Qaacidooyinka wareegga korantada iyo iska caabbinta korantada

Su'aalo iyo qaacidooyin tusaale ah oo ku saabsan wareegyada korantada iyo iska caabbinta korantada

Wareegyada korantada

1. Eeg jaantuska wareegga korantada ee soo socda. Cabbirka qulqulka korantada ee dhex mara iska caabinta 8 Ohm waa….
A. 1,8 Amperes
B. 1,2 Ampere
C. 0,8 Ampere
D. 0,6 Ampere

Dood
Waa la ogyahay in:
Iska caabbinta 1 (R)₁) = 12 Ω
Iska caabbinta 2 (R)₂) = 8 Ω
Iska caabbinta 3 (R)₃) = 10 Ω
Danabka korontada (V) = 12 Volts
Su'aal: Xoogga hadda jira iyada oo loo marayo R₁
Jawaab:

Korontada korontada waxay ka socotaa awood sare ilaa awood hoose. Jihada korontada ee wareegga kore waxay la mid tahay jihada gacmaha saacadda.

Korontada ka soo baxaysa baytariga:
Marka hore xisaabi iska caabinta beddelka ah (R). Intaa ka dib, xisaabi hadda adigoo isticmaalaya qaacidada sharciga Ohm: V = IR ama I = V / R halkaas oo V = danab, I = hadda, R = iska caabinta beddelka ah.

Iska-caabinta beddelka ah:
Iska caabbiye R₁ iyo iska caabinta R₂ Si taxane ah ayaa loo habeeyay. Resistor-ka beddelka ah waa:
R₁₂ =R₁ + R₂ = 12 + 8 = 20 Ω
Iska caabbiye R₁₂ iyo iska caabinta R₃ Isku barbar socda. Resistor-ka beddelka ah waa:
1/R = 1/R₁₂ + 1/R₃ = 1/20 + 1/10 = 1/20 + 2/20 = 3/20
R = 20/3 Ω

Korontada ka soo baxaysa baytariga:
I = V / R = 12: 20/3 = 12 x 3/20 = 36/20 = 1,8 Amperes
Korontada ka soo baxaysa baytariga waa 1,8 amperes.

Korontada ku socota iyada oo la adeegsanayo iska caabin 8 Ω ah
Vab = 12 Volt
R₁₂ = 20 Ω
R₃ = 10 Ω

Korontada ku socota iyada oo la adeegsanayo iska caabin 10 Ω ah
I₃ =V_ab /r₃ = 12 Volts / 10 Ω = 1,2 Ampere
Korontada ku socota iyada oo la adeegsanayo iska caabin 20 Ω ah
I₁₂ =V_ab /r₁₂ = 12 Volts / 20 Ω = 0,6 Ampere

Sharciga Koowaad ee Kirchhoff wuxuu dhigayaa in qulqulka korontada ee galaya laan uu la mid yahay kan korontada ee ka baxaya laantaas. Iyada oo ku saleysan... Sharciga ugu horreeya ee Kirchhoff waxaa lagu soo gabagabeeyay in korontada ay dhex marto iska caabin 20 Ω ah = korontada oo dhex marta iska caabin 12 Ω ah = korontada oo dhex marta iska caabin 8 Ω ah = 0,6 Amperes.
Jawaabta saxda ah waa D.

2.
Fiiro gaar ah u yeelo wareegga korontada ee xiran ee soo socda!
Haddii iska caabbinta isha danab la iska indho tiro, markaa farqiga u dhexeeya dhammaadka iska caabbinta 6 Ohm waa….
A. 3 Volt
B. 2 Volt
C. 2/3 Volt
D. 1/3 Volt
Dood
Waa la ogyahay in:
Iska caabbiye 1 = 4 Ω
Iska caabbiye 2 = 3 Ω
Iska caabbiye 3 = 6 Ω
Danab (V) = 6 volts
Su'aal: Farqiga suurtagalka ah ee u dhexeeya dhammaadka iska caabinta 6 Ohm
Jawaab:
Iska caabin 3 Ω ah iyo iska caabin 3 Ω ah ayaa isku xiran iyagoo is barbar socda. Iska caabinta beddelka ah waa:
1/R23 = 1/3 Ω + 1/6 Ω = 2/6 Ω + 1/6 Ω = 3/6 Ω
R23 = 6/3 Ω = 2 Ω

Iska-caabin 4 Ω ah iyo iska-caabin 2 Ω ah ayaa isku xiran taxane ahaan. Iska-caabinta beddelka ah waa:
R = 4 Ω + 2 Ω = 6 Ω

Xoogga korantada ee ka soo baxaysa isha danabka:
I = V / R = 6 volts / 6 ohms = 1 Ampere
Iska caabinta 4 Ω iyo iska caabinta 2 Ω ayaa isku xiran si taxane ah si sida waafaqsan sharciga koowaad ee Kirchooff, qulqulka korontada ee dhex mara iska caabinta 4 Ω uu la mid yahay qulqulka korontada ee dhex mara iska caabinta 2 Ω = 1 Ampere.

Farqiga suurtagalka ah ee u dhexeeya dhibcaha a iyo b:
V = IR = (1 Ampere)(2 Ohms) = 2 volts
Farqiga suurtagalka ah ee u dhexeeya dhibcaha a iyo b = farqiga suurtagalka ah ee u dhexeeya dhammaadka iska caabbinta 2 Ω = farqiga suurtagalka ah ee u dhexeeya dhammaadka iska caabbinta 3 Ω = farqiga suurtagalka ah ee u dhexeeya dhammaadka iska caabbinta 6 Ω = 2 volts.
Jawaabta saxda ah waa B.

Iska caabbinta korontada

3. Saddex iska caabin koronto, haddii si is barbar socda loo habeeyo, qiimaha iska caabintu waa 12/11 Ohm, haddii si taxane ah loo habeeyo qiimahu waa 12 Ohm, markaa qiimaha iska caabin kasta waa...
A. 1 Ohm, 2 Ohm, 3 Ohm
B. 2 Ohm, 4 Ohm, 6 Ohm
C. 1 Ohm, 3 Ohm, 5 Ohm
D. 3 Ohm, 4 Ohm, 5 Ohm
Dood
Haddii iska caabbinta korantada loo habeeyey si taxane ah, iska caabbinta u dhiganta waxaa lagu xisaabiyaa qaacidada:
R = R₁ + R₂ + R₃
12=R₁ + R₂ + R₃
Xulashooyinka suurtagalka ah waa jawaabaha B iyo D.
Haddii iska caabbinta korantada loo habeeyey si is barbar socda, iska caabbinta u dhiganta waxaa lagu xisaabiyaa qaacidada: 1/R = 1/R₁ + 1/R₂ + 1/R₃.
Haddii aad isticmaasho jawaabta B markaas:
1/R = 1/2 + 1/4 + 1/6
1/R = 6/12 + 3/12 + 2/12
1/R = 11/12
R = 12/11
Jawaabta saxda ah waa B.

Nooca caqabadda

4. Iska caabinta tungsten iyo tin waa 5,5 x 10 siday u kala horreeyaan.^-8 Mitirka Ohm iyo 22 x 10^-8 Mitirka Ohm. Siligga tungsten iyo siligga tin-ka ah midkiiba waa 5 mitir. Haddii labada siligba ay leeyihiin qiimo iska caabin isku mid ah oo dhexroorka tungsten-ku uu yahay 2 mm markaas dhexroorka siligga rasaasta waa... mmm
A. 2
B. 4
C. √2
D. ½ √2
Dood
Waa la ogyahay in:
Iska caabinta Tungsten (ρ)₁) = 5,5 x 10^-8 Mitirka Ohm
Iska caabinta Tin (ρ)₂) = 22 x 10^-8 Mitirka Ohm
Dhererka siligga tungsten (L)₁) = 5 mitir
Dhererka siligga rasaasta (L)₂) = 5 mitir
Dhexroorka Tungsten (D)₁) = 2 mm
Gacanka Tungsten (R)₁) = 1/2 (D) = 1/2 (2 mm) = 1 mm = 1 x 10^-3 meter
Su'aal: Dhexroorka rasaasta (D)₂)
Jawaab:
Bedka dusha sare ee siligga tungsten:
A₁ = π r² = π (10^-3)² = π 10^-6

Qiimaha iska caabbinta ee siligga tungsten iyo siligga tin waa isku mid.
R₁ =R₂
ρ₁ L₁ / TO₁ = ρ₂ L₂ / TO₂
(5,5x10^-8)(5) / π 10^-6 = (22 x 10^-8)(5) / π r²
5,5 / 10^-6 = 22 / r²
5,5 r² = 22x10^-6
r² = (22 x 10^-6 ) / 5,5
r² = 4x10^-6
r = 2 x 10^-3 meter
r = 2 milimitir
Jawaabta saxda ah waa A.

Isha su'aasha:

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