Figure shows a network of resistances connected to a 2V battery. If the internal resistance of the battery is negligible, current I in the circuit is

Because P/Q = R/S, the circuit is a balanced Wheatstone's bridge. Therefore, potential at B = potential at D. Hence no current flows through the 5$\mathrm{\Omega }$ resistor; it is, therefore, not effective. The equivalent resistance between A and C is the resistance of a parallel combination of P+R = 2+4 6$\mathrm{\Omega }$ and Q+ S= 4+8 12$\mathrm{\Omega }$, which is given by

$\begin{array}{l}\frac{1}{\mathrm{R}}=\frac{1}{6}+\frac{1}{12}\\ \mathrm{R}=4\mathrm{\Omega }.\text{ Therefore current }\mathrm{I}\text{ is }\\ \mathrm{I}=\frac{2\mathrm{V}}{4\mathrm{\Omega }}=0.5\mathrm{A}\end{array}$