Question

In the circuit shown in Figure below, E₁ and E₂ are batteries having emfs of 25V and 26V. They have an internal resistance of 1 Ω and 5 Ω respectively. Applying Kirchhoff’s laws of electrical networks, calculate the currents I₁ and I₂.

Answer 1

Applying Kirchhoff's voltage law to loop ABCDEF, we get

2(I₁ + I₂) + 3I₂ - 26 + 5I₂ = 0

or 2I₁ + 2I₂ + 3I₂ + 5I₂ = 26

or I₁ + 5I₂ = 13     ...(i)

Now, applying Kirchhoff’s voltage law to loop HJBCDEG.

2(I₁ + I₂) + 4I₁ + I₁ - 25 = 0

or 2I₁ + 2I₂ + 5I₁ = 25

or 7I₁ + 2I₂ = 25     ...(ii)

Equation (i) × 2 - equation (ii) × 5

2I₁ + 10I₂ = 26
35I₁ + 10I₂ = 125
-       -             -      
- 33I₁ = - 99

      I₁ = 3

From equation (i)

3 + 5I₂ = 13

or I₂ = 2

Hence I₁ = 3 A

I₂ = 2 A.