Mesh Analysis

For the network shown in the figure,we consider three loops.Let the current through the first loop be i₁ due to the voltage source of 10V.The current i₁ gets divided and a current i₂ flows through the 6Ω resistor.For the second loop in which the source of emf is 12V let the current be i₃ through the 2Ω resistor.

For the loop 1 the voltage equation is:

4i₁ + 6i₂ = 10

or 2i₁ +3i₂ = 5 --------(1)

For the loop 2 the voltage equation is :

2i₃ + (i₃ + i₂) x 6 = 12 or i₃ + 3(i₃ + i₂) = 6

or 3i₂ + 4i₃ = 6 ---------(2)

For the whole loop the voltage equation is:

10 + 4i₁ + (i₁ - i₃)2 + (-12) = 0 or 4i₁ + 2i₁ - 2i₃ = 2 or 6i₁ - 2i₃ = 2

or 3i₁ - i₃ = 1 ---------(3)

From equation (3), i₃ = 3i₁ -1 ----------(4)

Substituting the value of i₃ from equation (4) in equation (2),we get

3i₂ - 4(3i₁ -1) = 6 or 3i₂ +12i₁ - 4 = 6

or 3i₂ + 12i₁ = 10 ---------(5)

Subtracting equation(1) from equation(5),we get

10i₁ = 5 or i₁ = (5/10) =(1/2)ampere.

Substituting the value of i₁ in equation (4),we get

i₃ = 3 x (1/2) - 1 = (1/2)ampere.

Substituting the value of i₃ in equation (1),we get

2 x (1/2) + 3i₂ = 5 or 3i₂ = 4 or i₂ = 4/3ampere.

Therefore the current in the first loop is i₁= (1/2) = 0.5ampere,the current in the second loop is i₃=(1/2) = 0.5ampere and the current through the 6Ω resistor is i₂ and is equal to (4/3) = 1.33 

ampere.