Question

Find the currents flowing through the branches AB and BC in the network shown.

Answer 1

Applying Kirchhoffs second law in mesh ANMBA,

(I₁ + I₂)10 − 5(I₁) + 5 = 0

⇒ 15I₁ + 10I₂ = 5   ...(i)

Applying Kirchhoff's second law in mesh BCDMB,

−10(I₁ + I₂) − 5I₂ − 20I₂ + 10 = 0

⇒ 35I₂ + 10I₁ = 10   ...(ii)

Solving Eqs (i) and (ii), we get (Eq. (i) × 10 − Eq. (ii) × 15)

⇒ 150I₁ + 100I₂ = 50
    150I₁ + 525I₂ = 150
    −           −           −      
               −425I₂ = −100 
                       I₂ = 0.23 A

From Eq. (i)

`I_1 = (5 - 10 I_2)/15`

= `(5 - 235)/15`

= 0.17 A

Thus, I₁ = 0.17 A and I₂ = 0.23 A in the branches AB and BC.