If you raise the voltage it will raise the current and power in the device. Since V / R = I and your load is "probably" a fix resistance.
If you have a 100watt light bulb. at 12 volt.
you change that voltage to 24 volt... it will be 400 watt now. and blow up
(First lets find the load resistance
12² / 100 = 1.44 ohm for that bulb,
Now let's see what the wattage is at 24 volt.
24² / 1.44 = 400 watt
current draw @ 12V
12 / 1.44 = 8.333 Amp
current draw @ 24V
24 / 1.44 = 16,66 Amps, and 400 watt its going to burn up.. can not do that).
That is, if it's just a load.
For linear regulator if you raise the input voltage, current draw remain the same(and since watt = V * I the wattage go up), efficiency(Watt-in/Watt-out) go down since the extra power(watts) have to be dissipated as heat.
Lowering the input voltage will help alot, it will improve efficiency but not reduce current.
If its using some kind of switch mode power supply, usually efficiency is pretty good, raising the input voltage will reduce the input current draw. But this has limits(circuit design)
Now, if you take the bulb above and replace with 2.88 ohm 100 watt..
You now need about 17 volt and 5.9 amp to get it going at full power again. You reduced current and its running at the correct wattage"not going to blow up" but it has drawback since you have to power it at higher voltage and you can't change the resistance of your load... You can add serie resistance but that's another story for another day
.