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 n 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 Ω and Q + S = 4 + 8 = 12 Ω, which is given by

$$\begin{gathered} \frac{1}{\mathrm{R}}=\frac{1}{6}+\frac{1}{12} \mathrm{or} \mathrm{R} = 4\mathrm{\Omega }. \mathrm{Therefore} \mathrm{current} \mathrm{I} \mathrm{is} \\ \mathrm{I}=\frac{2\mathrm{V}}{4\mathrm{\Omega }}= 0.5 \mathrm{A} \\ \mathrm{Hence} \mathrm{the} \mathrm{current} \mathrm{choice} \mathrm{is} \left(2\right). \end{gathered}$$