The current supplied by a source is given by
\[ I=\frac{E}{R+r} \]
where,
For a low-voltage source, the emf \(E\) is small. To obtain a large current, the total resistance \((R+r)\) must be as small as possible.
If the internal resistance \(r\) is large, a significant part of the voltage is lost inside the source itself due to the drop \(Ir\). This reduces the current delivered to the external circuit.
Therefore, a low-voltage supply must have very low internal resistance so that the voltage drop inside the source is minimum and a large current can be supplied efficiently.