Figure shows a part of an electric circuit. The potentials at...

Let the potential be $xV$ at the junction.

By using $V = IR$ we get,

$30 - x = 10{i}_{1}x - 12 = 20{i}_{2}x - 2 = 30{i}_{3}$

By using Kirchhoff's law,

${i}_{1} = {i}_{2} + {i}_{3}$

$⇒ \dfrac{30 - x}{10} + \dfrac{12 - x}{20} + \dfrac{2 - x}{30} = 0 ⇒ \dfrac{90 - 3x + 18 - 1 . 5x + 2 - x}{30} = 0$

$⇒ x = 20V$

The current through the $20Ω$ resistor is

${i}_{2} = \left(\dfrac{30 - 12}{20}\right)A$

$⇒ {i}_{2} = \dfrac{8}{20}A = 0 . 4A$