A battery of 6 V is connected to the circuit as shown below. The current I drawn from the battery is :

Balanced wheat stone bridge in circuit so there is no current in $5\mathrm{\Omega }$ resistor so it can be removed from the circuit.

$\begin{array}{l}{\mathrm{R}}_{\mathrm{eq}}=\frac{6\times 12}{6+12}+2\\ =\frac{6\times 12}{18}+2\\ {\mathrm{R}}_{\mathrm{eq}}=6\mathrm{\Omega }\\ \mathrm{I}=\frac{\mathrm{V}}{{\mathrm{R}}_{\mathrm{eq}}}=\frac{6}{6}=1\mathrm{Amp}\end{array}$