Figure shows a part of an electric circuit. The potentials at points a, b and c are 30 V , 12 V and 2 V respectively. The current through the 20$\mathrm{\Omega }$ resistor will be
 

i₁ = i₂ + i₃
$\begin{array}{l}\frac{(\mathrm{\Delta V}{)}_1}{{\mathrm{R}}_1}=\frac{(\mathrm{\Delta V}{)}_2}{{\mathrm{R}}_2}+\frac{(\mathrm{\Delta V}{)}_3}{{\mathrm{R}}_3}\\ \frac{30-{\mathrm{V}}_0}{10}=\frac{{\mathrm{V}}_0-12}{20}+\frac{{\mathrm{V}}_0-2}{30}\\ \frac{30-{\mathrm{V}}_0}{1}=\frac{\left({\mathrm{V}}_0-12\right)3+\left({\mathrm{V}}_0-2\right)2}{6}\end{array}$
6(30 - V₀) = 3V₀ - 36 + 2V₀ - 4
180 - 6V₀ = 3V₀ - 36 + 2V₀ - 4
6V₀ + 3V₀ + 2V₀ = 180 + 36 + 4
$11{\mathrm{V}}_0=220\Rightarrow {\mathrm{V}}_0=\frac{220}{11}=20$
Current through the $\mathrm{i}=\frac{{\mathrm{\Delta V}}_2}{{\mathrm{R}}_2}$, Resistor of $20\mathrm{\Omega }, \mathrm{i}=\frac{20-12}{20}\Rightarrow \mathrm{i}=\frac{8}{20}=0.4\mathrm{A}$