Figure shows a part of an electric circuit. The potentials at points a ,   b and c are 30   V ,…

Let the potential be $x \mathrm{V}$ at the junction.

By using $V=IR$ we get,

$$\begin{gathered} 30-x=10i_1 \\ x-12=20i_2 \\ x-2=30i_3 \end{gathered}$$

By using Kirchhoff's law,

$i_1=i_2+i_3$

$$\begin{gathered} \Rightarrow \frac{30-x}{10}+\frac{12-x}{20}+\frac{2-x}{30}=0 \\ \Rightarrow \frac{90-3x+18-1.5x+2-x}{30}=0 \end{gathered}$$

$\Rightarrow x=20 \mathrm{V}$

The current through the $20 \mathrm{\Omega }$ resistor is 

$i_2=\left(\frac{30-12}{20}\right) \mathrm{A}$

$$$\Rightarrow i_2=\frac{8}{20}\mathrm{A}=0.4 \mathrm{A}$