Use Kirchhoff's rules to find the current through 3 Ω . . .

Question : Use Kirchhoff's rules to find the current through 3 Ω resistor in the circuit shown in the figure : 

Doubt by Riddhi & Ridhi

Solution : 

Applying Kirchhoff's Loop Rule in closed loop 1
-3(I₂-I₁)+4I₁=3
-3I₂+3I₁+4I₁=3
7I₁-3I₂=3 — (1)

Again, applying Kirchhoff's Loop Rule in closed loop 2

2I₂+3(I₂-I₁)=5
2I₂+3I₂-3I₁=5
-3I₁+5I₂=5 — (2)

On solving equation (1) and (2) 
[7I₁-3I₂=3 ] × 3
[-3I₁+5I₂=5] × 7

21I₁-9I₂=9
-21I₁+35I₂=35

Adding the above equations

0+26I₂=44
26I₂=44
I₂=44/26
I₂=22/13 A

Putting in equation (1) 
7I₁-3(22/13)=3
7I₁-66/13=3
7I₁=3+66/13
7I₁=(39+66)/13
7I₁=105/13
I₁=105/(13×7)
I₁=15/13 A

Current through 3 Ω resistor 
= I₂-I₁
= 22/13-15/13
= 7/13 A
= 0.54 A

Hence, the current through 3Ω resistor is (7/13) A or 0.54 A from E to B.