A battery of 6 volt is connected to the terminals of a three-metre long wire of uniform thickness and resistance of the order of 100 ohm. The difference of potential between two points separated by 50 cm on the wire will be
when a battery of 6 V is connected to the terminals of a three metre long wire, then the potential difference per metre will be 6 V/3 m = 2 volt per metre. Now fall of potential along 50 cm or 0.5 m wire = potential gradient x length
= 2 (volt/m) x (0.5 m) = 1 volt