In the figure shown, the current in the 10 V battery is close to :

$\begin{array}{l}KJR, i=i_1+i_2\text{}\\ KVL in left loop, \\ 20-5i-10i_1-2i=0\\ \text{}\Rightarrow i_1=\frac{20-7i}{10}\\ KVL in outermost loop, \\ 20-5i-10-4i_2-2i=0\text{}\\ \Rightarrow i_2=\frac{10-7i}{4}\\ u\sin g first equation , \\ \text{}i=\frac{20-7i}{10}+\frac{10-7i}{4}\\ \Rightarrow 40i=80-28i+100-70i\text{}\\ \Rightarrow i=\frac{180}{138}A\end{array}$

$\therefore i_2=\frac{10-7\left({\displaystyle \frac{180}{138}}\right)}{4}=0.21A$