In the circuit diagram shown in figure given below, the current flowing through resistance $3\mathrm{\Omega }\text{ is }\frac{\mathrm{x}}{3}\mathrm{A}$.
 

The value of x is.......

For given circuit effective emf is  ${\mathrm{\epsilon }}^1=8-4=4\mathrm{V}$
Total resistance $=0.5+1+4.5+\frac{3\times 6}{3+6}=8\mathrm{\Omega }$
current through the batteries, ${\mathrm{i}}^1=\frac{4}{8}=\frac{1}{2}\mathrm{A}$ 

Now the current through $3\mathrm{\Omega },$

${\mathrm{i}}_{3\mathrm{\Omega }}=\left(\frac{6}{6+3}\right)\frac{1}{2}\mathrm{A}=\frac{2}{3}\left(\frac{1}{2}\right)=\frac{1}{3}\mathrm{A}$
$\therefore \mathrm{x}=1$