In the electrical circuit shown in the figure, the current i through the side AB is :

The current drawn from battery 

$\begin{array}{l}\mathrm{I}=\frac{10}{10+3+\frac{20\times 30}{50}}=\frac{10}{25}\mathrm{A}\\ \mathrm{i}:\mathrm{I}-\mathrm{i}=3:2\\ \mathrm{i}=\frac{3}{5}\mathrm{I}=\frac{3}{5}\times \frac{10}{25}=\frac{6}{25}\mathrm{A}\end{array}$