β 5 β 1012 =
= 45,5 β 10β19ππ
π
π
πΈ
π
π1
π
πΉ
π
π2 πΉ
ABDUBANANOV AKRAMJON
15
π = πππ Javob: π = 4,55 β 10β18ππ
Topish kerak:
π =?
7-masala. B nuqtada turgan 2 β 10β8πΆ zaryadli zaryad A nuqtada turgan zaryadga
60 Β΅N kuch bilan taβsirlashmoqda. A nuqtadagi zaryadning B nuqtada hosil qilgan
maydon kuchlanganligini aniqlang.
Berilgan Formulasi Hisoblash
π = 2 β 10β8πΆ
πΉ = 60ππ = 60 β 10β6π
πΈ =πΉ
π
πΈ =60 β 10β6
2 β 10β8= 3 β 103π/πΆ
Javob: πΈ = 3000 π/πΆ
Chizmasi
Topish kerak:
E=?
II BOB. ELEKTR TOKI
6-mashq
1-masala. Elektr zanjiridagi lampochkadan maβlum vaqt davomida 25 C zaryad
oβtib, 75 J ish bajarildi. Lampochka qanday elektr kuchlanish asosida yongan?
Berilgan Formulasi Hisoblash
π = 25 πΆ
π΄ = 75 π½
π =π΄
π
Chizmasi
π =75
25= 3 π
Topish kerak:
U=?
Javob: π = 3 π
πΉ
π
πΈ πβππ§masi
π
π΅ π΄
πΉ
πΈ
U
ABDUBANANOV AKRAMJON
16
2-masala. Uyali aloqa 5 V kuchlanishli tok manbaiga ega. Maβlum vaqt davomida
undan 10 C zaryad oβtganida qancha ish bajariladi?
Berilgan Formulasi Hisoblash
π = 10 πΆ
π = 5 π
π =π΄
πβ π΄ = ππ
π΄ = 10 β 5 = 50 π½
Topish kerak:
U=?
Javob: π΄ = 50 π½
3-masala. Koβchma magnitafon 9 V kuchlanishli tok manbaiga ega. Maβlum vaqt
davomida 450 J ish bajarish uchun manba qancha zaryad berishi kerak?
Berilgan Formulasi Hisoblash
π = 9 πΆ
π΄ = 450 π½
π =π΄
πβ π =
π΄
π
π =450
9= 50 πΆ
Topish kerak:
q=?
Javob: π = 50 πΆ
4-masala. Elektr zanjirdagi lampochkaga parallel ulangan voltmeter 3 V ni
koβrsatmoqda. Maβlum vaqt davomida 24 J ish bajarilishi uchun lampochkadan
qancha elektron oβtishi kerak?
Berilgan Formulasi Hisoblash
π = 3 π
π΄ = 24 π½
π = 1,6 β 10β19πΆ
π =π΄
πβ π =
π΄
π
π =π
π=
π΄
ππ
Chizmasi
π =24
1,6 β 10β19 β 3= 5 β 1019
ABDUBANANOV AKRAMJON
17
Topish kerak:
N=?
Javob: π = 5 β 1019
7-mashaq
1-masala. Elektr zanjiridagi lampochkadan 5 minutda 30 C zaryad oβtgan boβlsa,
zanjirdagi tok kuchi nimaga teng?
Berilgan Formulasi Hisoblash
π = 30 πΆ
π‘ = 5 π = 300 π
πΌ =π
π‘
πΌ =30
300= 0,1π΄
Topish kerak:
I=?
Javob: πΌ = 0,1 π΄
2-masala. Zanjirdagi tok kuchi 0,3 A ga teng boβlsa, 0,5 minut davomida
oβtkazgichning koβndalang kesimidan qancha zaryad oβtadi?
Berilgan Formulasi Hisoblash
πΌ = 0,3 π΄
π‘ = 0,5 π = 30 π
πΌ =π
π‘β π = πΌπ‘
Chizmasi
π = 0,3 β 30 = 9 πΆ
Topish kerak:
π =?
Javob: π = 9 πΆ
3-masala. Elektr zanjiriga ulangan lampochkadan 0,1 A tok oβtmoqda. Lampochka
spirali orqali 8 minutda qancha zaryad oβtadi? Shu vaqt davomida lampochkadan
oβtgan elektronlar sonini hisoblang.
Berilgan Formulasi Hisoblash
V
ABDUBANANOV AKRAMJON
18
πΌ = 0,1 π΄
π‘ = 8 π = 480 π
π = 1,6 β 10β19πΆ
πΌ =π
π‘β π = πΌπ‘
π =π
π=
πΌπ‘
π
Chizmasi
π = 0,1 β 480 = 48 πΆ
π =48
1,6 β 10β19= 3 β 1020
Topish kerak:
π =? π =?
Javob: π = 48 πΆ, π = 3 β 1020
4-masala. Elektr lampochkadan 0,8 A tok oβtmoqda. Uning spirali koβndalang
kesimidan 10 minutda oβtgan elektronlarning massasini aniqlang.
Berilgan Formulasi Hisoblash
πΌ = 0,8 π΄
π‘ = 10 π = 600 π
ππ = 9,1 β 10β31πΆ
πΌ =π
π‘β π = πΌπ‘
π =π
π=
πΌπ‘
π
π = πππ =πππΌπ‘
π
π =9,1 β 10β31 β 0,8 β 600
1,6 β 10β19=
= 2730 β 10β12ππ
Javob:
Topish kerak:
π =?
π = 2730 β 10β12ππ = 2,7πππ
5-masala. Manbaga ulangan isteβmolchilardan 20 mA tok oβtib turibdi. Tok manbai
2 soat davomida zaryadni koβchirishda 720 J ish bajargan boβlsa, isteβmolchi
uchlariga qanday kuchlanish berilgan?
Berilgan Formulasi Hisoblash
πΌ = 20 ππ΄ = 20 β 10β3π΄
π΄ = 720 π½
π‘ = 2 π πππ‘ = 7200 π
π΄ = πΌππ‘
π =π΄
πΌπ‘
π =720
20 β 10β3 β 7200= 5 π
Javob:
Topish kerak:
π =?
π = 5 π
ABDUBANANOV AKRAMJON
19
6-masala. Elektr zanjiridagi lampochkadan oβtayotgan tok kuchi 0,3 A ga teng.
Lampochka spiralidan qancha vaqtda 360 C zaryad oβtadi?
Berilgan Formulasi Hisoblash
πΌ = 0,3 π΄
π = 360 πΆ πΌ =
π
π‘β π‘ =
π
πΌ
π‘ =360
0,3= 1200 π
Javob:
Topish kerak:
π‘ =?
π‘ = 1200 π = 2 ππππ’π‘
7-masala. Akkumulator 25 minut davomida 4 A tok berib tura oladi. Bunday
akkumulator qancha elektr zaryadi toβplay oladi?
Berilgan Formulasi Hisoblash
πΌ = 4 π΄
π‘ = 25 π = 1500π πΌ =
π
π‘β π = πΌπ‘
π = 4 β 1500 = 6000 πΆ
Javob:
Topish kerak:
π =?
π = 6000 πΆ
8-masala. Elektr zanjiridagi lampochkadan 0,4 A tok oβtmoqda. Lampochka spirali
orqali 3 minutda uning kesimidan oβtgan zaryad miqdori va oβtgan elektronlar sonini
hisoblang.
Berilgan Formulasi Hisoblash
πΌ = 0,4 π΄
π‘ = 3 π = 180π
π = 1,6 β 10β19 πΆ
πΌ =π
π‘β π = πΌπ‘
π =π
π=
πΌπ‘
π
π = 0,4 β 180 = 72 πΆ
π =72
1,6 β 10β19= 45 β 1019
Javob:
Topish kerak:
π =? π =?
π = 72 πΆ, π = 45 β 1019 π‘π
ABDUBANANOV AKRAMJON
20
9-masala. 12 V kuchlanishli akkumulator avtomobilni yurgizishda generatorga 50
A tok bermoqda. Agar avtomobil dvigateli 2 s oβtgach oβt olsa, akkumulator qanday
ish bajargan?
Berilgan Formulasi Hisoblash
πΌ = 50 π΄
π‘ = 2 π
π = 12 π
π΄ = πΌππ‘
π΄ = 50 β 12 β 2 = 1200 π½
Javob:
Topish kerak:
π΄ =?
π΄ = 1200 π½
10-masala. Elektr zanjirdagi lampochkadan maβlum vaqt davomida 25 C zaryad
oβtib, tok manbai 100 J ish bajardi. Lampochka qanday elektr kuchlanish ostida
yongan?
Berilgan Formulasi Hisoblash
π = 25 πΆ
π΄ = 100 π½ π΄ = ππ β π =
π΄
π
π =100
25= 4 π
Javob:
Topish kerak:
π =?
π = 4 π£
8-mashq
1-masala. Uzunligi 100 m va kondalang kesimining yuzasi 2 mm2 boβlgan mis
simning qarshiligini toping.
Berilgan Formulasi Hisoblash
π = 100 π
π = 2 ππ2 = 2 β 10β6π2
π = ππ
π
Chizmasi
π = 0,017 β 10β6100
2 β 10β6=
= 0,85 ππ
ABDUBANANOV AKRAMJON
21
π = 0,017 β 10β6ππ β π Javob:
Topish kerak:
π =?
π = 0,85 ππ
2-masala. Uzunligi 3 m koβndalang kesimining yuzasi 0,5 mm2 boβlgan simning
qarshiligi 2,4 Om ga teng. Sim qanday moddadan tayyorlangan?
Berilgan Formulasi Hisoblash
π = 3 π
π = 0,5 ππ2
π = 0,5 β 10β6π2
π = 2,4 ππ
π = ππ
π
π =π π
π
Chizmasi
π =2,4 β 0,5 β 10β6
3=
= 0,4 β 10β6 ππ β π
Javob:
Topish kerak:
π =?
π = 0,4 β 10β6 ππ β π
Nikelin
3-masala. Bir xil moddadan tayyorlangan ikkita oβtkazgich sim bor. Birinchi
simning uzunligi 5 m, koβndalang kesimining yuzasi 0,1 mm2, ikkinchi simning
uzunligi 0,5 m, koβndalang kesimining yuzasi 3 mm2. Simlarning qarshiliklarini
taqqoslang.
Berilgan Formulasi Hisoblash
π1 = π2
π1 = 5 π
π1 = 0,1 ππ2π = 0,1 β 10β6π2
π2 = 0,5 π
π2 = 3 ππ2π = 3 β 10β6π2
π 1 = π1
π1π1
π 2 = π2
π2π2
π 1
π 2=
π1 β π2
π2 β π1
π 1
π 2=
5 β 3 β 10β6
0,5 β 0,1 β 10β6= 300
Javob:
S
l
S
l
ABDUBANANOV AKRAMJON
22
Topish kerak:
π 1
π 2=?
π 1
π 2= 300
4-masala. Koβndalang kesimining yuzasi 0,5 mm2 boβlgan 2 Om qarshilikli spiral
tayyorlash uchun qanday uzunlikda nikelin sim kerak boβladi?
Berilgan Formulasi Hisoblash
π = 2 ππ
π = 0,5 ππ2 = 0,5 β 10β6π2
π = 0,4 β 10β6 ππ β π
π = ππ
π
π =π π
π
π =2 β 0,5 β 10β6
0,4 β 10β6= 2,5 π
Javob:
Topish kerak:
π =?
π = 2,5 π
5-masala. 6 m uzunlikdagi nixrom simdan tayyorlangan spiralning qarshiligi 13,2
Om ga teng. Simning koβndalang kesim yuzasini toping.
Berilgan Formulasi Hisoblash
π = 13,2 ππ
π = 6 π
π = 1,1 β 10β6 ππ β π
π = ππ
π
π =π β π
π
π =6 β 1,1 β 10β6
13,2= 0,5 β 10β6π2
Javob:
Topish kerak:
π =?
π = 0,5 β 10β6π2 = 0,5 ππ2
6-masala. Agar metall simning uzunligi va koβndalang kesim yuzasi 2 marta
orttirilsa uning qarshiligi qanday oβzgaradi?
Berilgan Formulasi Hisoblash
π = ππππ π‘ π 1 = π1
π1π1
π 2
π 1=
2π1 β π1
π1 β 2π1= 1
ABDUBANANOV AKRAMJON
23
π2 = 2π1
π2 = 2π1
π 2 = π2
π2π2
π 2
π 1=
π2 β π1
π1 β π2
Javob:
Topish kerak:
π 2
π 1=?
π 2
π 1= 1 πβ²π§ππππππ¦ππ
7-masala. Uzunligi 20 m, qarshiligi 16 Om boβlgan nixrom simning hajmi qancha
boβladi?
Berilgan Formulasi Hisoblash
π = 16 ππ
π = 20 π
π = 1,1 β 10β6 ππ β π
π = ππ
π
π =π β π
π
π = ππ
π =20 β 1,1 β 10β6
16= 1,375 β 10β6π2
π = 1,375 β 10β6 β 20 = 27,5 β 10β6π3
Javob:
Topish kerak:
π =?
π = 27,5 β 10β6π3 = 27,5 ππ3
9-mashq
1-masala. Elektr zanjirga ulangan rezistorning qarshiligi 100 Om rezistor uchlari
orasidagi kuchlanish 10 V boβlsa, undan qanday tok oβtadi?
Berilgan Formulasi Hisoblash
π = 100 ππ
π = 10 π πΌ =
π
π
πΌ =10
100= 0,1 π΄
Javob:
ABDUBANANOV AKRAMJON
24
Topish kerak:
πΌ =?
πΌ = 0,1 π΄
2-masala. Qarshiligi 110 Om boβlgan oβtkazgich orqali 2 A tok oβtkazish uchun
oβtkazgich uchlariga qanday kuchlanish qoβyish kerak?
Berilgan Formulasi Hisoblash
π = 110 ππ
πΌ = 2 π΄ πΌ =
π
π β π = πΌπ
π = 2 β 110 = 220 π
Javob:
Topish kerak:
π =?
π = 220 π
3-masala. Elektr zanjiridagi isteβmolchiga 2 V kuchlanish berilganda, undagi tok
kuchi 0,1 A ga teng boβldi. Shu isteβmolchida tok kuchi 0,3 A ga yetishi uchun unga
qanday kuchlanish berish kerak?
Berilgan Formulasi Hisoblash
π1 = 2 π
πΌ1 = 0,1 π΄
πΌ2 = 0,3 π΄
πΌ =π
π
π =π1
πΌ1
π =π2
πΌ2
π =2
0,1= 20 ππ
20 =π2
0,3β π2 = 6 π
Javob:
Topish kerak:
π2 =?
π2 = 6 π
4-masala. Uzunligi 12 m va koβndalang kesim yuzi 0,6 mm2 boβlgan nixrom
oβtkazgich uchlariga 4,4 V kuchlanish berilganda undan qanday tok oβtadi?
Berilgan Formulasi Hisoblash
π
π
πΌ
ABDUBANANOV AKRAMJON
25
π = 12 π
π = 4,4 π
π = 1,1 β 10β6 ππ β π
π = 0,6 ππ2 = 0,6 β 10β6π2
πΌ =π
π
π = ππ
π
πΌ =ππ
ππ
πΌ =4,4 β 0,6 β 10β6
1,1 β 10β6 β 12= 0,2 π΄
Javob:
Topish kerak: I=? πΌ = 0,2 π΄
5-masala. Qarshiligi 16 Om boβlgan reostat yasash uchun koβndalang kesim yuzasi
0,25 mm2 boβlgan nikelin simdan necha metr kerak boβladi?
Berilgan Formulasi Hisoblash
π = 0,4 β 10β6 ππ β π
π = 0,25 ππ2 = 0,25 β 10β6π2
π = 16 ππ
π = ππ
π
π =π π
π
π =16 β 0,25 β 10β6
0,4 β 10β6= 10 π
Javob:
Topish kerak: l=? π = 10 π
6-masala. Oβzgarmas kuchlanish manbaiga ulangan oβtkazgichdan 30 mA tok
oβtmoqda. Agar oβtkazgichning chorak qismi kesib olinsa, bu oβtkazgichdan qanday
tok oβtadi?
Berilgan Formulasi Hisoblash
π1 = π
π2 =3π
4
πΌ1 = 30ππ΄ = 30 β 10β3π΄
π = ππππ π‘
π 1 = ππ1π
π 2 = ππ2π
π = πΌ1 β π 1
π = πΌ2 β π 2
π = π β πΌ1 β π 1 = πΌ2 β π 2
πΌ1 β ππ1π
= πΌ2 β ππ2π
πΌ1 β π1 = πΌ2 β π2
30 β 10β3 β π = πΌ2 β3π
4
πΌ2 = 40 β 10β3π΄
Javob:
Topish kerak: πΌ2 =? πΌ2 = 40 β 10β3π΄ = 40 ππ΄
ABDUBANANOV AKRAMJON
26
10-mashq
1-masala. Oβtkazgich uchlariga 6 V kuchlanish berilganda undan 5 s da 20 C zaryad
oβtdi. Oβtkazgich qarshiligi qanday?
Berilgan Formulasi Hisoblash
π = 6 π
π‘ = 5 π
π = 20 πΆ
πΌ =π
π‘
π =π
πΌ
π =ππ‘
π
π =6 β 5
20= 3 ππ
Javob:
Topish kerak: R=? π = 3 ππ
2-masala. Uzunligi 12 m va koβndalang kesim yuzi 0,6 mm2 boβlgan nixrom
oβtkazgich uchlariga 4,4 V kuchlanish berilganda undan qanday tok oβtadi?
Berilgan Formulasi Hisoblash
π = 12 π
π = 4,4 π
π = 1,1 β 10β6 ππ β π
π = 0,6 ππ2 = 0,6 β 10β6π2
πΌ =π
π
π = ππ
π
πΌ =ππ
ππ
πΌ =4,4 β 0,6 β 10β6
1,1 β 10β6 β 12= 0,2 π΄
Javob:
Topish kerak: I=? πΌ = 0,2 π΄
3-masala. Qarshiligi 10 Om boβlgan oβtkazgich uchlariga 2,5 V kuchlanish berilgan.
Oβtkazgich koβndalang kesim yuzidan 8 s da qancha elektron oβtadi?
Berilgan Formulasi Hisoblash
π = 10 ππ
π = 2,5 π
π‘ = 8 π
πΌ =π
π
πΌ =π
π‘=
ππ
π‘
π =2,5 β 8
1,6 β 10β19 β 10= 1,25 β 1019
ABDUBANANOV AKRAMJON
27
π = 1,6 β 10β19πΆ π
π =
ππ
π‘
π =ππ‘
ππ
Javob:
Topish kerak: π =? π = 1,25 β 1019 π‘π
4-masala. Koβndalang kesim yuzi 0,1 mm2 boβlgan nixromdan elektr plitkasining
qizdirgichi yasalgan. Unng uchariga 220 V kuchlanish berilganda undan 4 A tok
oβtdi. Qizdirgich uchun qanday uzulikdagi sim oligan?
Berilgan Formulasi Hisoblash
πΌ = 4 π΄
π = 220 π
π‘ = 8 π
π = 1,1 β 10β6 ππ β π
π = 0,1 ππ2 = 0,1 β 10β6π2
πΌ =π
π
π =ππ
π
πΌ =ππ
ππ
π =ππ
πΌπ
π =220 β 0,1 β 10β6
4 β 1,1 β 10β6= 5 π
Javob:
Topish kerak: π =? π = 5 π
5-masala. Uzunligi 20 m, koβndalang kesim yuzi 0,8 mm2 boβlgan nixrom
oβtkazgichning koβndalang kesimidan 3 s ichida 18 C zaryad oβtgan boβlsa, uning
uchlariga qanday kuchlanish qoβyilgan?
Berilgan Formulasi Hisoblash
π = 20 π
π‘ = 3 π
π = 18 πΆ
π = 0,8 ππ2 = 0,8 β 10β6π2
π = 1,1 β 10β6 ππ β π
πΌ =π
π
π =ππ
π
πΌ =ππ
ππ=
π
π‘
π =πππ
π‘π
π =20 β 18 β 1,1 β 10β6
3 β 0,8 β 10β6=
= 220 π
Javob:
ABDUBANANOV AKRAMJON
28
Topish kerak: π =? π = 220 π
6-masala. Uzunligi 100 m, koβndalang kesim yuzi 0,5 mm2 boβlgan alyuminiy
simning uchlaridagi kuchlanish 14 V berilganda, shu simdan oβtayotgan tok kuchi
qanday boβladi?
Berilgan Formulasi Hisoblash
π = 100 π
π = 14 π
π = 0,5 ππ2 = 0,5 β 10β6π2
π = 0,028 β 10β6 ππ β π
πΌ =π
π
π =ππ
π
πΌ =ππ
ππ
πΌ =14 β 0,5 β 10β6
100 β 0,028 β 10β6= 2,5π΄
= 220 π
Javob:
Topish kerak: πΌ =? πΌ = 2,5 π΄
7-masala. Maxsus dastgohda simni choβzib, u ikki marta uzun va ingichka qilingan.
Buning natijasida simning qarshiligini qanday oβzgargan?
Berilgan Formulasi Hisoblash
π1 = π
π2 = 2π
π = ππππ π‘
π 1 = ππ1π1
π 2 = ππ2π2
π 2
π 1=
π2π1
π1π2
π ππ πβπβ²π§πππππππ π ππ βππππ πβ²π§ππππππ¦ππ.
π1 = π2 π1π1 = π2π2
ππ1 = 2ππ2 β π1 = 2π2
π 2
π 1=
2π β 2π2
ππ2= 4
Javob:
Topish kerak:
πΌ2 =?
π 2
π 1= 4 4 ππππ‘π πππ‘πππ.
11-mashq
1-masala. Ketma-ket ulangn ikkita oβtkazgichdan 0,4 A tok oβtmoqda.
Oβtkazgichlarning qarshiligi 5 Om va 10 Om boβlsa, har bir oβtkazgich uchlaridagi
kuchlanishni, zanjirning toβliq qarshiligini va toβliq kuchlanishini toping.
ABDUBANANOV AKRAMJON
29
Berilgan Formulasi Hisoblash
πΌ = 0,4 π΄
π 1 = 5 ππ
π 2 = 10 ππ
πΌ = πΌ1 = πΌ2 = 0,4 π΄
π1 = πΌ1 β π 1, π2 = πΌ2 β π 2
π = π1 + π2, π = π 1 + π 2
Chizmasi:
π1 = 0,4 β 5 = 2 π
π2 = 0,4 β 10 = 4 π
π = 2 + 4 = 6 π
π = 5 + 10 = 15 ππ
Topish kerak:
π1 =?, π2 =?
π =? , π =?
Javob: π1 = 2 π, π2 = 4 π
π = 6 π, π = 15 ππ
2-masala. Qarshiligi 4 Om, 10 Om va 16 Om boβlgan oβtkazgichlar ketma-ket
ulangan. Zanjir uchlariga 6 V kuchlanish berilganda, har bir oβtkazgichdan
oβtayotgan tok kuchi va har bir oβtkazgich uchlaridagi kuchlanish qanday boβladi?
Berilgan Formulasi Hisoblash
π = 6 π
π 1 = 4 ππ
π 2 = 10 ππ
π 3 = 16 ππ
πΌ = πΌ1 = πΌ2 = πΌ3
πΌ =π
π =
π
π 1 + π 2 + π 3
π1 = πΌ1 β π 1, π2 = πΌ2 β π 2
π3 = πΌ3 β π 3
Chizmasi:
πΌ =6
4 + 10 + 16= 0,2π΄
π1 = 0,2 β 4 = 0,8 π
π2 = 0,2 β 10 = 2 π
π3 = 0,2 β 16 = 3,2 π
Javob:
Topish kerak:
πΌ1 =? , πΌ2 =? , πΌ3 =?
π1 =? , π2 =? , π3 =?
πΌ1 = πΌ2 = πΌ3 = 0,2 π΄
π1 = 0,8π, π2 = 2π
π3 = 3,2π
π 1 π 2
π1
πΌ1
π2
πΌ2
π
π 1 π 2
π1
πΌ1
π2
πΌ2
π
π 2
π3
πΌ3
ABDUBANANOV AKRAMJON
30
3-masala. Ikkita elektr lampochka 220 V kuchlanishli tarmoqqa ketma-ket ulangan
boβlib, ulardan 0,5 A tok oβtmoqda. Agar birinchi lampochkaning qarshiligi
ikkinchisinikidan 3 marta katta boβlsa, har bir lampochkadagi kuchlanishni toping.
Berilgan Formulasi Hisoblash
π = 220 π
πΌ = 0,5 π΄
π 1 = 3π 2
πΌ = πΌ1 = πΌ2
πΌ =π
π =
π
π 1 + π 2=
π
4π 2
π = 4π 2πΌ β π 2 =π
4πΌ
π1 = πΌ1 β π 1
π2 = πΌ2 β π 2
π 2 =220
4 β 0,5= 110 ππ
π 1 = 3π 2 = 330 ππ
π1 = 0,5 β 110 = 55 π
π2 = 0,5 β 330 = 165 π
Topish kerak:
π1 =? , π2 =?
Javob:
π1 = 55 π, π2 = 165 π
4-masala. Sxemada berilgan A va B nuqtalar orasidagi kuchlanish qanday boβlgan?
Bunda: π 1 = 5 ππ, π 2 = 10 ππ, π 3 = 15 ππ, π2 = 15 π
Berilgan Formulasi Hisoblash
π 1 = 5 ππ
π 2 = 10 ππ
π 3 = 15 ππ
π2 = 15 π
πΌ = πΌ1 = πΌ2 = πΌ3
πΌ2 =π2
π 2
π = π 1 + π 2 + π 3
π = π β πΌ = π β πΌ2
πΌ2 =15
10= 1,5 π΄
π = 10 + 15 + 5 = 30 ππ
π = 1,5 β 30 = 45 π
π 1 π 2 π 3
π1 π2 π3
πΌ1 πΌ2 πΌ3
π
π΄ π΅
ABDUBANANOV AKRAMJON
31
Topish kerak:
π =?
Javob:
π = 45 π
12-mashq
1-masala. Qarshiliklari 3 Om va 6 Om boβlgan ikkita isteβmolchi parallel ulangan.
Isteβmolchilar ulaangan zanjir qismining toβliq qarshiligini toping.
Berilgan Formulasi Hisoblash
π 1 = 3 ππ
π 2 = 6 ππ
1
π =
1
π 1+
1
π 2
π =π 1 β π 2
π 1 + π 2
π =3 β 6
3 + 6= 2 ππ
Topish kerak:
π =?
Javob:
π = 2 ππ
2-masala. Qarshiliklari 10 Om, 15 Om va 30 Om boβlgan uchta isteβmolchi parallel
ulangan. Isteβmolchilar ulangan zanjir qismining qarshiligini toping.
Berilgan Formulasi Hisoblash
π 1 = 10 ππ
π 2 = 15 ππ
π 2 = 30 ππ
1
π =
1
π 1+
1
π 2+
1
π 3
1
π =
1
10+
1
15+
1
30=
=1 β 3
10 β 3+
1 β 2
15 β 2+
1
30=
6
30
π =30
6= 5 ππ
Topish kerak:
π =?
Javob:
π = 2 ππ
π2 πΌ2
π
π 1
π 2
πΌ1 π1
ABDUBANANOV AKRAMJON
32
3-masala. Uyungizdagi qandilda oβzaro parallel ulangan 5 ta bir xil lampochka
yonib turibdi. Qandil ulangan simda 4 A tok oβtayotgan boβlsa, har bir
lampochkadan oβtayotgan tok kuchini toping.
Berilgan Formulasi Hisoblash
π 1 = π 2 = π 3 = π 4 = π 5
πΌ = 4 π΄
πΌπ π‘πππππβππππ ππππππππ
π’ππππππππ βππ
πππ ππ π‘πβ²ππππβπππππ
ππ’πβππππ βπππ π‘πππ
ππβ²ππππ.
ππππ βπππππππ π‘πππ
ππβ²πππππ π’πβπ’π π‘ππ
ππ’πβπ π‘πππ 5 ππ
ππβ²ππππππ.
πΌ1 = πΌ2 = πΌ3 = πΌ4 = πΌ5 =
=πΌ
5=
4
5= 0,8 π΄
Topish kerak:
πΌ1 = πΌ2 = πΌ3 = πΌ4 = πΌ5 =?
Javob:
0,8 π΄
4-masala. Qarshiligi 40 Om va 60 Om boβlgan ikkita lampochka oβzaro parallel
ulangan. Zanjirning shu qismidagi toβliq qrshiligi qancha boβladi? Agar
lampochkalar uchlaridagi kuchlanish 36 V boβlsa, zanjirdagi toβliq tok kuchini
toping.
Berilgan Formulasi Hisoblash
π 1 = 40 ππ
π 2 = 60 ππ
π = 36 π
π =π 1 β π 2
π 1 + π 2
πΌ =π
π
π =40 β 60
40 + 60= 24 ππ
πΌ =36
24= 1,5 π΄
Topish kerak: Javob:
ABDUBANANOV AKRAMJON
33
π =? πΌ =? π = 24 ππ, πΌ = 1,5 π΄
5-masala. Sxemadagi qarshiligi π 1 = 30 ππ boβlgan oβtkazgchdan πΌ1 = 0,6 π΄ tok
oβtmoqda, qarshiligi π 2 = 10 ππ boβlgan oβtkazgichdan qanday toβk oβtadi?
Berilgan Formulasi Hisoblash
π 1 = 30 ππ
πΌ1 = 0,6 π΄
π 2 = 10 ππ
π1 = π2
πΌ1 β π 1 = πΌ2 β π 2
πΌ2 =πΌ1 β π 1
π 2
πΌ2 =0,6 β 30
10= 1,8 π΄
Topish kerak:
πΌ2 =?
Javob:
πΌ2 = 1,8 π΄
6-masala. Rasmda keltirilgan elektr oβlchov asboblaridan qaysi biri ampermet va
qaysi biri voltmeter. Javobingizni asoslang.
13-mashq.
π2 πΌ2
π
π 1
π 2
πΌ1 π1
π2 πΌ2
π
π 1
π 2
πΌ1 π1
1 2 3
1 voltmetr chunki u isteβmolchiga parallel ulangan.
2 ampermetr chunki u isteβmolchiga ketma-ket
ulangan.
3 voltmetr chunki u isteβmolchiga parallel ulangan.
ABDUBANANOV AKRAMJON
34
1-masala. Rasmda tasvirlangan elektr zanjirining A va B nuqtalar orasidagi toβla
qarshiligini hisoblang. Har bir rezistorning elektr qarshiligi 4 Om ga teng.
2-masala. Qarshiliklari 20 Om va 80 Om boβlgan ikkita oβtkazgich oβzaro parallel
ulangan boβlib, ularning uchlaridagi kuchlanish 48 V ga teng. Ularga ketma-ket
ulangan uchunchi 5 Om qarshilikdagi tok kuchi va kuchlanishni toping.
Berilgan Formulasi Hisoblash
π 1 = 20 ππ
π 2 = 80 ππ
π 3 = 5 ππ
π1 = π2 = 48 π
π1 = π2
πΌ1 =π1
π 1, πΌ2 =
π2
π 2
πΌ3 = πΌ1 + πΌ2
π3 = πΌ3 β π 3
πΌ1 =48
20= 2,4 π΄
πΌ2 =48
80= 0,6 π΄
πΌ3 = 2,4 + 0,6 = 3 π΄
π3 = 3 β 5 = 15 π
Topish kerak: Javob:
A B R R R
R
1-ish. Ketma-ket qoβshamiz
π 1 = π + π + π = 3π
Hosil boβldi
A B
3R
R
2-ish. Parallel qoβshamiz
π π’π =3π β π
3π + π =
3
4π =
3 β 4
4= 3 ππ
Sodda zanjir sxemasini tuzamiz
π 1
π 2
π 3
πΌ1
πΌ2
πΌ3 π1
π2
π3
ABDUBANANOV AKRAMJON
35
πΌ3 =?, π3 =?
πΌ3 = 3 π΄, π3 = 15 π
3-maala. Sxemada berilgan rezistorning elaktr qarshiligi π 1 = 4 ππ, π 2 = 10 ππ
va π 3 = 15 ππ ga teng. Agar zanjir uchlariga 12 V kuchlanish berilsa, ampermetr
qanday qiymatni koβrsatadi?
Berilgan Formulasi Hisoblash
π 1 = 4 ππ
π 2 = 10 ππ
π 3 = 15 ππ
π = 12 π
π2 = π3
π = π1 + π2 = π1 + π3
π 3 π£π π 2 ππππππππ
π =π 2π 3
π 2 + π 3
πππ‘πππ π , π 1 ππ πππ‘ππ β πππ‘
π π’π = π 1 + π
πΌπ΄ = πΌ1 =π
π π’π
π =10 β 15
10 + 15= 6 ππ
π π’π = 4 + 6 = 10 ππ
πΌπ΄ =12
10= 1,2 π΄
Javob:
πΌ3 = 3 π΄, π3 = 15 π
Topish kerak:
πΌπ΄ =?
14-mashq
1-masala. Yassi kondensator qoplamalari orasidagi dielektrik singdiruvchanligi π =
2,1 boβlgan dielektrik bilan toβdirilsa, uning sigβimi qanday oβzgaradi?
π 1
π 3 π 2 π
π3 π2
πΌ3 πΌ2 π1
πΌ1 π΄
ABDUBANANOV AKRAMJON
36
Berilgan Formulasi Hisoblash
π1 = 1
π2 = 2,1
πΆ1 =π1π0π
π, πΆ2 =
π2π0π
π
πΆ2
πΆ1=
π2
π1
πΆ2
πΆ1=
2,1
1= 2,1
2,1 marta ortadi
Topish kerak:
πΆ2
πΆ1=?
Javob:
πΆ2
πΆ1= 2,1
2-masala. 24 V kuchlanishli tok manbaiga ulangan kondensator 30 Β΅C zaryad olgan
boβlsa, kondensator sigβimini aniqlang.
Berilgan Formulasi Hisoblash
π = 24 π
π = 30 ππΆ = 30 β 10β6πΆ
πΆ =π
π
chizmasi
πΆ =30 β 10β6
24= 1,25 β 10β6πΉ
Topish kerak:
πΆ =?
Javob:
πΆ = 1,25 β 10β6πΉ
3-masala. Sigβimi 40 nF boβlgan kondensator qolamalaridagi tok manbaidan 30 V
kuchlanish berilganda u qanday miqdordagi zaryad oladi?
Berilgan Formulasi Hisoblash
π = 30 π
πΆ = 40 ππΆ = 40 β 10β9πΉ
πΆ =π
π
π = πΆπ
π = 30 β 40 β 10β9 = 1200 β 10β9πΆ
Topish kerak:
π =?
Javob:
π = 1200 ππΆ = 1,2 ππΆ
4-masala. Yuzasi 40 sm2 boβlgan kondensator qoplamalari bir-biridan 8 mm
qlinlikdagi havo bilan ajratilgan. Kondensatorning sigβimi nimaga ten
π1
πΆ1
π2
πΆ2
q
πΆ
ABDUBANANOV AKRAMJON
37
Berilgan Formulasi Hisoblash
π = 40 π π2 = 40 β 10β4π2
π = 8 ππ = 8 β 10β3π
π = 1, π0 = 8,85 β 10β12πΉ/π
πΆ =ππ0π
π
πΆ =1 β 8,85 β 10β12 β 40 β 10β4
8 β 10β3
= 44,25 β 10β13πΉ
Topish kerak:
πΆ =?
Javob:
πΆ = 4,425 ππΉ
5-masala. Sigβimi 3 Β΅F boβlgan kondensator qoplamalari olgan zaryad miqdori 42
Β΅C ga teng boβlsa, uning qoplamalari orasidagi kuchlanish nimaga teng?
Berilgan Formulasi Hisoblash
π = 42 ππΆ = 42 β 10β6πΆ
πΆ = 3 ππΆ = 3 β 10β6πΉ
πΆ =π
π
π =π
πΆ
π =42 β 10β6
3 β 10β6= 14 π
Topish kerak:
π =?
Javob:
π = 14 π
15-mashq
1-masala. Sigβimi 3 Β΅F, 5 Β΅F va 8 Β΅F boβlgan uchta kondenstorlar 12 V kuchlanishli
tok manbaiga oβzaro parallel ulangan. Zanjirning umumiy sigβimi qanday boβladi?
ularning har biri qanday zaryad oladi?
Berilgan Formulasi Hisoblash
πΆ1 = 3 ππΉ = 3 β 10β6πΉ
πΆ2 = 5 ππΉ = 5 β 10β6πΉ
πΆ3 = 8 ππΉ = 8 β 10β6πΉ
π = 12 π
πΆ = πΆ1 + πΆ2 + πΆ3
π = π1 = π2 = π3
π1 = πΆ1π1
π2 = πΆ2π2
πΆ = (3 + 5 + 8) β 10β6
= 16 β 10β6πΉ
π1 = 3 β 10β6 β 12
= 36 β 10β6πΆ
ABDUBANANOV AKRAMJON
38
π3 = πΆ3π3 π2 = 5 β 10β6 β 12
= 60 β 10β6πΆ
π3 = 8 β 10β6 β 12
= 96 β 10β6πΆ
Topish kerak:
πΆ =?, π1 =? , π2 =
? , π3 =?,
Javob:
πΆ = 16 ππΉ, π1 = 36 ππΆ,
π2 = 60 ππΆ, π3 = 96ππΆ
2-masala. Sigβimi 12 Β΅F, 20 Β΅F va 30 Β΅F boβlgan kondensatorni oβzaro ketma-ket
ulab qanday sigβim olish mumkin?
Berilgan Formulasi Hisoblash
πΆ1 = 12 ππΉ = 12 β 10β6πΉ
πΆ2 = 20 ππΉ = 20 β 10β6πΉ
πΆ3 = 30 ππΉ = 30 β 10β6πΉ
1
πΆ=
1
πΆ1+
1
πΆ2+
1
πΆ3
1
πΆ=
1
12+
1
20+
1
30=
=1 β 5
60+
1 β 3
60+
1 β 2
60
1
πΆ=
10
60β πΆ = 6 ππΉ
Topish kerak:
πΆ =?
Javob:
πΆ = 6 ππΉ
3-masala. Sigβimlari bir xil boβlgan ikkita kondensator avval ketma-ket, soβngra
parallel ulandi. Parallel ulangan holdagi umumiy sigβim ketma-ket ulangandagidan
necha marta farq qiladi?
Berilgan Formulasi Hisoblash
πΆ1 = πΆ2 = πΆ
πΆπ =πΆ1πΆ2
πΆ1 + πΆ2
πΆπ = πΆ1 + πΆ2
πΆπ
πΆπ=
(πΆ + πΆ)2
πΆ β πΆ= 4
Javob:
πΆ1
π1 π1 πΆ2
π2 π2 πΆ3
π3 π3
π
ABDUBANANOV AKRAMJON
39
πΆπ
πΆπ=
(πΆ1 + πΆ2)2
πΆ1πΆ2
Topish kerak:
πΆπ
πΆπ=?
πΆπ
πΆπ= 4
4 marta ortadi
4-masala. Sigβimi C1=4 Β΅F, C2=6 Β΅F va C3=10 Β΅F boβlgan kondensatorlarni bir-
biriga ulash orqali 5 Β΅F sigβim olish mumkinmi? Mumkin boβlsa qanday?
Berilgan Formulasi Hisoblash
πΆ1 = 4 ππΉ = 4 β 10β6πΉ
πΆ2 = 6 ππΉ = 6 β 10β6πΉ
πΆ3 = 10 ππΉ = 10 β 10β6πΉ
πΆ = 5 ππΉ = 5 β 10β6πΉ
πΆ1 π£π πΆ2 ππ ππππππππ
π’πππ¦πππ§
πΆ1π’π = πΆ1 + πΆ2
πΆ1π’π ππ πΆ3 ππ πππ‘ππ
β πππ‘ π’πππ¦πππ§
πΆ =πΆ1π’π β πΆ3
πΆ1π’π + πΆ3
πΆ1π’π = (4 + 6) = 10ππΉ
πΆ =10 β 10
10 + 10= 5ππΉ
Topish kerak:
πΆ =?
16-mashq
1-masala. Yuzalari 30 sm2 dan boβlgan yassi kondensator qoplmalari orasidagi
masofa 4 mm ga teng. Agar kondensatorning sigβimi 20 pF boβlsa, kondensator
qopolamalari orasidagi muhitning dielektrik singdiruvchanligi nimaga teng?
Berilgan Formulasi Hisoblash
π = 30 π π2 = 30 β 10β4π2
π = 4ππ = 4 β 10β3π πΆ =
ππ0π
π
π =20 β 10β12 β 4 β 10β3
8,85 β 10β12 β 30 β 10β4
= 3
ABDUBANANOV AKRAMJON
40
πΆ = 20 ππΉ = 20 β 10β12πΉ π =
πΆπ
π0π
chizmasi
Topish kerak:
π =?
Javob:
π = 3
2-masala. Yassi kondensatorning doira shaklidagi radiusi 4 sm boβlgan qoplamalari
bir-biridan 2 mm qalinlikdagi sluda bilan ajratilgan. Kondensator qoplamalariga 4
V kuchlanish berilsa, kondensator qanday zaryad oladi? π = 6
Berilgan Formulasi Hisoblash
π = 4 π π = 4 β 10β2π
π = 2 π = 4 β 10β3π
π = 2 ππ = 2 β 10β3π
π = 4 π
π = 6
πΆ =ππ0π
π
π = ππ2
π = πΆπ
π =ππ0ππ2π
π
π =8,85 β 10β12 β 6 β 3,14 β 16 β 10β6 β 4
2 β 10β3
= 5335 β 10β15πΆ
Topish kerak:
π =?
Javob:
π = 5335 β 10β15πΆ = 5,335ππΆ
3-masala. Sigβimi 370 pF boβlgan yassi kondensator qoplamalarining yuzasi 300
sm2 ga teng. Qoplamalar orasiga shisha plastina qoβyilgan boβlsa, uning qalinligi
qanday boβlgan? Shisha uchun π = 7
Berilgan Formulasi Hisoblash
πΆ = 370 ππΉ = 370 β 10β12πΉ
π = 300 π π2 = 300 β 10β4πβ4 πΆ =
ππ0π
π
π =7 β 8,85 β 10β12 β 300 β 10β4
370 β 10β12=
= 50 β 10β4π
S
πΆ
π
π
π
ABDUBANANOV AKRAMJON
41
π = 7 π =
ππ0π
πΆ
Topish kerak:
π =?
Javob:
π = 50 β 10β4π = 5 ππ
4-masala. Qutichada 30 pF va 70 pF sigβimli bir nechta kondensatorlar bor. har
qaysi sigβimli kondensatordan nechtadan olib, ularni parallel ulash orqali 330 pF
sigβimli kondensatorlar batareyasini hosil qilish mumkin?
Berilgan Formulasi Hisoblash
πΆ1 = 30 ππΉ = 30 β 10β12πΉ
πΆ2 = 70 ππΉ = 70 β 10β12πΉ
πΆ = 330 ππΉ = 330 β 10β12πΉ
70 ππΉ π ππβ²ππππ
πππππππ ππ‘ππππ
π = 3 π‘π
30 ππΉ π ππβ²ππππ
πππππππ ππ‘ππππ
π = 4 π‘π
ππππππ§
πΆ1π’π = ππΆ2
πΆ1π’π = 3 β 70 = 210 ππΉ
πΆ2π’π = ππΆ1
πΆ1π’π = 4 β 30 = 120 ππΉ
πΆ = 210 + 120 = 330 ππΉ
Topish kerak:
π =? , π =?
III BOB. ELEKTR TOKINING ISHI VA QUVVATI
17-mashq
1-masala. 220 V kuchlanish tarmogβiga ulangan dvigateldan 2 A tok oβtmoqda. Bu
dvigatelda 20 minut davomida tok qanday ish bajaradi?
Berilgan Formulasi Hisoblash
π = 220 π π΄ = πΌππ‘ π΄ = 2 β 220 β 1200 =
ABDUBANANOV AKRAMJON
42
πΌ = 2 π΄
π‘ = 20 ππππ’π‘ = 1200 π
= 528000 π½
Topish kerak:
π΄ =?
Javob:
π΄ = 528000 π½ = 528 ππ½
2-masala. 12 V kuchlnishga ulangan oβtkazgichdan 20 mA tok oβtmoqda. Tok 15
minut davomida qanday ish bajaradi?
Berilgan Formulasi Hisoblash
π = 12 π
πΌ = 20 ππ΄ = 20 β 10β3π΄
π‘ = 15 ππππ’π‘ = 900 π
π΄ = πΌππ‘
π΄ = 20 β 10β3 β 12 β 900 =
= 24 π½
Topish kerak:
π΄ =?
Javob:
π΄ = 24 π½
3-masala. Qarshiligi 200 Om boβlgan oβtkazgich uchlariga 42 V kuchlanish
berilgan. 20 minut davomida tok qanday ish bajaradi?
Berilgan Formulasi Hisoblash
π = 200 ππ
π = 42 π
π‘ = 20 ππππ’π‘ = 1200 π
π΄ = πΌππ‘
πΌ =π
π
π΄ =π2π‘
π
π΄ =422 β 1200
200= 10584 π½
Topish kerak:
π΄ =?
Javob:
π΄ = 10584 π½ = 10,584 ππ½
4-masala. Lampochkaadagi kuchlanish 4,5 V, tok kuchi 0,2 A boβlsa, 5 minut
davomida qanch elektr energiya sarflanadi?
ABDUBANANOV AKRAMJON
43
Berilgan Formulasi Hisoblash
πΌ = 0,2 π΄
π = 4,5 π
π‘ = 5 ππππ’π‘ = 300 π
π΄ = πΌππ‘
π΄ = 0,2 β 4,5 β 300 = 270 π½
Topish kerak:
π΄ =?
Javob:
π΄ = 270 π½
5-masala. Elektr dazmol 220 V kuchlanishli tok tarmogβiga ulanganda undan 3 A
tok oβtadi. Dazmol 10 minut ishlaganda qancha elektr energiya sarflanadi?
Berilgan Formulasi Hisoblash
πΌ = 3 π΄
π = 220 π
π‘ = 10 ππππ’π‘ = 600 π
π΄ = πΌππ‘
π΄ = 3 β 220 β 600 = 396000 π½
Topish kerak:
π΄ =?
Javob:
π΄ = 39600 π½ = 396 ππ½
18-mashq
1-masala. 220 V kuchlanish va 4 A tok kuchida ishlayotgan dvigatelning isteβmol
quvvatini toping.
Berilgan Formulasi Hisoblash
πΌ = 4 π΄
π = 220 π
π = πΌπ
π = 4 β 220 = 880 π
Topish kerak:
π =?
Javob:
π = 880 π
ABDUBANANOV AKRAMJON
44
2-masala. 40 W quvvatli avtomobil lampochkas 12 V kuchlanishga moβljallangan.
Lampochkaning qarshiligini aniqlang.
Berilgan Formulasi Hisoblash
π = 40 π
π = 12 π
π = πΌπ
πΌ =π
π
π =π2
π β π =
π2
π
π =122
40= 3,6 ππ
Topish kerak:
π =?
Javob:
π = 3,6 ππ
3-masala. Zanjirdan 5 A tok oβtganda elektr plitasi 30 minut davomida 1800 kJ
energiya sarflaydi. Plitaning qarshiligi qanday boβlgan?
Berilgan Formulasi Hisoblash
πΌ = 5π΄
πΈ = 1800ππ½ = 1800 β 103π½
π‘ = 30 ππππ’π‘ = 1800 π
πΈ = πΌππ‘
π = πΌπ
πΈ = πΌ2π π‘ β π =πΈ
πΌ2π‘
π =1800 β 103
52 β 1800=
= 40 ππ
Topish kerak:
π =?
Javob:
π = 40 ππ
4-masala. Xonadondagi elektrhisoblagich oy boshida koβrsatgan raqami 1450
kWΒ·h, oy oxirida esa 1890 kWΒ·h boβldi. Xonadonda bir oy davomida qancha elektr
energiyasi sarflangan?
Berilgan Formulasi Hisoblash
πΈ1 = 1450 ππ β β
πΈ2 = 1890 ππ β β
βπΈ = πΈ2 β πΈ1
βπΈ = 1890 β 1450 =
= 440 ππ β β
ABDUBANANOV AKRAMJON
45
Topish kerak:
βπΈ =?
Javob:
βπΈ = 440 ππ β β
5-masala. 220 V kuchlanishga ulangan lampochkadan 0,4 A tok oβtmoqda. Tok 10
minut davomida qancha ish bajaradi?
Berilgan Formulasi Hisoblash
π = 220 π
πΌ = 0,4 π΄
π‘ = 10 ππππ’π‘ = 600 π
π΄ = πΌππ‘
π = πΌπ
π΄ = 0,4 β 220 β 600 = 52800 π½
Topish kerak:
π΄ =?
Javob:
π΄ = 52800 π½ = 52,8 ππ½
6-masala. Quvvati 10 W va 15 W boβlgan ikkita lamopochka parallel ulanib, 220 V
kuchlanishli tarmoqqa ulangan. Har bir lampochka choβgβlanma tolasining
qarshiligini aniqlang. Lampochkalarning har bridan qanday tok oβtadi?
Berilgan Formulasi Hisoblash
π1 = 10 π
π2 = 15 π
π = 220 π
ππππππππ π’ππππππππ
ππππππβπππππππ
220 π ππ’πβπππππ β
ππβ²ππππ
π1 =π2
π 1β π 1 =
π2
π1β π 2 =
π2
π2
π 1 =2202
10= 4840 ππ
π 2 =2202
15= 3227 ππ
πΌ1 =π
π 1=
220
4840= 0,045 π΄
πΌ2 =π
π 2=
220
3227= 0,068 π΄
Topish kerak: Javob:
ABDUBANANOV AKRAMJON
46
π 1 =?, π 2 =?
πΌ1 =?, πΌ2 =?
π 1 = 4840 ππ, π 2 = 3227 ππ
πΌ1 = 0,045 π΄, πΌ2 = 0,068 π΄
7-masala. Zanjirning kuchlanish 220 V boβlgan qismida tok 176 kJ ish bajardi. Shu
vaqt davomida oβtkazgich koβndalang kesimidan qancha elektron oqib oβtgan?
Berilgan Formulasi Hisoblash
π = 220 π
π΄ = 176 ππ½ = 176 β 103π½
π = 1,6 β 10β19πΆ
π΄ = πΌππ‘
πΌ =π
π‘=
ππ
π‘
π΄ =ππππ‘
π‘
π =π΄
ππ
π =176 β 103
1,6 β 10β19 β 220=
= 0,5 β 1022
Topish kerak:
π =?
Javob:
π = 5 β 1021 π‘π
8-masala. Qarshiligi 120 Om va 160 Om boβlgan isteβmolchilar zanjirga parallel
ulangan. Ikkinchi isteβmolchi 15 W quvvat bilan ishlayotgan boβlsa, birinchi
isteβmolchi qanday quvvat bilan ishlaydi?
Berilgan Formulasi Hisoblash
π 1 = 120 ππ
π 2 = 160 ππ
π2 = 15 π
π2 =π2
2
π 2β π2 = βπ2 β π 2
π1 = π2
π1 =π1
2
π 1=
π22
π 1
π2 = β15 β 160 = β2400 π
π1 =2400
120= 20 π
Topish kerak:
π1 =?
Javob:
π1 = 20 π
ABDUBANANOV AKRAMJON
47
9-masala. Qarshiligi 30 Om va 75 Om boβlgan isteβmolchilar zanjirga parallel
ulangan. Ikkinchi isteβmolchi 25 W quvvat bilan ishlayotgan boβlsa, birinchi
isteβmolchi qanday quvvat bilan ishlaydi?
Berilgan Formulasi Hisoblash
π 1 = 30 ππ
π 2 = 75 ππ
π2 = 25 π
π2 =π2
2
π 2β π2 = βπ2 β π 2
π1 = π2
π1 =π1
2
π 1=
π22
π 1
π2 = β25 β 75 = β1875 π
π1 =1875
30= 62,5 π
Topish kerak:
π1 =?
Javob:
π1 = 62,5 π
10-masala. Ikki isitkich yordamida suvni isitish kerak. isitkichlar ketma-ket
ulanganda suv tezroq isiydimi yoki parallel ulangandami? Javobingizni izohlang.
Javob: isitkichlar ketma-ket ulanganda ulardagi kuchlanish 220 V dan
boβlmaydi shu sababli ular toβla quvvat bilan ishlamaydilar. Isitkichlar parallel
ulanganda ularning har birida 220 V dan kuchlanish boβladi. shu sababli ular toβla
quvvati bilan ishlaydilar. Parallel ulanganda tezroq isiydi.
19-mashq
1-masala. Qarshiligi 100 Om boβlgan sim spiraldan 10 A tok oβtmoqda. Shu
spiraldan 1 minut davomida qancha issiqlik ajralib chiqadi?
Berilgan Formulasi Hisoblash
π = 100 ππ
πΌ = 10 π΄
π‘ = 1 π = 60 π
π = πΌ2π π‘ π = 102 β 100 β 60 = 60 β 104π½
Topish kerak: Javob:
ABDUBANANOV AKRAMJON
48
π =? π = 60 β 104π½ = 600 ππ½
2-masala. 220 V kuchlanishli tarmoqqa ulangan 20 Om qarshilikli elektrisitkichdan
1 soatda qancha issiqlik ajralib chiqadi?
Berilgan Formulasi Hisoblash
π = 20 ππ
π = 220 π
π‘ = 1 π πππ‘ =
= 3600 π
π = πΌ2π π‘
πΌ =π
π
π =π2π‘
π
π =2202 β 3600
20= 8712000 π½
Topish kerak:
π =?
Javob:
π = 8712000 π½ = 8,712 ππ½
3-masala. Tok manbai zanjiriga koβndalang kesimi va uzunligi bir xil boβlgan
alyuminiy va nixrom sim ketma-ket ulangan. Ulardan qaysi biri koβproq qiziydi?
Berilgan Formulasi Hisoblash
π1 = π2
π1 = π2
ππ = 0,028 β 10β6ππ β π
ππ = 1,1 β 10β6ππ β π
π = πΌ2π π‘
π = ππ
π
π = πΌ2ππ
ππ‘
Ketma-ket ulanganda I=const
boβladi.
ππ
ππ=
ππ
ππ=
1,1 β 10β6
0,028 β 10β6= 39
Topish kerak:
ππ π£π ππ
Javob:
ππ = 39ππ πππ₯πππ
4-masala. Dazmolning spirali koβndalang kesimining yuzasi 0,2 mm2 va uzunligi
2,5 m li nixromdan tayyorlangan. Dazmol 220 V ga moβljallangan boβls, uning
quvvati qanchaga teng?
Berilgan Formulasi Hisoblash
ABDUBANANOV AKRAMJON
49
π = 0,2 ππ2 = 0,2 β 10β6π2
π = 2,5 π
π = 1,1 β 10β6ππ β π
π = 220 π
π = πΌπ
πΌ =π
π
π = ππ
π
π =π2π
ππ
π =2202 β 0,2 β 10β6
1,1 β 10β6 β 2,5=
= 3520 π
Topish kerak:
π =?
Javob:
π = 3520 π
5-masala. 50 Om qrshilikli oβtkazgich orqali 10 minut davomida qanday tok
oβtkazilganda, 120 kJ issiqlik ajralib chiqadi?
Berilgan Formulasi Hisoblash
π = 50 ππ
π‘ = 10 π = 600 π
π = 120ππ½ = 120 β 103π½
π = πΌ2π π‘
πΌ = βπ
π π‘
πΌ = β120 β 103
50 β 600= 2 π΄
Topish kerak:
πΌ =?
Javob:
πΌ = 2 π΄
20-mashq
1-masala. 220 V kuchlanish tarmogβiga ulangan elektr choynak 1,1 kW isteβmol
quvvatiga ega. Choynak tarmoqqa ulanganda undan qancha tok oβtadi?
Berilgan Formulasi Hisoblash
π = 220 π
π = 1,1 ππ = 1,1 β 103 π
π = πΌπ
πΌ =π
π
πΌ =1,1 β 103
220= 5 π΄
ABDUBANANOV AKRAMJON
50
Topish kerak:
πΌ =?
Javob:
πΌ = 5 π΄
2-masala. Qarshiligi 50 Om boβlgan sim spiralidan 4 A tok oβtmoqda. Shu spiraldan
2 soat davomida qancha issiqlik miqdori ajralib chiqadi?
Berilgan Formulasi Hisoblash
π = 50 ππ
πΌ = 4 π΄
π‘ = 2 π πππ‘ = 7200 π
π = πΌ2π π‘
π = 42 β 50 β 7200 =
= 5760000 π½
Topish kerak:
π =?
Javob:
π = 5,76 ππ½
3-masala. 220 V kuchlanishli tarmoqqa ulangan 60 Om qarshilikli elektrisitkichdan
1 soatda qancha issiqlik miqdori ajralib chiqadi?
Berilgan Formulasi Hisoblash
π = 220 π
π = 60 ππ
π‘ = 1 π πππ‘ = 3600 π
π = πΌ2π π‘
πΌ =π
π
π =π2π‘
π
π =2202 β 3600
60= 2904000 π½
Topish kerak:
π =?
Javob:
π = 2,904 ππ½
4-masala. 2,2 kW quvvatli elektrisitkich 220 V kuchlanishli tarmoqqa ulangan.
Undan qancha tok oβtadi?
Berilgan Formulasi Hisoblash
ABDUBANANOV AKRAMJON
51
π = 2,2 ππ = 2200 π
π = 220 π
π = πΌπ
πΌ =π
π
πΌ =2200
220= 10 π΄
Topish kerak:
πΌ =?
Javob:
πΌ = 10 π΄
5-masala. Dazmol spiralining koβndalang kesimining yuzasi 0,1 mm2 va uzunligi 2
m li nixromdan tayyorlangan. Dazmol 220 V ga moβljallangan boβlsa, uning quvvati
qanchaga teng?
Berilgan Formulasi Hisoblash
π = 0,1 ππ2 = 0,1 β 10β6π2
π = 220 π
π = 2 π
π = 1,1 β 10β6ππ β π
π = πΌπ
πΌ =π
π =
ππ
ππ
π =π2π
ππ
π =2202 β 0,1 β 10β6
1,1 β 10β6 β 2=
= 2200 π
Topish kerak:
π =?
Javob:
π = 2200 π = 2,2 ππ
6-masala. Qarshiligi 200 Om va 300 Om boβlgan ikkita elektrisitkichlar tok
tarmogβiga parallel ulangan. Ulardan aynan bir vaqtda ajralib chiqqan issiqlik
miqdorlarini taqqoslang.
Berilgan Formulasi Hisoblash
π 1 = 200 ππ
π 2 = 300 ππ
π‘1 = π‘2
π1 =π1
2π‘
π 1
π2 =π2
2π‘
π 2
ππππππππππ π = ππππ π‘
π1
π2=
π 2
π 1=
300
200= 1,5
Topish kerak: Javob:
ABDUBANANOV AKRAMJON
52
π1 π£π π2 π1 = 1,5π2
7-masala. 220 V ga mβljallangan elektr choynakning isteβmol quvvati 500 W ga
teng. Choynak tarmoqqa ulanganda undan qancha tok oβtadi va uning elektr
qarshiligi nimaga teng?
Berilgan Formulasi Hisoblash
π = 220 π
π = 500 π
π = πΌπ
πΌ =π
π
π =π
πΌ
πΌ =500
220= 2,27 π΄
π =220
2,27= 97 ππ
Topish kerak:
πΌ =? , π =?
Javob:
πΌ = 2,27 π΄, π = 97 ππ
8-masala. Elektr dvigateliga ulangan simdan 0,5 A tok oβtmoqda, undagi kuchlanish
20 V. dvigatel 1 soatda qancha ish bajaradi? Dvigatelning FIK 80 % ga teng.
Berilgan Formulasi Hisoblash
π = 20 π
πΌ = 0,5 π΄
π‘ = 1 π πππ‘ 3600 π
Ξ·=0,8 %
π =π΄
πΌππ‘
π΄ = ππΌππ‘
π΄ = 0,8 β 0,5 β 20 β 3600 =
= 28800 π½
Topish kerak:
π΄ =?
Javob:
π΄ = 28,8 ππ½
9-masala. Qarshiligi 50 Om va 16 Om boβlgan isteβmolchilar ketma-ket ulangan.
Ikkinchi isteβmolchidan maβlum vaqt ichida 400 J ish bajarganda birinchi
isteβmolchi qancha ish bajaradi?
ABDUBANANOV AKRAMJON
53
Berilgan Formulasi Hisoblash
π 1 = 50 ππ
π 2 = 16 ππ
π΄2 = 400 π½
π΄1 = πΌ12π 1π‘
π΄2 = πΌ22π 2π‘
π΄1
π΄2=
π 1
π 2
π΄1
400=
50
16β π΄1 = 1250 π½
Topish kerak:
π΄1 =?
Javob:
π΄1 = 1250 π½
21-mashq
1-masala. Xonadon elektr zanjiriga 2 ta 100 W quvvatli, 2 ta 150 W quvvatli
lampochkalar, 100 W quvvatli muzlatkich, 300 W quvvatli televizor, 1,5kW
quvvatli dazmol, 2kW quvvatli elektr isitkich bir vaqtda ulanishi mumkin. Shunday
quvvatli elektr asboblar oladigan tok kuchiga bardosh berishi uchun tarmoqqa
ulanadigan mis simning koβndalang kesimi yuzasi kamida qancha boβlishi kerak?
Javob: bu elektr asboblar barchasi parallel ulangan. Shu bois ularda 220 V
kuchlanish mavjud boβladi.
Umumiy quvvat: π = 2 β 100 + 2 β 150 + 100 + 300 + 1500 + 2000 = 4400 π
π =π2
π =
π2π
ππβ π =
πππ
π2
Hisoblash: π =4400β0,017β10β6βπ
2202 masala yechimi uchun sim uzunligi yetishmaydi.
2-masala. Odam tanasining oβrtacha qarshiligi taxminan 10 kOm. Agar odam nam
yerda turib 42 V kuchlanishli ochiq simni bexosdan uslab olsa, undan qancha tok
oβtadi?
Berilgan Formulasi Hisoblash
π = 10 πππ = 10 β 103 ππ πΌ =π
π πΌ =
42
10000= 42 β 10β4π΄
ABDUBANANOV AKRAMJON
54
π = 42 π
Topish kerak:
πΌ =?
Javob:
πΌ = 4,2 ππ΄
3-masala. 220 V kuchlanishga moβljallangan 400 W quvvatli televizorga qoβyilgan
eruvchan saqlagichga 2 A deb yozilgan. Baβzida tarmoqdagi kuchlanish 220 V dan
ortib ketadi? Tarmoqdagi kuchlanish qanchaga yetganda eruvchan saqlagich erib
ketadi?
Berilgan Formulasi Hisoblash
π = 220 π
π = 400 π
πΌ = 2 π΄
π =π2
π
π =π2
π
ππ₯ = πΌπ
π =2202
400= 121 ππ
ππ₯ = 2 β 121 = 242 π
Topish kerak:
ππ₯ =?
Javob:
ππ₯ = 242 π πππ πππ‘ππ πππ‘π π
4-masala. Qarshiligi 4,5 Om va 6 Om boβlgan isteβmolchilar oβzaro parallel
ulangan. Zanjirdagi birinchi isteβmolchidan maβlum vaqt davomida 30 J issiqlik
miqdori ajralganda, ikkinchi isteβmolchidan shu vaqt davomida qanday issiqlik
miqdori ajralib chiqadi?
Berilgan Formulasi Hisoblash
π 1 = 4,5 ππ
π 2 = 6 ππ
π1 = 30 π½
π‘1 = π‘2
π1 =π2π‘1
π 1
π2 =π2π‘2
π 2
π2
π1=
π 1
π 2
π2
30=
4,5
6
π2 = 22,5 π½
ABDUBANANOV AKRAMJON
55
Topish kerak:
π2 =?
Javob:
π2 = 22,5 π½
5-masala. Qarshiligi 12 Om va 15 Om boβlgan isteβmolchilar oβzaro ketma-ket
ulangan. Zanjirdagi birinchi isteβmolchidan 8 J issiqlik miqdori ajralganda, ikkinchi
isteβmolchidan qanday issiqlik miqdori ajralib chiqadi?
Berilgan Formulasi Hisoblash
π 1 = 12 ππ
π 2 = 15 ππ
π1 = 8 π½
π‘1 = π‘2
π1 = πΌ2π 1π‘
π2 = πΌ2π 2π‘
π2
π1=
π 2
π 1
π2
8=
15
12
π2 = 10 π½
Topish kerak:
π2 =?
Javob:
π2 = 10 π½
IV BOB. TURLI MUHITLARDA ELEKTR TOKI
22-mashq
1-masala. Mis kuprosining suvdagi eritmasidan iborat boβlgan elektrolitdan 12,5 C
zaryad oβtdi. Elektrolitga botirilgan katodda qancha miqdorda mis yigβilgan?
Berilgan Formulasi Hisoblash
π = 12,5 πΆ
π = 0,329ππ
πΆ= 329 β 10β9
ππ
πΆ
π = ππ
π = 329 β 10β9 β 12,5 =
= 4112,2 β 10β9ππ
Topish kerak:
π =?
Javob:
π = 4112,2 β 10β9ππ
ABDUBANANOV AKRAMJON
56
2-masala. Elektroliz vaqtida katodda 10 mg miqdorda kumush yigβilishi uchun
kumush ionlari boβlgan elektrolitdan qancha zaryad oβtishi kerak?
Berilgan Formulasi Hisoblash
π = 10 ππ = 10 β 10β6ππ
π = 1,118ππ
πΆ= 1118 β 10β9
ππ
πΆ
π = ππ
π =π
π
π =10 β 10β6
1118 β 10β9= 8,9 πΆ
Topish kerak:
π =?
Javob:
π = 30,39 πΆ
3-masala. 1,5 soat davom etgan elektrolizda katodda 15 mg nikel yigβildi.Elektroliz
vaqtida elektrolitdan oβtgan tok kuchini toping.
Berilgan Formulasi Hisoblash
π = 15 ππ = 15 β 10β6ππ
π = 0,304ππ
πΆ= 304 β 10β9
ππ
πΆ
π‘ = 1,5 π πππ‘ = 5400 π
π = ππ
π =π
π
πΌ =π
π‘
π =15 β 10β6
304 β 10β9= 49 πΆ
πΌ =49
5400= 0,009 π΄
Topish kerak:
πΌ =?
Javob:
πΌ = 0,009 π΄ = 9 ππ΄
4-masala. Elektrolitik vannadan 20 minut davomida kuchi 1,6 A boβlgan tok oβtib
turganda, katodda massasi 0,632 g mis ajralib chiqdi. Ushbu natijalar asosida
misning elektrokimiyoviy ekvivalentini hisoblang.
Berilgan Formulasi Hisoblash
ABDUBANANOV AKRAMJON
57
π = 0,632 π = 632 β 10β6ππ
π‘ = 20 π = 1200 π
πΌ = 1,6 π΄
π = ππ
π = πΌπ‘
π = ππΌπ‘
π =π
πΌπ‘
π =632 β 10β6
1,6 β 1200
= 0,329 β 10β6 ππ/πΆ
Topish kerak:
π =?
Javob:
π = 0,329 ππ/πΆ
23-mashq
1-masala. 2 soat davom etgan katodda 20 mg nikel yigβilgan boβlsa, elektroliz
vaqtida elektrolitdan oβtgan tok kuchi qanday boβlgan?
Berilgan Formulasi Hisoblash
π = 20 ππ = 20 β 10β6ππ
π‘ = 2 π πππ‘ = 7200 π
π = 0,304ππ
πΆ= 304 β 10β9
ππ
πΆ
π = ππ
π = πΌπ‘
π = ππΌπ‘
πΌ =π
ππ‘
πΌ =20 β 10β6
304 β 10β9 β 7200
= 0,009 π΄
Topish kerak:
πΌ =?
Javob:
πΌ = 0,009 π΄ = 9 ππ΄
2-masala. 12 V kuchlanishga moβljallangan 6 kW quvvatli elektroliz qurilmasida 2
soat davomida qancha kumush moddasi yigβiladi?
Berilgan Formulasi Hisoblash
π = 12 π
π‘ = 2 π πππ‘ = 7200 π
π = ππΌπ‘
πΌ =π
π
π =
=6 β 103 β 1118 β 10β9 β 7200
12
ABDUBANANOV AKRAMJON
58
π = 1,118ππ
πΆ= 1118 β 10β9
ππ
πΆ
π = 6 ππ = 6 β 103π
π =πππ‘
π
= 4,0248 ππ
Topish kerak:
π =?
Javob:
π = 4,025 ππ
3-masala. Buyumni nikellashda 3 soat davomida elektrolitdan 5 A tok oβtib
turganida nikel qatlamining qalinligi 0,1 mm boβlgan. Nikel qoplangan yuza qanday
boβlgan? Nikel zichligi 8900 kg/m3 ga teng.
Berilgan Formulasi Hisoblash
πΌ = 5 π΄
π‘ = 3 π πππ‘ = 10800 π
π = 0,304ππ
πΆ= 304 β 10β9
ππ
πΆ
π = 0,1 ππ = 0,1 β 10β3π
π = 8900 kg/π3
π = ππΌπ‘
π = ππ = πππ
ππΌπ‘ = πππ
π =ππΌπ‘
ππ
π
=304 β 10β9 β 5 β 10800
8900 β 0,1 β 10β3
= 18444 β 10β6 π2
Topish kerak:
π =?
Javob:
π = 184 π π2
4-masala. Mis kuprosi eritmasidagi elektrodlar orasidagi kuchlanish 24 V
boβlganda, elektr toki 192 kJ foydali ish bajarsa, qancha mis ajralib chiqqan?
Berilgan Formulasi Hisoblash
π = 24 π
π΄ = 192 ππ½ = 192 β 103π½
π = 0,329ππ
πΆ= 329 β 10β9
ππ
πΆ
π΄ = πΌππ‘
π = ππ = ππΌπ‘
π
π΄=
π
π
π
192 β 103=
329 β 10β9
24
π = 2632 β 10β6ππ
ABDUBANANOV AKRAMJON
59
Topish kerak:
π =?
Javob:
π = 2632 β 10β6ππ
5-masala. Sirt yuzasi 30 sm2 boβlgan temir qoshiqni qalinligi 0,05 mm boβlgan
kumush bilan qoplash uchun kumush tuzi eritmasi orqali qanday zaryad oβtishi
kerak? Kumush zichligi 10500 kg/m3 ga teng.
Berilgan Formulasi Hisoblash
π = 0,05 ππ = 0,05 β 10β3π
π = 30 π π2 = 30 β 10β4π2
π = 1,118ππ
πΆ= 1118 β 10β9
ππ
πΆ
π = 10500 ππ
π3
π = ππ
π = ππ
= πππ
ππ = πππ
π =πππ
π
π
=10500 β 30 β 10β4 β 0,05 β 10β3
1118 β 10β9
= 14,08 β 102πΆ
Topish kerak:
π =?
Javob:
π = 1408 πΆ
6-masala. Kumushning molyar massasi 108 g/mol, valentligi 1 va elektrokimyoviy
ekvivalenti 1,08 mg/C, oltinning molyar massasi 197 g/mol, valentligi 1 boβlsa,
oltinning elektrokimyoviy ekvivalenti qanday?
Berilgan Formulasi Hisoblash
ππ = 108π
πππ
π§π = 1
ππ = 1,08ππ
πΆ
ππ = 197π
πππ
π§π = 1
π =1
πΉβπ§
π
ππ
ππ=
ππ
ππ
ππ
1,08=
108
197
ππ = 0,592 ππ/πΆ
ABDUBANANOV AKRAMJON
60
Topish kerak:
ππ =?
Javob:
ππ = 0,592 ππ/πΆ
24-mashq
1-masala. Agar anod toki 8 mA boβlsa, anod sirtiga har sekundda qancha elektron
kelib tushadi?
Berilgan Formulasi Hisoblash
πΌ = 8 ππ΄
π‘ = 1 π
π = 1,6 β 10β19πΆ
π = πΌπ‘ = ππ
π =πΌπ‘
π
π =8 β 1
1,6 β 10β19= 5 β 1019
Topish kerak:
π =?
Javob:
π = 5 β 1019
2-masala. Diodda anod kuchlanishi 180 V ga teng. Agar elektr maydoni 4,8 J ish
bajargan boβlsa, anodga qancha elektron yetib kelgan?
Berilgan Formulasi Hisoblash
π = 180 π
π΄ = 4,8 π½
π = 1,6 β 10β19πΆ
π΄ = ππ = πππ
π =π΄
ππ
π =4,8
1,6 β 10β19 β 180
= 0,0166 β 1019
Topish kerak:
π =?
Javob:
π = 1,7 β 1021
3-masala. Diodda anod bilan katod orasidagi maydon kuchlanganligi 4 β 103π/πΆ
boβlsa, elektron qanday tezlanish oladi?
Berilgan Formulasi Hisoblash
ABDUBANANOV AKRAMJON
61
πΈ = 4 β 103π/πΆ
π = 9,1 β 10β31ππ
π = 1,6 β 10β19πΆ
πΉ = ππ
πΉ = πΈπ
π =πΈπ
π
π =1,6 β 10β19 β 4 β 103
9,1 β 10β31=
= 0,7 β 1015π
π 2
Topish kerak:
π =?
Javob:
π = 7 β 1014π
π 2
V BOB. MAGNIT MAYDON
25-mashq
1-masala. Uzunligi 50 sm boβlgan oβtkazgich magnit induksiyasi 1,2 T boβlgan
magnit maydonga joylashtirilgan. Magnit maydon induksiyasiga tik joylashgan
oβtkazgichdan 2 A tok oβtganda unga magnit maydon tomonidan qanday kuch taβsir
qiladi?
Berilgan Formulasi Hisoblash
π = 50 π π = 0,5 π
π΅ = 1,2 π
πΌ = 2 π΄
πΌ = 900
πΉ = πΌπ΅ππ πππΌ
πΉ = 2 β 1,2 β 0,5 = 1,2 π
Topish kerak:
πΉ =?
Javob:
πΉ = 1,2 π
2-masala. Induksiyasi 0,4 T boβlgan magnit maydon chiziqlariga tik qilib
joylashtirilgan 15 sm uzunlikdagi oβkazgichga 60 mN kuch taβsir qiladi.
Oβtkazgichdan oβtayotgan tok kuchi qanday boβladi?
Berilgan Formulasi Hisoblash
ABDUBANANOV AKRAMJON
62
π = 15 π π = 0,15 π
π΅ = 0,4 π
πΉ = 60 ππ = 60 β 10β3π
πΌ = 900
πΉ = πΌπ΅ππ πππΌ
πΌ =πΉ
π΅π
πΌ =60 β 10β3
0,4 β 0,15= 1 π΄
Topish kerak:
πΌ =?
Javob:
πΌ = 1 π΄
3-masala. Uzunligi 25 sm boβlgan va 5 A tok oβtayotgan oβtkazgichga magnit
maydon tomonidan 2,5 mN kuch taβsir qilgan. Oβtkazgich joylashgan magnit
maydonning induksiyasini aniqlang.
Berilgan Formulasi Hisoblash
π = 25 π π = 0,25 π
πΌ = 5 π΄
πΉ = 25 ππ = 25 β 10β3π
πΌ = 900
πΉ = πΌπ΅ππ πππΌ
π΅ =πΉ
πΌπ
π΅ =25 β 10β3
5 β 0,25= 20 β 10β3π
Topish kerak:
π΅ =?
Javob:
π΅ = 20 ππ
4-masala. Induksiyasi 0,4 T boβlgan magnit maydoni chiziqlariga tik joylashgan 5
sm uzunlikdagi oβtkazgichga maydonning taβsir kuchi 2 mN ga teng. Oβtkazgichdagi
tok kuchi qanday boβlgan?
Berilgan Formulasi Hisoblash
π = 5 π π = 0,05 π
π΅ = 0,4 π
πΉ = 2ππ = 2 β 10β3π
πΉ = πΌπ΅ππ πππΌ
πΌ =πΉ
π΅π
π΅ =2 β 10β3
0,4 β 0,05= 0,1 π
ABDUBANANOV AKRAMJON
63
πΌ = 900
Topish kerak:
πΌ =?
Javob:
π΅ = 0,1 π
5-masala. Bir jisnsli maydonda joylashgan uzunligi 40 sm boβlgan toβgβri
oβtkazgichdan 8 A tok oβtkazilsa, maydon tomonidan qanday kuch taβsir qiladi?
Maydon induksiyasi 0,5 T ga teng.
Berilgan Formulasi Hisoblash
π = 40 π π = 0,4 π
π΅ = 0,5 π
πΌ = 8 π΄
πΌ = 900
πΉ = πΌπ΅ππ πππΌ
πΌ =πΉ
π΅π
πΉ = 8 β 0,5 β 0,4 = 1,6 π
Topish kerak:
πΉ =?
Javob:
πΉ = 1,6 π
6-masala. 0,8 m uzunlikdagi oβtkazgich induksiyasi 2 mT boβlgan magnit
maydonning induksiyasi chiziqlariga tik joylashgan. Oβtkazgich koβndalang kesim
yuzasidan har 3 minutda 720 C zaryad oqib oβtmoqda. Magnit maydon tomonidan
oβtkazgichga qanday kuch taβsir qiladi?
Berilgan Formulasi Hisoblash
π = 0,8 π
π΅ = 2 ππ = 2 β 10β3π
π‘ = 180 π
π = 720 πΆ
πΌ = 900
πΉ = πΌπ΅ππ πππΌ
πΌ =π
π‘
πΉ =ππ΅ππ πππΌ
π‘
πΉ =720 β 2 β 10β3 β 0,8
180=
= 6,4 β 10β3π
ABDUBANANOV AKRAMJON
64
Topish kerak:
πΉ =?
Javob:
πΉ = 6,4 ππ
26-mashq
1-masala. Ichida temir oβzgi boβlgan gβaltak orqali 165-rasmda koβrsatilgan
yoβnalishda tok oβtkaziladi. Bunda hosil boβlgan elektromagnit qutblarini aniqlang.
bu elektromagnit qutblarini qanday oβzgartirish mumkin?
Izoh: sxemadagi tok yoβnalishi soat strelkasi harakati yoβnalishida.
Elektromagnitdagi tok yoβnalishi chizmada koβrsatilgan
Elektromagnitdagi tok yoβnalishini bilgan holda unga parma qoidasini qoβllaymiz:
2-masala. 166-rasmda gβaltakdan tok oβtayotganda hosil boβlgan elektromagnit
qutblari koβrsatlgan. Gβaltakdagi tokning yoβnalishini va tok manbaining qutblarini
aniqlang.
π΅
π π
Javob: chap tomon
shimoliy, oβng tomoni
janubiy qutb
Javob: elektromagnit
qutbalarini tok
yoβnalishini oβzgartirib
oβzgartirish mumkin.
π΅ πππππ: gaβaltakdagi tok
yoβnalishi
ABDUBANANOV AKRAMJON
65
Manba qutblari: oβnda musbat, chapda manfiy
3-masala. Taqasimon elektromagnit choβlgβamining oβramlaridagi tokning
yoβnalishi 167-rasmda strelkalar bilan koβrsatilgan. Elektromagnitning qutblarini
aniqlang.
27-mashq
1-masala. Elektron bir jinsli magnit maydonga tik ravishda 2 β 106π/π tezlik bilan
uchib kirdi. Induksiyasi 0,8 T boβlgan magnit maydon tominidan elektronga taβsir
qiladigan kuchni aniqlang.
Berilgan Formulasi Hisoblash
π£ = 2 β 106π/π
π΅ = 0,8 π
π = 1,6 β 10β19πΆ
πΌ = 900
πΉ = ππ΅π£π πππΌ
πΉ = 1,6 β 10β19 β 0,8 β 2 β 106 =
= 2,56 β 10β13π
Topish kerak:
πΉ =?
Javob:
πΉ = 0,256 ππ
2-masala. Tezligi 4 β 107π/π va zaryadi 3,2 β 10β19πΆ boβlgan zarra magnit maydon
kuch chiziqlari yoβnalishiga tik ravishda uchib kirdi. Agar zarraga maydon
tomonidan 6,4 pN kuch taβsir qilgan boβlsa, magnit maydon induksiyasi qanday
boβlgan?
Berilgan Formulasi Hisoblash
π΅ π΅
π
π
ABDUBANANOV AKRAMJON
66
π£ = 4 β 107π/π
πΉ = 6,4ππ = 6,4 β 10β12π
π = 3,2 β 10β19πΆ
πΌ = 900
πΉ = ππ΅π£π πππΌ
π΅ =πΉ
ππ£
π΅ =6,4 β 10β12
3,2 β 10β19 β 4 β 107= 0,5 π
Topish kerak:
π΅ =?
Javob:
π΅ = 0,5 π
3-masala. Induksiyasi 0,4 T boβlgan magnit maydonga induksiya chiziqlariga tik
ravishda elektron uchib kirdi. Unga taβsir etuvchi kuch 0,64 pN boβlsa, uning tezligi
qanday boβlgan?
Berilgan Formulasi Hisoblash
π΅ = 0,4 π
πΉ = 0,64ππ = 0,64 β 10β12π
π = 1,6 β 10β19πΆ
πΌ = 900
πΉ = ππ΅π£π πππΌ
π£ =πΉ
ππ΅
π΅ =0,64 β 10β12
1,6 β 10β19 β 0,4=
= 107π/π
Topish kerak:
π£ =?
Javob:
π£ = 107π/π
4-masala. Magnit maydon induksiya chiziqlariga tik yoβnalishda 2 β 108π/π tezlik
bilan harakatlanayotgan proton uchib kirdi. Agar magnit maydon induksiyasi 0,4 T
boβlsa, pratonga magnit maydon tomonidan qanday kuch taβsir qiladi?
Berilgan Formulasi Hisoblash
π£ = 2 β 108π/π
π΅ = 0,4 π
π = 1,6 β 10β19πΆ
πΉ = ππ΅π£π πππΌ
πΉ = 1,6 β 10β19 β 0,4 β 2 β 108 =
= 1,28 β 10β11π
ABDUBANANOV AKRAMJON
67
πΌ = 900
Topish kerak:
πΉ =?
Javob:
πΉ = 12,8 ππ
5-masala. Induksiyasi 0,3 T boβlgan magnit maydon induksiya chiziqlariga tik
yoβnalishda 2 β 106π/π tezlik bilan uchib kirgan ionga magnit maydon tomonidan
0,48 pN kuch taβsir qiladi. Ionning zaryadi qanday boβlgan?
Berilgan Formulasi Hisoblash
π΅ = 0,3 π
πΉ = 0,48ππ = 0,48 β 10β12π
π£ = 2 β 106π/π
πΌ = 900
πΉ = ππ΅π£π πππΌ
π =πΉ
π£π΅
π =0,48 β 10β12
2 β 106 β 0,3=
= 0,8 β 10β18πΆ
Topish kerak:
π =?
Javob:
π = 8 β 10β19πΆ
ABDUBANANOV AKRAMJON
68