Yeah - @JoeMomma gave you a good answer there.
Just to add one thing though, you can't directly "choose" your current (=amperage/amps). You were considering running "10 volts with half the current", it doesn't work like that. You don't decide how much current flows through a circuit, the voltage and resistance of the circuit determine the current.
Let's take your example of the LED: that LED will have a specific resistance. In other words, raising the voltage has no impact on the resistance . So, according to our I = V/R equation, if you double the voltage the end result will be double the current (amperage) flowing through your LED, which will likely burn it out. You can't choose to "halve" the current; if you have fixed resistance and double the voltage, the resulting circuit will have double the current flowing through it (unless or until something breaks)... that's the way electronics work.
To offset it however, you could double the resistance in the circuit. If you added a second resistor to the circuit (either a second identical LED in series, or some other resistor which matched the resistance of the LED and could cope with the current), then it would all work fine. If you double the voltage, but double the resistance to match it, the level of current will stay the same and everything will keep working fine. What's happening to the voltage in that case is that each resistor will produce a 5V drop in voltage. The first resistor in the circuit will see the voltage drop from 10V to 5V, (with 5V drop across that resistor - so the LED is happily within its spec), while the second resistor will see the voltage drop from 5V to ground (0V).