EMFs a cikin jerin da layi ɗaya

EMFs a cikin jerin da layi ɗaya

Idan akwai hanyoyi biyu ko fiye na electromotive (emf) da aka haɗa kamar yadda aka nuna a cikin hoton, an shirya emf a jere.

Daidai da irin ƙarfin lantarki tushe (ε) shine:

ε = ε₁ + ε₂ + ε_n

Daidaiton juriyar ciki (r) shine:

r = r₁ +r₂ +r_n

Wutar lantarki da ke gudana ta hanyar juriyar waje (R) ita ce:

Ni = ε / (r + R)

Matsalar samfurin:

A ce batura biyu kowanne emf yana da ƙarfin Volt 1.5 kuma ƙimar juriya ta ciki a cikin kowane baturi shine 0.1 Ω. Juriyar waje (R) = 10 Ω. Alkiblar wutar lantarki a kowane lokaci.

Yi amfani da dabarar da ta gabata:

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

Ina = 0.294 A

Yi amfani da dokar Kirchhoff ta biyu:

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

Ina = 0.294 A

Idan akwai hanyoyi biyu ko fiye na electromotive (emf) da aka haɗa kamar yadda aka nuna a cikin hoton, an haɗa emf a layi ɗaya.

Madogarar wutar lantarki daidai (ε) ita ce:

ε = ε₁ = da₂ = da_n

Daidaiton juriyar ciki (r) shine:

1/r = 1/r₁ + 1/r₂ + 1/r_n

Wutar lantarki da ke gudana ta hanyar juriyar waje (R) ita ce:

Ni = ε / (r + R)

Matsalar samfurin:

A ce batura biyu kowanne emf yana da ƙarfin Volt 1.5 kuma ƙimar juriya a cikin kowane baturi shine 0.1 Ω. Juriyar waje (R) = 10 Ω.

Yi amfani da dabarar da ta gabata:

ε = Volt 1.5

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

Ina = 0.149 A

Yi amfani da ƙa'idar Kirchhoff

Aiwatar KirchhoffDokar farko:

I₁ + Ni₂ = Ni ………. Daidaito 1

Yi nazarin madaurin Aefca. Alkiblar madaurin tana juyawa ne a hannun agogo. Yi amfani da ƙa'idar Kirchhoff ta biyu:

ε₂ - Ni₁ r₂ – IR = 0

1.5 – 0.1 I₁ – 10 I = 0

– 0.1 I₁ = 10 I – 1.5

I₁ = (10 I – 1.5) / – 0.1

I₁ = -100 I + 15 ……… 2 Daidaitawa

Yi nazarin madaurin Befdb. Alkiblar madaurin tana juyawa ne a hannun agogo. Yi amfani da dokar Kirchhoff ta biyu:

ε₁ - Ni₂ r₁ – IR = 0

1.5 – 0.1 I₂ – 10 I = 0

- 0.1 I₂ = 10 I – 1.5

I₂ = (10 I – 1.5) / – 0.1

I₂ = -100 I + 15 ……… 3 Daidaitawa

Sauya daidaito 2 da 3 zuwa lissafi 1:

I₁ + Ni₂ = Ni

-100 I + 15 – 100 I + 15 = I

– 200 I + 30 = I

30 = I + 200 I

30 = 201 I

I = 30/201

Ina = 0.149 A

Kawar da lissafi na 2 da 3:

I₁ = -100 I + 15

I₂ = -100 I + 15

———————– –

I₁ - Ni₂ = 0

I₁ = Ni₂ ………. Daidaito 4

Domin Ni₁ + Ni₂ = Ni, inda nake₁ = Ni₂ sai I₁ = Ni₂ = 1/2 I = 1/2 (0.149) = 0.0745 A.