Xoogga elektaroonigga ee taxanaha ah iyo kuwa is barbar socda (EMF)

Walax koronto oo taxane ah iyo mid is barbar socda (EMF)

Haddii ay jiraan laba ama in ka badan oo ilo koronto ah (emf) oo isku xiran sida sawirka dhinaca ku yaal, emf-ku wuxuu ku xiran yahay taxane.

Isha danabka beddelka ah ( ε ) waa:

ε = ε ₁ + ε ₂ + ε _n

Iska caabbinta beddelka (r) waa:

r = r ₁ + r ₂ + r _n

Korontada ku socota iska caabbinta dibadda (R) waa:

I = ε / (r + R)

Tusaale ahaan dhibaatooyinka:

Ka soo qaad in laba baytari midkiiba uu leeyahay emf ah 1,5 Volts, iska caabbinta gudaha ee baytari kastana ay tahay 0,1 Ω. Iska caabbinta dibadda (R) = 10 Ω. Jihada qulqulka korontada waa saacad-wareegga.

Isticmaal qaacidadii hore :

ε = 1,5 + 1,5 = 3 Volts

r = 0,1 + 0,1 = 0,2 Ω

I = ε / (r + R) = 3 / (0,2 + 10)

I = 3/10,2

I = 0,294 Amperes

Isticmaal Sharciga 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 Amperes 

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

Isha danabka beddelka ah ( ε ) waa:

ε = ε ₁ = ε ₂ = ε _n

Iska caabbinta beddelka (r) waa:

1/ r = 1/ r ₁ + 1/ r ₂ + 1/ r _n

Korontada ku socota iska caabbinta dibadda (R) waa:

I = ε / (r + R)

Tusaale ahaan dhibaatooyinka:

Ka soo qaad in laba baytari midkiiba uu leeyahay emf ah 1,5 Volts, iska caabbinta gudaha ee baytari kastana ay tahay 0,1 Ω. Iska caabbinta dibadda (R) = 10 Ω.

Isticmaal qaacidadii hore :

ε = 1,5 Volts

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 Amperes

Isticmaal sharciga Kirchhoff

Adeegso sharciga ugu horreeya ee Kirchhoff:

I ₁ + I ₂ = I ………. Isle'egta 1

Falanqaynta wareegga efca . Jihada wareeggu waa saacad-wareegga. Ku dabaq sharciga labaad ee Kirchhoff :

ε ₂ – I ₁ r ₂ – I R = 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 = -1 0 0 I + 15 ……. Isla'egta 2

Falanqaynta wareegga befdb . Jihada wareeggu waa saacad-wareegga. Ku dabaq sharciga labaad ee Kirchhoff:

ε ₁ – I ₂ r ₁ – I R = 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 = -1 0 0 I + 15 ……. Isla'egta 3

Isle'egyada 2 iyo 3 ku beddel isle'egyada 1:

I ₁ + I ₂ = I

-1 0 0 I + 15 – 1 0 0 I + 1 5 = I

– 200 I + 30 = I

30 = I + 200 I

30 = 201 I

I = 30/201

I = 0,149 Amperes

Ka saar isleegyada 2 iyo 3:

I ₁ = -1 0 0 I + 15

I ₂ = -1 0 0 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 Ampere