Calculate battery current (in A) of the network shown in figure.

 

 

 

 

 

 

 

 

All four resistors in the given circuit are in parallel because all one side and other side terminals of each of the resistors are connected on the two sides of battery so equivalent resistance of the combination is given as $\frac{1}{R}=\frac{1}{8}+\frac{1}{4}+\frac{1}{6}+\frac{1}{12}\Rightarrow R=\frac{8}{5}\Omega$