Increasing voltage, but same current
The load determines the ratio of current to voltage at its terminals.

If you control the voltage you supply, it determines the current.

If you control the current you supply, it determines the voltage.

It's possible to break a load by supplying, or letting it define, too much of either.

Let's compare a 5&nbsp;V, 2&nbsp;A adapter, and a 20&nbsp;V, 2&nbsp;A adapter, driving different loads.

We'll drive a 100 ohm load, which has a thermal limit of 1 watt. With 5&nbsp;V, it will draw 50&nbsp;mA and dissipate 250&nbsp;mW, and run happily. With 20&nbsp;V, it will draw 200&nbsp;mA and dissipate 4&nbsp;watts, and eventually overheat.

Now let's drive a 1 ohm load, which has a thermal limit of 10 watts. With a 2&nbsp;A supply, it will collapse the input voltage to 2&nbsp;V, and dissipate 4&nbsp;watts. It doesn't matter whether it's the 5&nbsp;V supply, or the 20&nbsp;V supply driving it; if they both put out a constant current of 2&nbsp;A, then the load will drop their voltage to 2&nbsp;V. Different adapters may behave differently to the current limit however; some will deliver a constant current, some will shutdown for a moment and try to restart, and repeat that cycle continuously, and some will deliver a lower current (so-called foldback limiting) to protect themselves.

Now let's connect the gate-source junction of a FET, that has a voltage limit of 15&nbsp;V. On 5&nbsp;V, it will draw essentially no current and survive. On 20&nbsp;V, it will draw essentially no current, punch through and be destroyed.