एकसर आणि समांतर जोडणीतील विद्युत दाब

एकसर आणि समांतर जोडणीतील विद्युत दाब

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

समतुल्य विद्युतदाब source (ε) is:

ε = ε₁ + ε₂ + ε_n

The equivalent internal resistance (r) is:

r = r₁ + आर₂ + आर_n

The electric current flowing through the external resistance (R) is:

मी = 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)

मी = १,०००,००० / ८३१.३६

I = ०.२९४ अ

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

५ – १० I = ०

५ = १० मी

मी = १,०००,००० / ८३१.३६

I = ०.२९४ अ

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

The equivalent internal resistance (r) is:

1/r = 1/r₁ + 1/r₂ + 1/r_n

The electric current flowing through the external resistance (R) is:

मी = 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 = ०.२९४ अ

Use Kirchhoff’s rule

लागू करा किर्चॉफ‘s first rule:

I₁ + मी₂ = मी ………. Equation 1

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

ε₂ - मी₁ r₂ – I R = 0

६ – १२ I₁ – 10 I = 0

– ६ मी₁ = 10 I – 1.5

I₁ = (10 I – 1.5) / – 0.1

I₁ = -100 I + 15 ………. समीकरण १

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

ε₁ - मी₂ r₁ – I R = 0

६ – १२ I₂ – 10 I = 0

- 0.1 मी₂ = 10 I – 1.5

I₂ = (10 I – 1.5) / – 0.1

I₂ = -100 I + 15 ………. समीकरण १

Substitute equation 2 and 3 to equation 1:

I₁ + मी₂ = मी

-100 I + 15 – 100 I + 15 = I

– २०० आय + ३० = आय

३० = १ + २०० १

५ = १० मी

मी = १,०००,००० / ८३१.३६

I = ०.२९४ अ

Eliminate equation 2 and 3:

I₁ = -100 I + 15

I₂ = -100 I + 15

——————– –

I₁ - मी₂ = 0

I₁ = मी₂ ………. Equation 4

कारण मी₁ + मी₂ = I, where I₁ = मी₂ मग मी₁ = मी₂ = 1/2 I = 1/2 (0.149) = 0.0745 A.