The total current supplied to the circuit as shown in figure by the $5 \mathrm{V}$ battery is _____ $\mathrm{A}$.

As in this case, $5 \mathrm{\Omega }$ & $5 \mathrm{\Omega }$ are in parallel hence, their combined resistance value will be $2.5 \Omega$. Now, this $2.5 \Omega$ with existing $2.5 \Omega$ are in series and hence can be combined to write $5 \Omega$.

In the final diagram, we can write equivalent resistance across the battery as $2.5 \Omega$.

Hence, current through the battery is, $i=\frac{5 \mathrm{V}}{2.5 \Omega }=2 \mathrm{A}$