In the arrangement shown in figure, the current through 5Ω resistor is

Using the Kirchhoff's voltage law for loop ABCDA, 2i₁ + 5(i₁ + i₂) = 12 or 7i₁ + 5i₂ = 12.. (1)
and for loop EBCFE, 2i₂ + 5(i₁ + i₂) = 12 or 5i₁ + 7i₂ = 12.. (2)
Now,(1) × 7, (2) × 5 ⇒ 49i₁ + 35i₂ = 84.. (3)
and 25i₁ + 35i₂ = 60.. (4)
(3) - (4) ⇒ 24i₁ = 24 or i₁ = 1A
Substituting the value of i₁ in (1) we get, i₂ = 12 - 7/5 = 1A
Thus the current through 5Ω = i₁ + i₂ = 1 + 1 = 2A