Question

E₁ and E₂ are two batteries having emfs of 3 V and 4 V and internal resistances of 2 Ω and 1 Ω respectively. They are connected as shown in Figure 2 below. Using Kirchhoff’s Laws of electrical circuits, calculate the currents I₁ and I₂.

Answer 1

Given: E₁ = 3 V

r₁ = 2 Ω

R₁ = 4 Ω

E₂ = 4 V

r₂ = 1 Ω

R₂ = 7 Ω

R₃ = 8 Ω

I₁ = ?

I₂ = ?

Current through R₃ = I₁ + I₂    ...(Kirchhoff’s current rule)

Applying the loop rule to loop ABEFA,

E₁ − [(I₁ + I₂)R₃ + I₁R₁ + I₁r₁] = 0

⇒ (I₁ + I₂)8 + 4I₁ + 2I₁ = 3

⇒ 14I₁ + 8I₂ = 3    ...(i)

Applying the loop rule to loop CBEDC,

E₂ − [(I₁ + I₂)R₃ + I₂R₂ + I₂r₂] = 0

⇒ (I₁ + I₂)8 + 7I₂ + 1I₂ = 4

⇒ 16I₂ + 8I₁ = 4    ...(ii)

Multiplying equation (i) by 2, we get,

28I₁ + 16I₂ = 6    ...(iii)

Subtracting equation (ii) from equation (iii), we get,

20I₁ = 2

I₁ = `2/20`

I₁ = 0.1 A

Putting this value of I₁ in the equation, we get,

14(0.1) + 8I₂ = 3

I₂ = `(3 - (14 xx 0.1))/8`

= `(3 - 1.4)/8`

= `1.6/8`

= 0.2 A