Explanation:
The current distribution is shown in the figure.
By writing Kirchhoff’s laws
\(10 - 10{i_1} - 10i = 0\)
\(1 - {i_1} - i = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1)\)
\(10V + 5V - 15\left( {i - {i_1}} \right) - 10i = 0\)
\(3 - 5i - 2{i_1} = 0\,\,\,\,\,\,\,(2)\)
From both equations \({i_1} = \frac{1}{4}A\)
