Question

In the circuit shown below in figure, E₁ and E₂ are batteries having emfs 4.0 V and 3.5 V respectively and internal resistance 1 Ω and 2 Ω. respectively. Using Kirchhoff’s laws, calculate currents : I₁ I₂ and I₃. 

Answer 1

Applying Kirchhoff's 1st law at the point S

I3 = I₁ + I₂

Applying second law to the mesh  PQSP

4I₁ + 10I₃ + I₁  = 4

5I₁ + 10 I₃ = 4

Applying second law to the mesh RQSR, we get

3I₂ + 10I₃ + 2I₂ = 3.5

5I₂ + 10I₃ = 3.5

From (i), I₂ = I₃- I₁, putting the value in (iii), we get

5I₃ - 5I₁ + 10I₃ = 3.5

15I_{3 -} 5I₁ = 3.5

Adding (ii) and (iv), we get