To determine the current through the battery in this circuit, we need to consider the equivalent resistance and the biasing of the components involved.

Both components in the circuit are forward-biased. Given this, the equivalent resistance ($R_{\text{eq}}$) is $10 \, \Omega$.

The current ($i$) through the circuit can be calculated using Ohm's law:

$ i = \frac{V}{R} = \frac{5 \, \text{V}}{10 \, \Omega} = 0.5 \, \text{A} $

Thus, the current through the battery is $0.5 \, \text{A}$.