The current through 10$\mathrm{\Omega }$ resister in the figure is approximately  

 

Acc to junction law i = i₁ + i₂
$\begin{array}{l}\frac{10-\mathrm{V}}{10}=\frac{\mathrm{V}-7}{20}+\frac{\mathrm{V}-5}{30};10-\mathrm{V}=\frac{\mathrm{V}-7}{2}+\frac{\mathrm{V}-5}{30}\\ =\frac{3\mathrm{V}-21+2\mathrm{V}-10}{6};\frac{10-\mathrm{V}}{1}=\frac{\mathrm{V}-31}{6}\end{array}$
∴ 60 - 6V=5V - 31
11v = 91 $\therefore \mathrm{V}=\frac{91}{11}=8.27\mathrm{V}$
∴ Current through 10Ω resistor
$\mathrm{i}=\frac{10-8.27}{10}=\frac{1.73}{10}=0.173\mathrm{A}$