Confusion between Kirchhoff's current law and voltage law
There are two basic ways you can do this.

By Ohm's law if you have a current of 2A from a voltage of 5V then the load must equal 2.5 Ohms.
In circuit A if we add an identical piece of wire (same material, cross-sectional area and length) between X and Y the current will split equally (2A --&gt; 1A + 1A) between the two wires at point X. At point Y the two currents will recombine back to the 2A (1A + 1A --&gt; 2A).
Alternatively (circuit B) you could double the load resistance to form two parallel loads of 5 Ohms. Two 5 Ohms in parallel giving the original load of 2.5 Ohms. Feed each 5 Ohm with a separate wire from point X. In which case the 2A current will split at point X (2A --&gt; 1A + 1A) and 1A will pass through each 5 Ohm load with the currents recombining at the common load connection before returning to the negative terminal of the battery.