Question

In the circuit shown in the figure below, E₁ and E₂ are two cells having emfs 2 V and 3 V respectively, and negligible internal resistance. Applying Kirchhoff’s laws of electrical networks, find the values of currents l₁ and I₂.

Answer 1

The distribution of current in the circuit is as shown in figure

Applying Kirchoff's laws (Loop law) to loop ABEFA

-2 (I₁ + I₂) - I₁ × 1 + 2 = 0 

2 - I₁ - 2(I₁ + I₂) = 0

⇒ 2 - 3I₁ - 2I₂ = 0     .....(i)

Applying to loop BCDEB

-3 + 6I₂ + 2(I₁ + I₂) = 0

⇒ 3 - 6I₂ - 2I₁ - 2I₂ = 0

⇒ 3 - 8I₂ - 2I₁ = 0       ....(ii)

Solving equations (i) and (ii), we can write

I₁ = `1/2`A , I₂ = `1/4` A