It's a law of nature. If it can deliver the voltage it can deliver the current. There is no way around that in this universe. If it cannot deliver the current the voltage will drop, 100% of the time!
Make a graph with voltage on the horizontal axis and current on the vertical axis. Draw a line from the upper left-hand corner to the lower right-hand corner; that's the boundary of clipping for that amplifier, as a function of voltage and current. It could be a straight line, or a rectangular box, or some arbitrary curve.
Now draw a point on the graph and a line connecting the origin to the point. The slope of the line is the load admittance, and the product of voltage and current at the point is instantaneous power. For any point inside (with respect to the origin) the boundary, the amplifier will deliver the required power without clipping. For a point outside the boundary, the amplifier will deliver neither the demanded current nor voltage, but will clip at the intersection of the boundary and the line.
(Note to self: why am I explaining this?)
Every amplifier will have some absolute maximum voltage it can produce, and some absolute maximum current. They may be relatively independent, in which case your operating area boundary looks like a box. Or they may be interrelated, giving some other curve. The shape of the boundary is determined by the design of the amplifier. As a general rule, there are both current- and voltage-limiting mechanisms in play. The lower the load impedance, the steeper the load line, and the more likely the amplifier will clip due to current limiting rather than voltage limiting. The opposite is true for high-impedance loads.
I assumed solderdude objected to the latter part of Ezees' comment, but I see below that's not the case. I think his complaint is that Ezees said the RME amp lacks power. I took that to mean in the context of driving a low-impedance planar load. But, of course, it's not true in general.
My own headphone amplifier will deliver over a Watt into 100 Ohms, but one-tenth that power into 10 Ohms. The reason is that its single-ended Class A output stage is biased by a 150 mA current source. It will swing 30V peak-to-peak into 100 Ohms. As Ezees says, "It doesn't have a lot of current capability, but has no problem with voltage."
Of course, 150 mA is still a good deal of current.
I am drawn to pissing matches like a moth to a flame. I have to get back to work.