**EMFs in series and parallel**

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

The equivalent voltage source (ε) is:

ε = ε_{1} + ε_{2} + ε_{n}

The equivalent internal resistance (r) is:

r = r_{1} + r_{2} + r_{n}

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

I = 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 A**

**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 A**

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 = ε_{1} = ε_{2} = ε_{n}

The equivalent internal resistance (r) is:

1/r = 1/r_{1} + 1/r_{2} + 1/r_{n}

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

I = 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 A**

**Use Kirchhoff’s rule**

Apply Kirchhoff‘s first rule:

I_{1} + I_{2} = I ………. Equation *1*

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

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} = -100 I + 15 ………. *Equation 2*

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

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} = -100 I + 15 ………. *Equation 3*

Substitute equation 2 and 3 to equation 1:

I_{1} + I_{2} = I

-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**

Eliminate equation 2 and 3:

I_{1} = -100 I + 15

I_{2} = -100 I + 15

——————– –

I_{1} – I_{2} = 0

I_{1 }= I_{2 }………. Equation *4*

Because I_{1} + I_{2} = I, where I_{1} = I_{2} then I_{1 }= I_{2} = 1/2 I = 1/2 (0.149) = 0.0745 A.