Цуваа болон зэрэгцээ холбогдсон EMF-үүд

Цуваа болон зэрэгцээ холбогдсон EMF-үүд

Хэрэв зурагт үзүүлсэн шиг хоёр ба түүнээс дээш цахилгаан хөдөлгөгч хүч (EMF) холбогдсон бол EMF-ийг цуваагаар байрлуулна.

Үүнтэй адил хүчдэлийн эх үүсвэр (ε) нь:

ε = ε₁ + ε₂ + ε_n

Дотоод эсэргүүцэлтэй тэнцүү (r) нь:

r = r₁ +r₂ +r_n

Гадаад эсэргүүцэл (R)-ээр урсах цахилгаан гүйдэл нь:

Би = ε / (r + R)

Sample problem:

Suppose that two batteries each emf is 1.5 Volt and the internal resistance value in each battery is 0.1 Ω. External resistance (R) = 10 Ω. The direction of the electric current clockwise.

Use the previous formula:

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 А

Use Kirchhoff’s second rule:

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 А

If there are two or more sources of electromotive (emf) connected as shown in the figure, the emf is connected in parallel.

The equivalent voltage source (ε) is:

ε = ε₁ = ε₂ = ε_n

Дотоод эсэргүүцэлтэй тэнцүү (r) нь:

1/r = 1/r₁ + 1/r₂ + 1/r_n

Гадаад эсэргүүцэл (R)-ээр урсах цахилгаан гүйдэл нь:

Би = ε / (r + R)

Sample problem:

Suppose that two batteries each emf is 1.5 Volt and the resistance value in each battery is 0.1 Ω. External resistance (R) = 10 Ω.

Use the previous formula:

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 А

Use Kirchhoff’s rule

түрхэнэ Кирхгоф‘s first rule:

I₁ + Би₂ = Би ………. Equation 1

Analyze Aefca loop. The direction of the loop is clockwise. Apply Kirchhoff’s second rule:

ε₂ - Би₁ r₂ – I R = 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 тэгшитгэл

Analyze the Befdb loop. The direction of the loop is clockwise. Apply Kirchhoff’s second law:

ε₁ - Би₂ r₁ – I R = 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 тэгшитгэл

Substitute equation 2 and 3 to equation 1:

I₁ + Би₂ = Би

-100 I + 15 – 100 I + 15 = I

– 200 I + 30 = I

30 = I + 200 I

30 = 201 I

I = 30 / 201

I = 0.149 А

Eliminate equation 2 and 3:

I₁ = -100 I + 15

I₂ = -100 I + 15

——————– –

I₁ - Би₂ = 0

I₁ = Би₂ ………. Equation 4

Учир нь би₁ + Би₂ = I, where I₁ = Би₂ Тэгээд би₁ = Би₂ = 1/2 I = 1/2 (0.149) = 0.0745 A.