This is pure curiousity.
Turning on a light (and keeping it on for that matter) takes up a certain amount of electricity right?
All right, the question: Would a burnt out bulb use up the same amount of power when turned on even though it doesn’t work? Or does the electricity just stop because it has nowhere to go, thereby using no electricity.
Thanks
The use of electricity stops because the circuit cannot be completed.
Correct. Technically, there is a very small current draw due to leakage through the gas in the bulb, but it’s measurable in nanoamperes and may as well be ignored.
Great minds post alike! sort of… 
Also, in very rare circumstances, the bulb can go “out” due to a high-resistance short via the metal and insulator. But for all practical purposes when the (incandescent) light is not shining, there isn’t any real current flowing.
I think you need to explain that a bit better, Anthracite.
Your statement implies that the bulb goes out because of the high resistance short.
Do you really mean that a high resistance short can remain after the filament fails (presumably because of tungsten deposition on the central glass insulator)?
And Q.E.D., leakage through the bulb should be negligible not only in absolute terms, but also relative to the dielectric losses in the cabling.
I meant that I have seen 10,000+ ohm or so shorts in the base of the bulb, before it gets to the filament, in a really crappy, heavily corroded bulb.