Question

The figure below shows two batteries, E1 and E2, having emfs of 18V and 10V and internal resistances of 1 Ω and 2 Ω, respectively. W₁, W₂ and W₃ are uniform metallic wires AC, FD and BE having resistances of 8 Ω, 6 Ω and 10 Ω respectively. B and E are midpoints of the wires W₁ and W₂. Using Kirchhoff's laws of electrical circuits, calculate the current flowing in the wire W₃:

Answer 1

Since B and E are the mid-points of AC and FD respectively, so resistance between AB, BC, DE and EF will be 4 Ω, 4 Ω, 3 Ω and 3 Ω respectively as shown in the diagram.

Using Kirchhoff's law in the closed-loop ABEFA, we have:

4 I₁ + 10 I₁ + 10 I₂ + 3 I₁ + 1 I₁ = 18

18 I₁ + 10 I₂ = 18

9 I₁ + 5 I₂ = 9   ...(i)

For the closed-loop BEDCB, we have.

4 I₂ + 10 I₁ +10 I₂ + 3 I₂ + 2I₂ = 10

10 I₁ + 19 I₂ = 10   ...(ii)

Solving equation (i) and (ii),

9 I₁ + 5 I₂ = 9 ...(i) × 10

10 I₁ + 19 I₂ = 10 ...(ii) × 9

 90 I₁ + 50 I₂ = 90

90 I₁ + 171 I₂ = 90

(-)     (-)           (-)  
`=> -121  I_2 = 0`

`=>` I₂ = 0

Putting I₂ = 0 in equation (i),

9 I₁ + 5 × 0 = 9

9 I₁ = 9 or I₁ = 1A

∴ I₁ = 1A, I₂ = 0

and I₃ = I₁ + I₂ 

= 1A