Circuit theory+nodal analysis

at node 1.The current i₂ flows in the upper loop.The current in the loop 2 be i₄ and the current in the loop 3 be i₅ and the current in the last loop be i₂ + i₅.

Now for loop 1 the voltage equation is:

2i₁ + 4i₃ = 10 -----(1)

For the loop 2 the voltage equation is :

i₄ + (i₄ - i₃) x 4 = 0 ------(2)

For the loop 3 the voltage equation is :

4i₅ +(i₅ - i₄) = 0 -------(3)

For the upper loop the voltage equation is :

i₂= 0 -------(4)

For the last loop the voltage equation is :

i₂ + i₅ = 1 -------(5)

From equation (4) and (5),we get

i₅= 1ampere

From equation (3),we get

5i₅ - i₄ = 0 or i₄= 5ampere

From equation (2),we get

5i₄ - 4i₃ = 0 or 4i₃= 25 or i₃=(25/4)ampere

From equation (1),we get

2i₁ + 4i₃= 10 or 2i₁= 10 - 4i₃ = 10 - 4 x (25/4) = -15 or i₁ = -(15/2)ampere.The negative sign shows that the current i₁ flows in the opposite direction.

The voltage at node 1 is :

v₁= i₃ x 4 = (25/4) x 4 = 25Volt.

The voltage at node 2 is :

v₂= i₄ x 1 = 5 x 1 = 5Volt.

The voltage at node 3 is :

v₃ = i₅ x 4 = 1 x 4 = 4Volt.