
Let the potential be xV at the junction.
By using V=IR we get,
30−x=10i1x−12=20i2x−2=30i3
By using Kirchhoff's law,
i1=i2+i3
⇒1030−x+2012−x+302−x=0⇒3090−3x+18−1.5x+2−x=0
⇒x=20V
The current through the 20Ω resistor is
i2=(2030−12)A
⇒i2=208A=0.4A