
As in this case, 5Ω & 5Ω are in parallel hence, their combined resistance value will be 2.5Ω. Now, this 2.5Ω with existing 2.5Ω are in series and hence can be combined to write 5Ω.

In the final diagram, we can write equivalent resistance across the battery as 2.5Ω.

Hence, current through the battery is, i=2.5Ω5V=2A
