The total current supplied to the circuit as shown in figure by the...

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, $i = \dfrac{5V}{2 . 5Ω} = 2A$