
The figure above shows the current and voltage at different points in circuit.
Applying Kirchhoff's Junction law at A,
1x−(y+5)+2x−2+2x−0=0...(1)
Applying Kirchhoff's Junction law at B,
1y+5−x+1y−0+1y−2=0...(2)
Solving equation (1) and (2), we get
x=3 and y=0
Now, at D junction
I1=i1+i2
⇒I1=1y−0+2x−0
=10−0+23−0
⇒I1=1.5A.
