EMF-yada taxanaha ah iyo kuwa is barbar socda

EMF-yada taxanaha ah iyo kuwa is barbar socda

Haddii ay jiraan laba ama in ka badan oo ilo koronto-qaadis ah (emf) ah oo isku xiran sida ku cad sawirka, emf waxaa loo habeeyey si taxane ah.

Isha danab ee u dhiganta (ε) waa:

ε = ε ₁ + ε ₂ + ε _n

Iska caabbinta gudaha (r) ee u dhiganta waa:

r = r ₁ + r ₂ + r _n

Korontada ku socota iska caabbinta dibadda (R) waa:

I = ε / (r + R)

Dhibaato tusaale ah:

Ka soo qaad in laba baytari midkiiba emf yahay 1.5 Volt qiimaha iska caabbinta gudaha ee baytari kastana yahay 0.1 Ω. Iska caabbinta dibadda (R) = 10 Ω. Jihada qulqulka korantada saacadda u jeeda.

Isticmaal qaacidadii hore:

ε = 1.5 + 1.5 = 3 Volt

r = 0.1 + 0.1 = 0.2 Ω

I = ε / (r + R) = 3 / (0.2 + 10)

I = 3/10.2

I = 0.294 A

Isticmaal xeerka labaad ee Kirchhoff:

1.5 – 0.1 I + 1.5 – 0.1 I – 10 I = 0

3 – 0.2 I – 10 I = 0

3 – 10.2 I = 0

3 = 10.2 I

I = 3/10.2

I = 0.294 A

Haddii ay jiraan laba ama in ka badan oo ilo koronto-ku-saleysan (emf) ah oo isku xiran sida ku cad sawirka, emf-ku wuxuu ku xiran yahay si is barbar socda.

Isha danab ee u dhiganta (ε) waa:

ε = ε₁ = ε₂ = ε_n

Iska caabbinta gudaha (r) ee u dhiganta waa:

1/r = 1/r ₁ + 1/r ₂ + 1/r _n

Korontada ku socota iska caabbinta dibadda (R) waa:

I = ε / (r + R)

Dhibaato tusaale ah:

Ka soo qaad in laba baytari oo emf kasta uu yahay 1.5 Volt qiimaha iska caabbinta ee baytari kastana uu yahay 0.1 Ω. Iska caabbinta dibadda (R) = 10 Ω.

Isticmaal qaacidadii hore:

ε = 1.5 Volt

1/r = 1/0.1 + 1/0.1 = 2 / 0.1

r = 0.1 / 2 = 0.05 Ω

I = ε / (r + R) = 1.5 / (0.05 + 10) = 1.5 / 10.05

I = 0.149 A

Isticmaal xeerka Kirchhoff.

Adeegso qaanuunka ugu horreeya ee Kirchhoff :

I ₁ + I ₂ = I ………. Isle'egta 1

Falanqee wareegga Aefca. Jihada wareeggu waa saacad-wareegga. Ku dabaq xeerka labaad ee Kirchhoff:

ε ₂ – I ₁ r ₂ – IR = 0

1.5 – 0.1 I ₁ – 10 I = 0

– 0.1 I ₁ = 10 I – 1.5

I ₁ = (10 I – 1.5) / – 0.1

I ₁ = -100 I + 15 ……. Isla'egta 2

Falanqee wareegga Befdb. Jihada wareeggu waa saacad-wareegga. Ku dabaq sharciga labaad ee Kirchhoff:

ε ₁ – I ₂ r ₁ – IR = 0

1.5 – 0.1 I ₂ – 10 I = 0

– 0.1 I ₂ = 10 I – 1.5

I ₂ = (10 I – 1.5) / – 0.1

I ₂ = -100 I + 15 ……. Isla'egta 3

Ku beddel isle'egyada 2 iyo 3 isla'egyada 1:

I ₁ + I ₂ = I

-100 I + 15 – 100 I + 15 = I

– 200 I + 30 = I

30 = I + 200 I

30 = 201 I

I = 30/201

I = 0.149 A

Ka saar isle'egta 2 iyo 3:

I ₁ = -100 I + 15

I ₂ = -100 I + 15

———————– –

I ₁ – I ₂ = 0

I ₁ = I ₂ ………. Isla'egta 4

Sababtoo ah I ₁ + I ₂ = I, halkaas oo I ₁ = I ₂ ka dibna I ₁ = I ₂ = 1/2 I = 1/2 (0.149) = 0.0745 A.