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 5Ω 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 6Ω and Q+ S= 4+8 12Ω, which is given by1R=16+112R=4Ω. Therefore current I is I=2V4Ω=0.5A