EMFan ann an sreath agus co-shìnte

EMFan ann an sreath agus co-shìnte

Ma tha dà no barrachd thùsan dealan-motair (emf) ceangailte mar a chithear san fhigear, tha an emf air a chur ann an sreath.

An co-ionann bholtaids Is e an stòr (ε):

ε = ε₁ + ε₂ + ε_n

Is e an aghaidh a-staigh co-ionann (r):

r = r₁ +r₂ +r_n

Is e an sruth dealain a tha a’ sruthadh tron ​​aghaidh taobh a-muigh (R):

Mise = ε / (r + R)

Eisimpleir de dhuilgheadas:

Ma tha dà bataraidh a’ cleachdadh emf de 1.5 Volt agus gu bheil an aghaidh a-staigh anns gach bataraidh 0.1 Ω. Tha an aghaidh a-muigh (R) = 10 Ω. Stiùireadh an t-sruth dealain deiseal.

Cleachd am foirmle roimhe:

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

Cleachd an dàrna riaghailt aig 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

Ma tha dà no barrachd stòran dealan-motair (emf) ceangailte mar a chithear san fhigear, tha an emf ceangailte ann am parail.

Is e an stòr bholtaids co-ionann (ε):

ε = ε₁ = tha₂ = tha_n

Is e an aghaidh a-staigh co-ionann (r):

1/r = 1/r₁ + 1/r₂ + 1/r_n

Is e an sruth dealain a tha a’ sruthadh tron ​​aghaidh taobh a-muigh (R):

Mise = ε / (r + R)

Eisimpleir de dhuilgheadas:

Ma tha dà bataraidh a’ tomhas le emf de 1.5 Volt agus gu bheil luach an aghaidh anns gach bataraidh 0.1 Ω. Tha an aghaidh taobh a-muigh (R) = 10 Ω.

Cleachd am foirmle roimhe:

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

Cleachd riaghailt Kirchhoff

Cuir a-steach Kirchhoffa’ chiad riaghailt:

I₁ + I.₂ = Mise ………. Co-aontar 1

Dèan mion-sgrùdadh air lùb Aefca. Tha stiùireadh a’ lùb deiseal. Cuir an dàrna riaghailt aig Kirchhoff an sàs:

ε₂ - I.₁ 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 ………. Co-aontar 2

Dèan mion-sgrùdadh air lùb Befdb. Tha stiùireadh na lùibe deiseal. Cuir an dàrna lagh aig Kirchhoff an sàs:

ε₁ - I.₂ 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 ………. Co-aontar 3

Cuir co-aontaran 2 agus 3 an àite co-aontar 1:

I₁ + I.₂ = Mise

-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

Cuir às do cho-aontar 2 agus 3:

I₁ = -100 I + 15

I₂ = -100 I + 15

———————– –

I₁ - I.₂ = 0

I₁ = Mise₂ ………. Co-aontar 4

Seach gu bheil mi₁ + I.₂ = Mise, far a bheil mi₁ = Mise₂ an uairsin mi₁ = Mise₂ = 1/2 I = 1/2 (0.149) = 0.0745 A.