Determine the current through the battery in the left and right...

In the left circuit:
The positive terminal of the $5\text{ V}$ battery is connected to the p-side of both diodes. Therefore, both diodes are forward-biased and act as closed switches (assuming ideal diodes).
The two $10\Omega$ resistors are in parallel.
Equivalent resistance, $R_{eq} = \dfrac{10 \times 10}{10 + 10} = 5\Omega$.
Current through the battery, $I_{left} = \dfrac{V}{R_{eq}} = \dfrac{5}{5} = 1\text{ A}$.

In the right circuit:
The positive terminal of the battery is connected to the p-side of the top diode and the n-side of the bottom diode. Thus, the top diode is forward-biased (acts as a closed switch) and the bottom diode is reverse-biased (acts as an open switch).
Current flows only through the upper branch.
Equivalent resistance, $R_{eq} = 10\Omega$.
Current through the battery, $I_{right} = \dfrac{V}{R_{eq}} = \dfrac{5}{10} = 0.5\text{ A}$.

The currents are $1\text{ A}$ and $0.5\text{ A}$ respectively.