The given circuit can be represented as
\begin{gathered} \frac{V_{ A }}{10}+\frac{V_{ A }}{4}+\frac{V_{ A }-5}{5}+\frac{V_{ A }-50}{7}=0 \\ \frac{140 V_{ A }+350 V_{ A }+280 V_{ A }-1400+200 V_{ A }-10000}{1400}=0 \\ 970 V_{ A }=11400 V_{ A }=11.75 \\ I=\frac{11.75}{10}=1.175 A \end{gathered}